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Alkyl Halides

Alkyl halides (R–X, where X = F, Cl, Br, I) are the workhorses of synthetic organic chemistry — they undergo two major reaction families: nucleophilic substitution (SN1, SN2) and elimination (E1, E2). The AAMC MCAT content outline expects you to predict mechanism, product, and reactivity for any given alkyl halide.

AAMC content categories. This chapter spans three subtopics — Nomenclature/Structure/Reactivity, Nucleophilic Substitution Reactions, and Elimination Reactions.

Nomenclature, Structure and Reactivity

An alkyl halide is an alkane in which one (or more) hydrogen has been replaced by a halogen. The C–X bond is polarized δ+C–Xδ — the carbon is electrophilic and is the center of all subsequent reactivity.

IUPAC nomenclature

Classification (1°, 2°, 3°)

Primary (1°)
Halogen on a carbon bonded to one other carbon. Example: CH3CH2Br.
Secondary (2°)
Halogen on a carbon bonded to two other carbons. Example: (CH3)2CHBr.
Tertiary (3°)
Halogen on a carbon bonded to three other carbons. Example: (CH3)3CBr.

Bond strength and reactivity trends

Preparation

From alcohols

R–OH + HX → R–X + H2O (HI > HBr > HCl).
R–OH + PCl3/PCl5 → R–Cl.
R–OH + SOCl2 → R–Cl + SO2↑ + HCl↑ (cleanest method — gaseous by-products).

From alkenes

CH2=CH2 + HBr → CH3CH2Br. With unsymmetrical alkenes, addition follows Markovnikov's rule: H goes to the C with more H's, X to the C with fewer H's (carbocation stability).

Wurtz reaction

2 R–X + 2 Na ⟶{dry ether} R–R + 2 NaX. Used to make symmetrical alkanes; doubles the carbon count. Example: 2 CH3Br + 2 Na → CH3–CH3 + 2 NaBr.

Nucleophilic Substitution Reactions

A nucleophile (Nu) replaces the halide leaving group. Two competing mechanisms: SN2 (one step, bimolecular) and SN1 (two step, via carbocation).

SN2 mechanism — bimolecular, one step

Nu + R–X → [Nu···R···X] → Nu–R + X.
· Rate = k[R–X][Nu] — second-order overall.
· Backside attack → inversion of configuration (Walden inversion).
· Favored by: 1° > 2° >> 3° (steric); strong nucleophile; polar aprotic solvent (DMSO, acetone).

SN1 mechanism — unimolecular, two steps

Step 1 (slow): R–X → R+ + X.
Step 2 (fast): R+ + Nu → R–Nu.
· Rate = k[R–X] — first-order in halide only.
· Carbocation intermediate is planar → racemic mixture.
· Favored by: 3° > 2° >> 1° (carbocation stability); weak nucleophile; polar protic solvent (water, alcohol); good leaving group.

Common nucleophilic substitutions

Common trap. A 3° alkyl halide does NOT undergo SN2 — the bulky alkyl groups block the back-side approach of the nucleophile. Instead, with a strong base it gives mostly E2; with a weak nucleophile in protic solvent it gives SN1.

Elimination Reactions

A β-hydrogen and the halide leave together to give an alkene. Like substitution, elimination has two flavours: E2 (one step, bimolecular) and E1 (two step, via carbocation).

E2 mechanism — concerted

R2CH–CHR′–X + B → R2C=CHR′ + BH + X.
· Rate = k[R–X][B] — second-order.
· H and X depart anti-periplanar in one step.
· Favored by: strong, hindered base (KOtBu, OEt); high temperature; 3° > 2° > 1°.

E1 mechanism — via carbocation

Step 1 (slow): R–X → R+ + X.
Step 2 (fast): R+ ⟶{−H+} alkene.
· Rate = k[R–X] — first-order.
· Same conditions as SN1 (3°, polar protic solvent, weak base); the two compete.

Saytzeff (Zaitsev) rule

The major elimination product is the more substituted (more stable) alkene. Example: 2-bromo-2-methylbutane ⟶{KOH/ethanol} 2-methylbut-2-ene (major) + 2-methylbut-1-ene (minor).

SN vs E — how to choose

Quick mnemonic. "One stays, two attack." SN1/E1 are unimolecular — only the substrate is in the rate law; SN2/E2 are bimolecular — substrate and nucleophile/base are both in the rate law. 1° loves SN2; 3° loves SN1/E1.
SN1 vs SN2 vs E1 vs E2 — the four mechanisms at a glance
PropertySN1SN2E1E2
Steps2 (carbocation intermediate)1 (concerted)2 (carbocation intermediate)1 (concerted)
MolecularityUnimolecularBimolecularUnimolecularBimolecular
Rate lawrate = k[RX]rate = k[RX][Nu]rate = k[RX]rate = k[RX][B]
Substrate preference > 2° > 1°1° > 2° > 3° never (CH3X fastest) > 2° > 1°3° > 2° > 1° (with bulky base)
Nucleophile / baseWeak Nu (H2O, ROH)Strong Nu (OH, CN)Weak base (H2O)Strong, often bulky base (KOtBu)
SolventPolar protic (water, alcohol)Polar aprotic (DMSO, DMF, acetone)Polar proticPolar protic / aprotic, hot
StereochemistryRacemization (planar carbocation)Inversion (Walden inversion)Mixture, usually Zaitsev productAnti-periplanar elimination, Zaitsev
RearrangementsPossible (carbocation)NonePossible (carbocation)None
Effect of heatModest increaseModest increaseFavoredStrongly favored

Worked MCQs

Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — that's where the deeper concept lives.

Q1. Which alkyl halide undergoes SN2 reaction the fastest?

  • CH3Br
  • (CH3)2CHBr
  • (CH3)3CBr
  • (CH3)3CCH2Br (neopentyl)

SN2 requires backside attack on the carbon bearing the leaving group. Bromomethane is the least hindered. Tertiary halides cannot undergo SN2; neopentyl is also strongly hindered by the bulky t-butyl group adjacent to it.

Q2. 2-bromo-2-methylpropane reacts with aqueous ethanol predominantly via:

  • SN2
  • SN1
  • E2
  • Free-radical substitution

A 3° alkyl halide in a polar protic solvent with a weak nucleophile (water/ethanol) ionizes to a stable 3° carbocation, which is then captured by the solvent — the textbook SN1 setup. (Some E1 alkene also forms.)

Q3. The Wurtz reaction of two molecules of bromoethane in dry ether with sodium gives:

  • Ethene
  • Propane
  • n-butane
  • Hexane

Wurtz couples two R–X to give R–R: 2 CH3CH2Br + 2 Na → CH3CH2–CH2CH3 (n-butane) + 2 NaBr.

Q4. Dehydrohalogenation of 2-bromo-2-methylbutane with hot alcoholic KOH gives mainly:

  • 2-methylbut-1-ene
  • 2-methylbut-2-ene
  • 3-methylbut-1-ene
  • Pentene

By Saytzeff's rule, the more highly substituted (more stable) alkene is the major product. Removing a β-H from C-3 gives 2-methylbut-2-ene (trisubstituted). Removing one from the methyl on C-2 would give the disubstituted 2-methylbut-1-ene as a minor product.

Q5. Which of the following has the strongest C–X bond?

  • CH3F
  • CH3Cl
  • CH3Br
  • CH3I

Bond strength decreases down the halogen group as the C–X bond gets longer (C–F > C–Cl > C–Br > C–I). For the same reason, fluoroalkanes are the least reactive in SN/E reactions and iodoalkanes the most.

Quick Recap

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