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MHT-CET Chemistry · Alkenes

Preparing Alkenes: Dehydrohalogenation and Saytzeff's Rule

Heating an alkyl halide with alcoholic KOH removes H and X from neighbouring carbons to give an alkene, and when more than one alkene can form the more substituted one is the major product.

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Why this matters

3 PYQs, none HARD, and two of them are the same question set in both 14 May 2024 shifts: the major product from 3-bromo-2-methylpentane. One card.

Concept 1 of 1: Dehydrohalogenation with Alcoholic KOH

Alcoholic KOH is a strong base in a solvent that does not favour substitution, so it pulls an H off a carbon next to the C–X carbon and the halide leaves: a double bond forms. If H can come from two different neighbours, the product with more alkyl groups on the double bond wins (Saytzeff). AQUEOUS KOH would substitute instead, giving an alcohol.

Definition

  • Alcoholic KOH, heat: elimination (E2) — alkene. Aqueous KOH: substitution — alcohol.
  • Saytzeff (Zaitsev) rule: the more substituted alkene is the major product.
  • 2-Bromopropane → propene; 3-bromo-2-methylpentane → 2-methylpent-2-ene (trisubstituted) over 4-methylpent-2-ene.

Dehydrohalogenation

R−CH2−CHX−R′→alc. KOH, ΔR−CH=CH−R′+KX+H2O\mathrm{R{-}CH_2{-}CHX{-}R' \xrightarrow{\text{alc. KOH},\ \Delta} R{-}CH{=}CH{-}R' + KX + H_2O}

Worked example

Major product of 3-bromo-2-methylpentane with alcoholic KOH on heating?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q75Moderate

Example 1 · Alkenes · Preparation of Alkenes by Elimination

Identify major product formed: 3-Bromo-2-methylpentane →Alc. KOH/Δ\xrightarrow{\text{Alc. KOH}/\Delta} Major product

Picking the alcohol

2-Methylpentan-3-ol is offered for alcoholic KOH. That is the AQUEOUS-KOH (substitution) product; alcoholic KOH eliminates.
Drill 2 more on dehydrohalogenation with alcoholic koh

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 (1)

  • Dehydrohalogenation with Alcoholic KOH

    Dehydrohalogenation

    R−CH2−CHX−R′→alc. KOH, ΔR−CH=CH−R′+KX+H2O\mathrm{R{-}CH_2{-}CHX{-}R' \xrightarrow{\text{alc. KOH},\ \Delta} R{-}CH{=}CH{-}R' + KX + H_2O}

Watch out for (1)

Test yourself on Alkenes

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Alkenes chapter testDrill all 3 questions