Organic Chemistry

The big picture

Organic chemistry is the chemistry of carbon. This lesson explains why one element out of the whole periodic table gets its own entire field: carbon's four bonds let it build endless chains, rings, and branches, so the number of possible carbon molecules is effectively unlimited. Everything else in the course rests on this foundation, so we start by counting electrons and bonds carefully.

Organic chemistry is the chemistry of carbon compounds. That one element supports millions of distinct molecules, from methane to DNA, and understanding why begins with carbon's place in the periodic table. Carbon sits in Group 14 with four valence electrons, the outer-shell electrons that actually take part in bonding. To reach a stable octet of eight outer electrons, carbon neither easily gains four nor loses four; instead it shares, forming four covalent bonds. A covalent bond is a shared pair of electrons that both atoms count as their own, like two neighbors sharing a fence that belongs to both yards.

Two features of that sharing are worth a whole semester. A shared pair is directional, pointing from one nucleus toward another, so a covalent molecule has a definite three-dimensional shape. And sharing is almost never perfectly equal, so every imbalance is a handle a reaction can grab. Nearly every mechanism you draw this term is a story of electrons flowing from an electron-rich place to an electron-poor one, and learning to see which is which is the skill this lesson begins to build.

Counting bonds from valence electrons

A quick rule predicts how many bonds a second-row atom makes: it equals the number of electrons needed to fill the octet. Carbon needs 4 and makes 4 bonds. Nitrogen, with 5 valence electrons, needs 3 and makes 3 bonds plus one lone pair. Oxygen, with 6, makes 2 bonds plus two lone pairs. Hydrogen, needing only 2 electrons total, makes 1 bond. A lone pair is a pair of valence electrons that is not shared in a bond; it still occupies space and shapes the molecule. These "normal" valences appear again and again, so memorizing them saves enormous time and lets you spot mistakes instantly.

Atom	Valence electrons	Typical bonds	Lone pairs
Hydrogen (H)	1	1	0
Carbon (C)	4	4	0
Nitrogen (N)	5	3	1
Oxygen (O)	6	2	2
Halogen (F, Cl, Br, I)	7	1	3

Key idea: For neutral second-row atoms, bonds plus lone pairs fill the octet, so bonds = 8 minus valence electrons (and hydrogen simply makes one bond).

Formal charge: the bookkeeping that keeps you honest

That table describes neutral atoms. When an atom carries more or fewer bonds than its normal count, it must carry a formal charge, and getting formal charges right is what separates a structure that means something from a drawing that is merely decorative. The arithmetic is one line: formal charge = (valence electrons of the free atom) - (lone-pair electrons) - (number of bonds).

Apply it to hydroxide, HO^-. Oxygen brings 6 valence electrons; in hydroxide it has one bond and three lone pairs, so 6 - 6 - 1 = -1. Now hydronium, H₃O⁺: three bonds and one lone pair, so 6 - 2 - 3 = +1. Same atom, opposite charge, and the only difference is how many pairs it is sharing rather than hoarding.

Carbon obeys the identical rule, and the results are the three reactive intermediates you will meet in nearly every mechanism ahead. A carbon with three bonds and one lone pair is a carbanion: 4 - 2 - 3 = -1. A carbon with three bonds and an empty orbital is a carbocation: 4 - 0 - 3 = +1. A carbon with three bonds and one unpaired electron is a neutral radical: 4 - 1 - 3 = 0.

Key idea: Formal charge = valence electrons minus lone-pair electrons minus bonds; an atom that departs from its normal valence must carry a charge, and drawing that charge is never optional.

Why carbon in particular

Many elements form covalent bonds, but carbon is special for three reasons. First, four bonds is the maximum for a small atom, so carbon can branch in four directions at once. Second, carbon-carbon bonds are strong and stable, so long chains and rings do not fall apart. Third, carbon bonds well to hydrogen, oxygen, nitrogen, sulfur, and the halogens, so it can carry a huge variety of reactive groups. The ability of an element to bond to itself in chains and rings is called catenation, meaning "chaining," and carbon is the champion of catenation. This is why a single element deserves a whole course.

Key idea: Carbon's four strong, versatile bonds and its talent for catenation are what make an unlimited variety of organic molecules possible.

Unequal sharing: electronegativity and polar bonds

Covalent bonds share electrons, but "share" does not mean "share equally." Electronegativity measures an atom's pull on the electrons of a bond it is part of. On the Pauling scale the values an organic chemist uses constantly are hydrogen 2.2, carbon 2.6, nitrogen 3.0, oxygen 3.4, and fluorine 4.0, with chlorine 3.2, bromine 3.0, and iodine 2.7 filling in the rest of the halogens. Electronegativity rises left to right across a period and falls going down a group.

When two bonded atoms differ, the more electronegative one holds the shared pair closer and takes on a partial negative charge, written delta-, while its partner takes a partial positive charge, delta+. A C-O bond, with a gap of about 0.8 units, is strongly polarized toward oxygen. A C-H bond, with a gap of only about 0.4, is close enough to nonpolar that we treat hydrocarbons as greasy and unreactive. That single comparison is why methane is inert enough to pipe into homes and methanol is a reactive solvent.

Polarity is what converts a static drawing into a prediction. The delta+ carbon of C-Cl, C-O, or C=O is an electrophile, a site hungry for electrons. A lone pair, a negative charge, or an exposed pi bond is a nucleophile, a site with electrons to give. Every polar mechanism in this course, without exception, is a nucleophile finding an electrophile.

Key idea: Electronegativity differences create delta+ and delta- sites, and those sites are the street addresses where reactions happen.

Single, double, and triple bonds

Because carbon has four bonds to distribute, it can attach to four different atoms with single bonds, or concentrate bonding between two atoms. A single bond shares one pair of electrons, a double bond shares two pairs, and a triple bond shares three. In ethane the two carbons share one pair (C-C); in ethene they share two (C=C); in ethyne they share three (C≡C). Notice that every carbon in all three molecules still has exactly four bonds total; a double bond simply uses two of carbon's four bonds on the same neighbor.

Key idea: More shared pairs between the same two atoms means a shorter, stronger connection, but each carbon always keeps four bonds in total.

Sigma and pi bonds

Not all shared pairs are alike. The first bond between two atoms is always a sigma bond, formed by orbitals overlapping head-on directly between the nuclei, like two flashlight beams meeting tip to tip. Any additional bonds in a double or triple bond are pi bonds, formed by orbitals overlapping side-by-side above and below the axis, like two boards laid flat and touching along their faces.

So a double bond is one sigma plus one pi, and a triple bond is one sigma plus two pi. Sigma bonds are strong and allow free rotation; pi bonds are weaker and lock the atoms rigid, which is why molecules with double bonds cannot freely twist. This single distinction explains a great deal of organic reactivity: pi bonds, being exposed and weaker, are where many reactions happen.

Key idea: The first bond is always sigma; extra bonds are pi, and those weaker, exposed pi bonds are the usual sites of reaction.

Bond length and bond strength: the numbers behind the picture

The claim that more shared pairs give a shorter, stronger bond is measurable rather than decorative. In ethane the C-C distance is about 154 pm; in ethene the C=C distance is about 134 pm; in ethyne the triple bond is about 120 pm. Bond dissociation energies rise in step: roughly 377 kJ/mol for the C-C bond of ethane, about 728 kJ/mol for the C=C of ethene, and about 965 kJ/mol for the C≡C of ethyne. Breaking one C-H bond of methane costs about 439 kJ/mol.

Read those numbers carefully, because they hide a trap. Going from 377 to 728 kJ/mol suggests the pi bond alone is worth about 351 kJ/mol, nearly as much as a sigma bond. It is not. The sigma bond in ethene is itself stronger than the one in ethane, because sp² orbitals carry more s character, sit closer to the nucleus, and make a shorter, tighter sigma bond. The cleaner measure of a pi bond is the barrier to twisting about a C=C, which for ethene is roughly 270 kJ/mol: rotating 90 degrees destroys the side-by-side p overlap completely, so that barrier is the pi bond and nothing else. Real, but clearly weaker than sigma, which is exactly why alkenes react readily and alkanes largely do not.

Key idea: Shorter means stronger, but the extra strength of a double bond is split between a tightened sigma bond and a genuinely weaker pi bond worth roughly 270 kJ/mol.

Worked example: auditing acetic acid atom by atom

Take acetic acid, written condensed as CH₃-COOH, and audit every atom the way you should audit every new structure.

• The methyl carbon. Three bonds to H, one to the neighbouring carbon: four bonds, no lone pairs, formal charge 4 - 0 - 4 = 0. It is sp³ and tetrahedral, and its C-H bonds are barely polar, so nothing interesting happens here.
• The carboxyl carbon. A double bond to one oxygen, a single bond to the other, a single bond to the methyl carbon: four bonds again, formal charge 0. But it is sp², trigonal planar, and bonded to two of the most electronegative atoms in the molecule, so it carries a large delta+ and is the electrophilic site.
• The carbonyl oxygen (C=O). Two bonds and two lone pairs, formal charge 6 - 4 - 2 = 0, strongly delta-. Those lone pairs are the molecule's most available electrons.
• The hydroxyl oxygen (O-H). Two bonds and two lone pairs, formal charge 0. Its hydrogen is the acidic one, because the anion left behind spreads its negative charge across both oxygens - a point Lesson 11 develops in full.

One pass has located the inert end, the electrophilic carbon, the nucleophilic lone pairs, and the acidic hydrogen. That workflow never changes; only the molecules get larger.

Key idea: Audit a new structure by checking each atom's bond count, lone pairs, formal charge, and polarity, and the reactive sites announce themselves.

Where people get stuck

• "A double bond is twice as strong as a single bond." It is shorter and stronger, but not double. The extra strength is split between a tighter sigma bond and a pi bond worth roughly 270 kJ/mol, and it is that weaker pi bond that reacts.
• "Carbon can make five or six bonds if it needs to." Carbon has only four valence orbitals in the second row and essentially never exceeds four bonds. A drawing showing five bonds to one carbon is an error, not an exotic species.
• "If atoms share electrons, the bond is neutral and nothing else matters." Sharing is unequal whenever electronegativities differ, and those delta+ and delta- partial charges are precisely what drives polar reactions.
• "Lone pairs are inert leftovers." Lone pairs are usually the most reactive electrons present, acting as the donors in acid-base chemistry, nucleophilic substitution, and almost every arrow you will push.
• "Formal charge is the real charge on that atom." Formal charge assumes perfectly equal sharing. It tells you where to draw a plus or minus sign; the actual electron density is smeared out and is better described by polarity and resonance.
• "Sigma and pi are just names for first and second bonds." They are different geometries. Sigma overlap lies along the internuclear axis and permits free rotation; pi overlap lies above and below it and locks rotation, which is why cis and trans alkenes are separable compounds.

Recap

• Carbon has four valence electrons and forms four covalent bonds to complete its octet.
• For neutral second-row atoms, bonds equal the electrons needed to reach eight, with N making 3, O making 2, and H making 1.
• Formal charge = valence electrons - lone-pair electrons - bonds, and it identifies carbocations, carbanions, and charged heteroatoms.
• Electronegativity differences make bonds polar, creating the delta+ electrophilic and delta- nucleophilic sites where reactions begin.
• Single, double, and triple bonds share one, two, and three electron pairs, but each carbon still keeps four bonds total.
• The first bond of any pair is a sigma bond; additional bonds are weaker pi bonds, worth about 270 kJ/mol in ethene, where much of organic reactivity occurs.
• Carbon's bond strength, versatility, and talent for catenation are why organic chemistry is its own field.

Sources

1. McMurry, J. (2023). Structure and Bonding: Development of Chemical Bonding Theory. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Describing Chemical Bonds: Valence Bond Theory. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Polar Covalent Bonds and Electronegativity. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Formal Charges. In Organic Chemistry (OpenStax). openstax.org
5. Flowers, P., et al. (2019). Covalent Bonding. In Chemistry 2e (OpenStax). openstax.org
6. LibreTexts Chemistry. Describing Chemical Bonds: Molecular Orbital Theory. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Clayden, J., Greeves, N., and Warren, S. (2012). Organic Chemistry (2nd ed.), Chapters 2 and 4. Oxford University Press. find source ↗

The big picture

Molecules are three-dimensional, and their shape decides almost everything about how they behave. This lesson gives you a fast, reliable way to predict the shape at any carbon, nitrogen, or oxygen: count the groups around it. Master this and you can sketch the geometry of a molecule at a glance, which you will need for stereochemistry and mechanisms later.

A neutral carbon atom by itself has electrons in one 2s and three 2p orbitals, but that does not match the four identical bonds we see in methane. To explain equal bonds pointing to equal angles, chemists use hybridization, which means mixing atomic orbitals to make new, equivalent hybrid orbitals aimed for bonding. Think of hybridization like blending one cup of one paint color with several cups of another to get several identical cups of a new color. Which orbitals mix depends on how many groups surround the carbon.

The three hybridizations of carbon

• sp³: one s and three p orbitals mix into four equal orbitals. Four groups spread to the corners of a tetrahedron at 109.5°. This is carbon with four single bonds, as in methane (CH₄).
• sp²: one s and two p orbitals mix into three orbitals in a plane at 120° (trigonal planar), leaving one unhybridized p orbital for a pi bond. This is carbon in a double bond, as in ethene (H₂C=CH₂).
• sp: one s and one p orbital mix into two orbitals pointing opposite at 180° (linear), leaving two p orbitals for two pi bonds. This is carbon in a triple bond, as in ethyne (HC≡CH).

Key idea: The number of hybrid orbitals equals the number of atomic orbitals mixed, and leftover unmixed p orbitals are exactly what form pi bonds.

Where 109.5 degrees comes from

The tetrahedral angle is not a convention; it falls out of geometry. Put four points on a sphere and push them as far apart as possible and they land on the vertices of a regular tetrahedron, with the angle between any two, measured at the centre, equal to arccos(-1/3) = 109.47 degrees. Three points on a sphere pushed apart land on a great circle 120 degrees apart, and two land at opposite poles, 180 degrees apart. The three geometries in the table below are simply the answers to "how do you space 4, 3, or 2 things as far apart as possible around a point."

That framing is VSEPR, valence-shell electron-pair repulsion, and it is the reason hybridization and shape track each other so faithfully. Hybridization is the orbital language; VSEPR is the electrostatic reason.

A reliable shortcut: count the groups

You rarely need to think about orbitals from scratch. Count the number of groups attached to an atom, where a group is any single-bonded atom, any multiple-bonded atom (counted once), or any lone pair. Then read off the hybridization and shape:

Groups	Hybridization	Shape	Angle
4	sp3	Tetrahedral	109.5°
3	sp2	Trigonal planar	120°
2	sp	Linear	180°

This works for nitrogen and oxygen too. The nitrogen in ammonia has three bonds and one lone pair, so four groups, sp³, roughly tetrahedral. The oxygen in water has two bonds and two lone pairs, four groups, also sp³. Note the difference between electron geometry (the arrangement of all groups including lone pairs) and molecular geometry (the shape traced by atoms only). Water's electron geometry is tetrahedral, but because two corners are lone pairs, the shape you see through the atoms is bent.

Key idea: Count groups (bonds of any order plus lone pairs) to get hybridization and electron geometry instantly, then hide the lone pairs to see the molecular shape.

Real angles deviate, and the deviation is informative

Methane's H-C-H angle is 109.5 degrees, the ideal value, because all four groups are identical. Ammonia's H-N-H angle is about 107 degrees and water's H-O-H angle is 104.5 degrees, both squeezed below the ideal. The reason is that a lone pair is held by only one nucleus rather than two, so it spreads out closer to the central atom and takes up more angular room than a bonding pair. Each lone pair therefore compresses the remaining bond angles by roughly two to three degrees. Water, with two lone pairs, is compressed twice as much as ammonia, with one.

Read the deviation as evidence rather than as an annoyance: if a measured angle is smaller than the ideal, look for lone pairs; if it is larger, look for bulky substituents pushing each other apart.

Key idea: Lone pairs demand more space than bonding pairs, so each one narrows the remaining angles by about two to three degrees.

Worked example: a molecule with several centers

Consider acetonitrile, CH₃-C≡N. The left carbon has four single bonds (three to H, one to C), so four groups: sp³, tetrahedral, 109.5°. The middle carbon has one single bond and one triple bond, so two groups: sp, linear, 180°. The nitrogen has one triple bond and one lone pair, so two groups: sp, and the C≡N unit is linear. Working center by center like this lets you describe the whole molecule with confidence.

Key idea: Analyze one atom at a time; a single molecule can contain sp³, sp², and sp centers side by side.

The important exception: when a lone pair joins a pi system

The group-counting rule has one systematic exception, and it is worth more than the rule itself because it shows up in every peptide and every amide drug. The nitrogen of an amide, R-CO-NH₂, has three bonds and one lone pair. Count naively and you get four groups, sp³, pyramidal. Measure it and you find a flat, trigonal planar nitrogen with angles near 120 degrees.

The reason is that the nitrogen lone pair is not a spectator. If nitrogen adopts sp², its leftover p orbital lines up with the p orbital of the adjacent C=O, and the lone pair delocalizes onto the carbonyl oxygen. That stabilization is worth more than the small penalty of the tighter angles, so the molecule pays it. The consequence is measurable: rotation about the amide C-N bond costs roughly 75 to 85 kJ/mol, because rotating breaks that overlap. That barrier is exactly why the peptide backbone of a protein is a chain of rigid planar units rather than a freely tumbling string, and it is what makes protein secondary structure possible at all.

The same logic explains why the nitrogen of pyrrole is sp² with its lone pair in a p orbital feeding the aromatic ring, while the nitrogen of pyridine is sp² with its lone pair in an in-plane sp² orbital, pointing outward, free to act as a base. Same element, same hybridization, opposite chemistry, and the difference is only where the lone pair lives.

Key idea: A lone pair next to a pi bond will flatten its atom to sp² so it can conjugate; count groups first, then check whether a neighbouring pi system overrides the count.

Why shape matters

More s-character in a hybrid orbital pulls electrons closer to the nucleus, so sp bonds are shorter and stronger than sp³ bonds. s-character is simply the fraction of the hybrid that came from the s orbital: sp is 50 percent s, sp² is about 33 percent, sp³ is 25 percent.

The trend is quantitative. The C-H bond length falls from about 109 pm at an sp³ carbon (ethane) to 108 pm at sp² (ethene) to 106 pm at sp (ethyne), and the bond dissociation energy rises the other way, from about 423 kJ/mol to 465 kJ/mol to 558 kJ/mol.

The most useful consequence is acidity. An orbital with more s-character holds a negative charge closer to the nucleus and therefore stabilizes it better. Remove a proton from ethane and the carbanion's lone pair sits in an sp³ orbital: pK_a about 51. From ethene, the anion's pair sits in sp²: pK_a about 44. From ethyne, it sits in sp: pK_a about 25. That is a span of 26 orders of magnitude produced by hybridization alone, with no change in the atoms present. It is why a terminal alkyne can be deprotonated by sodium amide and turned into a carbon nucleophile, while an alkane simply cannot.

Key idea: s-character stabilizes negative charge, so C-H acidity rises sharply along sp³ (pK_a 51) to sp² (44) to sp (25).

Geometry also decides whether a molecule is flat or three-dimensional, which controls how molecules pack, react, and fit into biological targets like enzymes and drug receptors. When you see a carbon, glance at its bonds: all singles means sp³ and tetrahedral, one double means sp² and flat, a triple or two cumulated double bonds means sp and linear.

Key idea: Higher s-character gives shorter, stronger bonds and larger angles, and geometry ultimately governs physical properties and biological fit.

What hybridization is, and what it is not

It is worth saying plainly what kind of thing hybridization is, because students often over-read it. Hybridization is a bookkeeping device inside valence bond theory: a linear recombination of atomic orbitals chosen so that localized two-centre bonds come out pointing in the directions we observe. It is not a physical event that an atom undergoes before bonding. Carbon does not "promote an electron and then hybridize" in time; the mathematics simply produces a convenient basis.

There is direct evidence that the four sp³ orbitals of methane are not the true one-electron states of the molecule. Photoelectron spectroscopy of methane shows two distinct ionization bands, near 12.7 and 23 electron volts, not one. Molecular orbital theory predicts exactly that pattern: a set of three degenerate bonding orbitals plus one lower-lying orbital of different symmetry. Four identical sp³ bonds would give a single band.

Both descriptions are correct within their own framework, because the localized and delocalized pictures are related by a change of basis and predict the same total electron density. Use hybridization for shape, sterics, and mechanism, where it is fast and reliable. Reach for molecular orbital theory when you need spectroscopy, conjugation, aromaticity, or pericyclic selection rules, where the delocalized picture is the one that explains the data.

Key idea: Hybridization is a localized-bond model that predicts geometry beautifully; molecular orbital theory is the picture that matches spectroscopy, and neither is "the truth" that replaces the other.

Where people get stuck

• "You must draw orbital diagrams every time." In practice you count groups. The orbital picture is the justification, not the daily tool.
• "Lone pairs do not count as groups." They count for hybridization and electron geometry, which is why the oxygen of water is sp³ and not sp.
• "A double bond counts as two groups." A double or triple bond to one neighbour counts once, because all of its electrons point toward the same atom.
• "sp² carbons are only roughly flat." The three groups on an sp² carbon lie in a genuine plane, and that flatness is what makes conjugation and aromaticity possible later.
• "Amide nitrogen has four groups, so it is sp³." Conjugation with the neighbouring C=O flattens it to sp². Always check whether an adjacent pi system can take the lone pair before trusting the raw count.
• "Hybridization is something the atom physically does." It is a choice of mathematical basis within valence bond theory, useful because it reproduces observed geometry, not a step in a reaction.
• "Electron geometry and molecular geometry are two names for the same thing." Electron geometry counts lone pairs; molecular geometry traces only atoms. Water is tetrahedral in the first sense and bent in the second.

Recap

• Hybridization mixes s and p orbitals to make equal hybrid orbitals aimed at bonding; leftover p orbitals form pi bonds.
• Count groups (bonds of any order plus lone pairs): 4 gives sp³ tetrahedral 109.5°, 3 gives sp² trigonal planar 120°, 2 gives sp linear 180°.
• The ideal angles are the geometric answers to spacing 4, 3, or 2 points around a centre; VSEPR supplies the electrostatic reason.
• Lone pairs occupy more room than bonding pairs, compressing real angles to about 107 degrees in ammonia and 104.5 degrees in water.
• Electron geometry includes lone pairs; molecular geometry traces atoms only, so water is bent though its electron geometry is tetrahedral.
• A lone pair adjacent to a pi bond flattens its atom to sp², which is why amide nitrogen is planar and the peptide bond is rigid.
• More s-character means shorter, stronger bonds and much greater carbanion stability: pK_a falls from about 51 (sp³) to 44 (sp²) to 25 (sp).

Sources

1. McMurry, J. (2023). sp3 Hybrid Orbitals and the Structure of Methane. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). sp2 Hybrid Orbitals and the Structure of Ethylene. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur. In Organic Chemistry (OpenStax). openstax.org
4. Flowers, P., et al. (2019). Hybrid Atomic Orbitals. In Chemistry 2e (OpenStax). openstax.org
5. Flowers, P., et al. (2019). Molecular Structure and Polarity (VSEPR). In Chemistry 2e (OpenStax). openstax.org
6. LibreTexts Chemistry. sp Hybrid Orbitals and the Structure of Acetylene. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Anslyn, E. V., and Dougherty, D. A. (2006). Modern Physical Organic Chemistry, Chapter 1. University Science Books. find source ↗

The big picture

Organic molecules are too large to spell out atom by atom every time, so chemists use shorthand drawings. This lesson teaches the three drawing styles and, most importantly, the skeletal (line-angle) style you will see on nearly every page from here on. Learning to read carbons and hidden hydrogens off a zig-zag is a survival skill for the rest of the course.

Organic molecules are large, so chemists use compact drawings. The same molecule can be shown three ways, and fluency means moving between them without effort.

Three ways to draw one molecule

• A Lewis (structural) formula shows every atom and every bond as a line. Butane is H₃C-CH₂-CH₂-CH₃ with every C-H bond drawn out. It is complete but slow.
• A condensed formula groups atoms without drawing most bonds: butane is CH₃CH₂CH₂CH₃, or even CH₃(CH₂)₂CH₃. Hydrogens are written next to the carbon they attach to.
• A skeletal (line-angle) formula is the fastest and most common. Draw only the carbon skeleton as a zig-zag of lines. Each line end and each vertex is a carbon. Hydrogens on carbon are not drawn at all.

Key idea: Lewis structures show everything, condensed formulas hide most bond lines, and skeletal formulas hide carbons and their hydrogens for maximum speed.

Reading condensed formulas without being fooled

Condensed formulas are read strictly left to right, and every hydrogen belongs to the atom written immediately before it. So CH₃CH₂OH is a two-carbon chain whose second carbon carries an -OH, and CH₃COCH₃ is acetone, in which the middle carbon is a C=O rather than a C-O-C. That second reading is the one beginners miss: a bare "CO" inside a condensed chain is a carbonyl, while an ether oxygen is written with carbons on both sides, as in CH₃OCH₃.

Branches are written in parentheses and hang off the atom immediately to their left. In CH₃CH(CH₃)CH₂CH₃, the parenthetical methyl is attached to the second carbon, giving 2-methylbutane and not pentane. Repeated units are collapsed with a subscript: CH₃(CH₂)₄CH₃ is hexane, six carbons in total, not four.

Key idea: In a condensed formula, hydrogens belong to the atom on their left, parentheses hang branches off the atom on their left, and a "CO" in the middle of a chain is a carbonyl, not an ether.

The two rules that make skeletal formulas work

1. Every corner and every line end is a carbon atom, unless another atom's symbol is written there.
2. Each carbon has enough hidden hydrogens to reach four bonds. A carbon at the end of one line has 3 hydrogens; a carbon between two lines has 2; a carbon at a three-line junction has 1; a carbon at a four-line junction has 0.

An atom that is not carbon is called a heteroatom, meaning a "different atom" such as O, N, S, or a halogen. Heteroatoms and the hydrogens attached to them (as in O-H or N-H) are always drawn explicitly. So an alcohol is drawn with its -OH shown, but the carbon chain stays as bare lines.

Key idea: Fill each carbon with implied hydrogens up to four bonds, but always draw heteroatoms and any H bonded to them.

Reading back the atoms

Given a skeletal drawing, you can always recover the molecular formula. Count the vertices and ends for carbons, then add the implied hydrogens so each carbon has four bonds, plus any hydrogens shown on heteroatoms. A three-segment zig-zag has four carbons and, filling to four bonds each, ten hydrogens, giving C₄H₁₀. This skill is worth practicing until it is automatic, because almost every structure you meet from here on will be drawn skeletally.

Key idea: To get the formula from a skeletal drawing, count carbons at every vertex and end, top up hydrogens to four bonds per carbon, then add hydrogens shown on heteroatoms.

Worked example: reading a small alcohol

Suppose a two-segment zig-zag ends in an -OH written at the right end. The three vertices and ends are three carbons, so the skeleton is propan-based. The left carbon (one line) carries 3 H, the middle carbon (two lines) carries 2 H, and the right carbon bonds to the -OH plus its line, so it carries 2 H. Adding the -OH hydrogen, the formula is C₃H₈O. Notice we never draw the carbon hydrogens, yet we can count them exactly by the four-bond rule.

Key idea: Even complex-looking skeletal structures reduce to simple bookkeeping: carbons at the corners, hydrogens filled to four bonds, heteroatoms spelled out.

Showing three dimensions: wedges and dashes

A flat drawing cannot express which groups point toward you and which point away, and from Lesson 9 onward that will matter enormously. The convention is minimal. A plain line lies in the plane of the paper. A solid wedge, drawn narrow at the atom and wide at the far end, points out of the page toward the reader. A hashed or dashed wedge points behind the page, away from the reader.

The practical rule is that at a single sp³ carbon you draw two plain lines, one wedge, and one dash, and you keep the two plain bonds adjacent to each other rather than opposite. Drawing a wedge and a dash on the same side, or using three wedges at one carbon, produces a picture that no real tetrahedron matches and it will lead you to the wrong stereochemical answer. This is exactly the notation you will use to distinguish two mirror-image molecules that may differ by the gap between a medicine and a poison.

Key idea: Wedge toward the reader, dash away, plain line in the plane, and at one sp³ carbon use exactly one of each wedge and dash.

When one drawing is not enough: resonance

Some molecules cannot be described honestly by any single structure. The classic case is the acetate ion, CH₃-COO^-. Draw it with the C=O on the left oxygen and the negative charge on the right, and you have a perfectly valid Lewis structure. Draw it the other way round and you have an equally valid one. Neither is correct on its own, because experiment says the two carbon-oxygen bonds in acetate are identical, both about 126 pm long, sitting neatly between a normal C=O of about 121 pm and a normal C-O of about 136 pm.

The resolution is resonance. The real molecule is a single species, the resonance hybrid, and the two drawings are two incomplete sketches of it joined by a double-headed arrow. The molecule does not flicker back and forth between them; it never was either one. The negative charge is genuinely spread over both oxygens, half a charge on each, which is why acetate is so much more stable than an alkoxide and why acetic acid is roughly ten billion times more acidic than ethanol.

Four rules keep resonance drawings honest:

1. Only electrons move. Nuclei stay exactly where they are. If two drawings differ in the position of an atom, they are different compounds (tautomers or isomers), not resonance forms.
2. Only lone pairs and pi electrons move. Sigma bonds never move in a resonance arrow.
3. Second-row atoms never exceed an octet. Carbon, nitrogen, oxygen, and fluorine can hold at most eight valence electrons in any contributing structure.
4. The total charge and the number of unpaired electrons are the same in every form.

Not all contributors count equally. The major contributor is the one with the most covalent bonds, with complete octets, with negative charge on the more electronegative atom, and with the least separation of opposite charges. Applied to the amide of Lesson 2, the neutral form with C=O is the major contributor and the charge-separated form with C=N⁺ and O^- is minor - yet that minor contributor is exactly what flattens the nitrogen and stiffens the peptide bond, which is why minor contributors are still worth drawing.

Key idea: Resonance forms are alternative electron bookkeeping for one real molecule; move only lone pairs and pi electrons, keep every octet legal, and weight contributors by bond count, octets, and where the charge sits.

Where people get stuck

• "The ends of the zig-zag are hydrogens." The ends are carbons, usually CH₃ groups. The hydrogens are implied, not drawn.
• "You can leave out an O-H hydrogen the way you leave out C-H hydrogens." Only hydrogens on carbon are hidden. Hydrogens on heteroatoms are always drawn, because they are usually the acidic or hydrogen-bonding ones.
• "A straight line and a zig-zag mean different molecules." The zig-zag is a readability convention. Bond angles in a drawing are near 120 degrees because that is easy to draw, not because the molecule is trigonal.
• "CH₃COCH₃ is an ether." It is acetone. An oxygen written between two carbons in a condensed chain with no other bonds shown is a carbonyl; an ether is written with the oxygen flanked as CH₃OCH₃.
• "Resonance means the molecule flips between two structures." It does not. There is one molecule with one electron distribution, and the two drawings are both incomplete descriptions of it.
• "If I can draw it, it is a resonance form." Move an atom and you have drawn a different compound. Exceed an octet on carbon or oxygen and you have drawn nothing at all.
• "All resonance forms contribute equally." Only when they are identical by symmetry, as in acetate. Otherwise weight them, because the major contributor is the better single description of the real molecule.

Recap

• Three equivalent styles: Lewis (all atoms and bonds), condensed (grouped atoms), and skeletal (carbon skeleton only).
• In condensed formulas, hydrogens and parenthetical branches attach to the atom on their left, and a mid-chain CO is a carbonyl.
• In skeletal drawings, every vertex and line end is a carbon unless another symbol is written there.
• Implied hydrogens fill each carbon to four bonds: 3 at a chain end, 2 mid-chain, 1 at a three-line junction, 0 at a four-line junction.
• Heteroatoms and the hydrogens bonded to them are always drawn explicitly.
• Wedges point toward the reader and dashes away, which is how a flat page carries three-dimensional information.
• Resonance forms describe one molecule with delocalized electrons; move only lone pairs and pi electrons, respect the octet, and rank the contributors.

Sources

1. McMurry, J. (2023). Drawing Chemical Structures. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Resonance. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Rules for Resonance Forms. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Drawing Resonance Forms. In Organic Chemistry (OpenStax). openstax.org
5. Flowers, P., et al. (2019). Formal Charges and Resonance. In Chemistry 2e (OpenStax). openstax.org
6. LibreTexts Chemistry. Drawing Chemical Structures. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Hardinger, S. A. Illustrated Glossary of Organic Chemistry. University of California, Los Angeles. chem.ucla.edu

The big picture

Functional groups are the vocabulary of organic chemistry. This lesson introduces the main groups so that, from now on, you can look at any structure and immediately predict how it behaves. Molecules that share a functional group react in similar ways regardless of the rest of the molecule, so learning about a dozen groups unlocks the chemistry of thousands of compounds.

A functional group is a specific arrangement of atoms that gives a molecule characteristic properties and reactions, like a standard plug that fits the same socket no matter which appliance it is attached to. This is the central organizing idea of the whole subject: molecules that share a functional group behave alike, no matter how large the rest of the molecule is. Learn the groups once and you can predict the chemistry of thousands of compounds.

Hydrocarbon groups (carbon and hydrogen only)

• Alkane: only C-C single bonds, as in ethane CH₃CH₃. It is saturated, meaning it holds the maximum number of hydrogens with no multiple bonds, and it is relatively unreactive.
• Alkene: contains a C=C double bond, as in ethene H₂C=CH₂. It is unsaturated because the double bond means fewer hydrogens than an alkane, and the pi bond is a reactive site.
• Alkyne: contains a C≡C triple bond, as in ethyne HC≡CH.
• Aromatic (arene): contains a benzene ring, a flat six-carbon ring with delocalized pi electrons that make it unusually stable.

Key idea: Hydrocarbon groups differ only in their carbon-carbon bonding, and the presence of a pi bond (alkene or alkyne) marks the reactive spot.

Groups with oxygen

• Alcohol: an -OH (hydroxyl) group on carbon, as in ethanol CH₃CH₂OH.
• Ether: an oxygen between two carbons, C-O-C, as in dimethyl ether CH₃OCH₃.
• Aldehyde: a C=O (carbonyl) with at least one H on the carbonyl carbon, at the end of a chain, written -CHO.
• Ketone: a carbonyl with carbons on both sides, C-CO-C, as in acetone (CH₃COCH₃).
• Carboxylic acid: a carbonyl bearing an -OH, written -COOH, as in acetic acid.
• Ester: a carboxylic-acid-derived group -COO-C, formed from an acid and an alcohol.

The carbonyl group is a carbon double-bonded to oxygen (C=O); it is the shared heart of aldehydes, ketones, carboxylic acids, esters, and amides. Because oxygen pulls electrons strongly, the carbonyl carbon carries a partial positive charge, which is why so many reactions attack it.

Key idea: Oxygen-containing groups mostly revolve around either an -OH (alcohols, acids) or a C=O carbonyl (aldehydes, ketones, acids, esters).

Groups with nitrogen or halogen

• Amine: nitrogen bonded to carbon, as in methylamine CH₃NH₂. Amines are basic because the nitrogen lone pair can accept a proton.
• Amide: a carbonyl bonded to nitrogen, -CO-N, the linkage that joins amino acids in proteins.
• Haloalkane (alkyl halide): a halogen on carbon, as in chloroethane CH₃CH₂Cl. The polar C-X bond makes it a workhorse in substitution and elimination reactions.
• Nitrile: a carbon triple-bonded to nitrogen, C≡N, as in acetonitrile CH₃CN.

Key idea: Nitrogen groups are typically basic (amines) or structural (amides), while alkyl halides are prized for their easily displaced halogen.

A summary you should memorize

Group	Key atoms	Example
Alcohol	C-OH	ethanol CH3CH2OH
Ether	C-O-C	dimethyl ether CH3OCH3
Aldehyde	-CHO	acetaldehyde CH3CHO
Ketone	C-CO-C	acetone CH3COCH3
Carboxylic acid	-COOH	acetic acid CH3COOH
Ester	-COO-C	methyl acetate CH3COOCH3
Amine	C-N	methylamine CH3NH2
Amide	-CO-N	acetamide CH3CONH2
Haloalkane	C-X	chloroethane CH3CH2Cl

Key idea: The oxidation state of a carbon rises as it bonds to more oxygens, which is why the series alcohol, aldehyde, carboxylic acid represents increasing oxidation of the same carbon.

Counting oxidation level: a ladder you can climb in both directions

That last idea deserves a rule you can apply mechanically. For any carbon, count its bonds to atoms more electronegative than carbon - oxygen, nitrogen, sulfur, halogen - and count a double bond as two. The total is that carbon's oxidation level, and it sorts the functional groups into rungs.

• Level 0: alkane. CH₃-CH₃, no bonds to heteroatoms.
• Level 1: alcohol, ether, alkyl halide, amine. One bond to a heteroatom, as in CH₃-CH₂-OH.
• Level 2: aldehyde, ketone, imine, acetal. Two bonds, as in CH₃-CHO.
• Level 3: carboxylic acid, ester, amide, acid chloride, nitrile. Three bonds, as in CH₃-COOH.
• Level 4: carbon dioxide and carbonates. Four bonds, the top of the ladder.

A reaction that raises the level is an oxidation and must be run with an oxidizing agent; one that lowers it is a reduction. Moving within a level - hydrolysing an ester to an acid, or converting an alcohol to an alkyl halide - is neither, and needs no redox reagent at all. That single check saves you from proposing impossible steps when you start designing syntheses.

Key idea: Count a carbon's bonds to heteroatoms to place it on the oxidation ladder; changing rungs requires a redox reagent, moving sideways does not.

Finding functional groups in the laboratory: infrared spectroscopy

Functional groups are not just a classification scheme; they are physically detectable. An infrared spectrometer shines a range of infrared frequencies through a sample, and a bond absorbs when the frequency matches its natural stretching or bending vibration. Because each functional group vibrates in a characteristic window, an IR spectrum is essentially a readout of which groups are present. Positions are quoted in wavenumbers, cm^-1, and higher wavenumber means a stiffer bond or lighter atoms.

Absorption (cm-1)	Bond	Appearance and meaning
3200-3600	O-H, alcohol	Strong and broad; the breadth comes from hydrogen bonding
2500-3300	O-H, carboxylic acid	Very broad, often sitting on top of the C-H region; unmistakable
3300-3500	N-H, amine or amide	Moderate and sharper than O-H; two peaks for a primary amine, one for a secondary
3300 (sharp)	≡C-H, terminal alkyne	Narrow spike, distinct from the broad O-H
3020-3100	=C-H, alkene or arene	Just above 3000; sp3 C-H sits just below at 2850-2960
2210-2260	C≡N nitrile	Sharp and medium; alkyne C≡C is nearby at 2100-2260 but weaker
1670-1780	C=O carbonyl	Very strong and narrow; the single most informative band in organic IR
1640-1680	C=C alkene	Weak to medium; easily missed in a symmetric alkene

The carbonyl window is narrow enough to distinguish the carbonyl-containing groups from one another: an acid chloride absorbs near 1800 cm^-1, an ester near 1735, an aldehyde near 1725, a ketone near 1715, a carboxylic acid near 1710, and an amide much lower, near 1650. The trend is not arbitrary. A neighbouring atom that donates electron density into the carbonyl by resonance weakens the C=O bond and lowers its frequency, and nitrogen donates far better than oxygen, which donates better than chlorine. The IR frequency is therefore a direct readout of how much the carbonyl has been tamed.

Worked example. An unknown of formula C₄H₈O₂ shows a very broad absorption running from about 2500 to 3300 cm^-1 and a strong sharp band at 1710 cm^-1. Work it in order. The broad low band is the carboxylic acid O-H, not an alcohol, because an alcohol O-H does not reach down to 2500. The strong 1710 band confirms a carbonyl in the acid range. Two oxygens in the formula are exactly what -COOH requires, leaving C₃H₇ as the rest of the molecule. The answer is butanoic acid, CH₃-CH₂-CH₂-COOH. Had the spectrum shown 1735 cm^-1 with no O-H at all, the same formula would have pointed to an ester such as ethyl acetate.

Key idea: Read an IR spectrum in a fixed order - look for O-H or N-H above 3000, then for a C=O near 1700, then use the exact carbonyl position to say which carbonyl group it is.

Why this matters

The carbonyl group (C=O) appears in aldehydes, ketones, carboxylic acids, esters, and amides, and its polarity drives much of their shared chemistry. The -OH group makes both alcohols and carboxylic acids capable of hydrogen bonding, raising their boiling points and their solubility in water. When you meet a new molecule, the first move is always the same: scan for functional groups, and let them tell you how the molecule will behave.

Key idea: Identifying the functional groups in a molecule is the first step of every problem, because the groups, not the carbon backbone, drive the chemistry.

Where people get stuck

• "Aldehydes and ketones are basically the same." Both contain a carbonyl, but an aldehyde has at least one H on the carbonyl carbon and sits at a chain end. That H makes aldehydes easier to oxidize and, being smaller, makes their carbonyl carbon more open to attack, so aldehydes are consistently more reactive than ketones.
• "An -OH always means an alcohol." An -OH on a carbonyl carbon is a carboxylic acid and an -OH on a benzene ring is a phenol. The three differ by roughly ten orders of magnitude in acidity, so the distinction is not cosmetic.
• "Ethers and esters are the same because both contain C-O-C." An ester also contains a C=O. That carbonyl is what makes an ester hydrolysable and an ether famously inert.
• "Amines and amides behave alike because both contain nitrogen." Amines are basic, with a pK_a near 10 for the conjugate acid. Amides are not appreciably basic at all, because the nitrogen lone pair is delocalized into the adjacent carbonyl.
• "A carbonyl is a carbonyl." The carbonyl-containing groups differ by orders of magnitude in reactivity, in the order acid chloride greater than anhydride, then aldehyde and ketone, then ester, then amide, then carboxylate. Both leaving-group ability and resonance donation change along the series, and IR frequency tracks it.
• "Converting an alcohol to an alkyl halide is a reduction." Both carbons sit at oxidation level 1, so it is neither oxidation nor reduction. Check the ladder before you reach for a redox reagent.
• "A broad peak near 3300 cm^-1 means an alcohol." Look at how far down it runs. An alcohol O-H stops around 3200; a carboxylic acid O-H smears all the way to 2500, and a terminal alkyne C-H is a narrow spike at 3300 rather than a broad hump.

Recap

• A functional group is a set of atoms that gives a molecule its characteristic reactions.
• Hydrocarbon groups (alkane, alkene, alkyne, arene) differ in carbon-carbon bonding; pi bonds are reactive.
• Oxygen groups center on -OH (alcohols, acids) or the C=O carbonyl (aldehydes, ketones, acids, esters).
• Nitrogen groups include basic amines and structural amides; alkyl halides carry a reactive C-X bond.
• Counting a carbon's bonds to heteroatoms gives its oxidation level, and only a change of level requires a redox reagent.
• Infrared spectroscopy detects functional groups directly: O-H and N-H above 3000 cm^-1, C=O near 1700 cm^-1, with the exact carbonyl position naming the group.
• Always identify functional groups first; they, not the backbone, predict behavior.

Sources

1. McMurry, J. (2023). Functional Groups. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Infrared Spectra of Some Common Functional Groups. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Interpreting Infrared Spectra. In Organic Chemistry (OpenStax). openstax.org
4. Flowers, P., et al. (2019). Aldehydes, Ketones, Carboxylic Acids, and Esters. In Chemistry 2e (OpenStax). openstax.org
5. Flowers, P., et al. (2019). Amines and Amides. In Chemistry 2e (OpenStax). openstax.org
6. LibreTexts Chemistry. Infrared Spectroscopy. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. National Institute of Standards and Technology. NIST Chemistry WebBook (searchable infrared spectra for named compounds). webbook.nist.gov

The big picture

Every organic compound has one systematic name that encodes its exact structure, so a chemist anywhere can read the name and draw the molecule. This lesson teaches the naming rules on alkanes, the simplest family. The same four-step logic, learned here, extends to every other functional group in later lessons.

Every organic compound has a systematic IUPAC name that any chemist can decode into a structure. IUPAC stands for the International Union of Pure and Applied Chemistry, the body that sets these rules. The system is built on a small set of steps applied in order, like following a recipe. Master them on alkanes and the same logic extends to every other family.

The stems that count carbons

The number of carbons in the main chain sets the name stem, and the ending -ane marks an alkane (all single bonds).

Carbons	Stem	Alkane
1	meth	methane
2	eth	ethane
3	prop	propane
4	but	butane
5	pent	pentane
6	hex	hexane
7	hept	heptane
8	oct	octane
9	non	nonane
10	dec	decane

Key idea: Memorize the first ten stems; they are the backbone of every organic name you will ever write.

The naming steps

1. Find the longest continuous carbon chain and name it as the parent. It may bend around corners; length, not straightness, is what counts.
2. Number the chain from the end that gives the substituents the lowest set of locants. A locant is simply the position number telling you which carbon a branch sits on, like a house number on a street.
3. Name each branch (substituent) as an alkyl group: a one-carbon branch is methyl, two is ethyl, three is propyl. A substituent is anything hanging off the main chain in place of a hydrogen.
4. Assemble the name: locant-substituent pieces first, in alphabetical order of substituent, then the parent. Use multiplier prefixes di, tri, tetra for repeats, but ignore those prefixes when alphabetizing.

Key idea: Longest chain sets the parent, lowest locants set the numbering, and substituents are listed alphabetically, ignoring di and tri prefixes.

Choosing the longest chain carefully

The longest chain is not always the one drawn horizontally. When two chains tie for length, pick the one with more substituents, because that gives the simpler set of branch names. Draw a light line tracing each candidate path and count; a branch you thought was a substituent may actually be part of a longer parent chain. This single habit prevents most beginner naming errors.

Key idea: Trace every path to find the true longest chain, and break ties by choosing the chain with the greater number of substituents.

The branched alkyl groups worth memorizing

Four small branched substituents appear so often that both their common and systematic names are worth knowing on sight.

Common name	Structure	Systematic (IUPAC 2013) name
isopropyl	(CH3)2CH-	propan-2-yl
isobutyl	(CH3)2CH-CH2-	2-methylpropyl
sec-butyl	CH3-CH2-CH(CH3)-	butan-2-yl
tert-butyl	(CH3)3C-	2-methylpropan-2-yl

Alphabetization treats these differently, which is a classic trap. The italicized structural prefixes sec- and tert- are ignored when alphabetizing, so sec-butyl files under "b". But "iso" is part of the word, so isopropyl files under "i", not "p". In a name containing an ethyl, an isopropyl, and a tert-butyl group, the citation order is tert-butyl, then ethyl, then isopropyl - that is, b, e, i.

Key idea: Ignore sec- and tert- when alphabetizing, but treat "iso" as part of the substituent name.

Primary, secondary, tertiary, quaternary

Nomenclature also gives you a vocabulary for how crowded a carbon is, and that vocabulary will do real work in every mechanism from Lesson 13 onward. Classify a carbon by how many other carbons it is attached to: a primary (1°) carbon touches one other carbon, a secondary (2°) carbon touches two, a tertiary (3°) carbon touches three, and a quaternary (4°) carbon touches four and therefore carries no hydrogen at all.

In 2-methylbutane, CH₃-CH(CH₃)-CH₂-CH₃, the three CH₃ groups are primary, the CH₂ is secondary, and the branch point CH is tertiary. In 2,2-dimethylpropane, (CH₃)₄C, the central carbon is quaternary. A hydrogen inherits the label of the carbon it sits on, so a tertiary hydrogen is one attached to a tertiary carbon.

Hold on to these labels. Whether a substitution goes by SN1 or SN2, whether an elimination gives the more or less substituted alkene, and how stable a carbocation or radical is all depend on exactly this count.

Key idea: A carbon is 1°, 2°, 3°, or 4° according to how many carbons it touches, and this single label predicts most of the reactivity you will study later.

Worked example

Consider CH₃-CH(CH₃)-CH₂-CH₂-CH₃. The longest chain has five carbons, so the parent is pentane. There is one methyl branch. Numbering from the left puts the methyl at carbon 2; numbering from the right would put it at carbon 4. Lower wins, so it is at position 2. The name is 2-methylpentane.

A trickier case: a six-carbon chain with a methyl at carbon 3 and a methyl at carbon 4 (numbered for lowest locants) is 3,4-dimethylhexane. Two identical branches take the prefix di, and both locants are listed and separated by a comma. If instead the branches were one methyl and one ethyl, they would be cited alphabetically (ethyl before methyl) regardless of position number, giving names like 4-ethyl-3-methylheptane. Careful, consistent application of these four steps handles the great majority of names you will encounter.

Key idea: Numbers are separated from words by hyphens and from each other by commas, and the whole name is written as one word ending in the parent alkane.

Breaking a tie: the first point of difference

When both numbering directions give the same number of substituents, compare the two locant sets term by term from the smallest and stop at the first point of difference. The set that is lower at that point wins outright, even if its later numbers are larger. So {2, 3, 5} beats {2, 4, 5}, because the first entries tie at 2 and the second entries differ with 3 below 4.

Work a full example. Take CH₃-CH(CH₃)-CH(CH₂CH₃)-CH₂-CH(CH₃)-CH₂-CH₃. Trace the longest chain: seven carbons, so the parent is heptane, and running out through the ethyl branch only reaches five, so it is not a longer path. Numbering from the left places substituents at 2 (methyl), 3 (ethyl), and 5 (methyl), giving the set {2, 3, 5}. Numbering from the right places them at 3, 5, and 6, giving {3, 5, 6}. The first entries already differ, 2 against 3, so we number from the left. Now cite alphabetically: ethyl before methyl. The name is 3-ethyl-2,5-dimethylheptane.

If the locant sets are still identical after that comparison, the final tie-break is to give the lower locant to whichever substituent is cited first alphabetically.

Key idea: Compare locant sets entry by entry and decide at the first difference; if still tied, favour the substituent that comes first alphabetically.

Rings: cycloalkanes

A ring takes the prefix cyclo- before the stem, so a six-carbon ring is cyclohexane. Three rules cover almost every case. If there is only one substituent, no locant is needed: methylcyclohexane, not 1-methylcyclohexane. If there are several, the carbon bearing the substituent cited first alphabetically becomes C1, and you then number around the ring in whichever direction gives the lower set of locants, so 1-ethyl-3-methylcyclohexane rather than 1-ethyl-5-methylcyclohexane.

The third rule catches people out: compare the ring against the chain and let the larger one be the parent. Methyl on cyclohexane is methylcyclohexane, but cyclopropane attached to a heptane chain is named as a cyclopropylheptane, with the ring demoted to a substituent because seven beats three.

Key idea: Rings take the cyclo- prefix, a lone substituent needs no locant, and whichever of ring or chain has more carbons becomes the parent.

Why anyone bothers, and what the rules are today

Systematic naming exists because ambiguity in chemistry is expensive. "Amyl alcohol" names eight different compounds; pentan-1-ol names exactly one. The current authority is the IUPAC Nomenclature of Organic Chemistry: Recommendations and Preferred Names 2013, usually called the Blue Book, which introduced the idea of a single preferred IUPAC name for regulatory and legal use while still permitting acceptable alternatives in general writing. That is why you will see both 2-methylpropan-2-yl and tert-butyl in current literature, and why the locant now sits inside the name (butan-2-ol rather than 2-butanol) in preferred style.

Names are not the only identifiers in use. A CAS Registry Number labels a substance uniquely but tells you nothing about its structure. SMILES strings encode structure compactly for software. The IUPAC International Chemical Identifier, InChI, published in 2005, is the open standard string that lets databases match the same compound across sources. A working chemist reads names and machines read InChI, and both describe the same molecule.

Key idea: IUPAC names exist to make one structure map to one name; the 2013 recommendations define preferred names, while InChI and SMILES do the same job for software.

Where people get stuck

• "The longest chain is whatever is drawn in a straight line." The parent chain can bend around corners. Trace every path before committing, because a "branch" is often part of a longer chain.
• "Always number from left to right." Number from whichever end gives the lower locant set. Direction is decided by the substituents, not by the page.
• "Alphabetize using the di and tri prefixes." Ignore multiplying prefixes, so diethyl files under e. But do alphabetize isopropyl under i, because "iso" is part of the name rather than a multiplier.
• "Two identical branches on the same carbon only need one locant." Every substituent gets its own locant even when they repeat, as in 2,2-dimethylbutane.
• "Lowest locants means the smallest first number, full stop." It means the lowest set compared at the first point of difference. {2, 3, 5} beats {2, 4, 5} even though both start at 2.
• "A ring is always the parent." Only when it has at least as many carbons as the competing chain. Otherwise the ring becomes a cycloalkyl substituent.
• "Primary and secondary are just descriptive words." They are the single best predictor of substitution and elimination behaviour, and you will use them constantly from Lesson 14 onward.

Recap

• The stem names the number of carbons and -ane marks a saturated alkane.
• Step 1: longest continuous chain is the parent; step 2: number for lowest locants; step 3: name substituents as alkyl groups; step 4: assemble alphabetically.
• Locants are position numbers; substituents are groups replacing a hydrogen on the chain.
• Ignore di, tri, tetra when alphabetizing, and separate numbers from words with hyphens and numbers from numbers with commas.
• Trace all paths to find the true longest chain, breaking ties by the greater number of substituents, then by the first point of difference in the locant set.
• Learn isopropyl, isobutyl, sec-butyl, and tert-butyl, and remember that sec- and tert- are skipped in alphabetizing while "iso" is not.
• Carbons are 1°, 2°, 3°, or 4° by how many carbons they touch, and this classification drives later mechanism choices.

Sources

1. McMurry, J. (2023). Naming Alkanes. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Alkyl Groups. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Alkanes and Alkane Isomers. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Naming Cycloalkanes. In Organic Chemistry (OpenStax). openstax.org
5. International Union of Pure and Applied Chemistry. Nomenclature and terminology (including the 2013 recommendations for organic chemistry). iupac.org
6. LibreTexts Chemistry. Naming Alkanes. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Favre, H. A., and Powell, W. H. (2014). Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013. Royal Society of Chemistry. find source ↗

The big picture

You already know how to name alkanes. This lesson shows how the same system stretches to alcohols, alkenes, ketones, acids, and more by swapping the ending and letting the main functional group control the numbering. Once you learn the priority order, you can name most molecules you will meet in this course.

Once you can name alkanes, naming other families is a matter of changing the ending and making sure the main functional group gets the lowest number. The IUPAC system uses a suffix, the ending of the name, to signal the principal group, much like a file extension tells you what kind of file you have.

Suffixes for common families

Family	Suffix	Example
Alkane	-ane	propane
Alkene	-ene	propene
Alkyne	-yne	propyne
Alcohol	-ol	propan-1-ol
Aldehyde	-al	propanal
Ketone	-one	propan-2-one
Carboxylic acid	-oic acid	propanoic acid

Key idea: Change the parent ending to match the principal functional group; the stem still counts carbons exactly as before.

Which group wins: the priority order

When a molecule has more than one functional group, only one can be the principal characteristic group, the one named by the suffix. The rest are cited as prefixes. From higher to lower priority for suffix status: carboxylic acid, then ester, then amide, then aldehyde, then ketone, then alcohol, then amine. A carbon-carbon double or triple bond is expressed in the parent ending (-ene or -yne) but ranks below these oxygen groups for controlling the numbering. So in a molecule with both an -OH and a C=C, the -OH wins the suffix and the double bond becomes an -en- infix.

Key idea: Rank the groups, give the highest-priority one the suffix and the lowest locant, and demote the others to prefixes.

Every group that loses the contest still has to appear in the name, and it does so with a prefix. Learning the prefix forms alongside the suffixes is what lets you name a molecule carrying three or four groups.

Priority	Group	Suffix if it wins	Prefix if it loses
1	Carboxylic acid	-oic acid	carboxy-
2	Ester	-oate	alkoxycarbonyl-
3	Amide	-amide	carbamoyl-
4	Nitrile	-nitrile	cyano-
5	Aldehyde	-al	oxo- or formyl-
6	Ketone	-one	oxo-
7	Alcohol	-ol	hydroxy-
8	Amine	-amine	amino-
9	Alkene / alkyne	-ene / -yne	alkenyl- / alkynyl-
-	Ether, halide	never a suffix	alkoxy-, fluoro-, chloro-, bromo-, iodo-

Ethers and halides never take a suffix at all, which is a useful simplification: a chlorine or a methoxy group is always just a prefix, no matter what else is present.

Numbering for the functional group

The main functional group now controls the numbering: choose the chain and direction that give the principal group the lowest locant, even ahead of branches. For a double bond or an -OH, you cite the locant of the group in the name.

• But-2-ene is a four-carbon chain with the double bond starting at carbon 2 (CH₃-CH=CH-CH₃).
• Propan-1-ol is a three-carbon chain with the -OH on carbon 1 (CH₃CH₂CH₂OH), while propan-2-ol has the -OH on carbon 2.
• Butan-2-one is a four-carbon ketone with the C=O at carbon 2 (CH₃-CO-CH₂-CH₃).

Key idea: The locant of the principal group is written directly into the name, so the reader knows exactly where the group sits.

Aldehydes and acids need no locant

Because an aldehyde (-CHO) and a carboxylic acid (-COOH) must sit at the end of a chain, their carbon is automatically carbon 1, so you usually do not write a locant for them. Propanal is unambiguous, and so is propanoic acid. This is a small but frequent shortcut worth remembering.

Key idea: Chain-terminal groups like aldehydes and carboxylic acids are always at carbon 1, so their locant is understood and omitted.

Worked example

Name CH₂=CH-CH₂-CH₂-OH. The parent chain has four carbons. Two functional features compete: a C=C and an -OH. The alcohol is the higher-priority group, so it takes the lowest locant and the -ol suffix. Numbering from the OH end puts the -OH at carbon 1 and the double bond starting at carbon 3. The name is but-3-en-1-ol.

Notice how the double bond keeps its own locant (3) but is now written as an "-en-" infix before the "-ol" suffix. This layered structure, parent + double-bond position + principal-group suffix, is how IUPAC packs a full three-dimensional description into a single readable name.

Key idea: A name can carry a parent chain length, a double-bond locant, and a principal-group suffix all at once, read left to right as stem, then -en-, then the suffix.

Two-word names: esters and amides

Esters break the one-word pattern. The alkyl group that came from the alcohol is written first as a separate word, then the acyl portion takes the -oate ending. So CH₃-COO-CH₂CH₃ is ethyl ethanoate, still widely called ethyl acetate, and CH₃CH₂-COO-CH₃ is methyl propanoate. Read the name backwards to draw the molecule: the second word tells you the acid half, the first word tells you what is hanging off its oxygen.

Amides use -amide, and any substituent on nitrogen is flagged with the italic locant N instead of a number, because nitrogen has no place in the carbon numbering. CH₃CH₂-CO-NH-CH₃ is N-methylpropanamide, and HCO-N(CH₃)₂ is N,N-dimethylformamide, the common polar aprotic solvent usually abbreviated DMF.

Key idea: An ester name is two words, alcohol part then acid part as -oate; an amide names substituents on nitrogen with an italic N locant.

Naming benzene rings

Aromatic compounds keep a handful of names that predate the system and were too entrenched to discard. IUPAC retains benzene, toluene, phenol, aniline, benzaldehyde, and benzoic acid as preferred names. Others are merely acceptable: styrene has the preferred name ethenylbenzene, and anisole is systematically methoxybenzene.

When a benzene ring is a substituent rather than the parent it is called phenyl, C₆H₅-, and the ring-plus-CH₂ unit C₆H₅CH₂- is benzyl. Confusing the two is a common slip, and it matters, because benzyl positions are famously reactive while phenyl positions are not.

For a ring carrying two substituents, positions can be given either as numbers or with the classical prefixes: 1,2 is ortho, 1,3 is meta, and 1,4 is para. So 1,4-dichlorobenzene and para-dichlorobenzene name the same compound. These three words are not just historical decoration; they are the vocabulary you will need in Lesson 17, where substituents on a ring direct an incoming group specifically to the ortho and para positions or specifically to the meta position.

Key idea: A benzene ring as a substituent is phenyl, ring-plus-CH₂ is benzyl, and 1,2-, 1,3-, and 1,4-disubstitution are ortho, meta, and para.

Naming the geometry of an alkene: E and Z

Saying an alkene is "cis" or "trans" works only when each double-bond carbon carries one hydrogen and one other group. As soon as a third or fourth substituent appears, the words break down, and IUPAC replaces them with a system that never fails.

Look at each alkene carbon separately and rank its two attached groups by Cahn-Ingold-Prelog priority, which at this stage means simply: higher atomic number wins at the first point of attachment, and if those tie, move outward and compare the next set of atoms. Then ask where the two winners sit. If the two higher-priority groups are on the same side of the double bond, the alkene is Z (from the German zusammen, together). If they are on opposite sides, it is E (entgegen, opposite).

Worked example. Take CH₃-CH=C(Cl)-CH₂CH₃, which is 3-chloropent-2-ene. On C2 the two groups are CH₃ and H, so carbon beats hydrogen and CH₃ is the winner. On C3 the two groups are Cl and CH₂CH₃, and chlorine at atomic number 17 easily beats carbon at 6, so Cl is the winner. Now look at the drawing: if the methyl and the chlorine are drawn on the same side of the C=C, the compound is (Z)-3-chloropent-2-ene; if they are on opposite sides it is the E isomer. Notice that the answer would be unreachable with cis and trans, because neither carbon carries a hydrogen pair to compare.

Key idea: Rank the two groups on each alkene carbon by atomic number; same side means Z, opposite sides means E, and this works when cis and trans cannot.

Where people get stuck

• "The double bond always controls numbering." An alcohol, ketone, or acid outranks a C=C for both the suffix and the lowest locant. The double bond then survives only as an -en- infix.
• "You always write a number for the functional group." Aldehydes and carboxylic acids must be terminal, so the locant 1 is understood and omitted. It is propanal, never propan-1-al.
• "Every group in the molecule gets a suffix." Exactly one group takes the suffix. All the rest become prefixes such as hydroxy-, oxo-, amino-, or chloro-.
• "Propan-1-ol and propan-2-ol are basically the same." They are different constitutional isomers, with different boiling points, different oxidation products, and, in Lesson 14, entirely different substitution behaviour.
• "Cis is the same as Z." Often, but not always. Cis and trans describe positions of identical or obvious groups; E and Z describe CIP priorities, and a high-priority group can sit cis while still producing an E label.
• "Phenyl and benzyl are interchangeable." Phenyl is C₆H₅-, attached through the ring itself. Benzyl is C₆H₅CH₂-, attached through an extra carbon, and that carbon is a hotspot for radical and cationic chemistry.
• "An ester is one word like every other name." It is two: the alkyl group from the alcohol, then the acid part as -oate. Reading it in the wrong order swaps which half of the molecule is which.

Recap

• Suffixes signal the principal group: -ene, -yne, -ol, -al, -one, -oic acid, -oate, -amide, -nitrile.
• Priority order for the suffix: carboxylic acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine, then C=C or triple bond; ethers and halides are always prefixes.
• Number the chain to give the principal group the lowest locant, and write that locant in the name.
• Aldehydes and carboxylic acids are terminal, so their locant is omitted.
• Names can layer a chain length, a double-bond position, and a suffix, as in but-3-en-1-ol.
• Esters take two-word names and amides use an italic N locant for substituents on nitrogen.
• Benzene rings keep retained names, appear as phenyl or benzyl substituents, and use ortho, meta, and para for 1,2-, 1,3-, and 1,4-disubstitution.
• Alkene geometry is specified by E and Z from CIP priorities, which succeeds where cis and trans fail.

Sources

1. McMurry, J. (2023). Naming Alkenes. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Alkene Stereochemistry and the E,Z Designation. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Naming Alcohols and Phenols. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Naming Aldehydes and Ketones. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). Naming Aromatic Compounds. In Organic Chemistry (OpenStax). openstax.org
6. LibreTexts Chemistry. Sequence Rules: The E,Z Designation. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Favre, H. A., and Powell, W. H. (2014). Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, Chapters 4 and 5. Royal Society of Chemistry. find source ↗

The big picture

The same handful of atoms can be assembled into different molecules, and telling those molecules apart is a core organic skill. This lesson covers constitutional isomers (same formula, different connectivity) and a fast counting trick, the degree of unsaturation, that tells you how many rings or multiple bonds a formula must contain before you draw anything.

Isomers are different compounds that share the same molecular formula, like two different words built from the same letters. Because carbon links in so many ways, isomerism is everywhere in organic chemistry, and telling isomers apart is a basic skill. The broadest split is between constitutional isomers and stereoisomers.

Constitutional isomers

Constitutional isomers (also called structural isomers) have the same atoms connected in a different order. Connectivity, meaning which atom is bonded to which, is what differs. They are genuinely different molecules with different names and often very different properties. The formula C₄H₁₀ has two constitutional isomers: butane, an unbranched chain, and 2-methylpropane (isobutane), a branched one. The formula C₂H₆O has two: ethanol (CH₃CH₂OH, an alcohol) and dimethyl ether (CH₃OCH₃, an ether), which behave nothing alike even though every atom count matches.

Key idea: Constitutional isomers share a formula but differ in the order the atoms are connected, so they are distinct compounds with distinct names.

The map of isomerism

It is worth fixing the whole taxonomy now, because the next two lessons live inside it. Start with two molecules of identical molecular formula and ask a single question: is the connectivity the same?

• Different connectivity gives constitutional isomers. Ethanol and dimethyl ether, butane and 2-methylpropane. Different compounds, different names, different boiling points.
• Same connectivity but a different arrangement in space gives stereoisomers. These divide again. Enantiomers are non-superimposable mirror images. Diastereomers are stereoisomers that are not mirror images, and cis-trans alkene pairs are the familiar example.

Two structures that differ only by rotation about a single bond are not isomers at all; they are conformers of one compound, and Lesson 8 takes them up. The test is whether you would have to break a bond to interconvert them: if a twist suffices, it is the same molecule.

Key idea: Same formula plus different connectivity gives constitutional isomers; same connectivity plus different spatial arrangement gives stereoisomers; free rotation gives only conformers of one compound.

Counting isomers without missing any

Drawing "all the isomers" of a formula is a task that rewards a system and punishes guessing. Work downward by chain length. Draw the longest unbranched chain first. Then shorten it by one carbon and place the leftover carbon as a methyl branch at every distinct position. Then shorten again and place two branches, and so on. Check each candidate against the ones already drawn by naming it: two drawings with the same IUPAC name are the same compound, however different they look on paper.

For C₅H₁₂ this gives pentane, then 2-methylbutane, then 2,2-dimethylpropane, and there is no fourth. (3-methylbutane is simply 2-methylbutane numbered from the wrong end, which is exactly the trap the naming check catches.) For C₆H₁₄ the same procedure yields five.

The count grows brutally fast: 2 for C₄H₁₀, 3 for C₅H₁₂, 5 for C₆H₁₄, 9 for C₇H₁₆, 18 for C₈H₁₈, 75 for C₁₀H₂₂, and 366,319 for C₂₀H₄₂. That explosion is the reason organic chemistry needed a systematic naming scheme in the first place, and it is also why the analytical tools below matter so much.

Key idea: Generate isomers by shortening the parent chain step by step and branching systematically, then name each one to weed out duplicates.

The degree of unsaturation

Before drawing isomers it helps to know how many rings or multiple bonds a formula requires. The degree of unsaturation (also called the index of hydrogen deficiency) counts them. It works because each ring and each pi bond removes two hydrogens compared with a fully saturated chain. A saturated molecule has the most hydrogens possible, so any shortfall in hydrogens signals a ring or a multiple bond. For a compound of carbon and hydrogen only, C_nH_m:

degrees of unsaturation = (2n + 2 - m) ÷ 2

Worked example. For C₄H₈: n = 4 and m = 8, so (2×4 + 2 - 8) ÷ 2 = (10 - 8) ÷ 2 = 1. One degree of unsaturation means the molecule has exactly one ring OR one double bond. So C₄H₈ could be but-1-ene (a double bond) or cyclobutane (a four-membered ring), among others.

Key idea: The formula (2n + 2 - m) / 2 turns a molecular formula into a count of rings plus pi bonds before you draw a single structure.

Handling nitrogen, oxygen, and halogens

The counting rule extends to other atoms with two small adjustments. Oxygen (and sulfur) do not change the count at all, because inserting a two-bond atom into a chain adds no net hydrogen change. Each halogen counts like a hydrogen, so add halogens to m. Each nitrogen adds one to the saturated hydrogen count. The general formula, with C carbons, H hydrogens, N nitrogens, and X halogens, is (2C + 2 + N - H - X) / 2. Oxygen simply does not appear.

Worked example. For C₄H₅N: (2×4 + 2 + 1 - 5 - 0) / 2 = (11 - 5) / 2 = 3. Three degrees suggest, for example, a nitrile (which alone is two, since a triple bond counts twice) plus one more ring or double bond.

Key idea: Oxygen is ignored, halogens count as hydrogens, and each nitrogen adds one to the top of the formula.

What each degree can be

Each degree of unsaturation could be a ring or a double bond; a triple bond counts as two degrees because it is two pi bonds. A benzene ring counts as four degrees: one for the ring and three for its three double bonds. Computing this number first tells you immediately whether to expect a ring, a double bond, or a fully saturated skeleton, which narrows the isomers you must consider and is a favorite first step on exams.

Key idea: A ring or a double bond is one degree, a triple bond is two, and a benzene ring is four, so the total steers you toward the right skeleton.

Choosing between the isomers: what proton NMR tells you

Once the degree of unsaturation has narrowed the field, proton nuclear magnetic resonance decides between the survivors. A ¹H NMR spectrum gives three independent pieces of information, and a structure determination is nothing more than reading all three in order.

• Number of signals tells you how many chemically distinct kinds of hydrogen the molecule has. Equivalent hydrogens give one signal no matter how many there are.
• Integration - the area under each signal - gives the ratio of hydrogens in each set. Scale the ratio up until it matches the molecular formula.
• Chemical shift, quoted in parts per million (ppm) relative to tetramethylsilane at 0, tells you what each set of hydrogens is attached to. Electron-withdrawing neighbours deshield a proton and push it downfield to higher ppm.
• Multiplicity - how many lines a signal is split into - tells you how many hydrogens sit on the neighbouring carbons, by the n + 1 rule: n equivalent neighbours split a signal into n + 1 lines. So a CH₂ next to a CH₃ appears as a quartet, and that CH₃ appears as a triplet.

Shift (ppm)	Type of hydrogen
0.9-1.8	Ordinary alkyl C-H: CH3 near 0.9, CH2 near 1.3
2.0-2.6	H on a carbon next to a C=O, or on a benzylic carbon
3.3-4.0	H on a carbon bearing an O or N (ether, alcohol, amine)
4.0-4.5	H on the O-CH2 of an ester
4.5-6.5	Vinyl H on an alkene carbon
6.5-8.0	Aromatic H
9.5-10.5	Aldehyde H
10-13	Carboxylic acid O-H, broad

Hydrogens on oxygen or nitrogen are the exception to almost everything: they appear anywhere from 1 to 5 ppm, are usually broad, normally show no splitting because they exchange rapidly, and vanish from the spectrum if the sample is shaken with D₂O. A signal that disappears on a D₂O shake is an O-H or N-H, and that trick alone often settles a structure.

Key idea: Signals count distinct hydrogen environments, integration counts how many are in each, shift says what they are attached to, and the n + 1 rule counts their neighbours.

Worked example: from a formula to a structure

An unknown liquid has the molecular formula C₄H₈O₂. Its infrared spectrum shows a strong sharp band at 1735 cm^-1 and no absorption at all above 3100 cm^-1. Its ¹H NMR spectrum shows three signals: a quartet at 4.12 ppm integrating for 2H, a singlet at 2.04 ppm integrating for 3H, and a triplet at 1.26 ppm integrating for 3H. Identify it.

1. Degree of unsaturation. Oxygen is ignored, so (2×4 + 2 - 8) / 2 = 1. Exactly one ring or pi bond.
2. Infrared. A strong band at 1735 cm^-1 is a carbonyl, and that accounts for the single degree of unsaturation - so there is no ring. The absence of anything above 3100 rules out an O-H and an N-H, so this is not a carboxylic acid or an alcohol. The position 1735 rather than 1710 or 1715 points specifically at an ester.
3. Integration. 2 + 3 + 3 = 8 hydrogens, matching the formula exactly, so no signal is hidden and no set is being counted twice.
4. Splitting. A quartet and a triplet, integrating 2H and 3H, is the signature of an isolated ethyl group: the CH₂ is split into four by the three hydrogens of the CH₃, and the CH₃ is split into three by the two hydrogens of the CH₂. The remaining 3H singlet has no neighbouring hydrogens at all, so it is a methyl attached to something with no C-H next to it.
5. Shift. The CH₂ at 4.12 ppm is far downfield, in the ester O-CH₂ window, so the ethyl group is bonded through oxygen. The isolated methyl at 2.04 ppm sits in the "next to a carbonyl" window, so it is CH₃-CO-.
6. Assemble. CH₃-CO- plus -O-CH₂-CH₃ gives CH₃-COO-CH₂-CH₃, ethyl ethanoate (ethyl acetate), C₄H₈O₂. Everything checks.

Test the reasoning against the main competitor. The isomer methyl propanoate, CH₃CH₂-COO-CH₃, has the same formula and a similar IR band, but its NMR would place the isolated 3H singlet near 3.7 ppm (a methyl on oxygen) and its quartet near 2.3 ppm (a CH₂ next to a carbonyl). The observed shifts are the other way round, so the data distinguish the two cleanly. This is the whole method: the formula sets the budget, IR names the functional group, and NMR assembles the pieces.

Key idea: Run structure problems in a fixed order - degree of unsaturation, then IR for the functional group, then NMR integration, splitting, and shift - and check your answer against the isomer you did not choose.

Where people get stuck

• "Constitutional isomers are just the same molecule drawn differently." They have different connectivity and are different compounds. Two drawings that differ only by rotation are conformers of one compound, not isomers.
• "Oxygen changes the degree of unsaturation." A two-bond atom inserted into a chain changes nothing. Leave oxygen and sulfur out of the calculation entirely.
• "A triple bond is one degree of unsaturation." It is two pi bonds and therefore two degrees. A nitrile is two, and a benzene ring is four.
• "Zero degrees means I made an arithmetic error." It means the molecule is a saturated acyclic compound, which is a perfectly good answer.
• "Integration gives the actual number of hydrogens." It gives a ratio. Scale it against the molecular formula: a 2:3:3 pattern in a C₄H₈O₂ compound is 2H, 3H, 3H, but the same ratio in a C₈H₁₆O₄ compound would be 4H, 6H, 6H.
• "A signal splits according to the hydrogens on its own carbon." It splits according to the hydrogens on the neighbouring carbons. Equivalent hydrogens on the same carbon do not split each other.
• "An -OH proton should appear as a clean multiplet." It normally appears broad and unsplit because it exchanges rapidly between molecules, and it disappears on shaking with D₂O, which is how you confirm it.

Recap

• Isomers share a molecular formula; constitutional isomers differ in how the atoms are connected, stereoisomers only in their arrangement in space.
• Generate isomers systematically by shortening the parent chain and branching, then name each candidate to eliminate duplicates.
• The degree of unsaturation counts rings plus pi bonds: (2n + 2 - m) / 2 for hydrocarbons.
• With heteroatoms, use (2C + 2 + N - H - X) / 2; oxygen is ignored and halogens count like hydrogen.
• A ring or double bond is one degree, a triple bond is two, and benzene is four.
• Proton NMR supplies the number of environments, their relative populations by integration, their surroundings by chemical shift, and their neighbours by the n + 1 rule.
• Solve structures in a fixed order: formula, degree of unsaturation, IR, then NMR, and always test the answer against the nearest rejected isomer.

Sources

1. McMurry, J. (2023). Calculating the Degree of Unsaturation. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). A Review of Isomerism. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Integration of 1H NMR Absorptions: Proton Counting. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Chemical Shifts in 1H NMR Spectroscopy. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). Uses of 1H NMR Spectroscopy. In Organic Chemistry (OpenStax). openstax.org
6. LibreTexts Chemistry. Spin-Spin Splitting in 1H NMR Spectra. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. National Institute of Standards and Technology. NIST Chemistry WebBook: ethyl acetate, infrared and mass spectra. webbook.nist.gov

The big picture

Molecules with the same connectivity can still differ in three-dimensional shape. Some of those differences vanish by a simple twist around a single bond (conformations), while others are locked in place and create genuinely different compounds (cis-trans isomers). This lesson teaches you to tell a passing shape from a permanent one, a distinction that matters from fats in your diet to drug design.

Two molecules can share the same connectivity yet still differ in three-dimensional arrangement. Some of those differences vanish by simple rotation (conformations); others are locked in and make truly different compounds (configurational isomers). Telling these apart is essential.

Conformations: rotating around single bonds

A conformation is a shape a molecule takes by rotating around its single (sigma) bonds, like turning one end of a twist-tie while holding the other. Because sigma bonds rotate freely, conformations interconvert constantly and are not separate compounds. In ethane (CH₃-CH₃), rotating one CH₃ relative to the other passes through two limiting shapes.

In the staggered conformation the hydrogens on the front carbon sit in the gaps between those on the back carbon; this is lowest in energy. In the eclipsed conformation the front and back hydrogens line up directly, which raises the energy by about 12 kJ/mol because of torsional strain, the resistance that arises when bonds on adjacent atoms are forced to line up. Ethane spends most of its time near the staggered arrangement but is never frozen there.

Key idea: Conformations arise from free rotation around single bonds, so they are fleeting shapes of one compound, not different compounds.

Seeing conformations: the Newman projection

Chemists visualize conformations with a Newman projection, a way of drawing the molecule looking straight down one carbon-carbon bond. The front carbon is drawn as a dot with three bonds meeting at it, and the back carbon as a circle with three bonds emerging from behind. Staggered means the front and back bonds are offset by 60 degrees; eclipsed means they overlap. You do not need to draw one to reason about it: just remember that offset bonds (staggered) are comfortable and overlapping bonds (eclipsed) are strained.

Key idea: A Newman projection is the standard tool for comparing conformations, and offset (staggered) is always lower in energy than overlapping (eclipsed).

Butane adds gauche and anti

Butane (CH₃CH₂CH₂CH₃) has a bulkier group on each end, so rotation around the central C-C bond gives named staggered forms. The anti conformation places the two methyl groups 180 degrees apart and is the most stable. A gauche conformation places them 60 degrees apart; it is still staggered but slightly higher in energy because the two methyls crowd each other, an effect called steric strain (the strain from bulky groups bumping into one another). The eclipsed forms are the highest-energy points along the rotation.

Key idea: Among staggered forms, anti (groups far apart) beats gauche (groups crowded), because steric strain penalizes bulky groups that are close.

The butane energy profile, with numbers

Rotate the central C-C bond of butane through a full 360 degrees and the energy traces a repeating landscape with four named stopping points. Taking the anti conformation as the zero point:

Dihedral angle	Conformation	Relative energy	Source of strain
180°	anti (staggered)	0 kJ/mol	none - the global minimum
120°	eclipsed, CH3 with H	about 16 kJ/mol	torsional plus modest steric
60°	gauche (staggered)	about 3.8 kJ/mol	steric crowding of the two methyls
0°	fully eclipsed, CH3 with CH3	about 19 kJ/mol	torsional plus severe steric - the global maximum

Put those numbers in context. Room-temperature thermal energy, RT, is about 2.5 kJ/mol, and a barrier of 19 kJ/mol is crossed millions of times per second. That is exactly why conformers cannot be isolated in a bottle, and it is also why the small 3.8 kJ/mol anti-gauche gap still matters: it does not stop the molecule from visiting gauche, but it does mean that at equilibrium the anti form is the more populated one. Roughly speaking, a gap of 3.8 kJ/mol corresponds to a population ratio of a few to one at 25 degrees Celsius, and since there are two equivalent gauche wells but only one anti well, real butane sits anti about 70 percent of the time.

Ethane's own 12 kJ/mol barrier looks like a settled textbook fact, but its origin is genuinely contested. The traditional explanation is steric or Pauli repulsion between eclipsing C-H bonds. A widely cited 2001 analysis by Pophristic and Goodman in Nature argued instead that the dominant term is hyperconjugation, the stabilizing donation of electron density from a filled C-H sigma orbital into the empty sigma* antibonding orbital of an anti-periplanar C-H bond, an interaction that is optimal in the staggered arrangement. Later energy-decomposition studies pushed back, arguing that the answer depends on how one partitions the energy and that Pauli repulsion remains the leading term. The honest summary for a first course is that both effects are present, the staggered preference is not in doubt, and the relative weighting is still argued over in the physical-organic literature.

Key idea: Butane's rotational landscape runs from 0 kJ/mol (anti) to about 19 kJ/mol (fully eclipsed), barriers this small are crossed constantly at room temperature, and the deeper origin of the ethane barrier is still debated.

Configurational isomers: locked arrangements

By contrast, a configurational isomer cannot interconvert without breaking a bond. The most important example here is cis-trans isomerism across a C=C double bond. Because a pi bond prevents rotation, the groups on each end of the double bond are held in fixed relative positions.

• Cis: the two like (or higher-priority) groups are on the same side of the double bond.
• Trans: they are on opposite sides.

Consider but-2-ene, CH₃-CH=CH-CH₃. In cis-but-2-ene both methyl groups are on the same side; in trans-but-2-ene they are on opposite sides. These are different compounds with different melting and boiling points, and one cannot become the other unless the pi bond is broken. Cis-trans isomerism requires that each double-bond carbon carry two different groups; if either carbon has two identical groups, no cis-trans distinction exists. This locked geometry is why fats described as cis or trans behave so differently in the body: the same atoms, held in a different fixed shape, are effectively different molecules.

Key idea: A pi bond blocks rotation, so cis and trans alkenes are separate compounds that can interconvert only by breaking the double bond.

Measuring the difference: heats of hydrogenation

How much more stable is trans than cis? The experiment that answers it is elegant. Hydrogenate several alkene isomers and each gives the identical alkane, so any difference in the heat released must reflect a difference in the starting alkenes. For the C₄H₈ series, all of which give butane:

• but-1-ene releases about 127 kJ/mol - the least stable
• cis-but-2-ene releases about 120 kJ/mol
• trans-but-2-ene releases about 115 kJ/mol - the most stable

Less heat released means the alkene started lower down, so trans-but-2-ene is about 5 kJ/mol more stable than cis and about 12 kJ/mol more stable than the terminal isomer. Two effects explain the ordering. The trans arrangement keeps the two methyl groups apart while cis forces them into each other, which is straightforward steric strain. And a more substituted double bond is stabilized by hyperconjugation from the adjacent C-H bonds into the pi system, which is why both internal isomers beat the terminal one.

Keep this ordering. It is the entire reason for Zaitsev's rule in Lesson 15, where an elimination will preferentially produce the more substituted alkene.

Key idea: Heats of hydrogenation put numbers on alkene stability - more substituted beats less substituted, and trans beats cis by roughly 5 kJ/mol.

Cis and trans without a double bond: rings

Rotation is blocked in a ring just as it is across a pi bond, so rings show cis-trans isomerism too. In 1,2-dimethylcyclohexane the two methyls can sit on the same face of the ring (cis) or on opposite faces (trans), and no amount of twisting converts one into the other. They are different compounds with different boiling points.

This is more than a curiosity. In Lesson 12 you will see that the ring itself flips between chair conformations, and the cis and trans isomers respond to that flip in different ways, which decides whether both methyls can occupy the roomy equatorial positions. In Lesson 15 the same geometry decides whether an E2 elimination is possible at all, because that reaction demands a specific anti-periplanar alignment that only one ring isomer can supply.

Key idea: A ring blocks rotation, so cis and trans substituents on a ring are configurational isomers, and their fixed relationship controls both conformation and reactivity.

Why locked geometry matters outside the classroom

The cis-trans distinction is not academic. A cis double bond puts a permanent kink of roughly 30 degrees into an otherwise straight hydrocarbon chain. That kink is why the cis-unsaturated fatty acids of vegetable and fish oils cannot stack neatly and are liquid at room temperature, while saturated chains pack tightly and are solid. Industrial partial hydrogenation converts some cis double bonds to trans, and a trans chain is nearly as straight as a saturated one, which is precisely why trans fats behave like saturated fat in the body and why many regulators have restricted them.

The same geometry drives vision. The retina's light-sensing pigment carries 11-cis-retinal bound to the protein opsin. Absorbing a photon isomerizes that one double bond to all-trans in a few hundred femtoseconds, the change of shape strains the protein, and the resulting conformational change begins the signal that reaches the brain. One double bond flipping from cis to trans is the first step of sight.

Key idea: A cis double bond kinks a chain and a trans one does not, which is the difference between an oil and a fat, and the flip between them is the primary event in vision.

Where people get stuck

• "Conformations are different compounds you can isolate." With barriers near 19 kJ/mol against thermal energy near 2.5 kJ/mol, conformers interconvert billions of times per second and cannot be bottled separately.
• "Cis and trans alkenes are just conformations." They are configurational isomers. Interconverting them requires breaking the pi bond, which costs roughly 270 kJ/mol.
• "Eclipsed is lower in energy because the atoms line up neatly." Lining bonds up raises the energy. Staggered is the minimum, whether the ultimate cause is hyperconjugation, Pauli repulsion, or both.
• "Every alkene has cis and trans forms." Only if each double-bond carbon carries two different groups. 2-methylbut-2-ene has no cis-trans pair on one of its carbons, so use E and Z instead.
• "Gauche is a strained, high-energy form." Gauche is a staggered minimum, only about 3.8 kJ/mol above anti. The high-energy points are the eclipsed maxima at 16 and 19 kJ/mol.
• "Rings cannot show cis-trans isomerism because there is no double bond." A ring blocks rotation just as effectively as a pi bond, so cis- and trans-1,2-dimethylcyclohexane are genuinely different compounds.
• "Trans is always more stable, so cis alkenes are rare." Trans is more stable in the free molecule, but enzymes routinely build cis double bonds on purpose, and nearly every unsaturated fatty acid in your cell membranes is cis.

Recap

• Conformations come from rotation around single bonds and are fleeting shapes of one compound.
• Staggered conformations are lower in energy than eclipsed ones due to torsional strain, about 12 kJ/mol in ethane.
• Newman projections show conformations; in butane, anti is the minimum, gauche costs about 3.8 kJ/mol, and the fully eclipsed maximum is about 19 kJ/mol.
• Whether the ethane barrier is mainly hyperconjugation or mainly Pauli repulsion is still argued over, though the staggered preference itself is not in doubt.
• Cis-trans (configurational) isomers are locked by a pi bond or by a ring and are genuinely different compounds.
• Heats of hydrogenation show trans-but-2-ene below cis by about 5 kJ/mol and both below but-1-ene, which is the basis of Zaitsev's rule later.
• A cis double bond kinks a chain, which is why unsaturated oils are liquid, trans fats are not, and the cis-to-trans flip of retinal starts vision.

Sources

1. McMurry, J. (2023). Conformations of Ethane. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Conformations of Other Alkanes. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Cis-Trans Isomerism in Alkenes. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Stability of Alkenes (heats of hydrogenation). In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). Cis-Trans Isomerism in Cycloalkanes. In Organic Chemistry (OpenStax). openstax.org
6. Pophristic, V., and Goodman, L. (2001). Hyperconjugation not steric repulsion leads to the staggered structure of ethane. Nature, 411(6837), 565-568. pubmed.ncbi.nlm.nih.gov
7. LibreTexts Chemistry. Conformations of Ethane. Organic Chemistry (Morsch et al.). chem.libretexts.org

The big picture

Some molecules come in left-handed and right-handed versions that are mirror images yet cannot be laid on top of each other, exactly like your two hands. This lesson explains this handedness, called chirality, and teaches the standard method for labeling each version R or S. Chirality matters enormously because living systems, including the targets of most drugs, respond to one hand and not the other.

Some molecules exist as a pair that relate like your left and right hands: mirror images that cannot be superimposed. This property, chirality (from the Greek for "hand"), is one of the most important ideas in organic chemistry because living systems are exquisitely sensitive to it.

Stereocenters and enantiomers

A carbon bonded to four different groups is a stereocenter (chirality center). A molecule with a single stereocenter is chiral: it and its mirror image are two distinct compounds called enantiomers, which are mirror-image isomers that cannot be superimposed. Enantiomers are non-superimposable, just as a left hand will not fit into a right glove.

They share almost every physical property (same melting point, same boiling point) but differ in how they rotate polarized light and, crucially, in how they interact with other chiral molecules such as enzymes. A quick test for chirality: if a molecule has an internal mirror plane of symmetry, it is achiral; a lone stereocenter with four different groups almost always makes it chiral.

Key idea: A carbon with four different groups is a stereocenter, and a molecule with one stereocenter is chiral, existing as a pair of non-superimposable enantiomers.

Assigning R and S

To name which enantiomer you have, use the Cahn-Ingold-Prelog (CIP) priority rules:

1. Rank the four groups by priority. Higher atomic number at the first point of difference wins. So I > Br > Cl > O > N > C > H. If two groups tie at the first atom, move outward to the next atoms until they differ.
2. Point the lowest-priority group (usually H) away from you.
3. Trace a path from priority 1 to 2 to 3. If that path curves clockwise, the center is R (from Latin rectus, right). If it curves counterclockwise, it is S (sinister, left).

For a double-bonded atom in the CIP rules, treat the double bond as if the atom were bonded to two copies of its partner. So a C=O carbon counts as bonded to two oxygens for ranking purposes, a duplication trick that resolves most ties.

Key idea: Rank the four groups by atomic number, put the lowest away from you, and read 1 to 2 to 3 as clockwise (R) or counterclockwise (S).

Worked example

Consider bromochlorofluoromethane, CHFClBr, a carbon bonded to H, F, Cl, and Br. Rank by atomic number: Br (35) is priority 1, Cl (17) is 2, F (9) is 3, and H (1) is 4. Orient the molecule so H points away. If Br → Cl → F sweeps clockwise, the molecule is the R enantiomer; if counterclockwise, it is S. A shortcut: if the lowest-priority group happens to point toward you instead of away, determine the rotation as seen and then flip the answer (clockwise seen actually means S).

Key idea: When the lowest-priority group points toward you rather than away, read the rotation and then reverse your R or S assignment.

Worked example: assigning alanine, where the first atoms tie

CHFClBr is easy because all four first atoms differ. Real molecules usually do not cooperate. Take the amino acid alanine, CH₃-CH(NH₂)-COOH, and work the stereocenter properly.

1. Identify the stereocenter. The central carbon carries -NH₂, -COOH, -CH₃, and -H: four different groups, so it is a stereocenter and alanine is chiral.
2. Rank the first atoms. Nitrogen is 7, the two attached carbons are 6, and hydrogen is 1. So -NH₂ is priority 1 and -H is priority 4 immediately. The two carbons are tied and must be broken open.
3. Explore outward. For each tied atom, list the three atoms attached to it in decreasing order. The carboxyl carbon is attached to one doubled-bonded O, which the duplication rule counts twice, plus the -OH oxygen: its set is (O, O, O). The methyl carbon is attached to three hydrogens: its set is (H, H, H). Compare the sets element by element - O against H at the very first comparison - and the carboxyl wins outright.
4. Final ranking. -NH₂ (1), -COOH (2), -CH₃ (3), -H (4).
5. Read the rotation. Orient the molecule with the hydrogen pointing away, then trace NH₂ to COOH to CH₃. Counterclockwise gives S; clockwise gives R. The alanine that your ribosomes install into proteins is the S enantiomer, conventionally written L-alanine.

Try a second case where the tie breaks one atom further out. In 2-chlorobutane, CH₃-CHCl-CH₂-CH₃, chlorine is priority 1 and hydrogen is 4, and the ethyl and methyl groups are again both carbon. Expand: the ethyl carbon's set is (C, H, H) and the methyl carbon's is (H, H, H). Carbon beats hydrogen at the first comparison, so ethyl is priority 2 and methyl is priority 3.

Two rules keep this reliable. Always compare the highest members of two sets first, then the second-highest, then the third, and stop at the first difference - never total the atomic numbers. And explore outward one whole sphere at a time rather than chasing one branch far ahead of another.

Key idea: When first atoms tie, compare the ordered sets of atoms one sphere out, count a double bond as a duplicated atom, and decide at the first point of difference.

More than one stereocenter: diastereomers and meso compounds

A molecule can have several stereocenters. Stereoisomers that are not mirror images of each other are diastereomers; unlike enantiomers, they have different physical properties such as melting point and solubility. A special case is a meso compound, a molecule that contains stereocenters but is achiral overall because an internal mirror plane makes one half the reflection of the other. Meso tartaric acid is the classic example: it has two stereocenters yet is superimposable on its mirror image, so it is optically inactive. As a rule of thumb, a molecule with n stereocenters has at most 2 to the power n stereoisomers, but meso forms reduce that count.

Key idea: Non-mirror-image stereoisomers are diastereomers with different properties, and a meso compound is achiral despite having stereocenters because of internal symmetry.

Optical activity

Chiral molecules are optically active: a pure enantiomer rotates the plane of polarized light. One enantiomer rotates it clockwise (called + or dextrorotatory) and the other by an equal amount counterclockwise (called - or levorotatory). A 50/50 mixture of both enantiomers, called a racemic mixture, shows no net rotation because the two effects cancel. Note that + and - describe measured rotation and are not the same labels as R and S, which come from structure alone.

Key idea: A single enantiomer rotates polarized light, but an equal (racemic) mixture cancels out, and the +/- sign of rotation is independent of the R/S label.

Putting numbers on rotation: specific rotation and enantiomeric excess

The raw angle a polarimeter reads depends on how much sample the light passed through, so it is normalized into a specific rotation, a genuine physical constant of the compound:

[alpha] = alpha(observed) / (l × c)

where alpha(observed) is the measured rotation in degrees, l is the sample tube length in decimetres, and c is the concentration in grams per millilitre. Reported values also carry the temperature and the wavelength used, almost always the sodium D line at 589 nm.

Worked example. Dissolve 1.20 g of a compound in enough solvent to make 10.0 mL, so c = 0.120 g/mL, and place it in a 1.00 dm tube. The polarimeter reads +6.24 degrees. Then [alpha] = 6.24 / (1.00 × 0.120) = +52.0 degrees. That number, not the +6.24, is what goes in a paper.

Now the more useful quantity. Real samples are rarely a single pure enantiomer, and the enantiomeric excess measures how far from racemic they are:

ee (%) = ([alpha] observed / [alpha] pure) × 100

Suppose the pure R enantiomer has [alpha] = +52.0 degrees and a sample measures +31.2 degrees. Then ee = 31.2 / 52.0 = 0.60, an enantiomeric excess of 60 percent. Interpret it carefully: 60 percent ee does not mean 60 percent R. It means 60 percent of the sample is pure R and the remaining 40 percent is racemic, splitting evenly, so the true composition is 80 percent R and 20 percent S. A racemic mixture has 0 percent ee and a single pure enantiomer has 100 percent ee.

Key idea: Specific rotation normalizes for path length and concentration; enantiomeric excess converts an observed rotation into composition, where x percent ee means (50 + x/2) percent of the major enantiomer.

Why the pharmaceutical industry cares

Enantiomers are identical in every achiral environment, and an enzyme or a receptor is about as chiral an environment as exists. The two hands of a molecule therefore routinely do different things in a body.

The mildest illustration is smell: (R)-carvone smells of spearmint and (S)-carvone smells of caraway, because the two olfactory receptors that bind them are themselves chiral. The commercially important one is ibuprofen, sold for decades as a racemate even though only the (S) enantiomer inhibits cyclooxygenase - the body converts much of the (R) form to (S) in vivo, which is why the racemate works at all.

The cautionary one is thalidomide. Marketed in the late 1950s as a racemate for morning sickness, it caused severe birth defects in thousands of children. It is often said that the (R) enantiomer is the sedative and the (S) enantiomer the teratogen, and that separating them would have averted the disaster. The first part is broadly right; the second is not. Pharmacokinetic work on the separated enantiomers showed that thalidomide undergoes rapid chiral inversion at physiological pH, so administering pure (R) simply produces a racemic mixture in the blood within hours. The real lesson is subtler and more useful than the popular version: enantiopurity in the bottle does not guarantee enantiopurity at the receptor.

Regulators responded anyway. The United States Food and Drug Administration issued its policy statement on the development of new stereoisomeric drugs in 1992, requiring sponsors to characterize each enantiomer separately, and single-enantiomer approvals have dominated ever since; surveys of new small-molecule approvals over 2013 to 2022 find that a clear majority of chiral drugs now reach the market as a single enantiomer.

Key idea: Because receptors are chiral, enantiomers can differ in smell, potency, and toxicity, which is why single-enantiomer drugs are now the regulatory norm - though thalidomide shows that in vivo racemization can undo enantiopurity.

Where people get stuck

• "R always means it rotates light clockwise." R and S are read off the structure; plus and minus are measured on an instrument. There is no general correlation, and the same compound can even change sign with solvent or wavelength.
• "Any molecule with a stereocenter is chiral." A meso compound has stereocenters yet is achiral, because an internal mirror plane makes it superimposable on its own reflection.
• "Enantiomers differ in melting and boiling point." They are identical in every achiral property. Only diastereomers differ in melting point, solubility, and chromatographic behaviour - which is exactly how enantiomers are separated, by first converting them to diastereomers.
• "You rank CIP groups by the first atom only." If the first atoms tie, expand one sphere at a time, compare ordered sets, and count a double bond as a duplicated atom.
• "Add up the atomic numbers of the attached atoms to break a tie." Never. (O, H, H) beats (C, C, C) because the comparison stops at the first difference, even though the totals say otherwise.
• "60 percent ee means 60 percent of one enantiomer." It means 80 percent, because the remaining 40 percent is racemic and splits evenly between the two.
• "Selling the single good enantiomer always solves the problem." Thalidomide racemizes in the body, and ibuprofen's inactive enantiomer is converted to the active one. What happens after dosing matters as much as what is in the tablet.

Recap

• A stereocenter is a carbon with four different groups; a single stereocenter makes a molecule chiral.
• Enantiomers are non-superimposable mirror images with identical physical properties but opposite optical rotation.
• Assign R or S with CIP: rank by atomic number, point the lowest group away, and read 1 to 2 to 3 clockwise (R) or counterclockwise (S).
• Diastereomers are non-mirror stereoisomers with different properties; a meso compound is achiral despite stereocenters.
• Pure enantiomers are optically active, but a racemic (50/50) mixture shows no net rotation.
• Break CIP ties by expanding one sphere at a time and comparing ordered sets, as in ranking -COOH (O,O,O) above -CH₃ (H,H,H) in alanine.
• Specific rotation [alpha] = alpha(observed) / (l × c) is the reportable constant, and enantiomeric excess converts a measured rotation into composition.
• Chiral receptors make enantiomers behave differently in the body, which is why single-enantiomer drugs are now the norm and why thalidomide's in vivo racemization is the instructive caveat.

Sources

1. McMurry, J. (2023). Sequence Rules for Specifying Configuration. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Optical Activity. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Meso Compounds. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Racemic Mixtures and the Resolution of Enantiomers. In Organic Chemistry (OpenStax). openstax.org
5. Eriksson, T., Bjorkman, S., Roth, B., Fyge, A., and Hoglund, P. (1995). Stereospecific determination, chiral inversion in vitro and pharmacokinetics in humans of the enantiomers of thalidomide. Chirality, 7(1), 44-52. pubmed.ncbi.nlm.nih.gov
6. McVicker, R. U., and O'Boyle, N. M. (2024). Chirality of New Drug Approvals (2013-2022): Trends and Perspectives. Journal of Medicinal Chemistry, 67(4), 2305-2320. pmc.ncbi.nlm.nih.gov
7. LibreTexts Chemistry. Sequence Rules for Specifying Configuration. Organic Chemistry (Morsch et al.). chem.libretexts.org
8. Khan Academy, Organic Chemistry, "Stereochemistry" and "Chirality." khanacademy.org ↗

The big picture

Reactions are not magic; they are step-by-step stories of electrons moving to break old bonds and make new ones. This lesson teaches the single most important skill in the whole course: reading and drawing curved arrows, the notation chemists use to show where electrons go. Once you can follow the electrons, most reactions become predictable rather than memorized.

A reaction mechanism is a step-by-step account of how bonds break and form as reactants turn into products, like the frames of an animation showing each move. The universal notation for a mechanism is the curved arrow, and learning to read and draw these arrows is like learning the grammar of the language: once you have it, reactions stop being facts to memorize and become stories you can follow.

Curved arrows move electrons

A curved arrow shows the movement of a pair of electrons, always from a source of electrons to where they end up. The tail of the arrow starts at the electrons (a lone pair or a bond), and the head points to where the new bond forms or where the electrons land. Two crucial habits: arrows track electrons, never atoms, and a normal (two-electron) arrow has a full double-barbed head, like a complete arrowhead. Electrons flow from electron-rich to electron-poor sites, like water flowing downhill from high to low.

Key idea: A curved arrow tracks a pair of electrons from an electron-rich source (its tail) to an electron-poor destination (its head), and never moves atoms.

Nucleophiles and electrophiles

Because electrons flow one way, every step has two roles.

• A nucleophile ("nucleus-loving") is electron-rich and donates a pair of electrons, like the negative end of a magnet seeking a positive partner. It has a lone pair, a negative charge, or a pi bond. Examples: hydroxide (OH^-), ammonia (NH₃), a C=C double bond.
• An electrophile ("electron-loving") is electron-poor and accepts a pair of electrons, like the positive end of that magnet. It has a positive charge or a partially positive atom. Examples: H⁺, a carbon bearing a good leaving group, a carbonyl carbon.

The arrow always runs from the nucleophile (source) to the electrophile (sink). Spotting which reactant is the nucleophile and which is the electrophile is usually the first and most important move in analyzing any reaction.

Key idea: Every polar step pairs a nucleophile (electron donor) with an electrophile (electron acceptor), and the arrow always points from the nucleophile to the electrophile.

Reading partial charges: where the arrow points

You can often predict the electrophilic site from polarity. When two bonded atoms differ in electronegativity (an atom's pull on shared electrons), the more electronegative atom takes a partial negative charge, written δ-, and its partner takes a partial positive charge, δ+. Nucleophiles attack the δ+ atom. In a carbonyl (C=O), oxygen is δ- and carbon is δ+, so nucleophiles attack the carbon. Recognizing these partial charges tells you where a reaction will happen before you draw anything.

Key idea: The more electronegative atom of a polar bond is δ- and the other is δ+, and nucleophiles head for the δ+ atom.

Making and breaking bonds

An arrow whose head points between two atoms forms a new bond there. An arrow whose tail sits on a bond breaks that bond, sending its electrons somewhere new (often onto a leaving group, an atom or group that departs taking a bonding pair, usually as an anion).

Many steps combine both: as a nucleophile forms a new bond to a carbon, a bond from that carbon to a leaving group breaks at the same time, so the carbon never exceeds four bonds. Keeping that "four bonds to carbon" limit in mind is the best check that your arrows make sense: if a step would give carbon five bonds, an arrow is missing.

Key idea: Arrow heads between atoms make bonds, arrow tails on bonds break them, and carbon must always end with exactly four bonds.

Almost every step is one of four moves

The reassuring truth about mechanisms is that the vocabulary is tiny. Nearly every polar step you will draw this year is one of four moves, and long mechanisms are just these four in sequence.

1. Nucleophilic attack. A lone pair or a pi bond forms a new bond to an electrophilic atom. One arrow, tail on the electrons, head between the two atoms.
2. Loss of a leaving group. A sigma bond breaks heterolytically and both electrons leave with the departing group. One arrow, tail on the bond, head on the leaving atom.
3. Proton transfer. A base takes a proton. Two arrows: one from the base's lone pair to the hydrogen, one from the H-X bond onto X.
4. Rearrangement. A neighbouring bond, usually C-H or C-C, shifts to a positively charged carbon. One arrow, tail on the migrating bond, head at the empty orbital.

Key idea: Learn four moves - attack, loss of leaving group, proton transfer, rearrangement - and every mechanism in this course is a sentence built from them.

Worked mechanism, arrow by arrow: hydroxide meets acetaldehyde

Take HO^- reacting with acetaldehyde, CH₃-CHO. Do not memorize a product; derive one.

1. Find the electrophile. Oxygen is far more electronegative than carbon, so the C=O bond is strongly polarized: the oxygen is delta- and the carbonyl carbon is delta+. That carbon is the target.
2. Find the nucleophile. Hydroxide has three lone pairs and a full negative charge. Any of the lone pairs will serve as the arrow's tail.
3. Draw arrow one. Tail on a lone pair of the hydroxide oxygen, head pointing to the carbonyl carbon. This forms the new C-O bond.
4. Check the octet. If only that arrow were drawn, the carbonyl carbon would now have five bonds - three plus the C=O counted as two. Illegal. So something must break.
5. Draw arrow two. Tail on the C=O pi bond, head on the oxygen. The pi electrons collapse onto oxygen, which now has three lone pairs, one bond, and a formal charge of 6 - 6 - 1 = -1.
6. Read off the product. A tetrahedral alkoxide intermediate: the former carbonyl carbon is now sp³, bonded to CH₃, H, the new OH, and the O^-.
7. Finish with a proton transfer. The alkoxide takes a proton from water: one arrow from an alkoxide lone pair to the hydrogen of H-OH, and a second from that O-H bond onto the water oxygen, releasing hydroxide. The final product is the hydrate, a 1,1-diol.

Notice what did the work. You never recalled a reaction name. You located a delta+ carbon, located a lone pair, made a bond, respected the octet, broke a bond, and cleaned up the charge with a proton transfer. That sequence covers an enormous fraction of organic chemistry.

Key idea: Derive mechanisms rather than recalling them: find the electrophile, find the nucleophile, make the bond, then break whatever must break to keep every octet legal.

Energy diagrams: transition states, intermediates, and rate

A mechanism tells you what happens; an energy diagram tells you how hard and how fast. Plot free energy on the vertical axis against reaction progress on the horizontal, and two kinds of feature appear.

A transition state is a maximum - the exact arrangement at the top of a hill, with bonds part-formed and part-broken. It exists for roughly the duration of a bond vibration, about 10^-13 seconds, and can never be isolated. An intermediate is a minimum between two maxima: a real species with a real lifetime, such as the alkoxide above or a carbocation, that can sometimes be detected or even trapped.

Two numbers govern everything. The activation energy, delta G double-dagger, is the height from reactants to the highest transition state, and it sets the rate. The overall delta G from reactants to products sets the equilibrium position. The two are independent: a reaction can be strongly favourable and still be immeasurably slow, which is exactly why a lump of sugar sits happily on a table despite its combustion being enormously downhill.

Attach numbers so the diagram means something. At 25 degrees Celsius, RT is about 2.5 kJ/mol, and lowering an activation barrier by 5.7 kJ/mol multiplies the rate by roughly ten. A barrier near 85 kJ/mol corresponds to a reaction that takes hours at room temperature; near 105 kJ/mol you must heat it; below about 60 kJ/mol it is over almost instantly. On the equilibrium side, a delta G of -5.7 kJ/mol gives K near 10, and -17 kJ/mol gives K near 1000, so quite modest energies produce lopsided equilibria.

When a mechanism has several steps, the slowest step - the one with the highest transition state - is the rate-determining step, and only changes that affect that step change the overall rate. This is why, in Lesson 14, adding more nucleophile speeds an SN2 reaction but does nothing at all to an SN1.

Key idea: Activation energy sets the rate and overall delta G sets the equilibrium; a 5.7 kJ/mol change in either is worth a factor of ten, and only the rate-determining step controls the observed speed.

The Hammond postulate: what does a transition state look like?

Transition states cannot be isolated, so how can anyone reason about them? George Hammond's 1955 proposal is the standard answer: a transition state resembles whichever species it is closer to in energy. In a strongly exothermic step the transition state comes early and looks like the reactants; in an endothermic step it comes late and looks like the product.

The payoff is that you can substitute a species you understand for one you cannot see. Forming a carbocation from an alkyl halide is endothermic, so its transition state resembles the carbocation. That single substitution licenses the argument used constantly from Lesson 13 onward: anything that stabilizes a carbocation also stabilizes the transition state that leads to it, and therefore speeds the reaction up. You never have to reason about the transition state directly.

Key idea: By the Hammond postulate a transition state resembles the nearer species in energy, so the stability of a carbocation intermediate can stand in for the stability of the transition state forming it.

Two kinds of arrows

Almost all of this course uses the double-barbed arrow for a pair of electrons. There is also a single-barbed "fishhook" arrow that moves just one electron, used only in radical reactions (reactions of species with a lone unpaired electron). For now, treat every arrow as a full two-electron arrow unless a problem specifically involves radicals. Do not confuse either curved arrow with the straight reaction arrow that simply separates starting materials from products.

Key idea: A double-barbed arrow moves two electrons (the default), a single-barbed fishhook moves one electron (radicals only), and the straight reaction arrow is not an electron arrow at all.

Where people get stuck

• "Curved arrows show atoms moving." They show electron pairs moving. Atoms relocate only as a consequence, and thinking otherwise makes every mechanism unreadable.
• "The arrow points from the electrophile to the nucleophile." The reverse. The tail sits on the electrons, so it always starts at the electron-rich species.
• "A full arrowhead and a fishhook mean the same thing." A double-barbed arrow moves two electrons; a fishhook moves one and appears only in radical steps.
• "You can give carbon five bonds mid-mechanism as long as it works out." Never. If a new bond forms at a saturated carbon, an old one must break in the same step - which is precisely why SN2 is concerted.
• "A transition state is just an unstable intermediate." They are different objects. A transition state is an energy maximum with no lifetime; an intermediate is a minimum and can in principle be observed or trapped.
• "A more favourable reaction is a faster reaction." Thermodynamics and kinetics are independent. Overall delta G sets the equilibrium; activation energy sets the rate, and a hugely favourable reaction with a 150 kJ/mol barrier simply does not happen.
• "Speeding up any step speeds up the reaction." Only the rate-determining step matters. Accelerating a fast step downstream changes nothing you can measure.
• "Arrows for a proton transfer can be drawn as one arrow." Proton transfer needs two: the base's lone pair to the hydrogen, and the H-X bond onto X. One arrow alone would leave hydrogen with two bonds.

Recap

• A mechanism is the step-by-step story of bond breaking and forming, shown with curved arrows.
• Curved arrows move electron pairs from an electron-rich tail to an electron-poor head, never atoms.
• Nucleophiles donate electrons; electrophiles accept them; the arrow runs nucleophile to electrophile.
• Polarity marks the electrophilic site: nucleophiles attack the partially positive (δ+) atom.
• Heads between atoms make bonds, tails on bonds break them, and carbon always keeps four bonds.
• Nearly every polar step is one of four moves: nucleophilic attack, loss of a leaving group, proton transfer, or rearrangement.
• Activation energy controls rate and overall delta G controls equilibrium; 5.7 kJ/mol is worth a factor of ten in either at 25 degrees Celsius.
• Transition states are maxima with no lifetime, intermediates are minima that can be trapped, and the Hammond postulate lets a nearby intermediate stand in for a transition state.

Sources

1. McMurry, J. (2023). How Organic Reactions Occur: Mechanisms. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Using Curved Arrows in Polar Reaction Mechanisms. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Describing a Reaction: Energy Diagrams and Transition States. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Describing a Reaction: Intermediates. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). The Hammond Postulate. In Organic Chemistry (OpenStax). openstax.org
6. Hammond, G. S. (1955). A correlation of reaction rates. Journal of the American Chemical Society, 77(2), 334-338. DOI: 10.1021/ja01607a027. find source ↗
7. LibreTexts Chemistry. Using Curved Arrows in Polar Reaction Mechanisms. Organic Chemistry (Morsch et al.). chem.libretexts.org

The big picture

Acid-base chemistry is the most useful single tool in this course, because a huge fraction of organic reactions are, at their core, a proton moving from one atom to another. This lesson shows how to rank acids and bases, predict which way a proton transfer goes, and explain acidity from the stability of the leftover anion. Master this and you can predict many reactions with just two numbers.

Organic chemists lean on the Bronsted-Lowry picture: an acid donates a proton (H⁺) and a base accepts one. A proton here just means a hydrogen nucleus, a hydrogen that has lost its electron.

Conjugate acid-base pairs

When an acid loses its proton, what remains is its conjugate base, the species left holding the electrons. When a base gains a proton, it becomes its conjugate acid. For example, acetic acid (CH₃COOH) donates a proton to become acetate (CH₃COO^-), its conjugate base. The acid and its conjugate base differ by exactly one proton, like the same person with and without a hat.

Key idea: An acid and its conjugate base differ by one proton; a strong acid has a stable, weak conjugate base.

pKa measures acid strength

The strength of an acid is captured by its pK_a, a number derived from the equilibrium for losing a proton. A lower pK_a means a stronger acid (more willing to give up its proton). Because the scale is logarithmic, each unit of pK_a is a factor of ten in acidity. The scale spans a huge range:

Acid	Approx. pKa	Strength
HCl	-7	Very strong
Carboxylic acid (CH3COOH)	4.8	Weak
Water (H2O)	15.7	Very weak
Alcohol (CH3CH2OH)	16	Very weak
Alkane (C-H)	about 50	Essentially not acidic

The difference is enormous: a carboxylic acid is roughly a trillion times more acidic than an alcohol, which is why carboxylic acids react with mild bases while alcohols barely do.

Key idea: Lower pK_a means a stronger acid, and because the scale is logarithmic, small pK_a differences mean large differences in strength.

What makes an acid strong: stabilize the conjugate base

The single best predictor of acid strength is the stability of the conjugate base. If the anion left behind after losing H⁺ is stable, the acid gives up its proton readily. Five factors stabilize a negative charge, and a handy way to remember them is the sequence Atom, Resonance, Induction, Orbital (with solvent as a fifth):

• Atom: charge on a more electronegative atom, or on a larger atom lower in the periodic table, is more stable. Electronegativity is an atom's pull on electrons; larger atoms spread the charge over more volume.
• Resonance: resonance is the spreading of electrons (and charge) over several atoms, shown by more than one valid Lewis structure. Charge shared over several atoms is far more stable, like weight spread across several people.
• Induction: nearby electron-withdrawing atoms (like electronegative halogens) pull negative charge away through the sigma bonds and stabilize it.
• Orbital (hybridization): charge in an orbital with more s-character (sp before sp³) is held closer to the nucleus and is more stable.

A carboxylic acid is much more acidic than an alcohol precisely because its conjugate base (a carboxylate) spreads the negative charge over two oxygens by resonance, while an alkoxide from an alcohol must bear it on a single oxygen.

Key idea: Anything that stabilizes the conjugate base (electronegativity, larger atom, resonance, induction, more s-character) makes the acid stronger.

The four factors, with real numbers

Qualitative rules are only convincing when the data agree, so work through each factor on measured pK_a values.

Atom, across a row. Compare CH₄ at about 50, NH₃ at 38, H₂O at 15.7, and HF at 3.2. The bond strengths hardly change; what changes is electronegativity, and putting the negative charge on a more electronegative atom is worth roughly 47 pK_a units from carbon to fluorine.

Atom, down a column. Now HF 3.2, HCl -7, HBr -9, HI -10. Here electronegativity runs the wrong way - fluorine is the most electronegative yet HF is the weakest of the four. Size wins: the larger anion spreads the same charge over a much bigger volume. The same reversal shows up in organic compounds, where ethanethiol (pK_a 10.6) is over five orders of magnitude more acidic than ethanol (pK_a 16).

Resonance. Acetic acid is 4.76 and ethanol is 16 - a difference of more than eleven units, or 10¹¹ in acidity, produced entirely by the carboxylate's ability to share the charge over two oxygens. Phenol at 10.0 against cyclohexanol at about 16 makes the same point on a ring: phenoxide delocalizes into the aromatic system, an alkoxide cannot.

Induction. Add electronegative atoms near the acidic site and watch the effect accumulate: acetic acid 4.76, chloroacetic acid 2.86, dichloroacetic acid 1.29, trichloroacetic acid 0.65. Three chlorines make the acid roughly ten thousand times stronger without touching the O-H bond itself.

Induction also fades sharply with distance, which is the cleanest evidence that it travels through sigma bonds rather than through space. Compare butanoic acid at 4.82 with its chlorinated relatives: 2-chlorobutanoic acid 2.86, 3-chlorobutanoic acid 4.05, 4-chlorobutanoic acid 4.52. Move the chlorine one carbon further away and most of the effect is gone; move it three carbons away and it has almost vanished.

Orbital. Ethane is about 51, ethene about 44, and ethyne 25. Nothing changes but the hybridization of the orbital holding the lone pair, and the sp orbital's 50 percent s character is worth 26 pK_a units against sp³.

Key idea: Across a row electronegativity dominates; down a column size dominates; resonance is worth about eleven units for a carboxylate; induction is strong but fades within two or three bonds; and s character is worth about twenty-six units from sp³ to sp.

Worked example: calculating an equilibrium constant from two pK_a values

The direction rule can be made quantitative. For any proton transfer,

K(eq) = 10^{(pK_a of product acid - pK_a of reactant acid)}

Try sodium hydroxide with acetic acid. The reactant acid is acetic acid, pK_a 4.76. The product acid, formed when hydroxide picks up the proton, is water, pK_a 15.7. So K(eq) = 10^{(15.7 - 4.76)} = 10^10.9, about 8 × 10¹⁰. The reaction goes essentially to completion, which is why titrating a carboxylic acid with NaOH is a reliable laboratory operation.

Now try sodium hydroxide with ethanol. The reactant acid is ethanol at 16 and the product acid is water at 15.7, so K(eq) = 10^{(15.7 - 16)} = 10^-0.3, about 0.5. Barely half deprotonated at equilibrium - hopeless if you actually need the alkoxide. To make ethoxide cleanly you must use a base whose conjugate acid is far weaker, such as sodium hydride, whose conjugate acid H₂ has a pK_a near 35. Then K(eq) = 10^{(35 - 16)} = 10¹⁹, and the deprotonation is complete and irreversible.

This calculation is how a synthetic chemist chooses a base. Line up the common ones by the pK_a of their conjugate acid and pick one comfortably above the proton you want to remove: triethylamine 10.7, pyridine 5.2, hydroxide 15.7, ethoxide 16, hydride 35, lithium diisopropylamide (LDA) 36, amide ion 38, butyllithium about 50. A useful working margin is three to four pK_a units, which puts the equilibrium beyond 99.9 percent.

Key idea: K(eq) = 10 to the power of (product-acid pK_a minus reactant-acid pK_a), so choosing a base is arithmetic, not intuition.

Lewis acids and bases: a broader view

The Bronsted picture is about protons, but a broader definition helps with many organic reactions. A Lewis base donates an electron pair and a Lewis acid accepts one. Every Bronsted base is a Lewis base, but the Lewis idea also covers electron-pair acceptors that are not protons, such as the boron in BF₃ or a metal cation. Notice that a Lewis base is just a nucleophile and a Lewis acid is just an electrophile, seen through the acid-base lens.

Key idea: Lewis acids accept electron pairs and Lewis bases donate them, generalizing acid-base chemistry to electrophiles and nucleophiles beyond the proton.

Which way does the equilibrium go?

An acid-base reaction always favors the side with the weaker acid and weaker base, that is, it runs toward the more stable, lower-energy species. To predict direction, compare the pK_a of the acid on each side: the proton ends up on whichever partner holds it more weakly (higher pK_a). This one rule lets you predict the outcome of a proton transfer just by comparing two numbers.

Key idea: Proton transfer favors formation of the weaker acid, so the proton settles on the atom that holds it least tightly (the higher-pK_a side).

Where people get stuck

• "A higher pK_a means a stronger acid." The opposite. Lower pK_a means stronger, and each unit is a factor of ten.
• "Acidity depends mostly on the H-X bond strength." The dominant term is nearly always the stability of the anion left behind. Compare HF and HI: HI has the weaker bond and the far more stable anion, and it wins on both counts.
• "Resonance and induction are the same effect." Resonance moves electron pairs through pi systems and is essentially distance-independent within the conjugated system. Induction pulls charge through sigma bonds and collapses within two or three bonds, as the chlorobutanoic acid series shows.
• "Only proton donors are acids." Under the Lewis definition, any electron-pair acceptor is an acid. BF₃, AlCl₃, and Mg²⁺ all qualify, which is why Lewis acids show up as catalysts throughout Lesson 17.
• "Fluorine is the most electronegative, so HF must be the strongest hydrohalic acid." Down a column, anion size beats electronegativity. HF is the weakest of HF, HCl, HBr, and HI by roughly ten orders of magnitude.
• "A stronger base is always the better choice." A base that is too strong destroys other functional groups or forces the wrong deprotonation. Match the base to the proton, aiming for three to four pK_a units of margin, not fifty.
• "pK_a values are absolute constants." They are measured in a solvent, usually water or DMSO, and the two scales can differ by many units for the same compound. Compare values only within one scale.

Recap

• Bronsted acids donate protons and bases accept them; conjugate pairs differ by one proton.
• pK_a measures acid strength on a logarithmic scale; lower pK_a means a stronger acid.
• Acid strength tracks conjugate-base stability: electronegativity, atom size, resonance, induction, and s-character all help.
• Real numbers back each factor: CH₄ 50 to HF 3.2 across a row, HF 3.2 to HI -10 down a column, ethanol 16 to acetic acid 4.76 for resonance, acetic acid 4.76 to trichloroacetic acid 0.65 for induction, and ethane 51 to ethyne 25 for hybridization.
• Lewis acids accept electron pairs and Lewis bases donate them, matching electrophiles and nucleophiles.
• A proton transfer favors the weaker acid and base, and K(eq) = 10^{(product pK_a - reactant pK_a)} makes that quantitative.
• Choose a base by the pK_a of its conjugate acid, with a few units of margin over the proton you intend to remove.

Sources

1. McMurry, J. (2023). Acid and Base Strength. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Predicting Acid-Base Reactions from pKa Values. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Acids and Bases: The Lewis Definition. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Substituent Effects on Acidity. In Organic Chemistry (OpenStax). openstax.org
5. Flowers, P., et al. (2019). Relative Strengths of Acids and Bases. In Chemistry 2e (OpenStax). openstax.org
6. LibreTexts Chemistry. Acid and Base Strength. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Bordwell, F. G. (1988). Equilibrium acidities in dimethyl sulfoxide solution. Accounts of Chemical Research, 21(12), 456-463. DOI: 10.1021/ar00156a004. find source ↗

The big picture

Alkanes are the simplest, least reactive organic molecules, but their shapes teach lessons that apply everywhere. This lesson covers why alkanes are stable, how their physical properties trend with size and branching, and how rings, especially the six-membered cyclohexane chair, arrange themselves to avoid strain. The chair conformation you learn here reappears in sugars, steroids, and drugs.

Alkanes are hydrocarbons with only single bonds, following the general formula C_nH_2n+2 for open chains. They are the least reactive organic family because their C-C and C-H bonds are strong and nonpolar, offering no electron-rich or electron-poor sites for reagents to attack. This is exactly why they make good fuels and lubricants: they are stable until deliberately burned. Their main reactions are combustion (burning in oxygen to give CO₂ and water) and, under light, halogenation.

The two reactions of alkanes

Combustion is the complete oxidation of an alkane by O₂ to carbon dioxide and water, releasing large amounts of energy; this is what powers engines and furnaces. Halogenation replaces a C-H bond with a C-X bond using a halogen (Cl₂ or Br₂) and light or heat. It proceeds by a radical mechanism, a chain of steps involving species with a single unpaired electron, rather than the electron-pair (polar) mechanisms that dominate the rest of the course. Because halogenation can occur at many positions, it often gives mixtures, which limits its usefulness.

Key idea: Alkanes mostly just burn (combustion) or, under light, undergo radical halogenation; otherwise they are chemically inert.

Radical halogenation, step by step, and why bromine is fussier than chlorine

Halogenation runs as a three-phase chain, and each phase uses fishhook arrows because single electrons move.

1. Initiation. Light or heat splits the halogen homolytically: Cl-Cl -> 2 Cl·. Two fishhook arrows, one electron to each chlorine.
2. Propagation, step one. A chlorine radical abstracts a hydrogen from the alkane: Cl· + CH₃-CH₃ -> HCl + CH₃-CH₂·. This step is the selective one, because it decides which C-H is attacked.
3. Propagation, step two. The carbon radical attacks Cl₂: CH₃-CH₂· + Cl₂ -> CH₃-CH₂-Cl + Cl·. A new chlorine radical is regenerated, so one initiation event can produce thousands of product molecules.
4. Termination. Any two radicals combine, which stops that chain.

Now the interesting part. Chlorination and bromination differ enormously in how carefully they choose which hydrogen to remove. Measured relative reactivities per hydrogen atom at room temperature are, for chlorination, primary : secondary : tertiary = 1 : 3.8 : 5.0, but for bromination they are roughly 1 : 82 : 1640. Bromine is hundreds of times more discriminating.

The Hammond postulate from Lesson 10 explains it exactly. Hydrogen abstraction by chlorine is exothermic, so the transition state is early and looks like the starting alkane, where the three kinds of C-H differ very little. Abstraction by bromine is endothermic, so the transition state is late and looks like the carbon radical - and radicals differ a great deal in stability, tertiary being far more stable than primary. A late transition state inherits that difference; an early one does not.

Worked example. Chlorinate propane, CH₃-CH₂-CH₃. It has six primary hydrogens and two secondary ones. Weight each count by its reactivity: primary gives 6 × 1.0 = 6.0 and secondary gives 2 × 3.8 = 7.6. So the product ratio is 6.0 to 7.6, about 44 percent 1-chloropropane and 56 percent 2-chloropropane. Statistics favours the primary position and reactivity favours the secondary, and the two nearly cancel - which is precisely why chlorination gives messy mixtures and is rarely used for synthesis. Run the same calculation with bromine and the secondary product dominates almost completely.

Key idea: Radical halogenation is an initiation-propagation-termination chain, and bromination is far more selective than chlorination because its endothermic abstraction step has a late, radical-like transition state.

Physical trends

Because alkanes are nonpolar, they attract each other only through weak London dispersion forces, the fleeting attractions between temporary dips in electron distribution. Longer chains have more surface contact and stronger attraction, so boiling point rises steadily with chain length. Branching makes a molecule more compact, reducing surface contact and lowering the boiling point relative to its straight-chain isomer, much as a crumpled ball touches its neighbors less than a stretched-out string. Alkanes are also less dense than water and do not dissolve in it, which is why oil floats.

Key idea: Boiling point rises with chain length (more dispersion contact) and falls with branching (less contact), and alkanes are nonpolar so they do not mix with water.

Three kinds of strain in rings

Cycloalkanes are alkanes closed into a ring (formula C_nH_2n). Rings can suffer three kinds of strain. Angle strain arises when bond angles are forced away from the ideal 109.5 degrees of an sp³ carbon. Torsional strain arises when bonds on adjacent carbons are forced to eclipse. Steric strain arises when atoms across the ring are pushed too close. Cyclopropane, a three-membered ring, is locked at 60-degree angles, far from ideal, so it is highly strained and unusually reactive. Cyclobutane (about 90 degrees) is also strained. The strain eases as rings grow toward six carbons.

Key idea: Ring strain comes from bad angles (angle strain), eclipsing bonds (torsional strain), and crowded atoms (steric strain), and small rings suffer the most.

Measuring ring strain by burning rings

Strain is not a metaphor; it is measured. Burn a cycloalkane and divide the heat released by the number of CH₂ groups. A completely unstrained CH₂ releases about 658.6 kJ/mol; anything above that is stored strain. Subtracting gives the total strain of each ring:

Ring	Total strain (kJ/mol)	Chief cause
cyclopropane	115	severe angle strain plus fully eclipsed C-H bonds
cyclobutane	110	angle strain plus torsional strain; puckers slightly to relieve it
cyclopentane	26	almost no angle strain; residual torsional strain, relieved by an envelope pucker
cyclohexane	0	none - the chair achieves ideal angles and full staggering
cycloheptane	26	transannular crowding
cyclooctane	41	transannular crowding across the ring

Cyclopropane deserves a closer look because its bonding is genuinely odd. Forcing three carbons into a triangle demands 60-degree internal angles, but sp³ orbitals cannot point at 60 degrees. Instead the orbitals overlap at an angle, outside the line joining the nuclei, producing what are usually called bent or banana bonds with an interorbital angle nearer 104 degrees. The overlap is poor, the bonds are weak, and on top of that all six C-H bonds are perfectly eclipsed. That 115 kJ/mol of stored strain is why cyclopropane rings open under conditions that leave ordinary alkanes untouched, and why cyclopropane-containing drugs are used deliberately as reactive handles.

Key idea: Heats of combustion put strain on a numerical scale - 115 kJ/mol for cyclopropane, essentially zero for cyclohexane - and small rings store that energy in bent, poorly overlapping bonds.

Cyclohexane's chair

Cyclohexane, the six-membered ring, is special: it can pucker into a three-dimensional shape that gives every carbon a perfect 109.5-degree angle, so it has essentially no ring strain. The lowest-energy shape is the chair conformation, which also staggers all its C-H bonds to avoid torsional strain. A higher-energy boat conformation exists but is disfavored because some hydrogens eclipse and two point inward and bump. In the chair, each carbon has one axial bond (pointing straight up or down, parallel to the ring's axis) and one equatorial bond (pointing outward around the ring's equator).

Key idea: Cyclohexane escapes strain by puckering into a chair, giving each carbon ideal 109.5-degree angles and one axial plus one equatorial position.

The ring flip and why equatorial wins

A cyclohexane chair can flip into an equivalent chair, a motion called the ring flip. During the flip, every group that was axial becomes equatorial and every equatorial group becomes axial. This matters because bulky groups prefer the roomier equatorial position; an axial bulky group suffers steric crowding with the other axial groups on the same face (called 1,3-diaxial interactions). So the more stable chair is the one with the largest substituent equatorial. The stability of the cyclohexane chair is why six-membered rings are so common throughout organic and biological chemistry, from sugars to steroids.

Key idea: A ring flip swaps axial and equatorial positions, and the favored chair places bulky groups equatorial to avoid 1,3-diaxial crowding.

How strong is the equatorial preference? A-values

The preference has a name and a number. The A-value of a substituent is the free-energy difference between its axial and equatorial forms on a cyclohexane ring, and it is essentially a measure of how bulky the group is.

Substituent	A-value (kJ/mol)	Percent equatorial at 25 °C
-F	0.6	about 56
-Cl, -Br	about 2.2	about 71
-OH	about 4.0	about 83
-CH3	7.6	about 95
-CH(CH3)2	about 9.2	about 98
-C(CH3)3	about 23	greater than 99.99

Worked example. Methylcyclohexane has an A-value of 7.6 kJ/mol. Convert that to a ratio with delta G = -RT ln K, taking RT = 2.48 kJ/mol at 298 K: K = e^{(7.6 / 2.48)} = e^3.06 = 21. Twenty-one molecules equatorial for every one axial, so about 95 percent equatorial. Put a tert-butyl group on instead and the A-value near 23 kJ/mol gives K above 10⁴, which is why tert-butylcyclohexane is treated as conformationally locked, effectively frozen with that group equatorial.

Where does 7.6 kJ/mol come from? An axial methyl group has exactly two 1,3-diaxial interactions with the axial hydrogens on the same face, and each is worth about 3.8 kJ/mol. That number should look familiar: it is the same 3.8 kJ/mol as a gauche butane interaction from Lesson 8, because it is the same physical situation - two groups held about 60 degrees apart. The cyclohexane ring is not a new phenomenon, only a rigid frame for an old one.

The flip itself is fast. The barrier is about 45 kJ/mol, so at room temperature a cyclohexane ring inverts on the order of 10⁵ times per second. Room-temperature NMR sees a single averaged signal; cool the sample enough and the two environments separate.

Key idea: An A-value is the axial-to-equatorial free-energy cost, built from about 3.8 kJ/mol per 1,3-diaxial interaction, and delta G = -RT ln K converts it directly into a population ratio.

Two substituents: which chair can put both equatorial?

With two substituents the cis or trans relationship decides whether a diequatorial chair even exists. Work it out once and the pattern is memorable.

• 1,2-disubstituted: the trans isomer can be diequatorial; the cis isomer is always one axial and one equatorial.
• 1,3-disubstituted: the cis isomer can be diequatorial; the trans isomer is always one axial and one equatorial.
• 1,4-disubstituted: the trans isomer can be diequatorial; the cis isomer is always one axial and one equatorial.

So trans-1,4-dimethylcyclohexane is more stable than its cis isomer by roughly one methyl A-value, about 7.6 kJ/mol, because trans can put both methyls equatorial while cis is stuck with one axial no matter which way the ring flips. Note that these are genuinely different compounds, not conformers - flipping the ring never converts cis into trans, since that would require breaking a bond.

This is not bookkeeping for its own sake. In Lesson 15 an E2 elimination will demand that the leaving group be axial, and whether a given ring isomer can put it there decides whether the reaction happens at all.

Key idea: For 1,2 and 1,4 substitution the trans isomer can be diequatorial, for 1,3 it is the cis isomer, and the diequatorial arrangement is more stable by roughly the sum of the A-values it saves.

Where people get stuck

• "Alkanes react with acids and bases like other organic molecules." They are essentially inert to acids, bases, nucleophiles, and electrophiles. They combust, and under light or heat they undergo radical halogenation. That is close to the whole list.
• "Branching raises the boiling point." Branching lowers it, by making the molecule more compact and reducing the surface available for dispersion contact.
• "Cyclohexane is strained like the smaller rings." The chair has ideal angles and fully staggered bonds, so its measured strain is zero.
• "Axial and equatorial are fixed properties of a substituent." A ring flip swaps them about 10⁵ times per second. What is fixed is the cis or trans relationship between two substituents.
• "A ring flip can turn cis into trans." It cannot. That would require breaking and remaking a bond, so cis and trans ring isomers are separate compounds.
• "Chlorination is fine for making a specific alkyl halide." With relative reactivities of only 1 : 3.8 : 5.0, chlorination gives mixtures. Bromination, at 1 : 82 : 1640, is the selective choice when you want the tertiary product.
• "The boat is the second-best cyclohexane conformation." The true second minimum is the twist-boat, several kJ/mol below the eclipsed boat, which is actually a maximum along the flip pathway.

Recap

• Alkanes (C_nH_2n+2) are unreactive nonpolar hydrocarbons whose main reactions are combustion and radical halogenation.
• Radical halogenation runs by initiation, propagation, and termination, and bromination is far more selective than chlorination because of its late, radical-like transition state.
• Boiling point rises with chain length and falls with branching; alkanes are insoluble in water.
• Ring strain comes from angle, torsional, and steric strain, measured at 115 kJ/mol for cyclopropane down to zero for cyclohexane.
• Cyclohexane adopts a nearly strain-free chair with ideal angles and staggered bonds, flipping about 10⁵ times per second.
• A-values quantify the equatorial preference: 7.6 kJ/mol for methyl gives about 95 percent equatorial, and about 23 kJ/mol for tert-butyl locks the ring.
• For 1,2 and 1,4 substitution the trans isomer can be diequatorial; for 1,3 it is the cis isomer.

Sources

1. McMurry, J. (2023). Stability of Cycloalkanes: Ring Strain. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Axial and Equatorial Bonds in Cyclohexane. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Conformations of Monosubstituted Cyclohexanes. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Conformations of Disubstituted Cyclohexanes. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). Preparing Alkyl Halides from Alkanes: Radical Halogenation. In Organic Chemistry (OpenStax). openstax.org
6. McMurry, J. (2023). Properties of Alkanes. In Organic Chemistry (OpenStax). openstax.org
7. LibreTexts Chemistry. Conformations of Cyclohexane. Organic Chemistry (Morsch et al.). chem.libretexts.org

The big picture

Alkenes react because their double bond is a ball of exposed, loosely held electrons that electrophiles find irresistible. This lesson teaches the signature alkene reaction, electrophilic addition, and the rule (Markovnikov's) that predicts which product forms. The key to all of it is the stability of the carbocation intermediate, an idea that will return in substitution and elimination.

Alkenes owe all their chemistry to the C=C double bond. The pi bond's electrons stick out above and below the molecular plane, exposed and loosely held, which makes the double bond electron-rich, so it behaves as a nucleophile (an electron-rich electron donor, like the negative end of a magnet). Alkenes therefore react with electrophiles, and the signature reaction is electrophilic addition, in which the pi bond opens and a new group adds to each carbon.

Addition of a hydrogen halide (HX): the mechanism in words

When an alkene meets HBr, the reaction happens in two steps. In step one, the pi bond acts as the nucleophile: an arrow runs from the C=C pi bond to the hydrogen of H-Br (the electrophile), and at the same time an arrow breaks the H-Br bond, sending those electrons onto bromine.

This puts H on one carbon and leaves the other carbon with only three bonds and a positive charge, a carbocation (a carbon bearing a positive charge and an empty orbital). In step two, an arrow runs from a lone pair on the bromide ion (Br^-) to the positive carbon, forming the new C-Br bond. The net result is that H and Br add across the former double bond, converting the alkene into a haloalkane.

Key idea: Electrophilic addition of HX goes in two steps: the pi bond grabs H to make a carbocation, then the halide fills the empty orbital.

The same mechanism, arrow by arrow, on 2-methylpropene

Words like "the pi bond attacks" only become useful when you can place every arrow. Run HBr onto 2-methylpropene, (CH₃)₂C=CH₂.

1. Arrow 1. Tail on the C=C pi bond, head on the hydrogen of H-Br. The pi electrons form a new sigma bond from the terminal CH₂ carbon to that hydrogen.
2. Arrow 2, drawn in the same step. Tail on the H-Br sigma bond, head on the bromine. Without this arrow, hydrogen would end up with two bonds. Bromine leaves with both electrons as Br^-, formal charge 7 - 6 - 1 = -1.
3. Read the intermediate. The terminal carbon now has three hydrogens and one bond to the central carbon: four bonds, neutral. The central carbon has lost its share of the pi bond and now holds only three bonds - to two methyls and to the new CH₃. Its formal charge is 4 - 0 - 3 = +1. It is sp², trigonal planar, with an empty p orbital perpendicular to that plane. This is the tertiary carbocation.
4. Why this step is favourable. It is the slow, rate-determining step, and it is uphill - which by the Hammond postulate means a late transition state that resembles the cation. Anything stabilizing the cation therefore lowers the barrier, and a tertiary cation is stabilized by nine adjacent C-H bonds able to hyperconjugate into the empty p orbital.
5. Arrow 3. Tail on a lone pair of Br^-, head at the positive carbon, into the empty p orbital. This forms the C-Br bond and neutralizes the charge. It is fast and strongly downhill, because a full positive and a full negative charge are being annihilated.
6. Product. 2-bromo-2-methylpropane, (CH₃)₃C-Br.

Note that bromide can approach the flat cation from either face with equal ease. If the cationic carbon were a stereocenter, that would give a racemic mixture - a point that returns in force in Lesson 14.

Key idea: Two arrows form the cation and one traps it; the first step is slow and uphill, so cation stability controls the rate, and the flat cation is attacked from both faces equally.

Markovnikov's rule

When the alkene is unsymmetrical, the two carbons are not equivalent, and one product dominates. Markovnikov's rule states that in the addition of HX, the hydrogen adds to the carbon that already has more hydrogens, and the halogen ends up on the more substituted carbon. The reason lies in the intermediate: adding H that way generates the more stable carbocation. Carbocation stability increases with substitution because neighboring carbon groups donate electron density to the electron-poor center, an effect called hyperconjugation (stabilization from adjacent C-H bonds sharing their electrons with the empty orbital):

tertiary (3°) > secondary (2°) > primary (1°) > methyl

Worked example. Add HBr to propene, CH₃-CH=CH₂. Placing H on the terminal CH₂ (which has more hydrogens) gives a secondary carbocation on the middle carbon; placing H on the middle carbon would give a less stable primary carbocation. The reaction takes the more stable route, so Br ends up on the middle carbon. The major product is 2-bromopropane, CH₃-CHBr-CH₃, not 1-bromopropane.

Key idea: Markovnikov's rule follows from carbocation stability: H adds so as to make the most substituted, most stable cation, and the halide follows.

How much more stable? Putting numbers on carbocations

Gas-phase measurements of the energy needed to pull a hydride ion off an alkane, R-H -> R⁺ + H^-, give a clean stability scale for the cations: methyl 1093 kJ/mol, ethyl (primary) 962, isopropyl (secondary) 862, tert-butyl (tertiary) 803. Lower cost means a more stable cation, so a tertiary cation is about 290 kJ/mol easier to form than a methyl cation, and roughly 60 kJ/mol easier than a secondary one.

Two effects produce the trend. Alkyl groups are weakly electron-donating through the sigma framework, and more importantly each C-H bond on a neighbouring carbon can align with the empty p orbital and donate electron density into it - hyperconjugation. Count the available bonds: a tert-butyl cation has nine such C-H bonds, isopropyl six, ethyl three, methyl none. The ladder of stability is a count of hyperconjugative donors.

Resonance beats hyperconjugation when it is available. An allylic cation, with the empty p orbital next to a C=C, and a benzylic cation, with it next to a benzene ring, delocalize the positive charge over several carbons and are about as stable as secondary or tertiary cations even when they are formally primary. Conversely, a cation directly on an sp² carbon of an alkene or a ring - a vinyl or aryl cation - is so unstable that reactions simply do not form it.

Key idea: Hydride affinities put the cation ladder at 803 (3°), 862 (2°), 962 (1°), and 1093 kJ/mol (methyl), and resonance in allylic or benzylic cations can outrank the substitution count entirely.

When the cation moves: rearrangements

A carbocation is not obliged to sit still. If shifting a neighbouring hydrogen or methyl group with its bonding pair produces a more stable cation, it happens - fast, and before the nucleophile can arrive.

Worked example. Add HBr to 3-methylbut-1-ene, CH₂=CH-CH(CH₃)₂.

1. Markovnikov protonation puts H on the terminal CH₂, generating a secondary cation at C2: CH₃-CH⁺-CH(CH₃)₂.
2. Look at the neighbour. C3 carries a hydrogen and two methyls. Shifting that hydrogen, with its bonding electron pair, from C3 to C2 is a 1,2-hydride shift. Draw one arrow: tail on the C3-H bond, head at C2.
3. The positive charge is now on C3, which bears three alkyl groups - a tertiary cation, about 60 kJ/mol more stable. The shift is strongly downhill and essentially instantaneous.
4. Bromide traps the new cation, giving 2-bromo-2-methylbutane as the major product rather than the 2-bromo-3-methylbutane a naive Markovnikov analysis would predict.

When no hydrogen is available in the right place, a whole methyl group can migrate instead, in a 1,2-methyl shift, by exactly the same single arrow from the C-C bond to the empty orbital. The practical rule is simple and worth internalizing: whenever a mechanism produces a secondary cation next to a tertiary carbon, check for a shift before writing the product. Rearrangement is also the strongest single piece of evidence that a free carbocation really forms, because concerted mechanisms never rearrange.

Key idea: A hydride or methyl group will migrate with its electron pair to convert a less stable cation into a more stable one, and unexpected rearranged products are the fingerprint of a genuine carbocation intermediate.

Other important additions

• Acid-catalyzed hydration: alkene plus water with an acid catalyst adds H and OH across the double bond, following Markovnikov's rule to give the more substituted alcohol.
• Halogenation: alkene plus Br₂ or Cl₂ adds a halogen to each carbon, giving a vicinal (neighboring) dihalide. This one goes through a bridged halonium ion rather than an open carbocation, so the two halogens add to opposite faces (anti addition).
• Hydrogenation: covered below, adds H to each carbon over a metal catalyst.

Key idea: Most alkene additions convert the pi bond into two new single bonds, and whether they follow Markovnikov depends on whether an open carbocation forms.

Hydrogenation

A different, cleaner addition is hydrogenation: an alkene plus H₂ gas over a metal catalyst (such as Pt, Pd, or Ni) adds one hydrogen to each carbon of the double bond, converting the alkene into an alkane. Ethene plus H₂ gives ethane. Because both hydrogens are delivered from the catalyst surface, they add to the same face (syn addition).

This reaction is how liquid vegetable oils (rich in C=C bonds) are turned into semisolid fats, and it is a workhorse for removing double bonds in synthesis. Across all these reactions the theme is constant: the electron-rich pi bond seeks out an electron-poor partner, and the double bond gives way to two new single bonds.

Key idea: Hydrogenation adds H2 across a double bond over a metal catalyst with syn stereochemistry, turning alkenes into alkanes.

Getting the other regiochemistry: two anti-Markovnikov routes

Markovnikov's rule is a consequence of one mechanism, not a law of nature. Change the mechanism and the regiochemistry inverts.

Radical addition of HBr. Add HBr to an alkene in the presence of a peroxide and the bromine ends up on the less substituted carbon. The mechanism is radical, not polar. A peroxide splits to give radicals, which generate Br·; the bromine radical adds to the alkene first, and it adds to the terminal carbon precisely because that leaves the unpaired electron on the more substituted carbon, where the radical is more stable (the radical stability ladder is 3° > 2° > 1°, the same order as for cations and for the same reasons). That carbon radical then abstracts H from HBr, regenerating Br·. Because the bromine attaches first, it lands on the opposite carbon from where the polar mechanism would put it. Notably this works only for HBr: with HCl the hydrogen-abstraction step is too endothermic and with HI the halogen-addition step is, so neither chain propagates.

Hydroboration-oxidation. Treat an alkene with borane, BH₃ in tetrahydrofuran, then with hydrogen peroxide and hydroxide, and you obtain the anti-Markovnikov alcohol. Boron has only six valence electrons, so BH₃ is an electrophile, and the addition is concerted through a four-centre transition state in which the B-H bond and the C=C interact simultaneously. Two consequences follow directly. Boron ends up on the less substituted, less hindered carbon, partly on steric grounds and partly because the partial positive charge that develops in the transition state is better carried by the more substituted carbon, which is where the hydrogen goes. And because everything happens on one face at once, the boron and the hydrogen add syn. The oxidation step then replaces boron with -OH with retention of configuration, so the net result is syn addition of H and OH with the OH on the less substituted carbon.

Compare the three routes on propene, CH₃-CH=CH₂: HBr alone gives 2-bromopropane; HBr with peroxides gives 1-bromopropane; and acid-catalysed hydration gives propan-2-ol while hydroboration-oxidation gives propan-1-ol. Same alkene, four products, chosen entirely by mechanism.

Key idea: Radical HBr addition and hydroboration-oxidation both place the new group on the less substituted carbon, because in each case the atom that adds first is not the hydrogen.

Stereochemistry: what the bromonium ion proves

Halogenation looks simple until you check the stereochemistry. Add Br₂ to cyclopentene and the product is exclusively trans-1,2-dibromocyclopentane; the cis isomer is not formed at all. An open carbocation cannot explain that, because a flat cation would be attacked from both faces and give a mixture.

The explanation is the bromonium ion. As the pi bond attacks Br₂, the bromine that has already bonded uses a lone pair to bridge across to the other carbon, forming a strained three-membered ring with a positive charge on bromine. One face of the alkene is now completely blocked. Bromide must therefore attack the back face, at either carbon, opening the ring in what is effectively an SN2 step. Backside attack forces the two bromines onto opposite faces, which is anti addition, and the trans product is the only possible outcome.

The same bridged intermediate explains halohydrin formation. Run the reaction in water and water rather than bromide opens the ring - still by backside attack, still anti, but now attacking the more substituted carbon, because that carbon carries more of the positive charge in the unsymmetrical bridged ion. So the OH ends up Markovnikov and the Br anti-Markovnikov, on opposite faces.

Key idea: Stereochemistry is evidence about mechanism: strict anti addition proves a bridged bromonium ion, strict syn addition proves a concerted or surface-delivered pathway, and a racemic mixture proves a free planar cation.

Where people get stuck

• "Markovnikov's rule is just about counting hydrogens." The counting is a shortcut. The cause is that the reaction takes whichever protonation gives the more stable carbocation, which is why anti-Markovnikov results appear the moment the mechanism changes.
• "A primary carbocation is fine if that is where the H lands." A primary cation costs about 160 kJ/mol more than a tertiary one. Reactions avoid it, rearrange past it, or take a different mechanism entirely.
• "Every alkene addition follows Markovnikov." Halogenation, hydrogenation, hydroboration, and radical HBr addition do not, because none of them makes an open carbocation.
• "The double bond is electron-poor because it is unsaturated." It is electron-rich and nucleophilic. Unsaturation means fewer hydrogens, not fewer electrons.
• "Rearrangements are rare exceptions." They are routine whenever a cation can improve by one shift. Check every secondary cation for an adjacent tertiary carbon before writing a product.
• "Anti and syn are just labels." They are experimental facts that discriminate between mechanisms. Anti addition rules out an open cation; syn addition rules out a bridged one.
• "Peroxides make HCl add anti-Markovnikov too." Only HBr. The radical chain fails on thermodynamic grounds for HCl and HI, so those still add by the polar, Markovnikov route.

Recap

• The alkene pi bond is electron-rich and acts as a nucleophile toward electrophiles.
• Electrophilic addition of HX is two steps: form a carbocation, then trap it with the nucleophile.
• Markovnikov's rule places H on the carbon with more hydrogens so the more stable carbocation forms.
• Carbocation stability runs tertiary (803 kJ/mol hydride affinity) > secondary (862) > primary (962) > methyl (1093), aided by hyperconjugation, with allylic and benzylic cations stabilized by resonance.
• Cations rearrange by 1,2-hydride or 1,2-methyl shifts whenever that produces a more stable cation.
• Radical HBr addition and hydroboration-oxidation deliver anti-Markovnikov products by changing which atom adds first.
• Other additions include acid-catalyzed hydration (Markovnikov), halogenation via a bromonium ion (anti), and hydrogenation (syn).

Sources

1. McMurry, J. (2023). Electrophilic Addition Reactions of Alkenes. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Orientation of Electrophilic Additions: Markovnikov's Rule. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Carbocation Structure and Stability. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). Hydration of Alkenes: Addition of H2O by Hydroboration. In Organic Chemistry (OpenStax). openstax.org
6. McMurry, J. (2023). Halogenation of Alkenes: Addition of X2. In Organic Chemistry (OpenStax). openstax.org
7. Brown, H. C. (1979). From Little Acorns to Tall Oaks: From Boranes through Organoboranes (Nobel Lecture). The Nobel Foundation. nobelprize.org

The big picture

Nucleophilic substitution is one of the most common reaction types in organic chemistry: a new group swaps in for a leaving group on a carbon. This lesson contrasts the two ways it can happen, SN2 and SN1, so that from the substrate, nucleophile, and solvent you can predict the rate, the mechanism, and the three-dimensional product. The carbocation idea from the alkene chapter returns here as the heart of SN1.

In nucleophilic substitution, a nucleophile replaces a leaving group on a carbon. A leaving group is an atom or group that departs taking a bonding pair of electrons, like a passenger stepping off a bus with their own belongings. Alkyl halides are the classic substrates: the halogen is a good leaving group because it departs as a stable halide ion. Two distinct mechanisms compete, and knowing which one operates lets you predict both the rate and the three-dimensional product.

The SN2 mechanism

SN2 stands for Substitution, Nucleophilic, Bimolecular. It happens in one concerted step, meaning bond-making and bond-breaking occur together: the nucleophile attacks the carbon from the side directly opposite the leaving group (a backside attack) at the same moment the leaving group departs. Because both the substrate and the nucleophile appear in the rate-determining step, the rate depends on both: rate = k[substrate][nucleophile].

The backside attack turns the carbon inside out like an umbrella flipping in the wind, causing inversion of configuration at a stereocenter (called Walden inversion). SN2 is fastest when the carbon is uncrowded, so it prefers methyl and primary substrates and is blocked at hindered tertiary carbons. It also wants a strong nucleophile and works best in polar aprotic solvents.

Key idea: SN2 is a one-step backside attack whose rate depends on both reactants, favors uncrowded (methyl and primary) carbons, and inverts the stereocenter.

SN2 arrow by arrow, and what the transition state looks like

Run hydroxide onto (S)-2-bromobutane, CH₃-CHBr-CH₂-CH₃.

1. Arrow 1. Tail on a lone pair of HO^-, head at the carbon bearing the bromine - but approaching along the line of the C-Br bond, from the face opposite bromine. The nucleophile must come in through the back lobe of the C-Br antibonding orbital, because that is the only place with room for electron density to enter.
2. Arrow 2, same step. Tail on the C-Br sigma bond, head on bromine. Bromide leaves with both electrons. This arrow is compulsory and simultaneous: without it the carbon would briefly hold five bonds, which is why SN2 cannot be split into two steps.
3. The transition state. At the top of the barrier the carbon is bonded to three unchanged groups arranged in a plane, with the incoming oxygen and the departing bromine on opposite sides, each about half-bonded. The carbon is momentarily five-coordinate and sp²-like, and the overall species carries the nucleophile's charge spread across two atoms. It is a maximum, not an intermediate - it has no lifetime.
4. The product. As bromide departs, the three unchanged groups snap through the plane, the way an umbrella inverts in the wind. The spatial arrangement is inverted, giving (R)-butan-2-ol.

One caution about labels. The spatial arrangement always inverts in an SN2 reaction, but the R/S descriptor only flips if the incoming group holds the same CIP rank as the one that left. Here bromine was priority 1 and the -OH that replaces it is also priority 1 against ethyl, methyl, and hydrogen, so S does become R. Replace bromine with a group of different rank and the letter can stay the same even though the molecule has genuinely turned inside out. Trust the geometry, not the letter.

Key idea: SN2 is one step with two simultaneous arrows through a five-coordinate transition state, and it always inverts the spatial arrangement even when the R/S letter happens not to change.

The SN1 mechanism

SN1 (Substitution, Nucleophilic, Unimolecular) goes in two steps. First, the leaving group departs on its own to form a carbocation (a positively charged carbon with an empty orbital); this slow step is rate-determining. Second, the nucleophile adds to the carbocation. Because only the substrate is present in the slow step, rate = k[substrate]; the nucleophile's concentration does not affect the rate.

Since the intermediate carbocation is flat (sp²), the nucleophile can attack either face, giving a mix of both configurations, that is, racemization (formation of both mirror-image products). SN1 needs a stable carbocation, so it prefers tertiary substrates, tolerates weak nucleophiles, and is helped by polar protic solvents that stabilize the ions.

Key idea: SN1 is a two-step reaction through a flat carbocation whose rate depends only on the substrate, favors tertiary carbons, and gives racemization.

One honest refinement. Pure 50:50 racemization is the idealization. Real SN1 reactions usually give a slight excess of the inverted product, often 5 to 20 percent, because the departing leaving group lingers briefly as an ion pair and partially shields the face it just left. Observing partial rather than complete racemization is itself evidence for the mechanism, and it is one of the standard experimental fingerprints separating SN1 from SN2.

How big are these effects? The rate data

The substrate rule is not a preference; it is a difference of many orders of magnitude, and both mechanisms respond to substitution in exactly opposite directions.

Substrate	Relative SN2 rate	Relative SN1 (solvolysis) rate
methyl, CH3-X	about 30	about 1
primary, CH3CH2-X	1	about 1
secondary, (CH3)2CH-X	about 0.02	about 12
tertiary, (CH3)3C-X	less than 0.001	about 1,200,000
neopentyl, (CH3)3CCH2-X	about 0.00001	about 1

Read the two columns against each other. Going from methyl to tertiary costs SN2 more than four orders of magnitude and buys SN1 six. The crossover happens at secondary carbons, which is precisely why secondary substrates are the ones that need real analysis rather than a rule.

The neopentyl row is the most instructive. Neopentyl halide is a primary substrate, yet it is the slowest SN2 substrate in the table by a wide margin - about a million times slower than ethyl. It also cannot do SN1, because ionizing it would give a primary cation. The reason is that the bulky tert-butyl group sits on the adjacent carbon and physically blocks the backside trajectory. This proves that the substrate effect on SN2 is steric, not electronic: what matters is whether a nucleophile can reach the back face, not how many carbons are attached to the reacting one.

Key idea: Substitution slows SN2 by 10⁴ and accelerates SN1 by 10⁶, and neopentyl halide - primary but hopeless at SN2 - shows the substrate effect is purely steric.

What makes a good leaving group

Both mechanisms need a good leaving group, and the rule is simple: a good leaving group is a weak base, because a weak base is stable on its own after it leaves. The halides I^-, Br^-, and Cl^- are good leaving groups (I best), while strong bases like hydroxide (OH^-) and amide are poor leaving groups. This is the same conjugate-base-stability idea from acid-base chemistry, now applied to a group that departs.

Key idea: Good leaving groups are weak, stable bases; the more stable the group is once it leaves, the better it works.

Because leaving-group ability tracks conjugate-base stability, the pK_a table from Lesson 11 doubles as a leaving-group table. Rank by the pK_a of the leaving group's conjugate acid, lowest first: triflate (TfOH, about -14), iodide (HI, -10), tosylate (TsOH, -2.8), bromide (HBr, -9), chloride (HCl, -7), water (H₃O⁺, -1.7), fluoride (HF, 3.2). Roughly, halide leaving rates run I : Br : Cl : F at about 30,000 : 10,000 : 200 : 1.

At the other extreme, hydroxide (water, pK_a 15.7), alkoxide (16), and amide (ammonia, 38) are so basic that they essentially never leave. That single fact explains two of the most-used tricks in organic synthesis. To make an alcohol react by substitution you either protonate the -OH with strong acid, converting a terrible leaving group (HO^-) into an excellent one (H₂O), or you convert it into a tosylate with TsCl, replacing it with a resonance-stabilized sulfonate that leaves readily. In both cases you have not changed the carbon at all - only the quality of what departs.

Nucleophile and solvent: how to read them

A strong nucleophile pushes toward SN2, while a weak nucleophile allows SN1. Nucleophile strength usually rises with negative charge and with a lone pair on a larger, more polarizable atom. Solvent matters too: a polar aprotic solvent (one that cannot hydrogen-bond to the nucleophile, such as acetone or DMSO) leaves the nucleophile bare and reactive, favoring SN2. A polar protic solvent (one with O-H or N-H, such as water or an alcohol) surrounds and stabilizes ions, favoring the ionization of SN1.

Key idea: Strong nucleophile plus polar aprotic solvent favors SN2; weak nucleophile plus polar protic solvent favors SN1.

The solvent effect deserves a number, because it is far larger than students expect. Take methyl iodide reacting with chloride ion. In methanol, a polar protic solvent, chloride is wrapped in a tight shell of hydrogen bonds that must be stripped away before it can attack, and the reaction is slow. Move the same reaction into dimethylformamide or dimethyl sulfoxide, which are polar enough to dissolve the salt but have no O-H to donate, and the rate rises by a factor of roughly a million. The cation is still solvated but the anion is left naked and furious. Changing nothing but the solvent buys six orders of magnitude.

The same effect reverses the halide nucleophilicity order, which is a favourite exam trap. In a protic solvent such as water or methanol, nucleophilicity runs I^- > Br^- > Cl^- > F^-, because the small, hard fluoride is the most tightly solvated and the large, polarizable iodide the least. In a polar aprotic solvent there is no hydrogen-bonding shell to escape, so nucleophilicity reverts to tracking basicity: F^- > Cl^- > Br^- > I^-. Fluoride in DMSO is a genuinely powerful nucleophile; fluoride in water is nearly useless. Always state the solvent before ranking nucleophiles.

Note also that nucleophilicity and basicity are related but not the same. Basicity is a thermodynamic equilibrium constant for grabbing a proton; nucleophilicity is a kinetic rate constant for attacking carbon. That is why bulky bases such as tert-butoxide are strongly basic yet poor nucleophiles - and, as Lesson 15 shows, exactly why they are chosen when you want elimination instead of substitution.

Side-by-side

Feature	SN2	SN1
Steps	One (concerted)	Two (via carbocation)
Rate law	k[substrate][nucleophile]	k[substrate]
Best substrate	Methyl, primary	Tertiary
Nucleophile	Strong	Weak is fine
Stereochemistry	Inversion	Racemization
Solvent	Polar aprotic	Polar protic

The single most useful predictor is the substrate: primary carbons lean strongly SN2, tertiary carbons lean SN1, and secondary carbons can go either way depending on the nucleophile and solvent.

Key idea: Start every substitution problem by classifying the carbon: primary points to SN2, tertiary to SN1, secondary depends on conditions.

Worked decision: three substrates, three answers

Run the analysis the way you would in an exam, in a fixed order: substrate, nucleophile or base, solvent, then temperature.

Case 1. 1-bromobutane plus sodium cyanide in DMSO. Substrate primary, so SN1 is impossible - a primary cation will not form. Cyanide is a strong, small nucleophile and a weak base, so it attacks carbon rather than removing a proton. DMSO is polar aprotic, leaving cyanide unsolvated and highly reactive. Verdict: clean SN2, giving pentanenitrile, with second-order kinetics.

Case 2. 2-bromo-2-methylpropane stirred in water. Substrate tertiary, so backside attack is blocked but the cation is excellent. Water is a weak nucleophile, so it cannot force SN2, but it is polar protic and stabilizes both ions as they separate. Verdict: SN1, giving 2-methylpropan-2-ol, first-order in substrate only, and adding more water changes the rate only as a solvent effect, not through the rate law.

Case 3. (S)-2-bromobutane plus sodium iodide in acetone. Substrate secondary, so both are conceivable. Iodide is an outstanding nucleophile and a very weak base, and acetone is polar aprotic. The strong nucleophile wins. Verdict: SN2, giving (R)-2-iodobutane with clean inversion. Swap the reagent for silver nitrate in ethanol and the same substrate goes SN1 instead, because Ag⁺ pulls the bromide off and the weak nucleophile cannot do otherwise - producing nearly racemic 2-ethoxybutane. Same molecule, opposite mechanism, decided entirely by the conditions.

Key idea: Classify the substrate first, then let the nucleophile and solvent decide any case that the substrate leaves open, and remember that a secondary carbon can be pushed either way.

Where people get stuck

• "SN1 and SN2 differ only in speed." They differ in mechanism, rate law, stereochemistry, substrate preference, sensitivity to solvent, and whether rearrangement is possible. Speed is a symptom, not a definition.
• "A strong nucleophile speeds up an SN1 reaction." The nucleophile is absent from the rate-determining step, so doubling its concentration changes the rate by nothing. It changes only which product forms.
• "Tertiary substrates react fastest by SN2." Tertiary substrates are the slowest SN2 substrates, by more than 10⁴. Crowding blocks the backside trajectory.
• "A primary substrate is always a fast SN2 substrate." Neopentyl halide is primary and about a million times slower than ethyl, because the branch on the adjacent carbon blocks the back face. Look at the whole neighbourhood, not just the reacting carbon.
• "Hydroxide is a fine leaving group because it is common." Its conjugate acid has pK_a 15.7, so it is far too basic to leave. Protonate it or tosylate it first.
• "Iodide is the best nucleophile, period." Only in protic solvents, where solvation dominates. In DMSO the order reverses and fluoride becomes the strongest of the halides.
• "SN1 gives exactly 50:50 racemization." Usually a small excess of inversion survives because of ion pairing, and observing that excess is itself evidence for the mechanism.
• "Inversion means the R/S letter must change." The geometry always inverts; the letter changes only if the incoming group has the same CIP rank as the one that left.

Recap

• Nucleophilic substitution swaps a nucleophile in for a leaving group on carbon.
• SN2 is one concerted backside-attack step: rate = k[substrate][nucleophile], inversion, favors methyl and primary.
• SN1 is two steps through a carbocation: rate = k[substrate], near-racemization with a small inversion excess from ion pairing, favors tertiary.
• Substitution slows SN2 by more than 10⁴ and accelerates SN1 by about 10⁶; the neopentyl case proves the SN2 effect is steric.
• Good leaving groups are weak, stable bases; rank them by the pK_a of their conjugate acid, and convert -OH to water or a tosylate before expecting it to leave.
• Strong nucleophile and polar aprotic solvent favor SN2; weak nucleophile and polar protic solvent favor SN1. Moving from methanol to DMSO can accelerate an SN2 reaction by roughly a million-fold and reverses the halide nucleophilicity order.
• Analyse in order - substrate, nucleophile or base, solvent, temperature - and remember that secondary substrates are decided by the conditions.

Sources

1. McMurry, J. (2023). The SN2 Reaction. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Characteristics of the SN2 Reaction. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). The SN1 Reaction. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Characteristics of the SN1 Reaction. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). The Discovery of Nucleophilic Substitution Reactions. In Organic Chemistry (OpenStax). openstax.org
6. LibreTexts Chemistry. Characteristics of the SN1 Reaction. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Hughes, E. D., and Ingold, C. K. (1935). Mechanism of substitution at a saturated carbon atom. Part IV. Journal of the Chemical Society, 244-255. DOI: 10.1039/JR9350000244. find source ↗

The big picture

Elimination reactions are the reverse of addition: instead of opening a double bond, they create one by kicking out a leaving group and a neighboring hydrogen. This lesson covers the two elimination mechanisms, E1 and E2, and then ties all four reactions of alkyl halides together so you can predict, for any substrate and reagent, whether you will get substitution or elimination. This decision is a capstone skill of the first organic course.

Elimination reactions do the opposite of addition: a leaving group and a neighboring hydrogen are removed from adjacent carbons, and a new C=C double bond forms between them. The hydrogen removed comes from the beta carbon, the carbon next to the one bearing the leaving group. Alkyl halides undergo elimination in competition with substitution, through two mechanisms that parallel SN1 and SN2.

The E2 mechanism

E2 (Elimination, Bimolecular) is a concerted, one-step reaction, meaning several bonds change at once. A base removes a hydrogen from the carbon next to the leaving group at the same instant the leaving group departs, and the electrons from the broken C-H bond swing over to form the double bond. Because the base and substrate both appear in the slow step, rate = k[substrate][base]. E2 is favored by a strong, often bulky base (such as hydroxide or an alkoxide) and works on primary, secondary, and tertiary substrates.

Key idea: E2 removes H and the leaving group in one concerted step; its rate depends on both substrate and base, and a strong base drives it.

E2 arrow by arrow, and the isotope experiment that proves it

Take 2-bromobutane with sodium ethoxide, and place all three arrows.

1. Arrow 1. Tail on a lone pair of the ethoxide oxygen, head at a hydrogen on C3, the beta carbon. The base is reaching for a proton, not for carbon.
2. Arrow 2. Tail on that C3-H sigma bond, head into the space between C2 and C3. Those electrons become the new pi bond. This is the arrow beginners forget, and without it the base would simply be making a carbanion.
3. Arrow 3. Tail on the C2-Br sigma bond, head onto bromine. Bromide departs with the pair. This arrow is compulsory in the same step, because otherwise C2 would end up with five bonds.
4. Product. But-2-ene, plus ethanol and bromide. Both carbons rehybridize from sp³ to sp² as the pi bond forms.

Why should anyone believe all three happen at once rather than in sequence? Because of a clean isotope experiment. Replace the beta hydrogens with deuterium and measure the rate. A C-D bond is harder to break than a C-H bond because deuterium is heavier and sits lower in the vibrational well, so if that bond is breaking in the rate-determining step the reaction must slow down measurably. E2 reactions show a primary kinetic isotope effect of k(H)/k(D) around 6 to 8: they run six to eight times slower with deuterium.

Now run the same test on an E1 reaction. The isotope effect is near 1, typically 1.0 to 1.5. Nothing changes, because in E1 the slow step is ionization and the C-H bond is not touched until afterwards. One experiment, two mechanisms, cleanly separated - and it is the strongest evidence that E2 really is concerted.

Key idea: E2 uses three simultaneous arrows, and a primary kinetic isotope effect of 6 to 8 - against roughly 1 for E1 - proves the C-H bond breaks in the rate-determining step.

E2 geometry: anti-periplanar

E2 has a strict geometric requirement: the C-H bond being broken and the C-leaving-group bond must be anti-periplanar, meaning they lie in the same plane pointing in opposite directions (180 degrees apart), like two hands reaching in opposite directions along a line. This alignment lets the electrons flow smoothly into the forming pi bond. The requirement matters most in rings, where the leaving group and the beta hydrogen must both be axial on a cyclohexane chair for E2 to occur.

Key idea: E2 needs the breaking C-H and C-leaving-group bonds anti-periplanar (180 degrees apart), which in rings means both must be axial.

The reason is orbital overlap. As the C-H bond breaks, its electrons must slide sideways into the p orbital that the departing leaving group leaves behind on the adjacent carbon. Only the anti-periplanar arrangement lines those orbitals up in the same plane, so only that arrangement lets a pi bond form without the electrons having to travel around a corner.

Worked example on a ring. Cyclohexane makes the requirement visible, because the chair fixes every bond as either axial or equatorial. Compare two menthyl-type chlorides that differ only in whether the chlorine sits axial or equatorial in the favoured chair.

• When the chlorine is axial, the two beta hydrogens that are also axial sit exactly anti-periplanar to it. E2 proceeds quickly, and it can choose the beta carbon that yields the more substituted Zaitsev alkene.
• When the chlorine is equatorial, no beta hydrogen is anti-periplanar to it at all. The ring must first flip into the far less stable chair - which forces the bulky substituents axial - before elimination can happen. That isomer reacts roughly two orders of magnitude more slowly, and once flipped it has only one anti-periplanar hydrogen available, so it gives the less substituted alkene exclusively, in direct violation of Zaitsev.

That is a striking result worth sitting with: two compounds differing only in stereochemistry give different products at very different rates, purely because of which bonds can line up. Geometry, not thermodynamics, is in charge.

The same requirement makes E2 stereospecific in open chains. If a substrate has stereocenters at both the alpha and beta carbons, each diastereomer has only one hydrogen that can reach the anti-periplanar position, and rotating the molecule to place it there fixes where the remaining groups end up. One diastereomer therefore gives the E alkene and the other gives the Z alkene, cleanly. A mechanism that produced a free carbanion or cation could not do that.

Key idea: The anti-periplanar rule is an orbital-overlap requirement, and it makes E2 both rate-sensitive to conformation and stereospecific in its alkene geometry.

The E1 mechanism

E1 (Elimination, Unimolecular) mirrors SN1. First the leaving group departs to make a carbocation (a positively charged carbon), the slow, rate-determining step; then a base removes a neighboring hydrogen to form the double bond. The rate is k[substrate] only. Like SN1, E1 prefers tertiary substrates that give stable carbocations, and it often accompanies SN1 under the same conditions (weak base, polar protic solvent, heat).

Key idea: E1 is two steps through a carbocation, its rate depends only on the substrate, and it favors tertiary carbons alongside SN1.

Zaitsev's rule

When more than one alkene can form, Zaitsev's rule predicts the major product: elimination favors the more substituted (more stable) alkene, the one with more carbon groups attached to the double-bond carbons. So removing HBr from 2-bromobutane gives mostly but-2-ene (more substituted) rather than but-1-ene. An exception: a very bulky base (like tert-butoxide) is too big to reach the crowded interior hydrogen and instead removes a less hindered one, giving the less substituted (Hofmann) alkene. Heat generally favors elimination over substitution because forming the alkene increases the number of molecules and is entropically favored.

Key idea: Zaitsev's rule favors the more substituted alkene, but a bulky base reverses this to give the less substituted (Hofmann) product.

Why Zaitsev, and why a bulky base overturns it

Zaitsev's rule is not a separate law; it is Lesson 8's alkene stability data reappearing. Heats of hydrogenation showed that a more substituted alkene sits lower in energy, and because the E2 transition state already has substantial double-bond character, the transition state leading to the more substituted alkene is the lower one. The product ratio follows the transition-state energies, not the product energies directly, but here the two point the same way.

Now change the base. Ethoxide is small and reaches the crowded interior hydrogen easily, so 2-bromo-2-methylbutane with sodium ethoxide gives mostly the trisubstituted Zaitsev alkene, on the order of 70 percent. Switch to potassium tert-butoxide, whose three methyl groups make it enormous, and the interior hydrogen becomes hard to reach while the exposed methyl hydrogens do not. The ratio inverts, and the terminal Hofmann alkene becomes the major product, typically around 70 to 75 percent.

Notice what changed and what did not. The thermodynamics of the products are identical in both experiments. Only the accessibility of the proton changed, which shifted which transition state is lower. This is regiochemical control by sterics, and it is a tool: choose ethoxide when you want the Zaitsev product and tert-butoxide when you want the Hofmann one.

Two further situations override Zaitsev. A quaternary ammonium leaving group, as in the Hofmann elimination, is bulky and positively charged and also gives the less substituted alkene. And an E1cB mechanism - base removes an unusually acidic beta hydrogen first, forming a carbanion, which then expels a poor leaving group - operates when the beta hydrogen sits next to a carbonyl. E1cB is common in biochemistry, where enzymes routinely deprotonate alpha to a carbonyl and then eliminate.

Key idea: Zaitsev follows from alkene stability acting on the transition state, and a bulky base, a bulky leaving group, or an E1cB pathway can each override it.

Temperature: why heat means elimination

The rule "heat favours elimination" has a thermodynamic reason worth stating. Elimination converts two particles into three - alkene, protonated base, and leaving group - while substitution converts two into two. Elimination therefore has the more positive entropy change. Since delta G = delta H - T delta S, the favourable -T delta S term grows with temperature, so raising the temperature always shifts the balance toward elimination.

In practice this is why the same alkyl halide with the same alkoxide gives mostly ether at room temperature and mostly alkene on reflux. Temperature is a genuine control variable, not a detail.

Key idea: Elimination increases the particle count and so has a more positive delta S, which is why the -T delta S term makes heat favour elimination over substitution.

Substitution versus elimination

All four pathways can compete, but a few guidelines usually settle which wins:

• Strong bulky base + heat pushes toward elimination (E2).
• Strong small nucleophile that is a weak base favors SN2.
• Weak nucleophile/base in a polar protic solvent on a tertiary substrate gives an SN1/E1 mixture.
• Primary substrates almost never do SN1 or E1 (a primary carbocation is too unstable).

Reading the substrate class and the strength and bulk of the reagent, together with temperature, will let you predict the dominant pathway in most textbook cases. These four reactions, SN1, SN2, E1, and E2, form a connected system, and understanding the competition among them is one of the central achievements of a first organic course.

Key idea: Classify the substrate, then judge the reagent: strong bulky base and heat favor E2, strong small nucleophiles favor SN2, and weak reagents on tertiary substrates give SN1/E1.

The complete decision table

Here is the whole system in one place. Read down the substrate column first, then across.

Substrate	Weak nucleophile/base (H2O, ROH)	Strong nucleophile, weak base (I-, CN-, N3-, RS-)	Strong small base (HO-, EtO-)	Strong bulky base (t-BuO-, LDA)
Methyl	no reaction	SN2	SN2	SN2 (slow)
Primary	no reaction	SN2	mostly SN2, some E2	E2
Secondary	slow SN1 and E1	SN2	E2 dominant	E2
Tertiary	SN1 and E1 mixture	E2 (SN2 is blocked)	E2	E2

Three cells are worth memorizing because they are the ones exams probe. Primary substrate plus bulky base gives E2, not SN2, because the base cannot reach carbon but can always reach a proton. Tertiary substrate plus any strong base or nucleophile gives E2, because backside attack is blocked but beta hydrogens are plentiful. And tertiary substrate plus a weak nucleophile in a protic solvent gives an SN1/E1 mixture, which is why solvolysis reactions are rarely clean.

Then apply two modifiers. Raising the temperature always shifts the balance toward elimination. And a primary or methyl substrate can never do SN1 or E1, because the required cation will not form.

Key idea: Four substrate classes crossed with four reagent types cover nearly every case; then adjust for temperature, and never allow a primary cation.

Where people get stuck

• "Elimination always beats substitution when a base is present." Only strong or bulky bases, and higher temperatures, clearly favour it. A strong nucleophile that is a weak base, such as cyanide or azide, still does SN2 on a primary substrate.
• "E1 rate depends on the base concentration." The base is absent from the slow ionization step, so the rate depends only on the substrate. Adding base changes what the cation becomes, not how fast it forms.
• "Zaitsev's rule has no exceptions." A bulky base, a bulky charged leaving group, an E1cB mechanism, or a ring that cannot reach the required anti-periplanar geometry all overturn it.
• "E2 works no matter how the atoms are arranged." It needs the breaking C-H and the C-leaving-group bond anti-periplanar. On a cyclohexane that means both must be axial, and an equatorial leaving group is close to unreactive.
• "Being a strong base is the same as being a strong nucleophile." Basicity is thermodynamic and aimed at protons; nucleophilicity is kinetic and aimed at carbon. tert-Butoxide is a very strong base and a very poor nucleophile, which is precisely why it is the reagent of choice for elimination.
• "E1 and E2 give the same product mixture." Both tend to obey Zaitsev, but only E2 is stereospecific. E1 goes through a flat cation that has forgotten the original geometry, so it gives whichever alkene is more stable rather than whichever alignment was available.
• "A deuterium label is just a tracer." A primary kinetic isotope effect near 7 is quantitative evidence that the C-H bond breaks in the rate-determining step, which is how E2 was distinguished from E1 in the first place.

Recap

• Elimination removes a leaving group and a beta hydrogen to form a C=C double bond.
• E2 is one concerted step with three arrows: rate = k[substrate][base], needs a strong base and anti-periplanar geometry, and shows a kinetic isotope effect of 6 to 8.
• E1 is two steps through a carbocation: rate = k[substrate], no isotope effect, favors tertiary substrates, accompanies SN1.
• The anti-periplanar requirement is an orbital-overlap condition, which makes E2 stereospecific and makes an equatorial leaving group on a ring nearly unreactive.
• Zaitsev's rule gives the more substituted alkene, except that a bulky base such as tert-butoxide gives the less substituted Hofmann product.
• Elimination raises the particle count, so its more positive delta S makes heat favour elimination over substitution.
• Predict SN1, SN2, E1, or E2 from substrate class, then reagent strength and bulk, then solvent and temperature.

Sources

1. McMurry, J. (2023). Elimination Reactions: Zaitsev's Rule. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). The E2 Reaction and the Deuterium Isotope Effect. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). The E2 Reaction and Cyclohexane Conformation. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). The E1 and E1cB Reactions. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). A Summary of Reactivity: SN1, SN2, E1, E1cB, and E2. In Organic Chemistry (OpenStax). openstax.org
6. LibreTexts Chemistry. A Summary of Reactivity: SN1, SN2, E1, E1cB, and E2. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. Clayden, J., Greeves, N., and Warren, S. (2012). Organic Chemistry (2nd ed.), Chapter 19: Elimination Reactions. Oxford University Press. find source ↗

The big picture

Alcohols sit at the crossroads of organic chemistry: they are useful in their own right and they convert into aldehydes, ketones, acids, alkenes, and alkyl halides. This lesson covers how to classify alcohols, why their hydroxyl group gives them high boiling points and water solubility, and their two signature reactions, oxidation and dehydration. These reactions connect back to the alkene and substitution chapters.

Alcohols carry the hydroxyl group (-OH) on a carbon, and they are among the most useful compounds in organic chemistry, serving as solvents, fuels, and stepping-stones to many other functional groups. Their chemistry follows directly from that polar O-H bond.

Classifying alcohols

Alcohols are labeled by the carbon that bears the -OH:

• Primary (1°): the OH carbon is attached to one other carbon, as in ethanol CH₃CH₂OH.
• Secondary (2°): the OH carbon is attached to two other carbons, as in propan-2-ol.
• Tertiary (3°): the OH carbon is attached to three other carbons, as in 2-methylpropan-2-ol.

This classification matters because primary, secondary, and tertiary alcohols behave differently, especially toward oxidation and substitution.

Key idea: Classify an alcohol by counting carbons on the OH-bearing carbon; the class predicts how it oxidizes and substitutes.

Hydrogen bonding and physical properties

The O-H bond is very polar, so alcohol molecules form hydrogen bonds with one another: a hydrogen bond is an especially strong attraction between an O-H (or N-H) hydrogen and a lone pair on a nearby oxygen or nitrogen, like tiny magnets clicking together. The partially positive H of one -OH is attracted to the lone pair on the oxygen of another.

This strong intermolecular attraction gives alcohols much higher boiling points than alkanes or ethers of similar size. Ethanol boils at 78 °C while propane, of comparable mass, boils at about -42 °C. Small alcohols also mix freely with water because they hydrogen-bond to it; as the carbon chain grows, the nonpolar part dominates and water solubility drops.

Key idea: Hydrogen bonding from the -OH group raises alcohol boiling points and makes small alcohols water-soluble, an effect that fades as the carbon chain grows.

The cleanest demonstration is a comparison of true isomers. Ethanol and dimethyl ether are both C₂H₆O, identical in molar mass at 46 g/mol. Ethanol boils at 78 °C; dimethyl ether boils at -24 °C. Both molecules are polar and both have oxygen lone pairs, but only ethanol has an O-H hydrogen to donate a hydrogen bond. That single difference is worth over 100 degrees of boiling point.

Alcohols as acids: how the -OH behaves

An alcohol is a weak acid, and the small differences among alcohols repay attention. In water: methanol 15.5, ethanol 16.0, propan-2-ol 17.1, and 2-methylpropan-2-ol 18.0, with water itself at 15.7 for comparison.

The trend looks like an electronic effect but is largely a solvation effect. In solution, a bulky alkoxide such as tert-butoxide is harder for water molecules to surround and stabilize, so it is a higher-energy anion and its alcohol is a weaker acid. The evidence is that in the gas phase, with no solvent at all, the order reverses completely and tert-butanol becomes the strongest of the four, because the larger alkyl group is more polarizable and stabilizes the charge better once solvation is removed. That inversion is a standing warning: much of what looks like intrinsic molecular behaviour in solution chemistry is really the solvent talking.

Put an electron-withdrawing group nearby and induction takes over, exactly as in Lesson 11. 2,2,2-Trifluoroethanol has a pK_a of 12.4, more than three units below ethanol, because three fluorines pull charge off the alkoxide oxygen.

Phenols are a different matter entirely. Phenol has a pK_a of 10.0, about a million times more acidic than cyclohexanol, because the phenoxide anion delocalizes its charge into the aromatic ring - onto the ortho and para carbons specifically. Add nitro groups at those positions and the effect compounds: 4-nitrophenol is 7.15, 2,4-dinitrophenol is 4.1, and picric acid, 2,4,6-trinitrophenol, is 0.4 - a stronger acid than acetic acid by four orders of magnitude. Practically, this means dilute NaOH deprotonates a phenol completely but barely touches an alcohol, which is a standard way to separate the two.

Key idea: Alcohol pK_a runs 15.5 to 18 and is governed mostly by how well the solvent can stabilize the alkoxide, while phenol at 10.0 is far more acidic because its anion delocalizes into the ring.

Oxidation of alcohols

A defining reaction of alcohols is oxidation, which removes hydrogen and forms a C=O bond. The outcome depends on the class:

Alcohol	Oxidizes to
Primary	Aldehyde, then carboxylic acid
Secondary	Ketone
Tertiary	No reaction (no H on the OH carbon)

A tertiary alcohol cannot be oxidized this way because its OH carbon has no hydrogen to remove; forming a C=O would require carbon to have five bonds. This is a common and useful way to distinguish the three classes in the lab. A strong oxidant (such as chromic acid) takes a primary alcohol all the way to a carboxylic acid, while a milder, specialized oxidant can be used to stop at the aldehyde.

Key idea: Primary alcohols oxidize to aldehydes then acids, secondary alcohols oxidize to ketones, and tertiary alcohols do not oxidize because their OH carbon has no hydrogen.

How the oxidation actually works, and what chemists use now

The classical chromium oxidation has a mechanism worth knowing, because it explains the whole pattern in one stroke. The alcohol oxygen attacks chromium to form a chromate ester, R₂CH-O-CrO₃H. A base then removes the hydrogen from the carbon bearing that oxygen while the Cr-O bond breaks, in what is essentially an E2-like step: the C-H electrons swing across to form the C=O bond and chromium leaves in a reduced oxidation state.

Everything follows. A tertiary alcohol forms the chromate ester perfectly well but has no hydrogen on that carbon, so the elimination step cannot happen and the reaction stops. A primary alcohol produces an aldehyde, but in water the aldehyde hydrates to a 1,1-diol, which still has a C-H on the oxygen-bearing carbon and is oxidized again straight through to the carboxylic acid. Stopping at the aldehyde therefore means excluding water - which is exactly what pyridinium chlorochromate (PCC) in dichloromethane does.

The field has moved on from chromium, and for good reason: Cr(VI) is a confirmed human carcinogen and its waste is a serious disposal problem. Current practice favours the Swern oxidation (oxalyl chloride and dimethyl sulfoxide, then triethylamine, at -78 °C), the Dess-Martin periodinane, a mild hypervalent-iodine reagent that works at room temperature, and catalytic TEMPO with household bleach as the stoichiometric oxidant, which is cheap and produces salt water as its main by-product. All three stop cleanly at the aldehyde and none of them uses a heavy metal. A modern synthesis paper reaching for Jones reagent would look distinctly dated.

Key idea: Oxidation goes through a chromate ester and an E2-like step, which is why tertiary alcohols cannot oxidize and why water carries a primary alcohol past the aldehyde - and modern chemistry replaces chromium with Swern, Dess-Martin, or TEMPO.

Dehydration to alkenes

Alcohols undergo dehydration (loss of water) with acid and heat to form alkenes, an elimination reaction that reverses the acid-catalyzed hydration of an alkene. The mechanism is essentially E1: the acid first protonates the -OH to turn it into water, a good leaving group; water then leaves to give a carbocation; finally a base removes a beta hydrogen to form the double bond. Because a carbocation forms, dehydration is easiest for tertiary alcohols and follows Zaitsev's rule, giving mainly the more substituted alkene.

Key idea: Acid-catalyzed dehydration converts an alcohol to an alkene by an E1-like path, so it favors tertiary alcohols and the more substituted (Zaitsev) alkene.

The mechanism, arrow by arrow. Dehydrate 2-methylbutan-2-ol with concentrated H₂SO₄ and heat.

1. Protonation. One arrow from a lone pair on the alcohol oxygen to the acidic proton, and a second from the H-O bond of the acid onto its oxygen. The alcohol becomes an oxonium ion, R-OH₂⁺. This step is fast and reversible, and it is the whole point of the acid: pK_a 15.7 has just become pK_a -1.7, so hydroxide has become water.
2. Ionization. One arrow from the C-O bond onto the oxygen. Water leaves; a tertiary carbocation remains. This is the slow, rate-determining step, which is why the reaction needs heat and why tertiary alcohols dehydrate far faster than primary ones.
3. Deprotonation. A weak base, usually another alcohol molecule or the bisulfate ion, takes a hydrogen from a beta carbon. Two arrows: base to hydrogen, and the C-H bond into the space between the two carbons to become the pi bond.
4. Product. 2-methylbut-2-ene, the more substituted Zaitsev alkene, dominates over 2-methylbut-1-ene.

Because a free carbocation is genuinely formed, the rearrangements of Lesson 13 apply in full. Dehydrating 3,3-dimethylbutan-2-ol gives mainly 2,3-dimethylbut-2-ene, because the secondary cation formed first undergoes a 1,2-methyl shift to a tertiary cation before losing a proton. If a dehydration product looks like it has the wrong skeleton, a shift is the reason.

Primary alcohols are a special case. They cannot form a primary cation, so their dehydration proceeds by an E2-like route in which the base removes the beta hydrogen while water leaves - and it needs harsher conditions. In practice, primary alcohols are usually dehydrated by other means, such as POCl₃ in pyridine, which works at low temperature and avoids rearrangement entirely.

Conversion to alkyl halides

The -OH of an alcohol is a poor leaving group on its own (hydroxide is a strong base), so alcohols do not readily undergo substitution directly. Chemists get around this by first converting the -OH into a better leaving group, for example by using reagents such as HX, SOCl₂, or PBr₃, which turn the alcohol into an alkyl halide. This is why alcohols are such valuable starting materials: they can be transformed into the alkyl halides that feed the substitution and elimination chemistry of earlier lessons.

Key idea: Because -OH is a poor leaving group, alcohols must be activated (for example to alkyl halides) before they can undergo substitution.

Which activation you choose determines the stereochemistry, so the choice is not arbitrary.

• HX with a tertiary alcohol. Protonation then ionization gives a carbocation, so the reaction is SN1 and a stereocenter racemizes. Rearrangement is possible.
• HX with a primary alcohol. Protonation then backside attack by halide, so the reaction is SN2 and the stereocenter inverts. No rearrangement.
• PBr₃ and SOCl₂. These convert the -OH into a phosphite or chlorosulfite ester, an excellent leaving group, and halide then attacks from the back face. The result is clean inversion, under mild conditions and without carbocations, which is why they are preferred over HX for secondary alcohols.
• Tosylation with TsCl in pyridine. This is the most elegant option. The alcohol oxygen attacks the sulfur of TsCl, so the C-O bond is never broken and the configuration is retained. What you now have is a substrate with a superb leaving group and an untouched stereocenter, which any nucleophile can then displace with inversion. Two steps, one predictable inversion, and no cation at any point.

Recognizing that -OH must be activated is arguably the single most useful practical idea in the whole lesson, because it is how alcohols connect to every other functional group in the course.

Key idea: Activation method sets stereochemistry: HX on a tertiary alcohol racemizes, PBr₃ and SOCl₂ invert, and tosylation retains configuration so that the following displacement inverts it once, cleanly.

Where people get stuck

• "All alcohols oxidize the same way." Primary gives aldehyde then acid, secondary gives ketone, tertiary gives nothing - because the elimination step needs a hydrogen on the carbon bearing the oxygen.
• "Alcohols undergo substitution as easily as alkyl halides." Hydroxide has a conjugate acid pK_a of 15.7 and does not leave. Protonate it, tosylate it, or convert it with PBr₃ or SOCl₂ first.
• "Alcohols boil high because they are polar." Polarity alone is not enough. Dimethyl ether is polar too and boils 100 degrees lower than its isomer ethanol. What matters is having an O-H to donate a hydrogen bond.
• "Dehydration gives the least substituted alkene." It goes through a carbocation and obeys Zaitsev, giving the more substituted alkene - and it can rearrange first.
• "tert-Butanol is the most acidic alcohol because alkyl groups donate electrons." In water it is the least acidic of the simple alcohols, at pK_a 18, because its bulky anion is poorly solvated. The order reverses in the gas phase, which shows the effect is solvation rather than induction.
• "Phenols are just aromatic alcohols." Phenol at pK_a 10.0 is about a million times more acidic than cyclohexanol, because the phenoxide charge delocalizes into the ring. Dilute NaOH deprotonates one and not the other.
• "Any oxidizing agent will stop at the aldehyde." Only under anhydrous conditions. In water the aldehyde hydrates and is oxidized straight on to the acid, which is why PCC, Swern, Dess-Martin, or TEMPO are used when the aldehyde is the target.

Recap

• Alcohols are classified primary, secondary, or tertiary by the carbons on the OH-bearing carbon.
• Hydrogen bonding raises alcohol boiling points and makes small alcohols water-soluble; ethanol boils 100 degrees above its isomer dimethyl ether.
• Alcohol pK_a runs 15.5 to 18 and is set largely by solvation of the alkoxide, while phenol at 10.0 is far more acidic through resonance.
• Oxidation goes via a chromate ester and an E2-like step, giving aldehydes then acids (primary), ketones (secondary), or nothing (tertiary).
• Modern practice replaces chromium with PCC, Swern, Dess-Martin, or catalytic TEMPO with bleach.
• Acid-catalyzed dehydration is an E1 elimination through a carbocation, favoring tertiary alcohols, obeying Zaitsev, and capable of rearrangement.
• Because -OH is a poor leaving group, alcohols are activated first, and the choice of activation fixes whether the product is racemized, inverted, or retained.

Sources

1. McMurry, J. (2023). Properties of Alcohols and Phenols. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Reactions of Alcohols. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Oxidation of Alcohols. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Preparing Alkyl Halides from Alcohols. In Organic Chemistry (OpenStax). openstax.org
5. Flowers, P., et al. (2019). Alcohols and Ethers. In Chemistry 2e (OpenStax). openstax.org
6. LibreTexts Chemistry. Oxidation of Alcohols. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. National Institute of Standards and Technology. NIST Chemistry WebBook: ethanol, thermophysical and spectral data. webbook.nist.gov

The big picture

Benzene and its relatives, the aromatic compounds, are among the most stable and widespread structures in chemistry, found in fuels, plastics, dyes, and half the drugs on the market. This lesson explains what makes a ring aromatic, why that stability makes benzene behave so differently from an ordinary alkene, and the reaction (electrophilic aromatic substitution) that results. It ties together resonance, hybridization, and mechanism from the whole course.

Aromatic compounds are built around the benzene ring, one of the most stable and important structures in all of chemistry. Benzene is a flat, six-membered ring of carbons with the formula C₆H₆. Each carbon is sp² hybridized, bonded to two neighboring carbons and one hydrogen, with 120° angles around a perfectly planar hexagon.

Delocalized electrons

Each of benzene's six carbons has one leftover p orbital perpendicular to the ring, and each holds one electron. Rather than pairing up into three fixed, alternating double bonds, these six electrons delocalize into a continuous ring of pi electron density above and below the plane. To delocalize means the electrons are spread over many atoms rather than stuck between two, like heat spread evenly through a room instead of concentrated at one radiator.

Benzene is best described as a resonance hybrid: a resonance hybrid is a single real structure that is the blend of two or more Lewis drawings, not a molecule flickering between them. Two equivalent structures with double bonds in different positions contribute equally, and the true molecule is the average. A telling piece of evidence is that all six carbon-carbon bonds in benzene are exactly the same length, intermediate between a single and a double bond, which no single Kekule structure with alternating bonds could explain. Benzene is often drawn as a hexagon with a circle inside to represent this shared, delocalized system.

Key idea: Benzene's six pi electrons are delocalized around the whole ring, so it is one resonance hybrid with six equal-length bonds, not alternating single and double bonds.

Aromatic stability and Huckel's rule

This delocalization makes benzene remarkably stable, far more so than three isolated double bonds would be. That extra stability is called aromatic stabilization (or resonance energy), and it is substantial, roughly 150 kJ/mol. A ring is aromatic when it meets four conditions: it is cyclic, planar, fully conjugated (a p orbital on every ring atom, so the pi system is unbroken), and contains 4n + 2 pi electrons where n is a whole number (2, 6, 10, and so on).

This last condition is Huckel's rule. Benzene has six pi electrons (n = 1), so it is aromatic. A ring with 4n pi electrons (like 4 or 8) that is otherwise conjugated is instead antiaromatic and unstable, which is why such systems avoid planarity.

Key idea: A ring is aromatic if it is cyclic, planar, fully conjugated, and has 4n + 2 pi electrons (Huckel's rule); benzene qualifies with six.

Measuring aromatic stabilization: the hydrogenation experiment

The 150 kJ/mol figure is not an estimate; it comes from a straightforward comparison. Hydrogenate cyclohexene, with one double bond, and 118 kJ/mol is released. If benzene were simply "cyclohexatriene" - three ordinary, independent double bonds - hydrogenating it should release three times that, about 354 kJ/mol. The measured value is 206 kJ/mol.

The arithmetic is the whole argument: 354 - 206 = 148 kJ/mol. Benzene starts about 148 kJ/mol lower in energy than three isolated double bonds would predict, and that shortfall is the aromatic stabilization energy. For comparison, 1,3-cyclohexadiene, which is merely conjugated and not aromatic, releases 230 kJ/mol against an expected 236, so ordinary conjugation is worth only about 6 kJ/mol. Aromaticity is roughly twenty-five times larger than plain conjugation, and that gap is why benzene behaves like an entirely different class of compound.

Key idea: Heats of hydrogenation put aromatic stabilization at about 148 kJ/mol - three-double-bond expectation of 354 minus the measured 206 - against only a few kJ/mol for ordinary conjugation.

Substitution, not addition

Here is the key chemical consequence. An ordinary alkene readily undergoes addition, because opening the pi bond costs little and gains two strong single bonds. Benzene does not, because addition would destroy the aromatic ring and forfeit that large stabilization energy. Instead, benzene undergoes electrophilic aromatic substitution (EAS): an electrophile replaces one of the ring hydrogens, and the aromatic system is preserved.

The EAS mechanism has two steps described in words. In step one, the ring's pi electrons act as a nucleophile and attack the electrophile, forming a new bond and breaking aromaticity to give a resonance-stabilized carbocation intermediate (called the arenium ion).

In step two, a base removes the hydrogen from the carbon that was attacked, and the electrons drop back into the ring, restoring the full aromatic system. The net result is that an H is swapped for the electrophile while the ring survives. For example, benzene reacts with a nitronium electrophile (NO₂⁺) to give nitrobenzene, swapping an H for an NO₂ group while keeping the ring intact.

Key idea: Benzene reacts by electrophilic aromatic substitution (add electrophile, then lose H) so it keeps its aromatic stability, whereas an alkene simply adds across its double bond.

The arrows, and the arenium ion in detail

Nitrate benzene with HNO₃ and H₂SO₄ and follow every arrow.

1. Making the electrophile. Sulfuric acid protonates nitric acid, and water leaves, generating the linear nitronium ion, NO₂⁺. Benzene is a poor nucleophile, so the electrophile must be exceptionally strong before anything happens.
2. Arrow 1. Tail on one of the ring's pi bonds, head at the nitrogen of NO₂⁺. A new C-N bond forms, that ring carbon becomes sp³, and the ring loses its aromaticity - which is exactly why this step is slow and uphill.
3. The intermediate. The result is the arenium ion, sometimes called the sigma complex or Wheland intermediate. It carries a positive charge spread over three ring carbons - the two ortho positions and the para position relative to the point of attack - and you should be able to draw all three resonance forms. The carbon that was attacked is sp³ and bears both the hydrogen and the new group; it is the only ring carbon not sharing the charge.
4. Arrow 2 and 3. A weak base, typically bisulfate, removes the hydrogen from that sp³ carbon: one arrow from the base to the hydrogen, one from the C-H bond back into the ring. Aromaticity is restored and the reaction is strongly downhill.

Notice what does not happen: a nucleophile never adds to the arenium ion. Losing a proton regains about 148 kJ/mol of aromatic stabilization, and nothing an addition could offer competes with that. This is the whole reason benzene substitutes rather than adds.

The same two-step pattern, differing only in how the electrophile is generated, runs the standard reactions: halogenation (Br₂ with FeBr₃, giving Br⁺), sulfonation (SO₃ in fuming H₂SO₄), Friedel-Crafts alkylation (RCl with AlCl₃, giving R⁺), and Friedel-Crafts acylation (RCOCl with AlCl₃, giving a resonance-stabilized acylium ion). Alkylation has two well-known flaws - the carbocation can rearrange, and the product is more reactive than the starting material so polyalkylation follows - which is why acylation, whose acylium ion cannot rearrange and whose ketone product is deactivated, is usually preferred.

Key idea: EAS is one uphill step forming a charge-delocalized arenium ion and one downhill step restoring aromaticity, and every EAS reaction differs only in how its electrophile is made.

Where does the second group go? Directing effects

Once one substituent is present, the ring is no longer symmetric and an incoming electrophile has a choice. Substituents sort into three groups.

Class	Examples	Effect on rate	Directs to
Strongly activating	-NH2, -NHR, -OH, -OR	much faster than benzene	ortho and para
Weakly activating	-CH3, alkyl, -C6H5	faster	ortho and para
Weakly deactivating	-F, -Cl, -Br, -I	slower	ortho and para
Strongly deactivating	-NO2, -CN, -SO3H, -CHO, -COR, -COOH, -CF3, -NR3+	much slower	meta

The resonance argument, done properly. The whole pattern comes from one question: in the arenium ion, does the substituent sit on a carbon that carries positive charge?

Recall that the arenium ion's positive charge sits on the carbons ortho and para to the point of attack. Work out where that puts things for an -OH group.

• Attack ortho or para to the -OH. One of the three resonance structures places the positive charge directly on the carbon bearing the -OH. Oxygen can then donate a lone pair into that empty position, giving a fourth resonance structure in which every atom has a complete octet and the positive charge sits on oxygen rather than carbon. That is a large extra stabilization of the intermediate, and by the Hammond postulate it stabilizes the transition state leading to it.
• Attack meta to the -OH. None of the three resonance structures places the charge on the substituted carbon, so the oxygen lone pair can do nothing at all. No fourth structure, no extra stabilization.

Ortho and para attack therefore have lower barriers, and the product is ortho and para substituted. The same reasoning applies to -NH₂, -OR, and -NHCOR, and nitrogen donates even better than oxygen, which is why aniline is one of the most reactive aromatic compounds known. Alkyl groups direct ortho and para for a weaker version of the same reason: they cannot donate a lone pair, but they stabilize adjacent positive charge by hyperconjugation and induction.

Now reverse it for a nitro group. Nitrogen in -NO₂ carries a formal positive charge, so the ring carbon it is attached to is strongly electron-poor.

• Attack ortho or para to the -NO₂. One resonance structure would place the arenium ion's positive charge on the very carbon bearing the positively polarized nitrogen - two positive charges on adjacent atoms. That structure is badly destabilized, so the intermediate as a whole is high in energy.
• Attack meta. No resonance structure places charge on the substituted carbon, so the clash is avoided entirely.

Meta attack is therefore the least bad option. Note the wording: a meta director does not make meta attack fast. Every position is slowed relative to benzene; meta is simply slowed least. Deactivation and meta direction are two faces of one effect.

The halogens are the instructive exception, and they are exceptional precisely because the two effects point in opposite directions. Chlorine is very electronegative, so inductively it withdraws electron density and slows every position - hence deactivating. But chlorine also has lone pairs, so when attack occurs ortho or para it can still donate one into the adjacent positive charge and supply that fourth resonance structure - hence ortho, para-directing. Induction wins on rate; resonance wins on regiochemistry.

Key idea: Ask whether the substituent sits on a charge-bearing carbon of the arenium ion; lone-pair donors stabilize ortho and para attack with a fourth all-octet resonance structure, electron-poor groups destabilize it, and halogens deactivate by induction while still directing ortho and para by resonance.

Worked example: order matters in synthesis

Suppose you need meta-bromonitrobenzene and, separately, para-bromonitrobenzene, from benzene. The reagents are the same in both cases; only the order changes.

For the meta isomer, nitrate first. HNO₃ with H₂SO₄ gives nitrobenzene. The nitro group is a strong deactivator and a meta director, so brominating with Br₂ and FeBr₃ - which will need forcing conditions, since the ring is now deactivated - places bromine meta. Result: 1-bromo-3-nitrobenzene.

For the para isomer, brominate first. Bromobenzene has a weakly deactivating but ortho, para-directing substituent, so subsequent nitration goes to the ortho and para positions, with para dominating because the ortho positions are sterically crowded. Separate the isomers and you have 1-bromo-4-nitrobenzene.

Same two reactions, two different products, decided entirely by which group is already on the ring when the second electrophile arrives. Planning that order is the core skill of aromatic synthesis.

Key idea: In a disubstituted aromatic synthesis, the group installed first directs the second, so choose the sequence that puts the right director on the ring first.

Where people get stuck

• "Benzene has three fixed alternating double bonds." All six C-C bonds are the same length, about 139 pm, between a single bond at 154 and a double at 134. It is one hybrid, not two alternating arrangements.
• "Resonance means the molecule flips between structures." A resonance hybrid is one unchanging structure. Nothing oscillates.
• "Benzene reacts like an alkene and adds reagents." Addition would forfeit about 148 kJ/mol of aromatic stabilization, so benzene substitutes instead.
• "Any flat ring with double bonds is aromatic." It must also be fully conjugated and carry 4n + 2 pi electrons. A 4n count makes it antiaromatic, and such rings distort out of planarity to escape.
• "Meta directors make the meta position fast." They slow every position. Meta is merely slowed the least, because it is the only attack that avoids putting positive charge next to an electron-poor substituent.
• "Halogens must be activating because they direct ortho and para." They deactivate by induction and direct by resonance. Rate and regiochemistry are controlled by different effects and can disagree.
• "Friedel-Crafts alkylation is a reliable way to add an alkyl group." The cation can rearrange, the product is activated so it alkylates again, and the reaction fails entirely on strongly deactivated rings. Acylation followed by reduction is the dependable route.
• "You can do the two steps of an aromatic synthesis in either order." Almost never. The first substituent decides where the second one lands.

Recap

• Benzene is a flat six-carbon ring of sp² carbons with six delocalized pi electrons.
• It is a resonance hybrid with six equal-length bonds of about 139 pm, not alternating single and double bonds.
• Aromaticity requires a cyclic, planar, fully conjugated ring with 4n + 2 pi electrons (Huckel's rule).
• Aromatic stabilization measures about 148 kJ/mol by heats of hydrogenation, roughly twenty-five times ordinary conjugation.
• Benzene undergoes electrophilic aromatic substitution through an arenium ion, keeping its ring, rather than addition.
• Lone-pair donors and alkyl groups activate and direct ortho and para; electron-withdrawing groups deactivate and direct meta; halogens deactivate yet still direct ortho and para.
• The directing pattern follows from whether the substituent sits on a positively charged carbon of the arenium ion, and it dictates the order of steps in aromatic synthesis.

Sources

1. McMurry, J. (2023). Structure and Stability of Benzene. In Organic Chemistry (OpenStax). openstax.org
2. McMurry, J. (2023). Aromaticity and the Huckel 4n + 2 Rule. In Organic Chemistry (OpenStax). openstax.org
3. McMurry, J. (2023). Electrophilic Aromatic Substitution Reactions: Bromination. In Organic Chemistry (OpenStax). openstax.org
4. McMurry, J. (2023). Substituent Effects in Electrophilic Substitutions. In Organic Chemistry (OpenStax). openstax.org
5. McMurry, J. (2023). Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction. In Organic Chemistry (OpenStax). openstax.org
6. LibreTexts Chemistry. An Explanation of Substituent Effects. Organic Chemistry (Morsch et al.). chem.libretexts.org
7. National Institute of Standards and Technology. NIST Chemistry WebBook: benzene, thermochemical data. webbook.nist.gov