Hydrocarbon A reacts with bromine by substitution to form an alkyl bromide, which by the Wurtz reaction is converted to a gaseous hydrocarbon containing less than four carbon atoms. A is:

Solution & Explanation

Substitution by bromine (free-radical) requires an alkane; ethene and ethyne would undergo addition, not substitution, so they are ruled out. Methane gives $\mathrm{CH}{}_4+\mathrm{Br}{}_2\stackrel{\mathrm{h}\nu }{\to }\mathrm{CH}{}_3\mathrm{Br}+\mathrm{HBr}$, and the Wurtz reaction $2\mathrm{CH}{}_3\mathrm{Br}+2\mathrm{Na}\to \mathrm{CH}{}_3\mathrm{CH}{}_3+2\mathrm{NaBr}$ gives ethane (2 carbons, $<4$). Starting from ethane would give butane (4 carbons), which violates the limit. Hence A is $\mathrm{CH}{}_4$.