Question

An equimolar mixture of alkyl bromide $(A)$ and ammonia gives $(B)$ which on treating with one equivalent of $CH _{3} I$ gives $(C) .(B)$ and $(C)$ on treating with $NaNO _{2}$ and $HCl$ gives $(D)$ and $(E)$ respectively. $(D)$ on oxidation followed by decarboxylation gives ethane. Structure of $(E)$ is

• A. $CH _{3}- CH _{2}- \underset{\overset{|}{CH_3}}{N} - N = O$
• B. $CH _{3}- \underset{\overset{|}{CH_3}}{CH} - CH _{2}- \underset{\overset{|}{CH_2-CH-3}}{N} - N =0$
• C. $CH _{3}- CH _{2}- CH _{2}- \underset{\overset{|}{CH_3}}{N} - N = O$
• D. $CH _{2}- \underset{\overset{|}{CH_3}}{C}H - \underset{\overset{|}{CH_2-CH_3}}{N} - N = O$

Answer 1

Answer: (C)

The reactions can be summarized as :



$R'- H$ is ethane, $C _{2} H _{6}$

Thus, $R- Br$ should be $C _{3} H _{7} Br$.