Identify the product ' B ' in the following reaction sequence. $\text { Alkyl halide } \frac{\mathrm{Mg}}{\text { Dry ether }} \mathrm{A} \xrightarrow{\mathrm{NH}_3} \mathrm{~B}$

Reaction sequence analysis:

Alkyl halide $\mathrm{R-X}$ reacts with magnesium in dry ether to form Grignard reagent $\mathrm{A = R-Mg-X}$.

Grignard reagents are strong bases that react with compounds containing acidic protons. Ammonia $\mathrm{NH_3}$ has acidic hydrogens, producing the corresponding hydrocarbon via proton transfer: $\mathrm{R-Mg-X + H-NH_2 \rightarrow R-H + Mg(X)NH_2}$.

The product $\mathrm{B}$ is therefore a hydrocarbon $\mathrm{R-H}$.

Among the options, the correct answer corresponds to hydrocarbon: $\boxed{C}$.