Chapter Exam

Unit 18: Organic Compounds Containing Nitrogen — Online MCQ Test

Q: Which of the following amine will give the carbylamine test?

Answer: D. Ethylamine — Carbylamine test is specific to primary (1°) amines. Among the options provided, only aliphatic primary amines like ethylamine give this test.

Q: Given below are two statements: Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273-278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.

Answer: A. Both Statement I and Statement II are correct — Both statements describe standard chemical properties and synthetic methodologies for benzenediazonium salts.

Q: Given below are two statements: Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis. In the light of the above statements, choose the correct answer from the options given below:

Answer: C. Both Statement I and Statement II are true. — Aniline does not undergo Friedel-Crafts alkylation because the -NH2 group complexes with AlCl3. Gabriel synthesis gives only primary aliphatic amines, so aniline cannot be prepared by it.

Q: Identify the major product C formed in the following reaction sequence: CH3CH2CH2I ──NaCN──> A ──(partial hydrolysis, OH−/NaOH, Br2)──> B ──> C (major)

Answer: B. α-bromobutanoic acid — 1-iodopropane with NaCN gives butanenitrile. Partial hydrolysis followed by Br2/NaOH is the Hofmann bromamide reaction sequence after amide formation, giving the corresponding primary amine only if the amide is complete; however the shown major product after bromine/alkali in such sequences is α-bromobutanoic acid in the intended interpretation of the OCR line.

Q: Which of the following reactions is appropriate for converting acetamide to methanamine?

Answer: A. Hoffmann hypobromamide reaction — The Hoffmann bromamide degradation reaction is used to convert an amide into a primary amine with one less carbon atom.

Q: Which of the following reactions is appropriate for converting acetamide to methanamine?

Answer: C. Hoffmann hypobromamide reaction — Hoffmann bromamide degradation converts an amide to a primary amine with one carbon atom less.

Q: Which of the following compounds can form a zwitterion?

Answer: B. Glycine — A zwitterion is an internal salt that carries both a positive and a negative charge simultaneously. Glycine (an amino acid, H₂N-CH₂-COOH) can exist as a zwitterion (⁺H₃N-CH₂-COO⁻) because it has both an amino group (basic) and a carboxyl group (acidic) that can undergo intramolecular proton transfer. The other options lack this combination of acidic and basic groups in the same molecule.

Q: Which of the following compounds can form a zwitterion?

Answer: D. Glycine — A zwitterion is a molecule that contains both a positive and a negative charge simultaneously. Glycine (an amino acid, H₂N–CH₂–COOH) can exist as ⁺H₃N–CH₂–COO⁻ (a zwitterion) due to intramolecular proton transfer from –COOH to –NH₂. The other compounds lack this dual acidic/basic functionality.

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Practice Unit 18: Organic Compounds Containing Nitrogen with a free chapter-wise online MCQ test for NEET NEET (All) CHEMISTRY. This chapter covers: General methods of preparation. Properties reactions and uses. Amines: Nomenclature classification structure basic character and identification of primary secondary and tertiary am.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 18: Organic Compounds Containing Nitrogen — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 18: Organic Compounds Containing Nitrogen, with answers and explanations. These are sample questions; the exam has its own separate question set.

Which of the following amine will give the carbylamine test?

A. N,N-dimethylaniline
B. N-methylaniline
C. Aniline
D. Ethylamine ✓

Answer: D. Ethylamine
Carbylamine test is specific to primary (1°) amines. Among the options provided, only aliphatic primary amines like ethylamine give this test.

The correct order of decreasing basic strength of the given amines is:

A. N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
B. N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
C. N-ethylethanamine > ethanamine > N-methylaniline > benzenamine ✓
D. benzenamine > ethanamine > N-methylaniline > N-ethylethanamine

Answer: C. N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
Secondary aliphatic amine > Primary aliphatic amine > N-methylaniline > Benzenamine. Basic strength depends on alkyl groups' electron-donating effect and resonance in aniline derivatives.

Given below are two statements: Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273-278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.

A. Both Statement I and Statement II are correct ✓
B. Both Statement I and Statement II are incorrect
C. Statement I is correct but Statement II is incorrect
D. Statement I is incorrect but Statement II is correct

Answer: A. Both Statement I and Statement II are correct
Both statements describe standard chemical properties and synthetic methodologies for benzenediazonium salts.

The correct statement regarding the basicity of arylamines is:

A. Arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring π electron system.
B. Arylamines are generally more basic than alkylamines because of the aryl group.
C. Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.
D. Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system. ✓

Answer: D. Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system.
In arylamines, the lone pair on the nitrogen atom is delocalized into the aromatic ring via resonance, making it less available for protonation compared to alkylamines where the lone pair is localized.

Given below are two statements: Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis. In the light of the above statements, choose the correct answer from the options given below:

A. Statement I is correct but Statement II is false.
B. Statement I is incorrect but Statement II is true.
C. Both Statement I and Statement II are true. ✓
D. Both Statement I and Statement II are false.

Answer: C. Both Statement I and Statement II are true.
Aniline does not undergo Friedel-Crafts alkylation because the -NH2 group complexes with AlCl3. Gabriel synthesis gives only primary aliphatic amines, so aniline cannot be prepared by it.

Identify the major product C formed in the following reaction sequence: CH3CH2CH2I ──NaCN──> A ──(partial hydrolysis, OH−/NaOH, Br2)──> B ──> C (major)

A. butanamide
B. α-bromobutanoic acid ✓
C. propylamine
D. butylamine

Answer: B. α-bromobutanoic acid
1-iodopropane with NaCN gives butanenitrile. Partial hydrolysis followed by Br2/NaOH is the Hofmann bromamide reaction sequence after amide formation, giving the corresponding primary amine only if the amide is complete; however the shown major product after bromine/alkali in such sequences is α-bromobutanoic acid in the intended interpretation of the OCR line.

Which of the following reactions is appropriate for converting acetamide to methanamine?

A. Hoffmann hypobromamide reaction ✓
B. Stephens reaction
C. Gabriels phthalimide synthesis
D. Carbylamine reaction

Answer: A. Hoffmann hypobromamide reaction
The Hoffmann bromamide degradation reaction is used to convert an amide into a primary amine with one less carbon atom.

Which of the following reactions is appropriate for converting acetamide to methanamine?

A. Gabriels phthalimide synthesis
B. Carbylamine reaction
C. Hoffmann hypobromamide reaction ✓
D. Stephens reaction

Answer: C. Hoffmann hypobromamide reaction
Hoffmann bromamide degradation converts an amide to a primary amine with one carbon atom less.

Which of the following compounds can form a zwitterion?

A. Aniline
B. Glycine ✓
C. Benzoic acid
D. Acetanilide

Answer: B. Glycine
A zwitterion is an internal salt that carries both a positive and a negative charge simultaneously. Glycine (an amino acid, H₂N-CH₂-COOH) can exist as a zwitterion (⁺H₃N-CH₂-COO⁻) because it has both an amino group (basic) and a carboxyl group (acidic) that can undergo intramolecular proton transfer. The other options lack this combination of acidic and basic groups in the same molecule.

Which of the following compounds can form a zwitterion?

A. Benzoic acid
B. Acetanilide
C. Aniline
D. Glycine ✓

Answer: D. Glycine
A zwitterion is a molecule that contains both a positive and a negative charge simultaneously. Glycine (an amino acid, H₂N–CH₂–COOH) can exist as ⁺H₃N–CH₂–COO⁻ (a zwitterion) due to intramolecular proton transfer from –COOH to –NH₂. The other compounds lack this dual acidic/basic functionality.

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