KCET 2021 Chemistry Solved Paper

From the given information question, addition of $\mathrm{HCl}$ occured on A. That means A is an alkene. $\underset{(A)}{\mathrm{C}_4\mathrm{H}_8}\xrightarrow{\mathrm{HCl}}\underset{(B)}{\mathrm{C}_4\mathrm{H}_9\mathrm{Cl}}$ Cl in compound B is substituted by $\mathrm{NH}_2$ to give compound $\mathrm{C}(\mathrm{C}_4\mathrm{H}_{11}\mathrm{N})$. $\underset{(B)}{\mathrm{C}_4\mathrm{H}_9\mathrm{Cl}}\xrightarrow{\mathrm{NH}_3}\underset{(C)}{\mathrm{C}_4\mathrm{H}_{11}\mathrm{N}}$ C gives a diazonium salt with $\mathrm{NaNO}_2 / \mathrm{HCl}$ which on further treatment with water give an optically active alcohol. That means C is an aliphatic amine. $\mathrm{C}\xrightarrow[(ii)\,\mathrm{H}_2\mathrm{O}]{(i)\,\mathrm{NaNO}_2/\mathrm{HCl}}\mathrm{D}$ [Alcohol (optically active)] From the given information it is concluded that the most probable structure of A is $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$. As only but-2-ene gives optically active alcohol $(D)$ as final product. The reactions involves are as follows.