Question
2-Bromobutane is treated with alcoholic KOH. Identify the major and minor elimination products and explain your choice using Zaitsev’s Rule.
Solution — Step by Step
In 2-bromobutane, the bromine is on C2. The β-carbons (adjacent to C2) are C1 and C3. Each gives a different alkene when we remove HBr.
Removing H from C1 gives but-1-ene (CH₂=CH-CH₂-CH₃) — a monosubstituted alkene.
Removing H from C3 gives but-2-ene (CH₃-CH=CH-CH₃) — a disubstituted alkene.
Zaitsev’s Rule says: the more substituted alkene is the major product. But-2-ene has two alkyl groups on the double bond carbons; but-1-ene has only one. So but-2-ene wins.
- Major product: but-2-ene (2-butene)
- Minor product: but-1-ene (1-butene)
Alcoholic KOH is a strong, moderately bulky base — ideal conditions for E2, which follows Zaitsev.
Why This Works
The key is hyperconjugation and alkyl group stabilisation. Every alkyl group attached to a double bond carbon donates electron density through hyperconjugation, lowering the energy of the π system. More alkyl groups = more stable alkene = lower energy product.
We always form the lower-energy product preferentially. The reaction is exothermic, and the transition state resembles the product (Hammond’s postulate for E2). So the more stable product has the lower-energy transition state — that’s why it forms faster.
A quick substitution count: count the number of carbon groups directly bonded to each carbon of the C=C bond. But-2-ene has methyl + ethyl = 2 groups. But-1-ene has only the propyl side = 1 group. Higher count → major product. This mental shortcut works in 10 seconds.
Alternative Method — Degree of Substitution Count
Instead of drawing full structures, use the degree of unsaturation approach directly.
Label the alkene carbons as C and C. For each possible alkene, count total carbon substituents on both double bond carbons combined:
Higher total → major product. This method is especially fast in MCQs where you need the answer in under 30 seconds. It appeared as a direct concept-check in JEE Main 2023.
Common Mistake
Confusing Zaitsev with Hofmann’s Rule. Students apply Zaitsev everywhere, but when the base is bulky (like t-BuOK / potassium tert-butoxide), Hofmann’s rule applies — the less substituted alkene becomes major because the bulky base cannot access the more hindered β-hydrogen.
In this question, KOH in alcohol is not bulky. Always check the base first: KOH/NaOEt → Zaitsev. t-BuOK → Hofmann. This distinction is a favourite JEE trap — at least one question in every cycle tests it.