Identify the final product [D] in the following sequence: $\mathrm{CH}{}_3\mathrm{CHO}\underset{\text{(ii) H3O+}}{\overset{\text{(i) LiAlH4}}{\to }}[\mathrm{A}]\underset{\mathrm{\Delta }}{\overset{\text{H2SO4}}{\to }}[\mathrm{B}]\stackrel{\mathrm{HBr}}{\to }[\mathrm{C}]\underset{\text{dry ether}}{\overset{\text{C6H5Br, Na}}{\to }}[\mathrm{D}]$

Solution & Explanation

\textbf{Answer:} (A) Ethylbenzene $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CH}{}_2\mathrm{CH}{}_3$ \textbf{Solution:} $\mathrm{CH}{}_3\mathrm{CHO}$ is reduced by $\mathrm{LiAlH}{}_4$ (then $\mathrm{H}{}_3\mathrm{O}{}^{+}$) to ethanol [A]. Dehydration with conc. $\mathrm{H}{}_2\mathrm{SO}{}_4$/heat gives ethene [B], which adds $\mathrm{HBr}$ to form bromoethane [C]. Wurtz-Fittig coupling of bromoethane with bromobenzene using $\mathrm{Na}$/dry ether gives ethylbenzene [D].