Ncert Exercises & Solutions

An alkyl halide $C_{5} H_{11} (A)$ 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 the sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D. Give the reactions involved.

The reaction scheme involved in the problem is

\underset{Alkyl halide}{\underset{(A)}{C_{5} H_{11} Br}} \overset{Alc . KOH}{\to} \underset{Alkene}{\underset{(B)}{C_{5} H_{10}}} \overset{{Br}_{2} / {CS}_{2}}{\to} \underset{Dilbromo alkane}{\underset{(C)}{C_{5} H_{10} {Br}_{2}}} \to_{- 2 HBr}^{Alc . KOH} \underset{Alkyne}{\underset{(D)}{C_{5} H_{8}}} \overset{H_{2}}{\to}

\underset{(straight chain alkene)}{C_{5} H_{8} \to_{Liq . {NH}_{3}}^{Na .} C_{5} H_{7}^{-} {Na}^{+} + \frac{1}{2} H_{2}}

Hydrogenation of alkyne (D) gives straight chain alkane hence all the compounds (A), (B), (C) and (D) must be straight chain compounds. Alkyne (D) form sodium salt which proves that it is terminal alkyne. Involved reactions are as follows

\underset{1 - bromopentane (A)}{{CH}_{3} {CH}_{2} {CH}_{2} {CH}_{2} {CH}_{2} Br} \to_{- HBr}^{Alc . KOH, ∆}

\underset{1 - pentene (B)}{{CH}_{3} {CH}_{2} {CH}_{2} CH = {CH}_{2}} \overset{{Br}_{2} in {CS}_{2}}{\to}

It is important point that alkyl halide (A) can not be 2-bromopentane because dehydrobromination of (A) would have given 2-pentene as the major product in accordance with Markownikoff's rule.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions