Question

The major product formed when 2-bromo-2-methylbutane is treated with alcoholic KOH is:

• A. 2-methylbut-1-ene
• B. 2-methylbut-2-ene
• C. 3-methylbut-1-ene
• D. 1-methylbut-1-ene

Hint:

In elimination reactions with tertiary alkyl halides, the major product is the more substituted alkene (Zaitsev's rule), formed via E2 mechanism.

Answer 1

The Correct Option is B

- The substrate is 2-bromo-2-methylbutane, a tertiary alkyl halide. - Treatment with alcoholic KOH typically causes dehydrohalogenation (elimination) via the E2 mechanism. - The elimination leads to the formation of the most stable alkene — the one with the most substituted double bond (Zaitsev's rule). - Possible alkenes: 1. 2-methylbut-1-ene (double bond between C1 and C2) — less substituted. 2. 2-methylbut-2-ene (double bond between C2 and C3) — more substituted. 3. 3-methylbut-1-ene (double bond between C1 and C2, methyl group at C3) — less favored. 4. 1-methylbut-1-ene — unlikely. - The most stable and major product is 2-methylbut-2-ene, the alkene with the double bond between C2 and C3, which is more substituted.