Aliphatic Amines General methods of preparation. Properties, reactions, and uses. Amines – Nomenclature: NEET MCQs with Answers

15 free questions on Aliphatic Amines General methods of preparation. Properties, reactions, and uses. Amines – Nomenclature for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Nomenclature, classification structure, basic character, and identification of primary, secondary, and tertiary amines and their basic character.

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Question 1

Previous-year question
The correct order of the basic strength of methyl-substituted amines in aqueous solution is [NEET 2019]
  1. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
  2. (CH₃)₃N > CH₃NH₂ > (CH₃)₂NH
  3. (CH₃)₃N > (CH₃)₂NH > CH₃NH₂
  4. CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
Show answer
Answer: (A) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
Explanation: In water the observed order results from a compromise between the inductive effect, which favours more methyl groups, and hydration of the ammonium ion together with steric hindrance, which oppose it. The secondary amine therefore comes out on top.

Question 2

Previous-year question
Which of the following reactions is appropriate for converting acetamide to methanamine? [NEET 2017]
  1. Hofmann bromamide reaction
  2. Stephen's reaction
  3. Gabriel phthalimide synthesis
  4. Carbylamine reaction
Show answer
Answer: (A) Hofmann bromamide reaction
Explanation: Methanamine has one carbon fewer than acetamide, which is exactly the carbon loss that accompanies the Hofmann bromamide degradation.

Question 3

Previous-year question
The number of structural isomers possible from the molecular formula C₃H₉N is [AIPMT 2015]
  1. 4
  2. 5
  3. 2
  4. 3
Show answer
Answer: (A) 4
Explanation: The isomers are propan-1-amine, propan-2-amine, N-methylethanamine and trimethylamine — two primary, one secondary and one tertiary amine.

Question 4

Which statement about methylamine is correct?
  1. It forms salts with sodium hydroxide.
  2. Its aqueous solution is acidic.
  3. It is a stronger base than ammonia.
  4. It is a weaker base than ammonia because the methyl group withdraws electrons.
Show answer
Answer: (C) It is a stronger base than ammonia.
Explanation: The methyl group releases electrons by its +I effect, so the nitrogen of methylamine has more electron density than that of ammonia and donates its lone pair to a proton more readily. The methylammonium ion is also stabilised by the same effect. Being a base, methylamine forms salts with acids, not with alkalis, and its aqueous solution is basic.

Question 5

Propanamide and propan-1-amine can be told apart by
  1. adding bromine water to each
  2. testing an aqueous solution of each with red litmus
  3. adding Tollens' reagent to each
  4. testing each with blue litmus
Show answer
Answer: (B) testing an aqueous solution of each with red litmus
Explanation: Propan-1-amine is a base, so its aqueous solution turns red litmus blue, while propanamide is practically neutral because the lone pair on its nitrogen is delocalised onto the carbonyl oxygen, so it leaves red litmus unchanged. Neither compound reduces Tollens' reagent or decolourises bromine water, and neither turns blue litmus red, so those tests do not separate them.

Question 6

A primary aliphatic amine, RNH₂, is treated with nitrous acid prepared from NaNO₂ and dilute HCl. The main product is
  1. a nitroalkane, RNO₂
  2. a secondary amine, R₂NH
  3. a stable alkyldiazonium chloride
  4. an alcohol, ROH, with evolution of nitrogen gas
Show answer
Answer: (D) an alcohol, ROH, with evolution of nitrogen gas
Explanation: Nitrous acid first converts the primary aliphatic amine into an alkyldiazonium salt. Unlike arenediazonium salts, this ion is not stabilised by resonance and breaks down at once, losing nitrogen gas. The fragment that remains reacts with water to give the alcohol, so the brisk evolution of N₂ is used to identify primary aliphatic amines.

Question 7

Compound P is reduced to Q. Q reacts with nitrous acid with evolution of nitrogen gas to give propan-1-ol. What is P?
  1. CH₃CH₂NC
  2. CH₃CH₂CH₂CN
  3. CH₃CH₂CN
  4. CH₃CH₂NO₂
Show answer
Answer: (C) CH₃CH₂CN
Explanation: A primary aliphatic amine reacts with nitrous acid to give the corresponding alcohol with evolution of N₂, so Q must be propan-1-amine, CH₃CH₂CH₂NH₂. Reduction of a nitrile adds the –CH₂NH₂ unit at the nitrile carbon, so propanenitrile gives propan-1-amine. Butanenitrile would give butan-1-amine, ethyl isocyanide would give a secondary amine, and nitroethane would give ethanamine.
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Question 8

Butanoic acid is treated with SOCl₂ to give X. X reacts with ammonia to give Y, and Y is warmed with Br₂ and aqueous NaOH to give Z. What is Z?
  1. CH₃CH₂CH₂NH₂
  2. CH₃CH₂CH₂CH₂NH₂
  3. CH₃CH₂NHCH₂CH₃
  4. CH₃CH₂CH₂CONH₂
Show answer
Answer: (A) CH₃CH₂CH₂NH₂
Explanation: Thionyl chloride converts the acid into butanoyl chloride (X), and ammonia converts this into butanamide (Y). Br₂ with aqueous NaOH brings about the Hofmann bromamide degradation, which removes the carbonyl carbon and gives a primary amine with one carbon fewer than the amide. Butanamide, with four carbons, therefore gives propan-1-amine.

Question 9

Which of the following compounds gives a secondary amine on reduction with LiAlH₄?
  1. Ethyl isocyanide, C₂H₅NC
  2. Propanenitrile, C₂H₅CN
  3. 1-Nitropropane, C₃H₇NO₂
  4. Propanamide, C₂H₅CONH₂
Show answer
Answer: (A) Ethyl isocyanide, C₂H₅NC
Explanation: In an isocyanide the nitrogen is already bonded to the alkyl group, and reduction turns the terminal carbon into a methyl group on that nitrogen. Ethyl isocyanide therefore gives C₂H₅NHCH₃, which is a secondary amine. A nitrile, an amide with an unsubstituted –NH₂ and a nitroalkane all have nitrogen at the end of the chain, so each gives propan-1-amine, a primary amine.

Question 10

Which amine reacts with benzenesulphonyl chloride to give a product that does not dissolve in aqueous KOH?
  1. Propan-2-amine
  2. Butan-1-amine
  3. N,N-Dimethylmethanamine
  4. N-Ethylethanamine
Show answer
Answer: (D) N-Ethylethanamine
Explanation: A secondary amine has only one hydrogen on nitrogen, and it is used up in forming the sulphonamide, so the product has no acidic N–H and does not dissolve in alkali. Primary amines such as propan-2-amine and butan-1-amine give sulphonamides that keep one N–H, which is acidic because of the strongly electron-withdrawing sulphonyl group, so these dissolve in KOH. A tertiary amine has no hydrogen on nitrogen and does not react with the reagent.

Question 11

Pentan-3-one is heated with methylamine, with loss of water, and the product is reduced with LiAlH₄ followed by aqueous work-up. The final organic product is
  1. Pentan-3-amine
  2. N,N-Dimethylpentan-3-amine
  3. N-Methylpentan-3-amine
  4. Pentan-3-ol
Show answer
Answer: (C) N-Methylpentan-3-amine
Explanation: The ketone condenses with methylamine, losing water, to give the imine (C₂H₅)₂C=N–CH₃. Lithium aluminium hydride reduces the carbon-nitrogen double bond, and work-up gives the secondary amine (C₂H₅)₂CH–NHCH₃. No alcohol forms because the carbonyl group has already been converted before the hydride is added.

Question 12

These days p-toluenesulphonyl chloride has replaced benzenesulphonyl chloride for distinguishing the classes of amines. Ethanamine (P), N-ethylethanamine (Q) and N,N-diethylethanamine (R) are treated separately with p-toluenesulphonyl chloride, and aqueous alkali is then added. Which observation is correct?
  1. P does not react, Q gives a sulphonamide soluble in alkali, and R gives one insoluble in alkali
  2. P and Q both give sulphonamides soluble in alkali, and R gives one insoluble in alkali
  3. P gives a sulphonamide insoluble in alkali, Q gives a sulphonamide soluble in alkali, and R does not react
  4. P gives a sulphonamide soluble in alkali, Q gives a sulphonamide insoluble in alkali, and R does not react
Show answer
Answer: (D) P gives a sulphonamide soluble in alkali, Q gives a sulphonamide insoluble in alkali, and R does not react
Explanation: The primary amine P gives a sulphonamide that still has a hydrogen on nitrogen. This hydrogen is strongly acidic because of the electron-withdrawing sulphonyl group, so the product dissolves in alkali. The secondary amine Q gives a sulphonamide with no hydrogen on nitrogen, so it is not acidic and stays insoluble in alkali. The tertiary amine R does not react.

Question 13

A 1.50 g sample of glycine, H₂N–CH₂–COOH (molar mass 75 g mol⁻¹), is treated with excess nitrous acid prepared in situ from NaNO₂ and a mineral acid, and its primary aliphatic amino group reacts completely. What gas is evolved, and what is its volume at 273 K and 1 atm? (Molar volume = 22.4 L mol⁻¹)
  1. 896 mL of N₂
  2. 448 mL of NO
  3. 448 mL of N₂
  4. 224 mL of N₂
Show answer
Answer: (C) 448 mL of N₂
Explanation: Moles of glycine = 1.50/75 = 0.020 mol. A primary aliphatic amino group forms an unstable aliphatic diazonium salt, which liberates nitrogen quantitatively, one mole of N₂ per mole of –NH₂. Volume of N₂ = 0.020 × 22.4 L = 0.448 L = 448 mL. This quantitative evolution of nitrogen is the basis of the estimation of amino acids and proteins.

Question 14

When aqueous methylamine is added to ferric chloride solution, a reddish-brown precipitate of hydrated ferric oxide separates. Which statement best explains this?
  1. Methylamine loses a proton to Fe³⁺, and the amide ion formed decomposes the salt
  2. Methylamine reduces Fe³⁺ to Fe²⁺, and the Fe²⁺ formed is oxidised by air directly to the oxide
  3. Methylamine accepts a proton from water and produces OH⁻ ions, which precipitate Fe³⁺ as the hydrated oxide
  4. Methylammonium chloride is insoluble, so removal of chloride leaves Fe³⁺ free to precipitate
Show answer
Answer: (C) Methylamine accepts a proton from water and produces OH⁻ ions, which precipitate Fe³⁺ as the hydrated oxide
Explanation: Methylamine is a stronger base than water, so CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻ gives a basic solution. The OH⁻ ions combine with Fe³⁺ to precipitate hydrated ferric oxide, Fe₂O₃·xH₂O. Methylammonium chloride itself stays in solution.

Question 15

Several drugs contain an amino group. Which one of the following descriptions is INCORRECT?
  1. Ephedrine contains a secondary amino group and is used to increase blood pressure
  2. Adrenaline contains a secondary amino group and is used to increase blood pressure
  3. Benadryl is an antihistaminic drug that contains a tertiary amino group
  4. Novocain is a quaternary ammonium salt that is used as a surfactant
Show answer
Answer: (D) Novocain is a quaternary ammonium salt that is used as a surfactant
Explanation: Novocain is a synthetic amino compound used as an anaesthetic in dentistry, so calling it a surfactant is wrong. Quaternary ammonium salts are the compounds used as surfactants. Adrenaline and ephedrine are biologically active secondary amines that raise blood pressure, and Benadryl is an antihistaminic with a tertiary amino group.

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