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CBSE Class 12 Chemistry Chapter 9 Amines — 20 MCQs with Answers

Chapter 9 Amines is a scoring chapter in CBSE Class 12 Chemistry, combining organic reaction mechanisms with industrial applications. Amines are organic derivatives of ammonia where one or more hydrogen atoms are replaced by alkyl or aryl groups. This chapter demands clarity on nomenclature, preparation methods (including reduction of nitro compounds and Gabriel phthalimide synthesis), chemical properties (basicity, alkylation, acylation, carbylamine test, Hinsberg test), and the entire chemistry of diazonium salts. The 2025 board paper expects you to recall specific reactions, distinguish isomers, and apply concepts in assertion-reason and case-based MCQs.

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Key takeaways

  • Amines are classified as primary (1°), secondary (2°), or tertiary (3°) based on the number of alkyl/aryl groups attached to nitrogen.
  • Gabriel phthalimide synthesis is specific for preparing primary aliphatic amines only, not aromatic amines.
  • The carbylamine test is positive only for primary amines (aliphatic and aromatic) and produces an offensive odour of isocyanide.
  • Hinsberg reagent (benzenesulphonyl chloride) differentiates 1°, 2°, and 3° amines based on solubility in aqueous KOH.
  • Diazonium salts are highly unstable at temperatures above 5°C and are prepared in situ by diazotisation of aromatic primary amines.
  • Azo coupling reactions of diazonium salts with phenols or amines produce azo dyes with brilliant colours.
  • In CBSE board exams Chapter 9 Amines typically carries 3-5 marks, with one MCQ and one short-answer or long-answer question.

Classification and Nomenclature of Amines — MCQs 1-3

Amines are classified based on the number of carbon atoms directly bonded to the nitrogen atom. A primary amine has one alkyl or aryl group (RNH₂), a secondary amine has two (R₂NH), and a tertiary amine has three (R₃N). IUPAC names use the suffix '-amine' with the longest carbon chain as the parent. Common names list alkyl groups alphabetically followed by '-amine'. Aromatic amines like aniline, N-methylaniline, and N,N-dimethylaniline are named with 'aniline' as the base. Understanding these distinctions is critical because reactivity (especially towards acylation and sulphonation) varies sharply among the three classes. NCERT Class 12 Chemistry MCQ questions often test your ability to identify the class of an amine from a given structure or IUPAC name.
  • MCQ 1: Which of the following is a secondary amine? (A) CH₃CH₂NH₂ (B) (CH₃)₂NH (C) (CH₃)₃N (D) C₆H₅NH₂ | Answer: (B) (CH₃)₂NH | Reason: Two methyl groups are attached to nitrogen, making it secondary.
  • MCQ 2: The IUPAC name of CH₃CH₂NHCH₃ is (A) N-methylethanamine (B) methylethylamine (C) N-ethylmethanamine (D) ethylmethylamine | Answer: (A) N-methylethanamine | Reason: Ethane is the parent chain; methyl is the N-substituent.
  • MCQ 3: Aniline is a (A) primary aromatic amine (B) secondary aromatic amine (C) tertiary aromatic amine (D) aliphatic amine | Answer: (A) primary aromatic amine | Reason: One phenyl group is attached to NH₂, making it a 1° aromatic amine.

Preparation of Amines — Gabriel Phthalimide Synthesis — MCQs 4-6

Amines can be prepared by reduction of nitro compounds, nitriles, and amides, or by nucleophilic substitution (alkylation of ammonia). The Gabriel phthalimide synthesis is a specific and elegant method for preparing pure primary aliphatic amines. In this method, phthalimide reacts with alcoholic KOH to form potassium phthalimide, which then undergoes nucleophilic substitution with an alkyl halide (preferably primary) to give N-alkyl phthalimide. Hydrolysis with aqueous NaOH or hydrazine (Gabriel reaction modification) releases the primary amine. This method avoids over-alkylation and is not suitable for aromatic amines because aryl halides do not undergo nucleophilic substitution under these conditions. CBSE Class 12 Chemistry chapter 9 quiz frequently includes MCQs on the scope and limitations of this synthesis.
  • MCQ 4: Gabriel phthalimide synthesis is used to prepare (A) primary aliphatic amines (B) primary aromatic amines (C) secondary amines (D) tertiary amines | Answer: (A) primary aliphatic amines | Reason: It specifically yields 1° aliphatic amines; aromatic halides do not react.
  • MCQ 5: In Gabriel synthesis, phthalimide is treated with (A) alcoholic KOH (B) aqueous NaOH (C) aqueous HCl (D) LiAlH₄ | Answer: (A) alcoholic KOH | Reason: Alcoholic KOH deprotonates phthalimide to form the nucleophile.
  • MCQ 6: Which reagent is used in the final step of Gabriel synthesis to release the primary amine? (A) HCl (B) H₂SO₄ (C) Hydrazine or aqueous NaOH (D) Alcoholic KOH | Answer: (C) Hydrazine or aqueous NaOH | Reason: These hydrolyse the N-alkyl phthalimide, liberating the amine.

Reduction of Nitro Compounds and Nitriles — MCQs 7-9

Reduction of nitro compounds is a versatile method for preparing both aliphatic and aromatic amines. Aromatic nitro compounds like nitrobenzene are reduced to aniline using Sn/HCl or Fe/HCl followed by neutralisation. Aliphatic nitroalkanes are reduced using LiAlH₄ or catalytic hydrogenation (H₂/Ni). Nitriles (R–C≡N) can be reduced to primary amines using LiAlH₄ or H₂/Ni, yielding an amine with one more carbon atom than the original nitrile. Amides can also be reduced to amines using LiAlH₄, producing amines with the same carbon count as the amide. These methods are staple questions in Class 12 Chemistry Amines multiple choice sets because they test understanding of reagent selection and the relationship between functional groups and product structure.
  • MCQ 7: Reduction of nitrobenzene with Sn and HCl gives (A) aniline (B) azobenzene (C) nitrosobenzene (D) phenylhydroxylamine | Answer: (A) aniline | Reason: Sn/HCl is a standard reducing agent for aromatic nitro → amine.
  • MCQ 8: Reduction of CH₃CH₂CN with LiAlH₄ gives (A) CH₃CH₂NH₂ (B) CH₃CH₂CH₂NH₂ (C) CH₃CONH₂ (D) CH₃CH₂COOH | Answer: (B) CH₃CH₂CH₂NH₂ | Reason: Nitrile reduction adds two H to C≡N, extending the chain by one carbon as –CH₂NH₂.
  • MCQ 9: Which reagent is used to reduce an aromatic nitro compound to aniline? (A) H₂/Ni (B) LiAlH₄ (C) Sn/HCl (D) NaBH₄ | Answer: (C) Sn/HCl | Reason: Sn/HCl or Fe/HCl in acidic medium is the classical reagent for aromatic nitro reduction.

Physical Properties and Basicity of Amines — MCQs 10-12

Amines exhibit hydrogen bonding due to the presence of a lone pair on nitrogen. Primary and secondary amines can form intermolecular H-bonds, resulting in higher boiling points than hydrocarbons of similar molecular mass. Tertiary amines cannot donate a hydrogen for H-bonding, so their boiling points are lower than corresponding 1° or 2° amines. Amines are basic because the lone pair on nitrogen can accept a proton. Aliphatic amines are generally more basic than ammonia due to the +I effect of alkyl groups, which increases electron density on nitrogen. Aromatic amines like aniline are weaker bases than ammonia because the lone pair on nitrogen is delocalised into the benzene ring by resonance, reducing its availability for protonation. NCERT Class 12 Chemistry MCQ questions often compare the basicity order of different amines or ask you to explain trends in boiling points.
  • MCQ 10: Which of the following is the most basic? (A) NH₃ (B) CH₃NH₂ (C) (CH₃)₂NH (D) C₆H₅NH₂ | Answer: (C) (CH₃)₂NH | Reason: Two alkyl groups provide maximum +I effect, increasing electron density on nitrogen.
  • MCQ 11: Aniline is less basic than methylamine because (A) aniline is aromatic (B) the lone pair on N is delocalised in the benzene ring (C) methyl group has +I effect (D) both B and C | Answer: (D) both B and C | Reason: Delocalisation in aniline reduces basicity; +I in methylamine increases it.
  • MCQ 12: The boiling point order is (A) 1° > 2° > 3° amine (B) 3° > 2° > 1° amine (C) 2° > 1° > 3° amine (D) all equal | Answer: (A) 1° > 2° > 3° amine | Reason: More N–H bonds mean stronger H-bonding and higher boiling point.

Chemical Reactions — Carbylamine Test and Hinsberg Test — MCQs 13-15

The carbylamine test is a qualitative test specific for primary amines. When a primary amine (aliphatic or aromatic) is heated with chloroform (CHCl₃) and alcoholic KOH, it forms an isocyanide (carbylamine) with a characteristic offensive smell. Secondary and tertiary amines do not give this test. The Hinsberg test uses benzenesulphonyl chloride (C₆H₅SO₂Cl) to distinguish among 1°, 2°, and 3° amines. A primary amine reacts to form N-alkylbenzenesulphonamide, which is soluble in aqueous KOH (due to acidic N–H). A secondary amine forms a sulphonamide that is insoluble in KOH (no acidic H). A tertiary amine does not react at all with Hinsberg reagent. These tests are high-yield for Amines important questions Class 12 and appear regularly in board MCQs and short-answer sections.
  • MCQ 13: Carbylamine test is given by (A) all amines (B) only primary amines (C) only secondary amines (D) only tertiary amines | Answer: (B) only primary amines | Reason: Primary amines react with CHCl₃ and alc. KOH to give isocyanides with a foul smell.
  • MCQ 14: In the Hinsberg test, a primary amine with C₆H₅SO₂Cl gives a product that is (A) soluble in KOH (B) insoluble in KOH (C) does not react (D) forms a gas | Answer: (A) soluble in KOH | Reason: The N-alkylbenzenesulphonamide has an acidic N–H, forming a salt in KOH.
  • MCQ 15: Which amine does NOT react with Hinsberg reagent? (A) CH₃NH₂ (B) (CH₃)₂NH (C) (CH₃)₃N (D) C₆H₅NH₂ | Answer: (C) (CH₃)₃N | Reason: Tertiary amines lack an N–H bond, so they cannot form a sulphonamide.

Diazonium Salts — Preparation and Stability — MCQs 16-18

Diazonium salts have the general formula Ar–N₂⁺X⁻ and are prepared by the diazotisation of aromatic primary amines with nitrous acid (NaNO₂ + HCl) at 0-5°C. The reaction must be carried out at low temperature because diazonium salts are highly unstable above 5°C and decompose explosively. Aliphatic diazonium salts are even more unstable and decompose immediately at room temperature, so only aromatic diazonium salts are practically useful. The nitrogen molecule (N₂) is an excellent leaving group, making diazonium salts versatile intermediates for introducing various functional groups (–OH, –Cl, –Br, –I, –CN, –F, –H) onto the benzene ring via Sandmeyer, Gattermann, or Balz-Schiemann reactions. CBSE Class 12 Chemistry chapter 9 quiz questions test both the conditions of diazotisation and the rationale behind temperature control.
  • MCQ 16: Diazonium salts are prepared by treating aniline with (A) HNO₃ (B) NaNO₂ + HCl at 0-5°C (C) NaNO₃ + H₂SO₄ (D) CH₃COOH | Answer: (B) NaNO₂ + HCl at 0-5°C | Reason: Nitrous acid (in situ) at low temperature converts ArNH₂ to ArN₂⁺Cl⁻.
  • MCQ 17: Diazonium salts are stable (A) above 10°C (B) at room temperature (C) below 5°C (D) at all temperatures | Answer: (C) below 5°C | Reason: Above 5°C they decompose, losing N₂ gas.
  • MCQ 18: Aliphatic diazonium salts are (A) more stable than aromatic (B) less stable than aromatic (C) equally stable (D) non-existent | Answer: (B) less stable than aromatic | Reason: Aromatic ring stabilises the cation by resonance; aliphatic salts decompose instantly.

Reactions of Diazonium Salts — Sandmeyer and Azo Coupling — MCQs 19-20

Diazonium salts undergo two main types of reactions: replacement reactions (where N₂ is replaced by another group) and coupling reactions (where the diazonium ion acts as an electrophile). In Sandmeyer reactions, diazonium salts react with CuCl, CuBr, or CuCN to introduce –Cl, –Br, or –CN groups respectively. The Gattermann reaction uses Cu powder with HCl or HBr. To introduce –F, the Balz-Schiemann reaction treats the diazonium salt with HBF₄, then heats the resulting diazonium fluoroborate. For hydroxyl groups, diazonium salts are warmed with water. Azo coupling occurs when diazonium salts react with phenols or aromatic amines in mildly alkaline solution, forming brightly coloured azo dyes (Ar–N=N–Ar'). These reactions are heavily tested in Class 12 Chemistry Amines multiple choice questions, especially in application and case-based formats, because they showcase the synthetic utility of diazonium chemistry in industrial dye manufacture.
  • MCQ 19: In Sandmeyer reaction, C₆H₅N₂⁺Cl⁻ + CuCN gives (A) benzonitrile (B) aniline (C) phenol (D) chlorobenzene | Answer: (A) benzonitrile | Reason: CuCN replaces the diazonium group with –CN, forming C₆H₅CN.
  • MCQ 20: Azo coupling of benzenediazonium chloride with phenol in NaOH gives (A) benzene (B) aniline (C) p-hydroxyazobenzene (D) phenyl cyanide | Answer: (C) p-hydroxyazobenzene | Reason: Diazonium ion couples at the para position of phenol, forming an orange-red azo dye.

How to Attempt MCQs in the CBSE Board Paper — Strategy Tips

In the 2025 CBSE Class 12 Chemistry board exam, Section A typically contains 16 MCQs of 1 mark each. Four of these are assertion-reason type. For Chapter 9 Amines, expect one MCQ on nomenclature or classification, one on preparation methods (Gabriel or reduction), one on tests (carbylamine or Hinsberg), and one on diazonium salts or their reactions. Read each question twice before marking. Eliminate obviously wrong options first. For assertion-reason MCQs, evaluate the assertion and reason independently, then check if the reason correctly explains the assertion. Watch for subtle differences: for example, Gabriel synthesis works only for aliphatic amines, not aromatic. Time management is crucial: allocate 20 minutes for all 16 MCQs, leaving ample time for the long-answer section. If unsure, flag the question and return after completing the rest. Never leave an MCQ unattempted — there is no negative marking in CBSE. Practice with previous years' question papers and NCERT exemplar problems. CBSETUTOR.ai offers a 24×7 AI tutor where you can upload photos of MCQ sets, get instant explanations, and clarify doubts on the spot at a flat ₹999/month for all classes 6-12, with a 3-day free trial. Regular practice sharpens pattern recognition and boosts speed, both essential for scoring full marks in Section A.
  • Read each MCQ stem carefully and underline keywords like 'only', 'not', 'most', 'least' to avoid traps.
  • For assertion-reason questions, mark A if both are true and reason explains assertion; B if both true but reason does not explain; C if assertion true, reason false; D if assertion false.
  • Use the process of elimination: cross out two obviously incorrect options to improve odds from 25% to 50%.
  • In nomenclature MCQs, quickly draw the structure if you are a visual learner — it prevents mixing up IUPAC and common names.
  • For reagent-based MCQs (like Sandmeyer or Hinsberg), recall the specific reagent name and condition; partial recall costs the mark.
  • Manage your time: aim for 1 minute per MCQ in the first pass, then revisit flagged questions in the final 5 minutes.

Frequently asked questions

How many MCQs from Chapter 9 Amines appear in the CBSE Class 12 Chemistry board exam?+
Typically one MCQ of 1 mark is set from Amines in Section A. Additionally, the chapter may feature in assertion-reason or case-based MCQs in Section B, bringing the total weightage to 3-5 marks across the paper.
What is the difference between Gabriel phthalimide synthesis and reduction of nitriles?+
Gabriel synthesis produces primary aliphatic amines only and avoids over-alkylation. Reduction of nitriles (R–CN) with LiAlH₄ also yields primary amines but extends the carbon chain by one carbon atom. Gabriel cannot prepare aromatic amines; nitrile reduction can if you start with an aromatic nitrile.
Why is the carbylamine test specific for primary amines?+
Primary amines have an –NH₂ group that reacts with chloroform and alcoholic KOH to form isocyanides (R–N≡C), which have a foul smell. Secondary and tertiary amines lack the necessary N–H bonds for this reaction, so they do not produce isocyanides.
How do I remember the reagents for Sandmeyer and Gattermann reactions?+
Use the mnemonic 'Sandmeyer uses Copper salts (CuCl, CuBr, CuCN), Gattermann uses Copper powder + HX'. For fluorine, remember Balz-Schiemann uses HBF₄. Write these out on a flashcard and revise daily until automatic recall is achieved.
What is the role of temperature in the preparation of diazonium salts?+
Diazonium salts are prepared at 0-5°C because they are highly unstable and decompose above this temperature, releasing nitrogen gas. Low temperature slows decomposition, allowing the salt to be isolated and used in subsequent coupling or substitution reactions.
Why is aniline less basic than ammonia?+
In aniline, the lone pair on nitrogen is delocalised by resonance into the benzene ring, reducing its availability to accept a proton. In ammonia, the lone pair is localised, making it more available for protonation and hence more basic.
Can diazonium salts be stored for long periods?+
No. Aromatic diazonium salts are stable only at low temperatures (below 5°C) and decompose on warming or storage. They are prepared in situ and used immediately in further reactions. Aliphatic diazonium salts are so unstable they cannot be isolated at all.
How does CBSETUOR.ai help with Chapter 9 Amines MCQ practice?+
CBSETUTOR.ai provides a 24×7 AI tutor where you can upload photos of MCQ sets, get instant step-by-step solutions, and clarify doubts on nomenclature, reaction mechanisms, and tests like Hinsberg or carbylamine. At ₹999/month for all classes 6-12 with a 3-day free trial, it is an affordable way to practise unlimited questions and track your progress before the board exam.
What is the major product when benzenediazonium chloride reacts with phenol in alkaline medium?+
The major product is p-hydroxyazobenzene (para isomer), an orange-red azo dye. The diazonium ion acts as an electrophile and couples at the para position of phenol, which is activated by the –OH group. This reaction is called azo coupling.
Are there any common mistakes students make in Amines MCQs?+
Yes. Students often confuse Gabriel synthesis applicability (only aliphatic 1° amines), misidentify the class of an amine from structure, forget that tertiary amines have no N–H bond (so no Hinsberg or carbylamine reaction), and mix up Sandmeyer reagents (CuCl vs. CuBr vs. CuCN). Careful reading and diagram-drawing prevent these errors.

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