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.
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An alkyl halide is an alkane in which one (or more) hydrogen has been replaced by a halogen.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
- Treat the halogen as a substituent (fluoro-, chloro-, bromo-, iodo-).
- Choose the longest chain containing the halogen-bearing carbon; number it so that the halogen has the lowest locant.
- Examples: CH3Cl = chloromethane, CH3CH2Br = bromoethane, (CH3)2CHCl = 2-chloropropane, (CH3)3CBr = 2-bromo-2-methylpropane.
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
- C–X bond strength: C–F > C–Cl > C–Br > C–I (longer bond, weaker).
- Reactivity in SN/E reactions: R–I > R–Br > R–Cl > R–F (weaker bond breaks more easily).
- Polarity: C–F most polar, but its bond strength keeps fluoroalkanes inert.
Preparation
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).
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).
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).
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).
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
- R–X + OH− → R–OH (alcohol).
- R–X + CN− → R–CN (nitrile, extends chain by 1 C).
- R–X + R′O− → R–O–R′ (Williamson ether synthesis).
- R–X + NH3 (excess) → R–NH2 (amine).
- R–X + AgNO2 → R–NO2 (nitroalkane); R–X + KNO2 → R–O–N=O (alkyl nitrite). Ambident-nucleophile MCQ favourite.
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).
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°.
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.
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
- 1° halide + strong nucleophile: SN2.
- 1° halide + bulky strong base (KOtBu): E2.
- 2° halide + strong base/nucleophile: SN2 vs E2 — mixture; high temperature favours E2.
- 3° halide + strong base: E2 dominates.
- 3° halide + weak nucleophile / polar protic solvent: SN1 + E1 mixture.
| Property | SN1 | SN2 | E1 | E2 |
|---|---|---|---|---|
| Steps | 2 (carbocation intermediate) | 1 (concerted) | 2 (carbocation intermediate) | 1 (concerted) |
| Molecularity | Unimolecular | Bimolecular | Unimolecular | Bimolecular |
| Rate law | rate = k[RX] | rate = k[RX][Nu] | rate = k[RX] | rate = k[RX][B] |
| Substrate preference | 3° > 2° > 1° | 1° > 2° > 3° never (CH3X fastest) | 3° > 2° > 1° | 3° > 2° > 1° (with bulky base) |
| Nucleophile / base | Weak Nu (H2O, ROH) | Strong Nu (OH−, CN−) | Weak base (H2O) | Strong, often bulky base (KOtBu) |
| Solvent | Polar protic (water, alcohol) | Polar aprotic (DMSO, DMF, acetone) | Polar protic | Polar protic / aprotic, hot |
| Stereochemistry | Racemisation (planar carbocation) | Inversion (Walden inversion) | Mixture, usually Zaitsev product | Anti-periplanar elimination, Zaitsev |
| Rearrangements | Possible (carbocation) | None | Possible (carbocation) | None |
| Effect of heat | Modest increase | Modest increase | Favoured | Strongly 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?
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:
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:
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:
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?
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
- Reactivity in SN/E: R–I > R–Br > R–Cl > R–F (bond strength inverse).
- SN2: 1°, strong Nu, polar aprotic, inversion, second-order rate.
- SN1: 3°, weak Nu, polar protic, racemisation, first-order rate.
- E2: strong/bulky base, second-order, anti-periplanar, follows Saytzeff.
- E1: 3°, weak base, polar protic, competes with SN1.
- SOCl2 = cleanest alcohol → chloride conversion (gaseous SO2 + HCl).
- Wurtz: 2 R–X + 2 Na → R–R + 2 NaX (symmetrical alkanes).