Organic Chemistry

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. To reach a stable octet of eight outer electrons, it neither easily gains four nor loses four; instead it shares, forming four covalent bonds.

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. These "normal" valences appear again and again, so memorizing them saves enormous time.

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

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).

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. Any additional bonds in a double or triple bond are pi bonds, formed by orbitals overlapping side-by-side above and below the axis. 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.

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: mixing atomic orbitals to make new, equivalent hybrid orbitals aimed for bonding. 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).

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.

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. Geometry also decides whether a molecule is flat or three-dimensional, which controls how molecules pack, react, and fit into biological targets. 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 double bonds means sp and linear.

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.

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.

Atoms other than carbon 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.

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.

A functional group is a specific arrangement of atoms that gives a molecule characteristic properties and reactions. 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. Saturated and relatively unreactive.
• Alkene: contains a C=C double bond, as in ethene H₂C=CH₂. 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.

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.
• 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.

Groups with nitrogen or halogen

• Amine: nitrogen bonded to carbon, as in methylamine CH₃NH₂. Amines are basic.
• Amide: a carbonyl bonded to nitrogen, -CO-N, the linkage in proteins.
• Haloalkane (alkyl halide): a halogen on carbon, as in chloroethane CH₃CH₂Cl.

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. 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.

Every organic compound has a systematic IUPAC name that any chemist can decode into a structure. The system is built on a small set of steps applied in order. 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.

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

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 (position numbers).
3. Name each branch (substituent) as an alkyl group: a one-carbon branch is methyl, two is ethyl, three is propyl.
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.

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. Careful, consistent application of these four steps handles the great majority of names you will encounter.

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 to signal the principal group.

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

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).

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.

Isomers are different compounds that share the same molecular formula. 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. 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.

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 index of hydrogen deficiency) counts them. Each ring and each pi bond removes two hydrogens compared with a fully saturated chain. 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.

Each degree could be a ring or a double bond; a triple bond counts as two degrees. Nitrogen adds one hydrogen per atom to the saturated count and halogens count like hydrogen, but for pure hydrocarbons the simple formula above is all you need. 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.

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. 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. Ethane spends most of its time near the staggered arrangement but is never frozen there.

Configurational isomers: locked arrangements

By contrast, a configurational isomer cannot interconvert without breaking a bond. The most important example 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.

Some molecules exist as a pair that relate like your left and right hands: mirror images that cannot be superimposed. This property, chirality, 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. 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.

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).

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).

Chirality has real consequences. Many drugs are chiral, and often only one enantiomer is active while the other is inert or harmful, which is why modern synthesis works so hard to make a single pure enantiomer.

A reaction mechanism is a step-by-step account of how bonds break and form as reactants turn into products. 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. Electrons flow from electron-rich to electron-poor sites.

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. 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. 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.

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 atom, which departs 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.

Acid-base chemistry is the most useful tool in the whole course because so many reactions are, at heart, a proton moving from one place to another. Organic chemists lean on the Bronsted-Lowry picture: an acid donates a proton (H⁺) and a base accepts one.

Conjugate acid-base pairs

When an acid loses its proton, what remains is its conjugate base. 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.

pKa measures acid strength

The strength of an acid is captured by its pK_a. A lower pK_a means a stronger acid (more willing to give up its proton). 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.

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. Factors that stabilize a negative charge, and so strengthen the acid, include: the charge sitting on a more electronegative atom, the charge being spread out by resonance, and the atom bearing the charge being larger. 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.

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.

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.

Physical trends

Because alkanes are nonpolar, they attract each other only through weak London dispersion forces. 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. Alkanes are also less dense than water and do not dissolve in it.

Ring strain in cycloalkanes

Cycloalkanes are alkanes closed into a ring (formula C_nH_2n). Small rings suffer ring strain because their bond angles are forced away from the ideal 109.5° of an sp³ carbon. Cyclopropane, a three-membered ring, is locked at 60° angles, far from ideal, so it is highly strained and unusually reactive. Cyclobutane (about 90°) is also strained. The strain eases as rings grow.

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° 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) and one equatorial bond (pointing outward); bulky groups prefer the roomier equatorial position. The stability of the cyclohexane chair is why six-membered rings are so common throughout organic and biological chemistry, from sugars to steroids.

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 - a nucleophile. 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)

When an alkene meets HBr, the pi bond acts as the nucleophile and attacks the H of H-Br (the electrophile). This breaks the pi bond and puts H on one carbon, leaving the other carbon as a positively charged carbocation. The bromide ion (Br^-) then adds to the carbocation. The result is that H and Br add across the former double bond, converting the alkene into a haloalkane.

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:

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.

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. 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.

In nucleophilic substitution, a nucleophile replaces a leaving group on a carbon. 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: 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 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.

The SN1 mechanism

SN1 (Substitution, Nucleophilic, Unimolecular) goes in two steps. First, the leaving group departs on its own to form a carbocation - 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. SN1 needs a stable carbocation, so it prefers tertiary substrates, tolerates weak nucleophiles, and is helped by polar protic solvents that stabilize the ions.

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.

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. 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. 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.

The E1 mechanism

E1 (Elimination, Unimolecular) mirrors SN1. First the leaving group departs to make a carbocation (slow, rate-determining), 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).

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. Heat generally favors elimination over substitution because forming the alkene increases the number of molecules and is entropically favored.

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, E2 - form a connected system, and understanding the competition among them is one of the central achievements of a first organic course.

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.

Hydrogen bonding and physical properties

The O-H bond is very polar, so alcohol molecules form hydrogen bonds with one another: 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.

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. Alcohols also undergo dehydration (loss of water) with acid and heat to form alkenes, an elimination reaction that reverses the hydration of an alkene, showing again how these reactions connect into one web.

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. Benzene is best described as a resonance hybrid: 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.

Aromatic stability

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 is cyclic, planar, fully conjugated (a p orbital on every ring atom), and contains the right number of pi electrons (for benzene, six). This stability is the defining feature of aromatic compounds and governs how they react.

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: an electrophile replaces one of the ring hydrogens, and the aromatic system is preserved. For example, benzene reacts with a nitronium electrophile to give nitrobenzene, swapping an H for an NO₂ group while keeping the ring intact. So while alkenes add and lose their double bond, aromatic rings substitute and keep their special stability, a difference that comes down entirely to protecting the delocalized aromatic system.