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(D) The product is $m$-bromonitrobenzene. $1$. Bromination of benzene using $Br_2$ and $FeBr_3$ gives bromobenzene. $2$. The $-Br$ group is ortho/para

Vedclass · Accepted Answer

$(1) \, Br_2, FeBr_3; 	ext{ then } (2) \, HNO_3, H_2SO_4$

Vedclass · Answer

(D) The product is $m$-bromonitrobenzene. $1$. Bromination of benzene using $Br_2$ and $FeBr_3$ gives bromobenzene. $2$. The $-Br$ group is ortho/para-directing,but it is deactivating. $3$. Nitration of bromobenzene using $HNO_3$ and $H_2SO_4$ yields a mixture of $o$-bromonitrobenzene and $p$-bromonitrobenzene. $4$. However,if we perform nitration first,the $-NO_2$ group is meta-directing. $5$. Nitration of benzene gives nitrobenzene. $6$. Subsequent bromination of nitrobenzene using $Br_2$ and $FeBr_3$ gives $m$-bromonitrobenzene as the major product because the $-NO_2$ group is meta-directing. Therefore,the correct sequence is $(1) \, HNO_3, H_2SO_4; 	ext{ then } (2) \, Br_2, FeBr_3$.

Suggest the best reaction conditions for the synthesis shown below.

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