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JEE Mains Chemistry · Haloalkanes and Haloarenes

Elimination Versus Substitution

Hydroxide or alkoxide can attack carbon (substitution) or remove a β-hydrogen (elimination); water favours substitution, alcoholic KOH with heat, a bulky base or a tertiary halide favours the alkene, and the major alkene is the more substituted one unless another is conjugated.

All Haloalkanes and Haloarenes notesJEE Mains Chemistry strategy

Why this matters

Fifteen PYQs, three numerical, one from 2026. Ten decide between substitution and elimination from the reagent, the solvent and the halide, often across a two- or three-step sequence; five ask which alkene forms, or how many.

Concept 1 of 2: Substitution or elimination: reading the reagent, solvent and halide

The same reagent can do two jobs. As a nucleophile, OH−\mathrm{OH^-} attacks the carbon that holds the halogen and gives an alcohol. As a base, it pulls off a hydrogen from the next carbon (the β-carbon) and the halide leaves, giving a C=C. Water surrounds hydroxide and keeps it acting as a nucleophile; ethanol and heat let it act as a base.

Definition

  • Aqueous KOH or NaOH: substitution, giving the alcohol.
  • Alcoholic KOH, heat: β-elimination (dehydrohalogenation) by E2, giving the alkene.
  • A strong nucleophile that is a weak base (CN−\mathrm{CN^-}, I−\mathrm{I^-}, RS−\mathrm{RS^-}) substitutes, by SN2 on a primary or secondary halide.
  • A tertiary halide with a strong base (C2H5O−\mathrm{C_2H_5O^-}, (CH3)3CO−\mathrm{(CH_3)_3CO^-}) eliminates; the same halide in water with no base substitutes by SN1.
  • A bulky base such as (CH3)3COK\mathrm{(CH_3)_3COK} eliminates even where substitution is possible.
  • No β-hydrogen, no elimination: (CH3)3C−CH2Br\mathrm{(CH_3)_3C{-}CH_2Br} cannot give an alkene by E2.
  • Moving a halogen along a chain: alcoholic KOH gives the alkene, HBr adds by Markovnikov's rule, and aqueous KOH then gives the new alcohol.

Dehydrohalogenation with alcoholic KOH

R−CH2−CH2−X+KOH→ethanol, ΔR−CH=CH2+KX+H2O\mathrm{R{-}CH_2{-}CH_2{-}X + KOH \xrightarrow{\text{ethanol},\ \Delta} R{-}CH{=}CH_2 + KX + H_2O}

Worked example

Give the major product in each case: (i) 1-bromobutane with NaCN in DMSO; (ii) bromocyclohexane with sodium ethoxide in ethanol, heated; (iii) 2-bromo-2-methylpropane in water at room temperature.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q106Moderate

Example 1 · Haloalkanes and Haloarenes · Elimination Versus Substitution

Given below are two statements : Statement (I) : Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction. Statement (II) : In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the β\beta-carbon. In the light of the above statements, choose the most appropriate answer from the options given below:

A bulky alkoxide gives the alkene, not the ether

Potassium tert-butoxide is an alkoxide, but it is too crowded to attack carbon. With a tertiary or secondary halide it removes a β-hydrogen and gives the alkene.

No β-hydrogen, no elimination

A halide such as (CH3)3C−CH2Br\mathrm{(CH_3)_3C{-}CH_2Br} has no hydrogen on the carbon next to the C–Br carbon. Alcoholic KOH cannot give an alkene from it by E2.
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Concept 2 of 2: Zaitsev rule and counting the alkenes from dehydrohalogenation

When hydrogens can be removed from more than one β-carbon, several alkenes can form. The major one is the most stable, which is usually the one with more alkyl groups on the double-bond carbons. A C=C that can conjugate with a benzene ring or another C=C is more stable still, and wins even with fewer alkyl groups.

Definition

  • Zaitsev (Saytzeff) rule: the more substituted alkene is the major product.
  • Conjugation wins: if one alkene is conjugated with a ring or another C=C, it is the major product.
  • A very bulky base such as (CH3)3CO−\mathrm{(CH_3)_3CO^-} gives more of the less substituted alkene.
  • Counting: list each different β-carbon that carries H, write the alkene each gives, then check each for cis and trans isomers. Ignore rearrangement unless the question allows it.
  • Excess alcoholic KOH removes two HX from a dihalide and gives a diene, conjugated where possible.
  • In yield problems, carry moles through each step: moles of product = moles of starting material × each fractional yield.

Stability order of alkenes (Zaitsev)

tetrasubstituted>trisubstituted>disubstituted>monosubstituted\text{tetrasubstituted} > \text{trisubstituted} > \text{disubstituted} > \text{monosubstituted}

Worked example

2-Bromo-2-methylbutane is heated with alcoholic KOH. Name the two alkenes that can form and say which is major.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q124Moderate

Example 2 · Haloalkanes and Haloarenes · Elimination Versus Substitution

The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement)

Count cis and trans separately

A question asking for isomeric alkenes counts geometrical isomers. An alkene such as hept-3-ene counts as two, cis and trans.

Conjugation beats the Zaitsev count

When one possible C=C lies next to a benzene ring, it is the major product even if another alkene carries more alkyl groups. Check for conjugation before counting substituents.

A dihalide with excess base gives a diene

Excess alcoholic KOH removes both HX molecules. The two new double bonds form in conjugation, with each other and with any ring, wherever the structure allows.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (5)

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