The reaction of 1-^{14}C-chlorobenzene with N | vedclass

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(A) The reaction of $1-^{14}C$-chlorobenzene with $NaNH_2/NH_3$ proceeds via the benzyne mechanism:$1$. The amide ion $(NH_2^-)$ abstracts an ortho-hy

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(A) The reaction of $1-^{14}C$-chlorobenzene 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: $1-^{14}C$-aniline (ipso), $2-^{14}C$-aniline (ortho), and $3-^{14}C$-aniline (meta).$3$. Based on the statistical probability of the benzyne formation and the subsequent nucleophilic attack, the products are formed in a $1:2:1$ ratio, where the ortho-substituted aniline ($2-^{14}C$-aniline) is the major product ($50$% of the total mixture).$4$. The image corresponding to the ortho-substituted aniline ($2-^{14}C$-aniline) is $(A)$.

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