Question

A hydrocarbon $A\left(C_4{H}_8\right)$ on reaction with$HCl$ gives a compound $B\left(C_4{H}_{9}Cl\right)$ which on reaction with $1$ mole of $N{H}_3$ gives a compound $C\left(C_4{H}_{11}N\right)$. On reacting with $NaN{O}_2$ and $HCl$ (low temperature) followed by treatment with water, compound $C$ yields optically active alcohol(s). Ozonolysis of $A\left(1\text{mole}\right)$ gives $2$ moles of acetaldehyde.
The compound C is:

• A. Butan-2-amine
• B. N, N-dimethylethanamine
• C. Diethylamine
• D. Butylamine

Answer 1

The correct option is A Butan-2-amine

Ozonolysis is the clue here. On ozonolysis, A gives 2 moles of ethanal, so A is $C{H}_3-CH=CH-C{H}_3$.
The reactions are as follows:
$\underset{\left(A\right)}{C{H}_3-CH=CH-C{H}_3}\stackrel{HCl}{\to }\underset{\left(B\right)}{C{H}_3-C{H}_2-\stackrel{Cl}{\stackrel{|}{C}}H-C{H}_3}$

$\underset{\text{1 mole}}{\overset{N{H}_3}{\to }}\underset{\left(C\right)}{C{H}_3-C{H}_2-\stackrel{N{H}_2}{\stackrel{|}{C}}H-C{H}_3}\underset{(NaN{O}_2+HCl)}{\overset{HN{O}_2}{\to }}\underset{}{C{H}_3-C{H}_2-\stackrel{N_{2}Cl}{\stackrel{|}{C}}H-C{H}_3}$

$\stackrel{H_{2}O}{\to }\underset{\left(D\right)}{C{H}_3-C{H}_2-\stackrel{OH}{\stackrel{|}{C}}H-C{H}_3}$

(D) is optically active due to presence of chiral centre.