Identify product ' $B$ ' in the following sequence of reactions. $\text { 1,2 Dibromoethane } \xrightarrow[\text { alcohol }]{\mathrm{KOH}} \mathrm{~A} \xrightarrow{\text { sodamide }} \mathrm{B}$

1,2-Dibromoethane undergoes dehydrohalogenation with alcoholic KOH, producing bromoethene (vinyl bromide) through elimination of one molecule of HBr: $\mathrm{CH_2 Br-CH_2 Br \rightarrow[\text{alc. KOH}] CH_2=CHBr + KBr + H_2 O}$. Bromoethene then reacts with sodamide ($\mathrm{NaNH_2}$), a strong base, to undergo further dehydrohalogenation, yielding ethyne (acetylene): $\mathrm{CH_2=CHBr \rightarrow[\mathrm{NaNH_2}] HCCH + NaBr + NH_3}$.

Product B is ethyne. Among the options, D) Ethyne is correct.