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NEET Organic Compounds Containing Nitrogen MCQs with answers

223 questions on Organic Compounds Containing Nitrogen for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. Which of the following represents a secondary (2°) allylic amine?

  1. CH₂=CH-NH-CH₂CH₃
  2. CH₂=CH-CH(CH₃)-NH₂
  3. CH₂=CH-CH₂-NH-CH₃
  4. CH₃-CH=CH-CH₂-NH₂
Show answer and explanation

Answer: (C) CH₂=CH-CH₂-NH-CH₃

An allylic amine has the nitrogen atom attached directly to an sp³ hybridized allylic carbon (adjacent to a C=C double bond). In CH₂=CH-CH₂-NH-CH₃, the nitrogen is bonded to one allylic carbon and one methyl carbon, making it a secondary (2°) allylic amine.

medium

2. Given below are two statements: Statement I: Gabriel phthalimide synthesis is preferred for preparing pure aliphatic primary amines. Statement II: Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with potassium phthalimide. In light of the above statements, choose the correct answer:

  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
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Gabriel phthalimide synthesis exclusively gives pure primary aliphatic amines without any contamination of secondary or tertiary amines. Statement II is also correct because aryl halides do not undergo nucleophilic substitution with the nucleophile formed by phthalimide.

hard

3. Heating anilinium hydrogensulphate (from aniline and conc. H2SO4) at 453-473 K gives:

  1. Aniline back unchanged
  2. p-Aminobenzenesulphonic acid (sulphanilic acid), which exists as a zwitterion
  3. m-Aminobenzenesulphonic acid
  4. Benzenesulphonic acid
Show answer and explanation

Answer: (B) p-Aminobenzenesulphonic acid (sulphanilic acid), which exists as a zwitterion

Sulphonation of aniline forms anilinium hydrogensulphate, which on heating at 453-473 K gives p-aminobenzenesulphonic acid (sulphanilic acid), existing mainly as a zwitterion.

NCERT Class 12

medium

4. Consider the synthesis of 4-hydroxyazobenzene. Why is it essential to maintain the pH between 9 and 10 during the coupling of benzenediazonium chloride with phenol?

  1. To convert phenol into the more nucleophilic phenoxide ion without hydrolysing the diazonium ion
  2. To protonate the diazonium ion and increase its electrophilicity
  3. To prevent benzenediazonium chloride from converting into non-reactive phenyl cation
  4. To neutralise the hydrochloric acid so that aniline does not precipitate
Show answer and explanation

Answer: (A) To convert phenol into the more nucleophilic phenoxide ion without hydrolysing the diazonium ion

In alkaline solution (pH 9-10), phenol is converted to the phenoxide ion (C₆H₅O⁻), whose ring is far more nucleophilic and strongly activated towards electrophilic substitution. If the pH is too high (>11), diazonium ions are converted into unreactive diazotate ions (C₆H₅-N=N-O⁻).

easy

5. Which of the following gas(es) is/are quantitatively evolved when an aqueous solution of benzenediazonium chloride is warmed to 323 K?

  1. Only Cl₂
  2. Only NO₂
  3. N₂ only
  4. A mixture of N₂ and Cl₂
Show answer and explanation

Answer: (C) N₂ only

When an aqueous solution of benzenediazonium chloride is warmed (above 283 K), it undergoes hydrolysis to produce phenol, releasing nitrogen gas quantitatively along with HCl: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂↑ + HCl.

hard

6. The conversion of benzenediazonium chloride to fluorobenzene via benzenediazonium fluoroborate involves which type of intermediate or reaction mechanism during thermal decomposition?

  1. Nucleophilic bimolecular displacement (SN2)
  2. Aryne intermediate pathway
  3. Phenyl cation or free-radical pathway (SN1-like)
  4. Electrophilic addition pathway
Show answer and explanation

Answer: (C) Phenyl cation or free-radical pathway (SN1-like)

Thermal decomposition of dry benzenediazonium tetrafluoroborate (Balz-Schiemann reaction) proceeds via an unimolecular pathway where N₂ leaves to generate an extremely reactive aryl cation (or radical pair depending on conditions) which captures fluoride ion (F⁻) from BF₄⁻ to give ArF, BF₃, and N₂.

easy

7. What is the primary reason why arenediazonium salts are synthesized at low temperatures (273-278 K) and generally not isolated in the dry crystalline state?

  1. They are explosive in the dry solid state and decompose easily on warming.
  2. They sublime rapidly at room temperature into toxic fumes.
  3. They dimerize spontaneously at temperatures above 273 K.
  4. They rapidly reduce to primary aromatic amines in dry air.
Show answer and explanation

Answer: (A) They are explosive in the dry solid state and decompose easily on warming.

Benzenediazonium chloride and related salts are unstable at temperatures above 5 °C (278 K) in aqueous solution and readily decompose by releasing N₂ gas. In the dry solid state, many diazonium salts are sensitive and explode violently upon percussion or warming.

easy

8. When aniline is treated with chloroform and ethanolic KOH, an extremely foul-smelling compound is formed. This reaction is known as:

  1. Sandmeyer reaction
  2. Gattermann reaction
  3. Carbylamine reaction
  4. Coupling reaction
Show answer and explanation

Answer: (C) Carbylamine reaction

The carbylamine reaction (or isocyanide test) is given by primary amines upon heating with chloroform and ethanolic KOH to form foul-smelling isocyanides (carbylamines). Aniline produces phenyl isocyanide (C₆H₅NC).

easy

9. Identify the reagent or sequence of reagents used in the Gattermann reaction to synthesize chlorobenzene from benzenediazonium chloride:

  1. Cu₂Cl₂ in concentrated HCl
  2. Copper foil in aqueous NaCl
  3. CuCl and KClO₃ at high temperature
  4. Copper powder in concentrated HCl
Show answer and explanation

Answer: (D) Copper powder in concentrated HCl

In the Gattermann reaction, benzenediazonium chloride is treated with copper powder in the presence of halogen acid (HCl for chlorobenzene, HBr for bromobenzene). The use of cuprous halide (Cu₂Cl₂) instead corresponds to the Sandmeyer reaction.

medium

10. In the Sandmeyer reaction, the replacement of the diazo group by a halogen atom is generally understood to proceed through:

  1. Carbocation intermediate
  2. Benzyne intermediate
  3. Carbanion intermediate
  4. Aryl free radical intermediate
Show answer and explanation

Answer: (D) Aryl free radical intermediate

The Sandmeyer reaction catalyzed by Cu(I) salts involves a single-electron transfer mechanism generating aryl free radicals (Ar•), which then capture the halogen from copper(II) species.

hard

11. In a reaction, 1 mole of methylamine (CH₃NH₂) reacts with excess of methyl iodide (CH₃I) in the presence of base until no further reaction occurs. How many moles of methyl iodide are consumed per mole of methylamine to convert it completely into the quaternary ammonium salt?

  1. 3
  2. 2
  3. 1
  4. 4
Show answer and explanation

Answer: (A) 3

Starting from methylamine (CH₃NH₂), 1 mole of CH₃I converts it to dimethylamine (CH₃)₂NH, the 2nd mole of CH₃I converts it to trimethylamine (CH₃)₃N, and the 3rd mole of CH₃I converts it to tetramethylammonium iodide [(CH₃)₄N]⁺I⁻. Hence, 3 moles of CH₃I are consumed per mole of primary amine.

medium

12. When the temperature of a benzenediazonium chloride solution is allowed to rise to about 283 K, the salt is hydrolysed to:

  1. Benzene
  2. Chlorobenzene
  3. Aniline
  4. Phenol
Show answer and explanation

Answer: (D) Phenol

If the temperature of a diazonium salt solution rises to 283 K, the salt gets hydrolysed to phenol.

NCERT Class 12

easy

13. Aniline reacts with concentrated H₂SO₄ at 453-473 K to yield predominantly which of the following compounds?

  1. o-Aminobenzenesulphonic acid
  2. Benzenesulphonic acid
  3. p-Aminobenzenesulphonic acid (Sulphanilic acid)
  4. m-Aminobenzenesulphonic acid
Show answer and explanation

Answer: (C) p-Aminobenzenesulphonic acid (Sulphanilic acid)

Heating aniline with concentrated sulphuric acid at 453-473 K produces sulphanilic acid (p-aminobenzenesulphonic acid) as the major product, which exists in equilibrium as a dipolar zwitterion.

hard

14. The correct decreasing order of basic strength among the following substituted pyridines/anilines is: (I) Pyridine (II) Piperidine (III) Pyrrole (IV) Aniline

  1. II > I > IV > III
  2. II > IV > I > III
  3. I > II > IV > III
  4. II > I > III > IV
Show answer and explanation

Answer: (A) II > I > IV > III

Piperidine (II) is a non-aromatic secondary aliphatic amine with an sp³-hybridized nitrogen, making it the most basic (pKb ≈ 2.8). Pyridine (I) has an sp²-hybridized nitrogen where the lone pair is not involved in aromaticity (pKb ≈ 8.8), making it more basic than aniline (IV, pKb ≈ 9.4), whose lone pair is delocalized. In pyrrole (III), the lone pair is an integral part of the aromatic 6π-electron sextet, making it an extremely weak base (pKb ≈ 13.6).

hard

15. Three isomeric amines A, B, and C with formula C₄H₁₁N were subjected to reaction with p-toluenesulphonyl chloride: - Isomer A reacts to give a product that is soluble in KOH. - Isomer B reacts to give a product that is completely insoluble in KOH. - Isomer C does not react with p-toluenesulphonyl chloride at all. When treated with NaNO₂ and HCl at 0 °C, which of these isomers will form a clear solution that does not evolve gas until warmed?

  1. None of the isomers
  2. Isomer B only
  3. Isomer C only
  4. Isomer A only
Show answer and explanation

Answer: (A) None of the isomers

Isomers A, B, and C are primary, secondary, and tertiary aliphatic amines, respectively. Aliphatic primary amines (like A) decompose immediately even at 0 °C to evolve N₂ gas. Secondary amines (B) separate out as an insoluble yellow oil (nitrosoamine). Tertiary aliphatic amines (C) form soluble nitrite salts that remain in solution without decomposing, but they do not liberate N₂ on warming—they decompose into nitrosoamines and aldehydes/alcohols. Therefore, none of these aliphatic isomers form diazonium salts stable at 0 °C that evolve N₂ only on warming (a property characteristic of aromatic primary amines).

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