The reaction of 2,6-dideutero-fluorobenzene w | vedclass

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(C) The reaction of $2,6-$dideutero-fluorobenzene with $NaNH_2$ in liquid $NH_3$ proceeds via the benzyne mechanism.$1$. The amide ion $(NH_2^-)$ abst

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$2-$deuteroaniline and $3-$deuteroaniline

Vedclass · Answer

(C) The reaction of $2,6-$dideutero-fluorobenzene with $NaNH_2$ in liquid $NH_3$ proceeds via the benzyne mechanism.$1$. The amide ion $(NH_2^-)$ abstracts a proton from the ortho position relative to the fluorine atom.$2$. This leads to the elimination of $F^-$ and the formation of a benzyne intermediate, specifically $3-$deuterobenzyne.$3$. The nucleophilic attack of $NH_2^-$ on the benzyne intermediate can occur at two non-equivalent positions due to the presence of the deuterium atom.$4$. This results in a mixture of $2-$deuteroaniline and $3-$deuteroaniline as the final products.

The reaction of $2,6-$dideutero-fluorobenzene with $NaNH_2$ in liquid $NH_3$ gives:

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