1. Tardigrade
  2. Question
  3. Chemistry
  4. An alkyl halide C5H11 Br (A) reacts with ethanolic KOH to give an alkene B which reacts with Br2 to give a compound ‘C, which on dehydrobromination gives an alkyne D On treatment with sodium metal in liquid ammonia, one mole of 'D' gives one mole of sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify D.

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Q. An alkyl halide $C_5H_{11} Br (A)$ reacts with ethanolic $KOH$ to give an alkene $B$ which reacts with $Br_2$ to give a compound ‘$C$, which on dehydrobromination gives an alkyne $D$ On treatment with sodium metal in liquid ammonia, one mole of '$D$' gives one mole of sodium salt of '$D$' and half a mole of hydrogen gas. Complete hydrogenation of ‘$D$’ yields a straight chain alkane. Identify $D$.

Hydrocarbons

Solution:

Correct answer is (b) Pent-$1$-yne

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