Match List I with List II.

List I List II

Reaction Reagents

(A) Hoffmann
Degradation (I) Conc. $KOH, \Delta$

(B) Clemmensen
reduction (II) ${CHCl}_{3}, NaOH / {H}_{3}{O}^{+}$

(C) Cannizzaro reaction (III) ${Br}_{2}, NaOH$

(D) Reimer-Tiemann
reaction (IV) $Zn - Hg / HCl$

Question

Match List I with List II.

List I		List II
Reaction		Reagents
(A)	Hoffmann Degradation	(I)	Conc. $KOH, \Delta$
(B)	Clemmensen reduction	(II)	${CHCl}_{3}, NaOH / {H}_{3}{O}^{+}$
(C)	Cannizzaro reaction	(III)	${Br}_{2}, NaOH$
(D)	Reimer-Tiemann reaction	(IV)	$Zn - Hg / HCl$

Quantrex Academy · Accepted Answer

Correct answer: C. (A) - III, (B) - IV, (C) - I, (D) - II

(A) Hoffmann degradation reaction: Hoffmann developed a method for the preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.

$O ∥ R - C - {NH}_{2} + {Br}_{2} + 4NaOH \to R - {NH}_{2} + {Na}_{2}{CO}_{3} + 2NaBr + 2{H}_{2}O$

(B) Clemmensen reduction reaction: The Clemmensen reduction is a reaction that is used to reduce aldehydes or ketones to alkanes using hydrochloric acid and zinc amalgam.

(C) Cannizzaro reaction: Aldehydes which do not have an $\alpha$ -hydrogen atom, undergo self-oxidation and reduction (disproportionation) reaction on heating with concentrated alkali.

(D) Reimer-Tiemann reaction: On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring. This reaction is known as Reimer-Tiemann reaction.

So, the correct matching is given in the following table:

Reactions	Reagent used
(A) Hoffmann degradation	${Br}_{2} / NaOH$
(B) Clemmensen reduction	$Zn - Hg / HCl$
(C) Cannizzaro reaction	conc.KOH/ $KOH / ∆$
(D) Reimer-Tiemann reaction	${CHCl}_{3}, NaOH / {H}_{3}{O}^{+}$

Match List I with List II. List I List II Reaction Reagents (A)...

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