250+ TOP MCQs on Preparation of Amines and Answers

Chemistry Multiple Choice Questions on “Preparation of Amines – 2”.

1. Which of the following reagents cannot be used to convert ethanenitrile to ethylamine?
a) H₂/Ni
b) LiAlH₄
c) Na(Hg), ethanol
d) Sn, HCl
Answer: d
Clarification: Nitriles on reduction with lithium aluminium hydride or catalytic hydrogenation or with Na(Hg)-ethanol produce primary amines.

2. How many more carbon atoms are present in the amine formed from the reduction of a nitrile, than in the nitrile itself?
a) 0
b) 1
c) 2
d) 1 or 2
Answer: a
Clarification: The -CN group of the nitrile is reduced to -CH₂-NH₂ group. This means that the parent carbon chain of the amine has one more carbon than the parent chain of the nitrile. However, the total number of carbons in both compounds is same as one of the carbon in nitriles is present in the cyanide group.

3. The reduction of phenyl isocyanide with H₂ and Ni catalyst gives a/an _________
a) primary amine
b) secondary amine
c) tertiary amine
d) arylalkyl amine
Answer: b
Clarification: The reduction of isocyanide compounds (where the CN group is attached through N atom) with H₂/Ni gives secondary amines or N-alkyl amines. For example, phenyl isocyanide gives N-methylaniline.

4. What is the product of the following reaction?

a) Phenylmethanamine
b) 1-Phenylethan-1-amine
c) 2-Phenylethan-1-amine
d) Toluene
Answer: c
Clarification: Benzyl cyanide reacts with LiAlH₄ to undergo reduction of the CN group to a CH₂NH₂ group and form a primary aromatic amine. The formula of this product will be C₆H₅CH₂CH₂NH₂, which is 2-phenylethan-1-amine.

5. Which of the following amines cannot be formed from the reduction of amides with LiAlH₄?
a) Ethylamine
b) Benzenamine
c) Benzylamine
d) Ethylmethylamine
Answer: b
Clarification: The -CONH₂ group of amides is reduced to a -CH₂NH₂ group and results in a compound having the same number of carbon atoms. The simplest aromatic amide is benzamide, which is reduced to give benzylamine. Hence, aniline cannot be formed as there is no corresponding amide that can be reduced to it with LiAlH₄. However, it can be achieved by Hoffmann bromamide degradation.

6. Only primary amines can be obtained from the reduction of amides with LiAlH₄.
a) True
b) False
View Answer

Answer: b
Clarification: Primary, secondary or tertiary amines can be formed depending on whether the amide used is primary, secondary or tertiary. This is because irrespective of the number of alkyl groups, the CO group of amides is directly reduced to CH₂.

7. Ethyldimethylamine is obtained from the reduction of _______ with LiAlH₄.
a) acetamide
b) benzamide
c) N-methylacetamide
d) N,N-dimethylacetamide
Answer: d
Clarification: Ethyldimethylamine is a tertiary amine with two methyl substituents directly attached to N atom. It can be obtained only from the reduction of a corresponding tertiary amide that has two methyl substituents attached to N atom.

8. Which path gives propan-1-amine as the product in the reaction below?

a) Only A
b) Only B
c) Both A and B
d) Not A nor B
Answer: a
Clarification: Propanamine has three carbon atoms and so does the parent amide. So, the reduction of CO to CH₂ will give the required compound. Whereas in path B, the amide undergoes Hoffmann bromamide degradation to give ethanamine.

9. Which of the following is not a by product of Hoffmann bromamide degradation of acetamide with alcoholic KOH?
a) KBr
b) KCN
c) K₂CO₃
d) H₂O
Answer: b
Clarification: Acetamide (CH₃CONH₂) reacts with Br₂ and 4 molecules of KOH to form methylamine (main product) along with 2 molecules KBr, 2 molecules of H₂O and one molecule of K₂CO₃.

10. Hoffmann bromamide degradation reaction is used for preparing _______ amines.
a) primary
b) secondary
c) tertiary
d) mixed
Answer: a
Clarification: Mixed amines are a classification of secondary and tertiary amines. The Hoffmann bromamide degradation is only possible with 1° amides so as to produce primary amines.

11. The best reagent for converting 2-phenylpropanamide to 1-phenylethanamine is ________
a) H₂/Ni
b) Na(Hg)/C₂H₅OH
c) LiAlH₄
d) NaOH/Br₂
Answer: d
Clarification: 2-Phenylpropanamide has 9 carbon atoms while 1-phenylethamine has 8 carbon atoms. This depicts stepping down of carbon from amide to amine conversion and this takes place in Hoffmann degradation reaction.

12. Which of the following compounds undergoes Hoffmann bromamide degradation reaction?
a) C₆H₅NH₂
b) C₆H₅NO₂
c) C₆H₅CONH₂
d) C₆H₅CH₂NH₂
Answer: c
Clarification: Only primary amides undergo this reaction with bromine in NaOH to form primary amines. Benzamide undergoes this reaction to form benzenamine or aniline.

13. Which of the following amines can be prepared from Gabriel phthalimide synthesis?
a) Benzylamine
b) Aniline
c) o-Toluidine
d) N-Methylbenzenamine
Answer: a
Clarification: Since Gabriel phthalimide synthesis is a nucleophilic substitution reaction, aryl halides cannot be prepared form this. This is because of the sp² hybridised aryl carbon and stable benzene ring.

14. Gabriel phthalimide synthesis can be used for preparing phenylmethanamine.
a) True
b) False
Answer: a
Clarification: Potassium phthalimide can undergo nucleophilic substitution with a benzyl halide to form N-phenylphthalimide, which on hydrolysis with NaOH gives benzylamine or phenylmethanamine.

15. Which of the following steps is not present in Gabriel phthalimide synthesis?
a) Treating phthalimide with alcoholic KOH
b) Heating potassium phthalimide with alkyl halide
c) Alkaline hydrolysis of N-alkylphthalimide
d) Heating phthalic acid with NaOH
Answer: d
Clarification: The chemical compounds involved in the Gabriel phthalimide synthesis are phthalimide, potassium phthalimide and N-alkyl phthalimide. These are formed in order when reacted with different reagents like KOH and haloalkanes.