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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.

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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