JAM Haloalkanes and Haloarenes — Reaction Mechanisms
The SN1/SN2 substitution comparison — substrate, nucleophile, solvent, leaving group and stereochemistry — is worked through in full detail in a companion article, linked below, and is not repeated here. This article covers the ground that comparison leaves out: how haloalkanes and haloarenes are actually prepared, what happens when elimination competes with substitution, and why an aryl halide behaves nothing like an alkyl one.
Preparing haloalkanes
| Method | Reagent | Note |
|---|---|---|
| From an alcohol | SOCl₂ (Darzens procedure) | Gives R–Cl with only gaseous by-products (SO₂, HCl), so purification is easy — the cleanest laboratory route to a chloroalkane |
| From an alcohol | PCl₅, PCl₃ or PBr₃ | Standard alternative routes to chloro- and bromoalkanes |
| From an alcohol (test + prep) | Conc. HCl + anhydrous ZnCl₂ (Lucas reagent) | Reacts fastest with 3° alcohols (SN1), slower with 2°, needs heat for 1° — the basis of the Lucas test for distinguishing them |
| From an alkane | X₂ / hν (free-radical halogenation) | Selectivity of Br₂ over Cl₂ for the most substituted C–H is covered in the companion article on aliphatic hydrocarbons |
| From an alkene | HX or X₂ addition | Markovnikov or anti-Markovnikov depending on conditions; see the reaction-mechanisms article for the full addition table |
Haloarenes come mainly from direct halogenation of an arene (X₂ with a Lewis-acid catalyst such as FeX₃, an ordinary electrophilic aromatic substitution) or from a diazonium salt via the Sandmeyer or Gattermann reactions, both detailed in the companion article on amines.
Worked example 1 — preparation yield with SOCl₂. 9.20 g of ethanol (C₂H₅OH, M = 46.07 g/mol) is treated with excess SOCl₂. The isolated chloroethane (C₂H₅Cl, M = 64.51 g/mol) weighs 11.6 g. Find the percentage yield.
Moles of ethanol = 9.20 ÷ 46.07 = 0.200 mol.
The reaction is 1 : 1, so theoretical moles of chloroethane = 0.200 mol.
Theoretical mass = 0.200 × 64.51 = 12.90 g.
% yield = (11.6 ÷ 12.90) × 100 = 89.9%.
E2 vs E1 elimination — Zaitsev and Hofmann products
A haloalkane treated with a base does not always substitute; elimination is always a competing pathway, and predicting which alkene forms is a distinct skill from predicting SN1/SN2.
E1 — two steps, through the same carbocation intermediate as SN1; the base (often just the solvent) removes a proton from whichever position gives the more stable alkene. Rate = k[substrate] — first order, independent of base concentration.
Zaitsev's rule: with a small, unhindered base (ethoxide, hydroxide), the major product is the more substituted, more stable alkene — this is also always the outcome of E1, since the more stable alkene's transition state is lower in energy. Hofmann's rule: with a bulky base (tert-butoxide), steric hindrance makes the hydrogen needed for the Zaitsev product hard to reach, so the base instead removes a more accessible hydrogen, giving the less substituted alkene as the major product.
Worked example 2 — Zaitsev vs Hofmann on the same substrate. 2-Bromo-2-methylbutane, (CH₃)₂CBr–CH₂CH₃, is treated with two different bases.
Two sets of β-hydrogens are available: on the two equivalent methyl groups, and on the CH₂ of the ethyl side.
With sodium ethoxide (small base): elimination removes a hydrogen from the ethyl CH₂, giving the more substituted, trisubstituted alkene 2-methylbut-2-ene as the major (Zaitsev) product.
With potassium tert-butoxide (bulky base): the base cannot easily reach the more hindered ethyl CH₂ hydrogen, so it removes a methyl hydrogen instead, giving the less substituted, disubstituted alkene 2-methylbut-1-ene as the major (Hofmann) product.
Why haloarenes resist nucleophilic substitution
An aryl C–X bond is shorter and stronger than an alkyl C–X bond for two reasons: the carbon is sp²-hybridised (more s-character pulls the bonding electrons closer to carbon), and the halogen's lone pair is partly delocalised into the ring by resonance, giving the C–X bond some double-bond character. Neither SN1 nor SN2 can proceed easily: SN1 would need an extremely unstable aryl cation, and SN2 backside attack is blocked by the ring itself. Ordinary nucleophilic substitution of chlorobenzene with NaOH therefore needs genuinely forcing conditions — high temperature and high pressure — rather than the mild room-temperature reaction an alkyl halide undergoes.
Two routes that do work — addition-elimination and benzyne
Addition-elimination (nucleophilic aromatic substitution): works readily when a strong electron-withdrawing group (typically –NO₂) sits ortho or para to the leaving group. The nucleophile first adds to the ring carbon bearing the halogen, forming a resonance-stabilised anionic intermediate in which the negative charge is delocalised onto the ortho/para nitro group's oxygens; the halide then leaves, restoring aromaticity. 2,4-Dinitrochlorobenzene reacts with aqueous NaOH under mild conditions to give 2,4-dinitrophenol for exactly this reason, while plain chlorobenzene does not react at all under the same conditions.
Worked example 3 — the benzyne mechanism. Chlorobenzene, which carries no activating group, is treated with NaNH₂ in liquid NH₃ — a very strong base with no resonance stabilisation available. Explain how substitution still occurs.
1. NH₂⁻ removes a proton from the carbon ortho to the chlorine (an
E2-like step), forming a carbanion.
2. That carbanion immediately expels chloride, generating benzyne — a
highly strained ring carrying an extra, in-plane π bond between the two carbons that had
borne the H and the Cl.
3. NH₂⁻ (or NH₃) adds back to either carbon of that strained triple bond,
because both are similarly electrophilic, and a proton transfer completes the aniline
product.
Because addition in step 3 is not selective for one particular carbon, classic isotopic-labelling experiments (using a chlorobenzene labelled at the carbon bearing chlorine) have confirmed that the product amine ends up labelled at both the original position and the adjacent one, in roughly equal amounts — direct evidence for the symmetric benzyne intermediate rather than a simple one-step displacement.
Common mistakes
- Applying Zaitsev's rule regardless of the base. A bulky base flips the outcome to the Hofmann product — always check the base's size, not just its strength.
- Forgetting the anti-periplanar requirement for E2. In a cyclohexane ring this means the leaving group and the β-hydrogen must both be axial; an equatorial leaving group cannot undergo a normal E2 elimination.
- Treating aryl and vinyl halides as ordinary alkyl halides. Neither undergoes SN1 or SN2 under normal conditions — the mechanism has to be addition-elimination or benzyne instead.
- Assuming nucleophilic aromatic substitution works on any haloarene. The addition-elimination route needs a strong electron-withdrawing group positioned ortho or para to the leaving group; without one, only the much harsher benzyne route (or forcing high-temperature/pressure conditions) will work at all.
- Expecting a single, clean product from a benzyne reaction on an unsymmetrically substituted substrate. Non-selective addition to the strained triple bond is exactly what benzyne mechanisms predict.
Exam relevance
| Question style | What to check first |
|---|---|
| Predict the major elimination product | Size of the base — Zaitsev for a small base, Hofmann for a bulky one |
| Explain a stereospecific elimination | The anti-periplanar requirement of E2 |
| Explain why a haloarene fails to react with NaOH at room temperature | sp² C–X bond strength and resonance delocalisation of the halogen lone pair |
| Predict the product of NaNH₂/NH₃(l) on an unactivated aryl halide | The benzyne mechanism, and its lack of positional selectivity |
| Numerical (NAT) | Preparation stoichiometry and percentage yield |
Check molar masses before balancing a preparation reaction. The molar mass tool accepts any formula and is the fastest way to catch a stoichiometry error in a haloalkane or haloarene synthesis problem.
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