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

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

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

Column-$I$	Column-$II$
$A$. Ammonolysis	$1$. Amine with lesser number of carbon atoms
$B$. Gabriel phthalimide synthesis	$2$. Detection test for primary amine
$C$. Hoffmann Bromamide reaction	$3$. Reaction of phthalimide with $KOH$ and $R-X$
$D$. Carbylamine reaction	$4$. Reaction of alkyl-halides with $NH_3$

The correct sequence of reagents for the following transformation is:

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