Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum, and also by reduction with metals in acidic medium; nitroalkanes are similarly reduced to alkanamines. Reduction with iron scrap and hydrochloric acid is preferred because the FeCl2 formed gets hydrolysed to release hydrochloric acid, so only a small amount of acid is needed to initiate the reaction.
-- NCERT Class 12 Chemistry, Ch. 9, p. 262Amine Preparation
8 MCQs9-step worked example
Source: NCERT AminesPYQ coverage: NEET 2023, 2024, 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Amine Preparation, explained for NEET
Amines are prepared by reducing nitrogen-containing compounds (nitro compounds, nitriles, amides), by replacing the halogen of an alkyl halide with nitrogen (ammonolysis, Gabriel synthesis), or by Hoffmann bromamide degradation of an amide. NCERT Class 12 Chemistry, Chapter 9, page 262 opens section 9.4 with the words "Amines are prepared by the following methods" and lists six.
Roadmap: nitro reduction → ammonolysis of halides → nitrile reduction → amide reduction → Gabriel phthalimide → Hoffmann bromamide. For each method, fix three things: the starting material, the reagent, and what happens to the carbon count.
1. Reduction of nitro compounds. Hydrogen gas over finely divided nickel, palladium or platinum, or a metal in acidic medium, reduces a nitro compound to the amine; nitroalkanes are reduced the same way to alkanamines (NCERT Class 12 Chemistry, Chapter 9, page 262). Iron scrap with hydrochloric acid is preferred because the FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction, so only a small amount of acid is required to initiate it (NCERT Class 12 Chemistry, Chapter 9, page 262). This is the standard route to aniline from nitrobenzene. Bridge: nitro reduction needs a nitro compound; the next route starts from an alkyl halide instead.
2. Ammonolysis of alkyl/benzyl halides. An alkyl or benzyl halide heated with an ethanolic solution of ammonia in a sealed tube at 373 K undergoes nucleophilic substitution: the halogen is replaced by -NH2 (NCERT Class 12 Chemistry, Chapter 9, page 262). The primary amine formed also behaves as a nucleophile, so it reacts with more halide to give secondary and tertiary amines and finally a quaternary ammonium salt (NCERT Class 12 Chemistry, Chapter 9, page 262). The result is a mixture; a large excess of ammonia makes the primary amine the major product, and the free amine is released from its ammonium salt by a strong base (NCERT Class 12 Chemistry, Chapter 9, page 263). Reactivity of the halides: RI > RBr > RCl (NCERT Class 12 Chemistry, Chapter 9, page 263). Bridge: ammonolysis keeps the carbon count of the halide; the next route adds a carbon by going through a nitrile.
3. Reduction of nitriles. Nitriles are reduced to primary amines by lithium aluminium hydride (LiAlH4) or by catalytic hydrogenation, R-C≡N → R-CH2-NH2 (NCERT Class 12 Chemistry, Chapter 9, page 263). The nitrile carbon stays in the chain as the CH2 beside nitrogen, which is why NCERT calls this the ascent of the amine series: the amine has one carbon more than the starting amine (NCERT Class 12 Chemistry, Chapter 9, page 263). So benzonitrile, C6H5CN, gives benzylamine C6H5CH2NH2, never aniline. Bridge: the same reagent, LiAlH4, also reduces amides.
4. Reduction of amides. Amides on reduction with lithium aluminium hydride yield amines (NCERT Class 12 Chemistry, Chapter 9, page 263): R-CONH2 → R-CH2-NH2, same carbon count as the amide. Bridge: LiAlH4 keeps the carbonyl carbon; route 6 treats the same amide differently and removes it.
5. Gabriel phthalimide synthesis. Phthalimide with ethanolic KOH gives potassium phthalimide, which is heated with an alkyl halide and then hydrolysed with alkali to give the primary amine (NCERT Class 12 Chemistry, Chapter 9, page 264). It prepares primary amines only. It fails for aromatic primary amines because aryl halides do not undergo nucleophilic substitution with the phthalimide anion (NCERT Class 12 Chemistry, Chapter 9, page 264), so aniline cannot be made this way. Bridge: to reach aniline from an amide instead, use route 6.
6. Hoffmann bromamide degradation. An amide treated with bromine in an aqueous or ethanolic solution of sodium hydroxide gives a primary amine; an alkyl or aryl group migrates from the carbonyl carbon to the nitrogen, and the amine contains one carbon less than the amide (NCERT Class 12 Chemistry, Chapter 9, page 264). NCERT's Example 9.3 works both directions: propanamine (3 C) needs butanamide (4 C), and benzamide (7 C) gives aniline (6 C) (NCERT Class 12 Chemistry, Chapter 9, page 265). Watch-out: "Hoffmann bromamide degradation" is not "Hofmann elimination", which the sibling nomenclature lesson covers. A second watch-out: hydrolysing acetanilide back to aniline (NCERT Class 12 Chemistry, Chapter 9, page 272) only removes a protecting acetyl group that was put on an existing amine; section 9.4 does not list it as a preparation method.
| Method | Starting material | Reagent (NCERT) | Amine class | Carbon change | Gives an aryl amine (ArNH2)? |
|---|---|---|---|---|---|
| Nitro reduction | R-NO2 / Ar-NO2 | H2 + Ni/Pd/Pt; metal + acid (Fe + HCl preferred) | 1° | same | Yes (nitrobenzene → aniline) |
| Ammonolysis | R-X (alkyl or benzyl) | ethanolic NH3, sealed tube, 373 K | mixture 1°/2°/3° + quaternary salt | same | No |
| Nitrile reduction | R-CN | LiAlH4 or H2/catalyst | 1° | keeps the nitrile carbon (ascent) | No (ArCN → ArCH2NH2) |
| Amide reduction | R-CONH2 | LiAlH4 | amine, same skeleton | same | No (ArCONH2 → ArCH2NH2) |
| Gabriel phthalimide | R-X (not Ar-X) | potassium phthalimide, then alkaline hydrolysis | 1° only | same | No |
| Hoffmann bromamide | R-CONH2 / Ar-CONH2 | Br2 + NaOH (aq or ethanolic) | 1° | one carbon LESS | Yes (benzamide → aniline) |
Can you answer these Amine Preparation MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Nitrobenzene can be reduced to aniline by a metal in acidic medium. NCERT says reduction with iron scrap and hydrochloric acid is preferred. What reason does it give?
Show answer and why every option is right or wrong
Answer: D. D is correct: the FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction, so only a small amount of HCl is required to initiate the reaction (NCERT Class 12 Chemistry, Chapter 9, page 262).
Why A is wrong: A is wrong because NCERT does not rank Fe as inherently the strongest reducing metal here; the stated reason is acid economy through hydrolysis, not raw reducing power.
Why B is wrong: B is wrong because HCl does not reduce a nitro group by itself; the metal supplies the electrons for the reduction.
Why C is wrong: C is wrong because reducing a nitro compound gives the primary amine (aniline from nitrobenzene), not a secondary amine.
MCQ 2Easy RecallPractice
Which of the following, using only the reagents named, CANNOT prepare aniline (C6H5NH2)?
Show answer and why every option is right or wrong
Answer: A. A is correct: Gabriel synthesis needs an SN2 attack of the phthalimide anion on the halide, and aryl halides such as chlorobenzene do not undergo that substitution, so aromatic primary amines cannot be made this way (NCERT Class 12 Chemistry, Chapter 9, page 264).
Why B is wrong: B is wrong to pick because Fe scrap + HCl does reduce nitrobenzene to aniline; it is a valid route (page 262).
Why C is wrong: C is wrong to pick because Hoffmann bromamide degradation of benzamide does give aniline, since one carbon (the carbonyl carbon) is removed from the amide.
Why D is wrong: D is wrong to pick because catalytic hydrogenation of nitrobenzene over Ni is a standard route to aniline (page 262).
MCQ 3Easy RecallPractice
Which reagent converts propanenitrile (CH3CH2CN) directly into propan-1-amine (CH3CH2CH2NH2)?
Show answer and why every option is right or wrong
Answer: B. B is correct: nitriles are reduced to primary amines by LiAlH4 or by catalytic hydrogenation, keeping the nitrile carbon in the chain (NCERT Class 12 Chemistry, Chapter 9, page 263).
Why A is wrong: A is wrong because Br2/NaOH degrades an amide (Hoffmann bromamide); there is no amide in this stem, only a nitrile.
Why C is wrong: C is wrong because ethanolic NH3 in a sealed tube is the ammonolysis reagent for an alkyl halide, not for a nitrile.
Why D is wrong: D is wrong because aqueous NaOH alone does not reduce a nitrile to an amine.
MCQ 4Direct ApplicationPractice
1-Bromopropane is heated with ethanolic ammonia in a sealed tube at 373 K. Which statement correctly describes the outcome?
Show answer and why every option is right or wrong
Answer: C. C is correct: the primary amine formed is itself a nucleophile and keeps reacting with more halide, giving a mixture up to the quaternary salt; a large excess of NH3 favours the 1° amine (NCERT Class 12 Chemistry, Chapter 9, page 263).
Why A is wrong: A is wrong because the 1° amine formed is itself a nucleophile and reacts with more halide, so the product is a mixture, not a single amine.
Why B is wrong: B is wrong because propan-1-amine IS nucleophilic — that is exactly why the reaction does not stop at one stage.
Why D is wrong: D is wrong because ammonolysis is a direct nucleophilic substitution on the halide; no alkene intermediate is involved.
MCQ 5Direct ApplicationPractice
For ammonolysis of methyl halides with ethanolic NH3 at a given temperature, which reactivity order does NCERT give?
Show answer and why every option is right or wrong
Answer: A. A is correct: NCERT gives the order of reactivity of halides with amines as RI > RBr > RCl (NCERT Class 12 Chemistry, Chapter 9, page 263); the C-I bond is the easiest of the three to break.
Why B is wrong: B is wrong because it reverses the true order — a weaker C-X bond, as in C-I, reacts faster, not slower.
Why C is wrong: C is wrong because it places bromide ahead of iodide; iodide is the best leaving group of the three.
Why D is wrong: D is wrong because leaving-group ability differs sharply between Cl, Br and I even though the substitution mechanism is the same, so the rates are not equal.
MCQ 6Direct ApplicationPractice
Butanamide (CH3CH2CH2CONH2) is treated with Br2 and aqueous NaOH (Hoffmann bromamide degradation). How does the carbon count of the amine product compare with the starting amide?
Show answer and why every option is right or wrong
Answer: B. B is correct: in Hoffmann bromamide degradation, the group on the carbonyl carbon migrates to nitrogen and the carbonyl carbon is lost, so butanamide (4 carbons) gives propan-1-amine (3 carbons) (NCERT Class 12 Chemistry, Chapter 9, page 264; Example 9.3 on page 265).
Why A is wrong: A is wrong because this is the exact trap the topic flags: the carbon count is NOT preserved across Hoffmann degradation, unlike amide reduction with LiAlH4.
Why C is wrong: C is wrong because Br2 only brominates nitrogen during the mechanism; it does not add a carbon to the final product.
Why D is wrong: D is wrong because only the single carbonyl carbon is lost, not an additional adjacent carbon.
MCQ 7CalculationPractice
A student proposes four routes to propan-1-amine, each a single NCERT preparation method (CH3CH2CH2NH2): (i) propanenitrile + LiAlH4; (ii) butanamide + Br2/NaOH; (iii) 1-bromopropane + a large excess of ethanolic NH3, sealed tube, 373 K; (iv) propanamide + LiAlH4. Which of these give propan-1-amine as the major or sole organic product?
Show answer and why every option is right or wrong
Answer: C. C is correct: propanenitrile (3 carbons) reduces straight to propan-1-amine; butanamide (4 carbons) loses one carbon via Hoffmann degradation to give propan-1-amine; a large excess of NH3 makes propan-1-amine the major ammonolysis product; and LiAlH4 reduces propanamide to propan-1-amine with the same carbon count (NCERT Class 12 Chemistry, Chapter 9, page 263 for routes (i), (iii) and (iv); page 264 for (ii)).
Why A is wrong: A is wrong because it drops (iii) and (iv), both of which also give propan-1-amine as the major product under the stated conditions.
Why B is wrong: B is wrong because it drops (i) and (iii); nitrile reduction and excess-ammonia ammonolysis both reach propan-1-amine too.
Why D is wrong: D is wrong because it drops (ii) and (iv); Hoffmann degradation of butanamide and LiAlH4 reduction of propanamide both reach propan-1-amine.
MCQ 8Concept TrapPractice
Consider two statements. Statement I: Aniline can be prepared by Gabriel phthalimide synthesis starting from chlorobenzene. Statement II: When an alkyl halide is heated with a large excess of ethanolic ammonia, the primary amine is the major product. Which option correctly evaluates them?
Show answer and why every option is right or wrong
Answer: B. B is correct: Statement I is false because aromatic primary amines cannot be prepared by Gabriel synthesis, since aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide (NCERT Class 12 Chemistry, Chapter 9, page 264). Statement II is true: ammonolysis gives a mixture, but taking a large excess of ammonia makes the primary amine the major product (NCERT Class 12 Chemistry, Chapter 9, page 263).
Why A is wrong: A is wrong on both counts: it accepts the Gabriel route to aniline, which fails because chlorobenzene does not undergo the substitution, and it rejects the excess-ammonia statement that NCERT states.
Why C is wrong: C is wrong because Statement I is false: Gabriel synthesis cannot use an aryl halide such as chlorobenzene.
Why D is wrong: D is wrong because Statement II is true: excess ammonia makes the primary amine the major ammonolysis product, even though a mixture forms.
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How do you solve a Amine Preparation question? A worked example
- 1
Given
Benzamide, C6H5CONH2 (6 ring carbons + 1 carbonyl carbon = 7 carbons), is treated with Br2 and aqueous NaOH. Separately, benzonitrile, C6H5CN (6 ring carbons + 1 nitrile carbon = 7 carbons), is treated with LiAlH4.
- 2
Required
Identify which of the two product amines is aniline, C6H5NH2 (6 carbons, ring only, no exocyclic carbon).
- 3
Concept
Hoffmann bromamide degradation moves the group attached to the carbonyl carbon onto nitrogen and expels that carbonyl carbon; the amine ends up with one carbon fewer than the amide. Nitrile reduction adds hydrogen across the C≡N triple bond without removing any carbon; the nitrile carbon survives as the CH2 next to NH2.
- 4
Formula
Amide (n carbons) --Hoffmann--> amine (n - 1 carbons). Nitrile (n carbons) --LiAlH4 or H2/catalyst--> amine (n carbons, as R-CH2-NH2).
- 5
Substitution
Benzamide: n = 7 (6 ring + 1 carbonyl). Hoffmann degradation removes the carbonyl carbon: 7 - 1 = 6 carbons, matching aniline's 6 ring carbons. Benzonitrile: n = 7 (6 ring + 1 nitrile carbon). Reduction keeps all 7 carbons, giving C6H5CH2NH2 (benzylamine), not aniline.
- 6
Calculation
Benzamide route: 7 - 1 = 6 carbons = aniline. Benzonitrile route: 7 + 0 = 7 carbons = benzylamine. The "1" carbon removed in Hoffmann degradation is a counting integer from the reaction mechanism, not a measured quantity, so it carries no sig-fig or measurement uncertainty of its own.
- 7
Final answer
Option B: benzamide + Br2/NaOH gives aniline (C6H5NH2); benzonitrile + LiAlH4 gives benzylamine (C6H5CH2NH2), not aniline.
- 8
Common trap
Assuming an amide degradation and a nitrile reduction land on the same amine because both routes formally involve one nitrogen-bearing carbon. They move the carbon count in opposite directions: Hoffmann degradation loses a carbon, nitrile reduction keeps it — so a nitrile with the same number of carbons as an arylamine's parent ring will always overshoot that arylamine by one carbon.
- 9
Similar NEET-style question
Acetamide (CH3CONH2, 2 carbons) is degraded with Br2 and NaOH. What amine forms, and how many carbons does it have? (Answer: Hoffmann degradation removes the carbonyl carbon, leaving 2 - 1 = 1 carbon, so the product is methanamine, CH3NH2.)
What to remember before solving Amine Preparation questions
An alkyl or benzyl halide reacts with an ethanolic solution of ammonia by nucleophilic substitution, the halogen being replaced by an amino (-NH2) group; this cleavage of the C-X bond by ammonia is ammonolysis, carried out in a sealed tube at 373 K. The primary amine formed is itself a nucleophile and can react further with the alkyl halide to give secondary and tertiary amines and finally a quaternary ammonium salt.
-- NCERT Class 12 Chemistry, Ch. 9, p. 262The free amine is obtained from the ammonium salt by treatment with a strong base. Ammonolysis has the disadvantage of yielding a mixture of primary, secondary and tertiary amines and a quaternary ammonium salt; the primary amine is obtained as the major product by taking a large excess of ammonia. The order of reactivity of halides with amines is RI > RBr > RCl.
-- NCERT Class 12 Chemistry, Ch. 9, p. 263Nitriles on reduction with LiAlH4 or catalytic hydrogenation produce primary amines; this is used for ascent of the amine series, giving amines containing one carbon atom more than the starting amine.
-- NCERT Class 12 Chemistry, Ch. 9, p. 263Amides on reduction with lithium aluminium hydride (LiAlH4) yield amines.
-- NCERT Class 12 Chemistry, Ch. 9, p. 263Gabriel synthesis is used for the preparation of primary amines: phthalimide on treatment with ethanolic potassium hydroxide forms the potassium salt of phthalimide, which on heating with an alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.
-- NCERT Class 12 Chemistry, Ch. 9, p. 264Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
-- NCERT Class 12 Chemistry, Ch. 9, p. 264Hoffmann's bromamide degradation converts an amide (treated with bromine in aqueous/ethanolic NaOH) into a primary amine that contains one carbon atom less than the starting amide.
-- NCERT Class 12 Chemistry, Ch. 9, p. 264In Hoffmann bromamide degradation (amide + bromine in aqueous or ethanolic sodium hydroxide) an alkyl or aryl group migrates from the carbonyl carbon of the amide to the nitrogen atom, which is why the primary amine formed contains one carbon less than the amide.
-- NCERT Class 12 Chemistry, Ch. 9, p. 264Example 9.3: (i) Propanamine contains three carbons, so the amide that gives it by Hoffmann bromamide reaction must contain four carbon atoms: butanamide. (ii) Benzamide is an aromatic amide containing seven carbon atoms, so the amine formed from it is an aromatic primary amine containing six carbon atoms: aniline (benzenamine).
-- NCERT Class 12 Chemistry, Ch. 9, p. 265The -NH2 group can be protected by acetylation with acetic anhydride before a substitution reaction, after which hydrolysis of the substituted amide gives back the substituted amine.
-- NCERT Class 12 Chemistry, Ch. 9, p. 272Where do students lose marks on Amine Preparation?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Organic Reaction Conditions
Student keeps the carbon count unchanged across Hoffmann bromamide degradation (or across nitrile reduction), so picks the amide-length amine, or the aromatic amine from a nitrile.
When it triggers
An amide with Br2/NaOH (or KOH), or a nitrile with LiAlH4, and options that differ by one carbon.
How to avoid
Hoffmann bromamide: the amine has one carbon LESS than the amide (the group migrates from the carbonyl carbon to N). Nitrile reduction: the amine keeps the nitrile carbon, one carbon MORE than the amine series it ascends from, so C6H5CN gives C6H5CH2NH2, not aniline. NCERT Class 12 Chemistry Ch 9, PDF pp. 5-7.
Root cause: concept gap
Correction
Gabriel synthesis needs an SN2 attack of the phthalimide anion on the halide; aryl halides do not undergo that substitution, so aromatic primary amines cannot be made this way (NCERT Class 12 Chemistry Ch 9, PDF p. 6). Make aniline by reducing nitrobenzene (Fe scrap + HCl) or by Hoffmann degradation of benzamide.
Root cause: concept gap
Correction
The primary amine formed is itself a nucleophile, so ammonolysis gives a mixture of 1°, 2° and 3° amines and the quaternary ammonium salt; a large excess of ammonia makes the primary amine the major product (NCERT Class 12 Chemistry Ch 9, PDF pp. 4-5).
Mistake
Keeps the carbon count unchanged through Hoffmann bromamide degradation or nitrile reduction.
Root cause: concept gap
Correction
Hoffmann bromamide: amine = amide minus one carbon (butanamide gives propanamine; benzamide gives aniline). Nitrile reduction keeps every carbon, so C6H5CN gives benzylamine C6H5CH2NH2, not aniline (NCERT Class 12 Chemistry Ch 9, PDF pp. 5-7).
More in Amines: 6 exam traps and mistakes · 1 formula · 2 question patterns from its other lessons.
Amine Preparation questions from past NEET papers
4 questions from NEET 2023, 2024, 2026. Answers verified against NTA official keys.
NEET 2026
1A and B only
2B and D only
3A and C only
4C and D only
NTA Answer: Option 2(final)
NEET 2024Revised key
1Both statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is correct but Statement II is false
4Statement I is incorrect but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2024Revised key
1propylamine
2butylamine
3butanamide
4α–bromobutanoic acid
NTA Answer: Option 1(revised_final)
NEET 2023
Which of the following reactions will NOT give primary amine as the product?
1CH₃CN —[(i) LiAlH₄ (ii) H₃O⊕]→ Product
2CH₃NC —[(i) LiAlH₄ (ii) H₃O⊕]→ Product
3CH₃CONH₂ —[(i) LiAlH₄ (ii) H₃O⊕]→ Product
4CH₃CONH₂ —[Br₂/KOH]→ Product
NTA Answer: Option 2(final)
How does NEET ask about Amine Preparation?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Given reagents or a short sequence (nitrile, amide, nitro compound, halide), decide which routes give a primary (or aromatic) amine and track the carbon count: Hoffmann bromamide removes one carbon, nitrile reduction keeps it, Gabriel fails for aryl halides.
InterpretationMedium
Common distractors
keeps carbon count through hoffmann
Amide and amine look like the same chain
gabriel for aniline
Gabriel is 'the' primary-amine synthesis
Sources
NCERT refs: Class 12 Chemistry Chapter 9, p.262 | Class 12 Chemistry Chapter 9, p.263 | Class 12 Chemistry Chapter 9, p.264 | Class 12 Chemistry Chapter 9, p.265 | Class 12 Chemistry Chapter 9, p.272
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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