$A$ hydrocarbon $'A'$ $(C_4H_8)$ on reaction with $HCl$ gives a compound $'B'$ $(C_4H_9Cl)$,which on reaction with $1 \ mol$ of $NH_3$ gives compound $'C'$ $(C_4H_{11}N)$. On reacting with $NaNO_2$ and $HCl$ followed by treatment with water,compound $'C'$ yields an optically active alcohol,$'D'$. Ozonolysis of $'A'$ gives $2 \ mols$ of acetaldehyde. Identify compounds $'A'$ to $'D'$. Explain the reactions involved.

(A) The reactions are as follows:
$1$. Ozonolysis of $'A'$ $(C_4H_8)$ gives $2 \ mols$ of acetaldehyde $(CH_3CHO)$,which indicates that $'A'$ is $CH_3-CH=CH-CH_3$ (but$-2-$ene).
$2$. $CH_3-CH=CH-CH_3 + HCl \rightarrow CH_3-CH_2-CHCl-CH_3$ (Compound $'B'$,$2$-chlorobutane).
$3$. $CH_3-CH_2-CHCl-CH_3 + NH_3 \rightarrow CH_3-CH_2-CH(NH_2)-CH_3$ (Compound $'C'$,butan-$2$-amine).
$4$. $CH_3-CH_2-CH(NH_2)-CH_3 + NaNO_2/HCl$ $\rightarrow [CH_3-CH_2-CH(N_2^+)-CH_3]$ $\xrightarrow{H_2O} CH_3-CH_2-CH(OH)-CH_3$ (Compound $'D'$,butan-$2$-ol,which is optically active).
The structures are:
$'A': CH_3-CH=CH-CH_3$
$'B': CH_3-CH_2-CHCl-CH_3$
$'C': CH_3-CH_2-CH(NH_2)-CH_3$
$'D': CH_3-CH_2-CH(OH)-CH_3$