1. Tardigrade
  2. Question
  3. Chemistry
  4. Primary alkyl halide C4H9Br (A) on reaction with alcoholic KOH gives compound (B). Compound (B) is reacted with HBr to give (C) which is an isomer of (A). When (A) is reacted with sodium metal, it gives compound (D), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. The compound (D) is

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Q. Primary alkyl halide $C_4H_9Br$ (A) on reaction with alcoholic $KOH$ gives compound (B). Compound (B) is reacted with $HBr$ to give (C) which is an isomer of (A). When (A) is reacted with sodium metal, it gives compound (D), $C_8H_{18}$ which is different from the compound formed when $n$-butyl bromide is reacted with sodium. The compound (D) is

Haloalkanes and Haloarenes

Solution:

There can be only two primary alkyl bromides with molecular formula, $C_4H_9Br$. These are
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(A) must be isobutyl bromide as, when treated with sodium it gave a compound different than that produced by $n$-butyl bromide, which reacts with sodium to give n-octane
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