Identify the reagents X,Y,and Z in the follow | vedclass

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(A) For reaction $(i)$:Benzene reacts with conc. $H_2SO_4$ (oleum) to form benzenesulfonic acid $(X = conc. H_2SO_4)$.Benzenesulfonic acid reacts with

Vedclass · Accepted Answer

$X = 	ext{conc. } H_2SO_4, Y = NaOH, Z = H_3O^+$; $X = NaNO_2/HCl, Y = H_2O/\Delta$; $X = NaOH, Y = H_3O^+$

Vedclass · Answer

(A) For reaction $(i)$:Benzene reacts with conc. $H_2SO_4$ (oleum) to form benzenesulfonic acid $(X = conc. H_2SO_4)$.Benzenesulfonic acid reacts with $NaOH$ (fusion) to form sodium phenoxide $(Y = NaOH)$.Sodium phenoxide reacts with dilute acid $(H_3O^+)$ to form phenol $(Z = H_3O^+)$.For reaction $(ii)$:Aniline reacts with $NaNO_2/HCl$ at $0-5 ^\circ C$ to form benzenediazonium chloride $(X = NaNO_2/HCl)$.Benzenediazonium chloride reacts with water ($H_2O$ or $H_3O^+$) upon warming to form phenol $(Y = H_3O^+)$.For reaction $(iii)$:Chlorobenzene reacts with $NaOH$ at high temperature and pressure $(623 K, 300 atm)$ to form sodium phenoxide $(X = NaOH)$.Sodium phenoxide reacts with dilute acid $(H_3O^+)$ to form phenol $(Y = H_3O^+)$.Thus,the correct reagents are $X = conc. H_2SO_4, Y = NaOH, Z = H_3O^+$ for $(i)$; $X = NaNO_2/HCl, Y = H_3O^+$ for $(ii)$; and $X = NaOH, Y = H_3O^+$ for $(iii)$.

Column-$I$ (Name)	Column-$II$ ($C-O$ bond length) and Column-$III$ ($C-O-H$ angle)
$A$. Methanol	$1$. $141 \ pm$,$i$. $109^o$
$B$. Phenol	$2$. $136 \ pm$,$ii$. $111.7^o$
$C$. Methoxymethane	$3$. $142 \ pm$,$iii$. $108.9^o$

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