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JEE MainChemistry

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Organic Compounds Containing Nitrogen: Chemistry | JEE Main

Match List I with List II.

List I		List II
Reaction		Reagents
(A)	Hoffmann Degradation	(I)	$\text{Conc.}\mathrm{KOH}, \Delta$
(B)	Clemenson reduction	(II)	$\mathrm{CHCl}_3, \mathrm{NaOH}/\mathrm{H}_3\mathrm{O}^{\oplus}$
(C)	Cannizaro reaction	(III)	$\mathrm{Br}_2, \mathrm{NaOH}$
(D)	Reimer-Tiemann Reaction	(IV)	$\mathrm{Zn}-\mathrm{Hg}/\mathrm{HCl}$

Choose the correct answer from the options given below:

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A-III, (B) - IV, (C) - I, (D) - IIB(A) - II, (B) -I, (C) - III, (D) - IVC(A) -III, (B) -IV, (C) - II, (D) - ID(A) -II, (B) - IV, (C) - I, (D) - III

Step-by-step solutionView

Correct answer

The correct matching is (A) - III, (B) - IV, (C) - I, (D) - II.

Option analysis

Why each option works or fails

A · -III, (B) - IV, (C) - I, (D) - II

None. This option correctly matches each named reaction with its characteristic reagent. Hoffmann degradation uses $\mathrm{Br}_2/\mathrm{NaOH}$ , Clemmensen reduction uses $\mathrm{Zn-Hg}/\mathrm{HCl}$ , Cannizzaro reaction uses concentrated alkali like $\text{Conc. }\mathrm{KOH}$ , and Reimer-Tiemann reaction uses $\mathrm{CHCl}_3/\mathrm{NaOH}$ .

B · (A) - II, (B) -I, (C) - III, (D) - IV

Confusing the halogen-containing base combinations across Hoffmann degradation and Reimer-Tiemann reaction, and swapping Clemmensen reduction with Cannizzaro reaction conditions. Remember that Hoffmann degradation specifically requires a halogen with base ( $\mathrm{Br}_2/\mathrm{NaOH}$ ) to convert amides to amines, not chloroform.

C · (A) -III, (B) -IV, (C) - II, (D) - I

Inverting the roles of concentrated base (Cannizzaro) and chloroform with aqueous base (Reimer-Tiemann). Cannizzaro reaction is a disproportionation of non-enolizable aldehydes driven solely by concentrated alkali ( $\text{Conc. }\mathrm{KOH}$ ), whereas Reimer-Tiemann specifically incorporates a formyl group using chloroform ( $\mathrm{CHCl}_3$ ) in base.

D · (A) -II, (B) - IV, (C) - I, (D) - III

Mistaking the electrophilic source in the Reimer-Tiemann reaction for molecular bromine rather than dichlorocarbene from chloroform. Reimer-Tiemann introduces a formyl group using $\mathrm{CHCl}_3/\mathrm{NaOH}$ , while Hoffmann degradation degrades primary amides with $\mathrm{Br}_2/\mathrm{NaOH}$ .

Reviewed route

Solution

StepWorking

01concept

Recall the characteristic reagent combinations for each named organic reaction: - Hoffmann bromamide degradation converts primary amides to amines using $\mathrm{Br}_2$ and aqueous $\mathrm{NaOH}$ (III). - Clemmensen reduction reduces carbonyl groups of aldehydes/ketones to methylene groups using zinc amalgam and concentrated hydrochloric acid, $\mathrm{Zn}\text{-}\mathrm{Hg}/\mathrm{HCl}$ (IV). - Cannizzaro reaction is the disproportionation of non-enolizable aldehydes under strong basic conditions, such as $\text{Conc. }\mathrm{KOH},\, \Delta$ (I). - Reimer-Tiemann reaction formylates phenols using chloroform and alkali, $\mathrm{CHCl}_3, \mathrm{NaOH}/\mathrm{H}_3\mathrm{O}^+$ (II).

02option_verdict

Matches (A) with III, (B) with IV, (C) with I, and (D) with II correctly.

03option_verdict

Incorrectly assigns $\mathrm{CHCl}_3/\mathrm{NaOH}$ to Hoffmann degradation and $\text{Conc. }\mathrm{KOH}$ to Clemmensen reduction.

04option_verdict

Swaps the reagents for Cannizzaro reaction and Reimer-Tiemann reaction.

05option_verdict

Matches Hoffmann degradation with $\mathrm{CHCl}_3/\mathrm{NaOH}$ and Reimer-Tiemann with $\mathrm{Br}_2/\mathrm{NaOH}$ .

✓discriminator

Knowing that Hoffmann degradation uniquely uses bromine in alkali ((A) -> III) immediately eliminates options (1) and (3). Checking Clemmensen reduction as $\mathrm{Zn}\text{-}\mathrm{Hg}/\mathrm{HCl}$ ((B) -> IV) and Cannizzaro as concentrated alkali ((C) -> I) uniquely confirms option (0).

Hints that build this answer step by step

Which reagent pair is characteristic of the Hoffmann bromamide degradation of primary amides?
$\mathrm{Br}_2, \mathrm{NaOH}$ (III)
What are the characteristic reagents for Clemmensen reduction and the Cannizzaro reaction?
Clemmensen: $\mathrm{Zn-Hg}/\mathrm{HCl}$ (IV); Cannizzaro: $\text{Conc. }\mathrm{KOH}, \Delta$ (I)
Given that Reimer-Tiemann reaction is paired with $\mathrm{CHCl}_3, \mathrm{NaOH}/\mathrm{H}_3\mathrm{O}^\oplus$ (II), which overall match sequence is correct?
(A) - III, (B) - IV, (C) - I, (D) - II

✓ Source and academic review

Question type: Match the following
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is heating with concentrated KOH specific to Cannizzaro here?

Cannizzaro reaction requires a strongly alkaline medium (e.g., 50% or concentrated KOH) to initiate hydride transfer from the hydrate intermediate of an aldehyde lacking alpha-hydrogens.

Answer

The correct matching is (A) - III, (B) - IV, (C) - I, (D) - II.

Why each option works or fails

A · correct: -III, (B) - IV, (C) - I, (D) - II - None. This option correctly matches each named reaction with its characteristic reagent. Hoffmann degradation uses $\mathrm{Br}_2/\mathrm{NaOH}$ , Clemmensen reduction uses $\mathrm{Zn-Hg}/\mathrm{HCl}$ , Cannizzaro reaction uses concentrated alkali like $\text{Conc. }\mathrm{KOH}$ , and Reimer-Tiemann reaction uses $\mathrm{CHCl}_3/\mathrm{NaOH}$ .
B: (A) - II, (B) -I, (C) - III, (D) - IV - Confusing the halogen-containing base combinations across Hoffmann degradation and Reimer-Tiemann reaction, and swapping Clemmensen reduction with Cannizzaro reaction conditions. Remember that Hoffmann degradation specifically requires a halogen with base ( $\mathrm{Br}_2/\mathrm{NaOH}$ ) to convert amides to amines, not chloroform.
C: (A) -III, (B) -IV, (C) - II, (D) - I - Inverting the roles of concentrated base (Cannizzaro) and chloroform with aqueous base (Reimer-Tiemann). Cannizzaro reaction is a disproportionation of non-enolizable aldehydes driven solely by concentrated alkali ( $\text{Conc. }\mathrm{KOH}$ ), whereas Reimer-Tiemann specifically incorporates a formyl group using chloroform ( $\mathrm{CHCl}_3$ ) in base.
D: (A) -II, (B) - IV, (C) - I, (D) - III - Mistaking the electrophilic source in the Reimer-Tiemann reaction for molecular bromine rather than dichlorocarbene from chloroform. Reimer-Tiemann introduces a formyl group using $\mathrm{CHCl}_3/\mathrm{NaOH}$ , while Hoffmann degradation degrades primary amides with $\mathrm{Br}_2/\mathrm{NaOH}$ .

Step-by-step solution

concept: Recall the characteristic reagent combinations for each named organic reaction: - Hoffmann bromamide degradation converts primary amides to amines using $\mathrm{Br}_2$ and aqueous $\mathrm{NaOH}$ (III). - Clemmensen reduction reduces carbonyl groups of aldehydes/ketones to methylene groups using zinc amalgam and concentrated hydrochloric acid, $\mathrm{Zn}\text{-}\mathrm{Hg}/\mathrm{HCl}$ (IV). - Cannizzaro reaction is the disproportionation of non-enolizable aldehydes under strong basic conditions, such as $\text{Conc. }\mathrm{KOH},\, \Delta$ (I). - Reimer-Tiemann reaction formylates phenols using chloroform and alkali, $\mathrm{CHCl}_3, \mathrm{NaOH}/\mathrm{H}_3\mathrm{O}^+$ (II).
option_verdict: Matches (A) with III, (B) with IV, (C) with I, and (D) with II correctly.
option_verdict: Incorrectly assigns $\mathrm{CHCl}_3/\mathrm{NaOH}$ to Hoffmann degradation and $\text{Conc. }\mathrm{KOH}$ to Clemmensen reduction.
option_verdict: Swaps the reagents for Cannizzaro reaction and Reimer-Tiemann reaction.
option_verdict: Matches Hoffmann degradation with $\mathrm{CHCl}_3/\mathrm{NaOH}$ and Reimer-Tiemann with $\mathrm{Br}_2/\mathrm{NaOH}$ .
discriminator: Knowing that Hoffmann degradation uniquely uses bromine in alkali ((A) -> III) immediately eliminates options (1) and (3). Checking Clemmensen reduction as $\mathrm{Zn}\text{-}\mathrm{Hg}/\mathrm{HCl}$ ((B) -> IV) and Cannizzaro as concentrated alkali ((C) -> I) uniquely confirms option (0).