The correct sequence of reagents for the foll | vedclass

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(D) The transformation involves converting nitrobenzene to $3$-chlorophenol. Step $1$: Chlorination of nitrobenzene using $Cl_{2} / FeCl_{3}$ gives $m

Vedclass · Accepted Answer

$i. Cl_{2}, FeCl_{3}; ii. Fe, HCl; iii. NaNO_{2}, HCl, 0^{\circ} C; iv. H_{2}O / H^{+}$

Vedclass · Answer

(D) The transformation involves converting nitrobenzene to $3$-chlorophenol. Step $1$: Chlorination of nitrobenzene using $Cl_{2} / FeCl_{3}$ gives $m$-chloronitrobenzene because the $-NO_{2}$ group is meta-directing. Step $2$: Reduction of the nitro group using $Fe / HCl$ yields $m$-chloroaniline. Step $3$: Diazotization of $m$-chloroaniline using $NaNO_{2} / HCl$ at $0^{\circ} C$ produces $m$-chlorobenzenediazonium chloride. Step $4$: Hydrolysis of the diazonium salt using $H_{2}O / H^{+}$ results in the formation of $3$-chlorophenol. Thus,the correct sequence is $i. Cl_{2}, FeCl_{3}; ii. Fe, HCl; iii. NaNO_{2}, HCl, 0^{\circ} C; iv. H_{2}O / H^{+}$.

The correct sequence of reagents for the following transformation is:

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