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57 questions on this topic of Organic Compounds Containing Nitrogen (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following amines is completely insoluble in water due to high steric hindrance and predominantly hydrophobic nature despite having an amino group?

  1. Ethanamine
  2. Propan-1-amine
  3. N,N-Diethylethanamine (Triethylamine)
  4. Methanamine
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Answer: (C) N,N-Diethylethanamine (Triethylamine)

Lower aliphatic amines like methanamine and ethanamine are readily miscible with water because they form intermolecular hydrogen bonds with water molecules. In triethylamine, the large hydrophobic hydrocarbon parts (three ethyl groups) strongly hinder hydrogen bonding, making it practically insoluble compared to the smaller lower amines.

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2. An organic compound 'A' on reaction with 2-chloro-2-methylpropane gives a single substitution product in negligible yield and predominantly forms 2-methylpropene. Compound 'A' is most likely:

  1. Ammonia
  2. Potassium hydroxide
  3. Ethanol
  4. Water
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Answer: (A) Ammonia

Ammonolysis of tertiary alkyl halides does not yield tertiary amines as major products; instead, tertiary halides predominantly undergo elimination in the presence of basic ammonia to give alkenes (2-methylpropene). Thus, 'A' is ammonia acting as a base rather than a nucleophile.

hard

3. Direct nitration of aniline gives a substantial amount of m-nitroaniline (about 47 per cent) because:

  1. In strongly acidic medium aniline is protonated to the anilinium ion, which is meta-directing
  2. -NH2 is a meta-directing group
  3. The meta position is the most electron-rich
  4. Nitric acid reduces the para product
Show answer and explanation

Answer: (A) In strongly acidic medium aniline is protonated to the anilinium ion, which is meta-directing

In strongly acidic conditions aniline becomes the anilinium ion, which deactivates the ring and directs meta. So nitration gives p-nitroaniline (51%), m-nitroaniline (47%) and o-nitroaniline (2%), along with tarry oxidation products.

NCERT Class 12

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4. Match List-I (Reaction/Reagents) with List-II (Name of Reaction): List-I: (A) R-CONH₂ + Br₂ + 4NaOH → R-NH₂ (B) C₆H₅N₂⁺Cl⁻ + CuCl / HCl → C₆H₅Cl + N₂ (C) R-NH₂ + CHCl₃ + 3KOH (alc.) → R-NC (D) C₆H₅N₂⁺Cl⁻ + Cu / HCl → C₆H₅Cl + N₂ List-II: (I) Gattermann reaction (II) Carbylamine reaction (III) Sandmeyer reaction (IV) Hofmann bromamide degradation Choose the correct option:

  1. A-III, B-IV, C-I, D-II
  2. A-IV, B-I, C-II, D-III
  3. A-II, B-III, C-IV, D-I
  4. A-IV, B-III, C-II, D-I
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Answer: (D) A-IV, B-III, C-II, D-I

Conversion of an amide to an amine with one less carbon using Br₂/NaOH is Hofmann bromamide degradation (A-IV). Replacement of diazonium by halogen using CuCl/HCl is Sandmeyer reaction (B-III). Conversion of primary amine to isocyanide using CHCl₃/KOH is Carbylamine reaction (C-II). Use of copper powder with halogen acid is Gattermann reaction (D-I).

easy

5. Among the following compounds, which one does NOT react with benzenesulphonyl chloride (Hinsberg's reagent) to form any product?

  1. CH₃-NH-CH₃
  2. CH₃-CH₂-NH₂
  3. (CH₃)₃N
  4. C₆H₅-NH-CH₃
Show answer and explanation

Answer: (C) (CH₃)₃N

Hinsberg's reagent (benzenesulphonyl chloride) reacts with primary amines to form sulfonamides soluble in alkali and with secondary amines to form insoluble sulfonamides. Tertiary amines like trimethylamine, (CH₃)₃N, lack any replaceable hydrogen atom directly attached to the nitrogen atom and therefore do not react with benzenesulphonyl chloride.

hard

6. In a reaction, 1 mole of an aromatic compound A (C₇H₉N) reacts with 1 mole of acetyl chloride to give B (C₉H₁₁NO). Oxidation of A with alkaline KMnO₄ followed by acidification gives benzoic acid. What is the IUPAC name of compound A?

  1. N-Methylaniline
  2. 2-Methylaniline
  3. 4-Methylaniline
  4. Benzylamine
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Answer: (D) Benzylamine

Since A on oxidation with alkaline KMnO₄ gives benzoic acid, the nitrogen-containing substituent must be on a side chain attached to the ring, or an alkyl side chain is present. A reacts with acetyl chloride to give a monoacetyl derivative, meaning it is a primary or secondary amine. However, nuclear methyl groups in toluidines oxidize to give aminobenzoic acids (or get destroyed), whereas benzylamine (C₆H₅CH₂NH₂) has formula C₇H₉N and oxidizes at the side chain to yield benzoic acid. Moreover, benzylamine is a primary amine and acetylates to N-benzylacetamide (C₉H₁₁NO).

easy

7. An aliphatic primary amine reacts with nitrous acid (HNO₂ generated in situ) at 273-278 K. What are the expected major observation and organic product?

  1. Quantitative liberation of N₂ gas and formation of alcohol
  2. Formation of a stable diazonium salt that precipitates out
  3. Formation of a yellow oily N-nitrosoamine layer
  4. Evolution of NO₂ gas with formation of a nitroalkane
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Answer: (A) Quantitative liberation of N₂ gas and formation of alcohol

Primary aliphatic amines react with nitrous acid to form highly unstable aliphatic diazonium salts, which decompose spontaneously and quantitatively to liberate nitrogen gas and form alcohols along with other secondary products. This quantitative evolution of N₂ is utilized in amino acid estimation.

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8. Consider the following statements regarding diazonium salts: Statement I: Arenediazonium fluoroborates are generally insoluble in water and stable at room temperature. Statement II: Aqueous solutions of benzenediazonium chloride can be stored safely at room temperature for several days. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Benzenediazonium fluoroborate is insoluble in water and reasonably stable at room temperature, making Statement I correct. Benzenediazonium chloride decomposes readily in aqueous solution above 273-278 K releasing nitrogen gas and cannot be stored, making Statement II incorrect.

easy

9. Which of the following methods CANNOT be used to prepare a primary amine?

  1. Reduction of ethanenitrile with LiAlH₄
  2. Hofmann bromamide degradation of propanamide
  3. Reduction of nitrobenzene with Fe and HCl
  4. Gabriel phthalimide synthesis using chlorobenzene
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Answer: (D) Gabriel phthalimide synthesis using chlorobenzene

Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines like aniline because aryl halides (such as chlorobenzene) do not undergo nucleophilic substitution with the phthalimide anion under normal conditions due to partial double bond character.

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10. An organic compound 'X' (C₆H₇N) is treated with NaNO₂ and HCl at 0 °C to yield 'Y'. Compound 'Y' when warmed with aqueous ethanol (CH₃CH₂OH) gives 'Z', an aldehyde 'W', and nitrogen gas. Identify 'Z' and 'W':

  1. Z = Phenol, W = Methanal
  2. Z = Benzene, W = Ethanal
  3. Z = Chlorobenzene, W = Ethanal
  4. Z = Nitrobenzene, W = Acetic acid
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Answer: (B) Z = Benzene, W = Ethanal

Aniline (X, C₆H₇N) reacts with NaNO₂/HCl at 0 °C to give benzenediazonium chloride (Y). Diazonium salts are reduced by mild reducing agents such as ethanol to yield benzene (Z), while ethanol is oxidised to ethanal (W, CH₃CHO), alongside the liberation of N₂ and HCl.

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11. Aniline does not undergo Friedel-Crafts alkylation or acylation reaction. What is the main reason for this?

  1. The phenyl ring of aniline is heavily deactivated towards electrophilic attack
  2. The amino group undergoes preferential substitution over the ring
  3. Aniline gets oxidised by anhydrous AlCl₃ to nitrobenzene
  4. Aniline forms a salt complex with anhydrous AlCl₃, deactivating the benzene ring
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Answer: (D) Aniline forms a salt complex with anhydrous AlCl₃, deactivating the benzene ring

Aniline acts as a Lewis base and forms a salt/complex with anhydrous AlCl₃ (the Lewis acid catalyst). Because of the positive charge acquired by nitrogen in this complex, it strongly deactivates the aromatic ring towards electrophilic attack.

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12. Reduction of amides with lithium aluminium hydride gives:

  1. Nitriles
  2. Carboxylic acids
  3. Alcohols
  4. Amines
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Answer: (D) Amines

Amides on reduction with LiAlH4 yield amines. This is another method of preparing amines besides reduction of nitro compounds and nitriles.

NCERT Class 12

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