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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 PMDC MDCAT 2026 syllabus expects you to predict mechanism, product, and reactivity for any given alkyl halide. Expect 1–2 high-yield MCQs.

PMC Table of Specifications. This chapter spans three PMDC 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An alkyl halide is an alkane in which one (or more) hydrogen has been replaced by a halogen.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 polarised δ+C–Xδ — the carbon is electrophilic and is the centre 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).
· Favoured 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.
· Favoured 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.
· Favoured 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
StereochemistryRacemisation (planar carbocation)Inversion (Walden inversion)Mixture, usually Zaitsev productAnti-periplanar elimination, Zaitsev
RearrangementsPossible (carbocation)NonePossible (carbocation)None
Effect of heatModest increaseModest increaseFavouredStrongly favoured

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) ionises 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

Test yourself. Take a timed practice test or browse topic-wise MCQs to lock these concepts in.