Match each organic reaction involving an amine or amide on the left with its corresponding… — JAMB UTME

Match each organic reaction involving an amine or amide on the left with its corresponding principal product on the right.

Reduction of propanamide ( $CH_3CH_2CONH_2$ ) using $LiAlH_4$ in dry etherPropylamine ( $CH_3CH_2CH_2NH_2$ )
Reaction of propanamide ( $CH_3CH_2CONH_2$ ) with bromine ( $Br_2$ ) in aqueous $KOH$ Ethylamine ( $CH_3CH_2NH_2$ )
Alkaline hydrolysis of propanamide ( $CH_3CH_2CONH_2$ ) by boiling with aqueous $NaOH$ Sodium propanoate ( $CH_3CH_2COONa$ ) and ammonia ( $NH_3$ )
Acylation of methylamine ( $CH_3NH_2$ ) using ethanoyl chloride ( $CH_3COCl$ ) $N$ -methylethanamide ( $CH_3CONHCH_3$ )

Answer

Reduction of propanamide with

LiAlH_4

pairs with propylamine; Reaction of propanamide with

Br_2/KOH

pairs with ethylamine; Alkaline hydrolysis of propanamide pairs with sodium propanoate and ammonia; Acylation of methylamine with ethanoyl chloride pairs with

N

-methylethanamide.

Each reaction pair is determined by its specific mechanistic pathway:

LiAlH_4

reduces the carbonyl group to methylene (retaining carbon count to form propylamine);

Br_2/KOH

undergoes Hofmann degradation to lose the carbonyl carbon (forming ethylamine); basic hydrolysis cleaves the

C-N

bond (yielding sodium propanoate and ammonia); and acylation of methylamine with ethanoyl chloride produces the substituted amide (

N

-methylethanamide).

Step-by-Step Solution

Analyze the reduction reaction of primary amides.

Reducing

CH_3CH_2CONH_2

with

LiAlH_4

reduces the

C=O

bond to a

-CH_2-

group without altering the total carbon count, yielding

CH_3CH_2CH_2NH_2

(propylamine).

Amide reduction retains the full carbon skeleton.

Identify the reaction of primary amides with

Br_2

and

KOH

This is Hofmann degradation, which removes the carbonyl carbon (

C=O

) as carbonate, reducing the carbon length by 1. Propanamide (3 carbons) yields ethylamine (2 carbons).

Hofmann degradation shortens the carbon chain by one atom.

Examine the basic hydrolysis of amides.

Nucleophilic attack of

OH^-

on the carbonyl carbon of propanamide cleaves the amide bond to generate propanoate anion (forming sodium propanoate with

Na^+

) and ammonia gas.

Base hydrolysis of amides yields a carboxylate salt and ammonia.

Examine the nucleophilic substitution between methylamine and ethanoyl chloride.

The nitrogen lone pair of methylamine attacks ethanoyl chloride, releasing

HCl

to form a secondary amide,

N

-methylethanamide (

CH_3CONHCH_3

Primary amines undergo acylation to form secondary amides.

Key Concept

Chemical Reactions and Interconversions of Amines and Amides

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic