C_6H_6 xrightarrow[text{Conc. } H_2SO_4]{text | vedclass

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(A) The reaction sequence is as follows:$1$. $C_6H_6$ (benzene) reacts with conc. $HNO_3$ and conc. $H_2SO_4$ to form $B$ (nitrobenzene,$C_6H_5NO_2$).

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(A) The reaction sequence is as follows:$1$. $C_6H_6$ (benzene) reacts with conc. $HNO_3$ and conc. $H_2SO_4$ to form $B$ (nitrobenzene,$C_6H_5NO_2$).$2$. Nitrobenzene $(B)$ is reduced by $Sn + HCl$ to form $C$ (aniline,$C_6H_5NH_2$).$3$. Aniline $(C)$ reacts with $HNO_2$ at $0-5^\circ C$ to form $D$ (benzenediazonium chloride,$C_6H_5N_2^+Cl^-$).$4$. Benzenediazonium chloride $(D)$ reacts with $HBF_4$ to form $E$ (benzenediazonium fluoroborate,$C_6H_5N_2^+BF_4^-$).$5$. Heating $(E)$ leads to the Balz-Schiemann reaction,yielding $F$ (fluorobenzene,$C_6H_5F$).Thus,$F$ is fluorobenzene.

$C_6H_6$ $\xrightarrow[\text{Conc. } H_2SO_4]{\text{Conc. } HNO_3} B$ $\xrightarrow{\text{Sn + HCl}} C$ $\xrightarrow[0-5^\circ C]{HNO_2} D$ $\xrightarrow{HBF_4} E$ $\xrightarrow{\Delta} F$
$F$ will be

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$C_6H_6$ $\xrightarrow[\text{Conc. } H_2SO_4]{\text{Conc. } HNO_3} B$ $\xrightarrow{\text{Sn + HCl}} C$ $\xrightarrow[0-5^\circ C]{HNO_2} D$ $\xrightarrow{HBF_4} E$ $\xrightarrow{\Delta} F$$F$ will be