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(A) The reaction of chlorobenzene labeled with $^{14}C$ at the $C-1$ position with $NaNH_2/NH_3$ proceeds via the benzyne mechanism.$1$. The amide ion

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(A) The reaction of chlorobenzene labeled with $^{14}C$ at the $C-1$ position with $NaNH_2/NH_3$ proceeds via the benzyne mechanism.$1$. The amide ion $(NH_2^-)$ abstracts an ortho-hydrogen,leading to the formation of two benzyne intermediates: $1,2$-dehydrobenzene (with $^{14}C$ at the $C-1$ position) and $2,3$-dehydrobenzene (with $^{14}C$ at the $C-2$ position).$2$. Nucleophilic attack by $NH_2^-$ on these intermediates can occur at two positions for each,resulting in three products: ortho-labeled aniline,meta-labeled aniline,and ipso-labeled aniline.$3$. Based on the statistical probability of the benzyne formation and the subsequent nucleophilic attack,the products are formed in a $1:2:1$ ratio.$4$. The major product is the one where the amino group is attached to the carbon adjacent to the $^{14}C$ label (ortho position) or the meta position,but specifically,the question asks for the product $(A)$ which corresponds to the ortho-substituted aniline.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A)$ $p$-Nitrochlorobenzene $\xrightarrow[(ii) H^+]{(i) NaOH, 443 \ K}$	$(i)$ Phenol
$(B)$ Chlorobenzene $\xrightarrow[(ii) H^+]{(i) NaOH, 623 \ K, 300 \ atm}$	$(ii)$ $2,4,6$-Trinitrophenol
$(C)$ $2,4$-Dinitrochlorobenzene $\xrightarrow[(ii) H^+]{(i) NaOH, 368 \ K}$	$(iii)$ $2,4$-Dinitrophenol
$(D)$ $2,4,6$-Trinitrochlorobenzene $\xrightarrow{Warm \ H_2O}$	$(iv)$ $p$-Nitrophenol

Two benzyne intermediates are likely to be formed equally. Reaction with amide ion can occur in two different directions with each benzyne,giving three possible products. They are formed in a $1 : 2 : 1$ ratio. Asterisk $(*)$ refers to $^{14}C$. Product $(A)$ is:

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