Question

Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	Hydrolysis of benzene diazonium chloride	(i)	p- Cresol
(B)	Phenol + methyl chloride in presence of an, $AlC{l}_3$	(ii)	Salicylic acid
(C)	Reaction of sodium phenoxide with $C{O}_2$	(iii)	Picric acid
(D)	Phenol + Conc. $HN{O}_3$	(iv)	Phenol

• A. (A) $\to$ (i), (B) $\to$ (iii), (C)$\to$(ii), (D)$\to$ (iv)
• B. (A) $\to$ (ii), (B) $\to$ (iii), (C)$\to$ (iv), (D)$\to$ (i)
• C. (A) $\to$ (iv), (B) $\to$ (i), (C)$\to$ (ii), (D)$\to$ (iii)
• D. (A) $\to$ (iii), (B) $\to$ (iv), (C)$\to$ (i), (D)$\to$ (ii)

Answer 1

Answer: (C)

(A) $\to$ (iv), (B) $\to$ (i), (C)$\to$ (ii), (D)$\to$ (iii)

A. Hydrolysis of benzene diazonium chloride gives phenol as:

$C_6{H}_5{N}_{2}Cl\stackrel{H^{+}/H_{2}O}{\to }{C}_6{H}_{5}OH+N_2+HCl$

B. Phenol + methyl chloride in presence of anhy. $AlC{l}_3$ undergoes Friedel craft alkylation at ortho and para position. The major product is p-methyl phenol or p-cresol as:

$C_6{H}_{5}OH+C{H}_{3}Cl\stackrel{AlC{l}_3}{\to }p-\left(C{H}_3\right)C_6{H}_{4}OH$

C. Reaction of sodium phenoxide with $C{O}_2$ is a Kolbe's reaction and gives COOH at ortho to OH on benzene ring. The product is called salicylic acid as:

$C_6{H}_{5}ONa+C{O}_2\to o-\left(COOH\right)C_6{H}_{4}OH$

D. Phenol + Conc. $HN{O}_3$ gives trinitro product called picric acid as:

$C_6{H}_{5}OH+conc.HN{O}_3\to o,o^{\prime },p-(N{O}_2{)}_3{C}_6{H}_{2}OH$

Thus (A) - (iv), (B) - (i), (C)- (ii), (D)- (iii)