ChemistryISC Class 12 · Organic Compounds Containing Nitrogen

Preparation of Amines: Methods and Conversions for ISC Class 12

Updated 15 October 2026 · Fact-checked

Amines are made by ammonolysis of alkyl halides (a mixture of amines), reduction of nitriles, amides or nitro compounds, Hofmann bromamide degradation of amides (one carbon fewer), and Gabriel phthalimide synthesis (pure primary amines). The choice of method controls the number of carbon atoms and the class of amine obtained.

Match the starting compound to the result: nitrile adds one carbon, Hofmann removes one, Gabriel gives pure primary amine, ammonolysis gives a mixture.

Illustration drawn for QuestionBuddy.

Understand Preparation of Amines

Amines are derivatives of ammonia in which alkyl or aryl groups replace hydrogen atoms: primary RNH₂, secondary R₂NH, tertiary R₃N. The CISCE syllabus lists six ways of making them.

1. From alkyl halides (ammonolysis). An alkyl halide heated with ethanolic ammonia in a sealed tube (about 373 K) undergoes nucleophilic substitution: $C H_{3} C H_{2} C H_{2} C H_{2} C l + N H_{3} \to C H_{3} C H_{2} C H_{2} C H_{2} N H_{2} + H C l$ (the HCl forms the ammonium salt; a strong base frees the amine). The primary amine formed is itself a nucleophile and attacks more alkyl halide, giving secondary and tertiary amines and finally a quaternary ammonium salt. Ammonolysis therefore gives a mixture; using a large excess of ammonia favours the primary amine. Aryl halides do not undergo ammonolysis under these conditions.

2. From alcohols. Alcohol vapour and ammonia passed over heated alumina also give a mixture of amines.

3. From cyanides (nitriles). Reduction with LiAlH₄, with H₂/Ni, or with sodium and ethanol (Mendius reduction) gives a primary amine with one more carbon than the alkyl halide used to make the nitrile: $C_{2} H_{5} C N + 4 [H] \to C H_{3} C H_{2} C H_{2} N H_{2}$ . This is the standard way to step up a carbon chain. Isocyanides (from alkyl halides with AgCN) are reduced to secondary amines: $C_{2} H_{5} N C + 4 [H] \to C_{2} H_{5} N H C H_{3}$ .

4. From amides. (a) Hofmann bromamide degradation: an amide warmed with bromine and aqueous or ethanolic NaOH gives a primary amine with one carbon fewer (the carbonyl carbon is lost as carbonate): $C H_{3} C H_{2} C O N H_{2} + B r_{2} + 4 N a O H \to C H_{3} C H_{2} N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$ . It steps a chain down. (b) Reduction of amides with LiAlH₄ gives an amine with the same number of carbons: $C_{2} H_{5} C O N H_{2} + 4 [H] \to C_{3} H_{7} N H_{2} + H_{2} O$ .

5. From nitro compounds. Reduction with tin or iron and HCl, or H₂ over Ni, Pd or Pt: $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ . This is the main route to aniline and other aromatic amines. Iron scrap and HCl is preferred industrially because the FeCl₂ formed is hydrolysed and releases HCl, so only a little acid is needed to start the reaction.

6. Gabriel phthalimide synthesis. Phthalimide is converted by ethanolic KOH into potassium phthalimide; this N-nucleophile reacts with a primary alkyl halide (SN2) to give an N-alkylphthalimide, which on hydrolysis with alkali gives a pure primary amine: $C_{6} H_{4} (C O)_{2} N K + C_{3} H_{7} B r \to C_{6} H_{4} (C O)_{2} N C_{3} H_{7} + K B r$ ; $C_{6} H_{4} (C O)_{2} N C_{3} H_{7} + 2 N a O H \to C_{6} H_{4} (C O O N a)_{2} + C_{3} H_{7} N H_{2}$ . Because the nitrogen carries only one alkyl group and has no further H to replace, no secondary or tertiary amine forms. Aromatic primary amines cannot be made this way, because aryl halides do not react with the phthalimide anion.

Choosing a method. Same number of carbons: reduce a nitro compound, an amide, or use Gabriel. One carbon more than RX: RX → RCN → reduce. One carbon fewer than the acid: amide → Hofmann. Pure primary amine: Gabriel (aliphatic) or reduction. Aniline: reduce nitrobenzene.

Preparation of Amines: formulas and key facts

Hofmann bromamide degradation: $R C O N H_{2} + B r_{2} + 4 N a O H \to R N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$; Product has one carbon fewer than the amide.
Reduction of nitriles: $R C N + 4 [H] L i A l H_{4} or Na/C_{2} H_{5} OH R C H_{2} N H_{2}$; One carbon more than RX used to make RCN.
Reduction of nitro compounds: $A r N O_{2} + 6 [H] S n or Fe, HCl A r N H_{2} + 2 H_{2} O$; Route to aniline.
Gabriel synthesis: $C_{6} H_{4} (C O)_{2} N K + R X \to C_{6} H_{4} (C O)_{2} N R N a O H R N H_{2}$; Pure primary aliphatic amines only.
Ammonolysis: $R X + N H_{3} \to R N H_{2} \to R_{2} N H \to R_{3} N \to R_{4} N^{+} X^{-}$; Mixture; excess NH₃ favours RNH₂.

How to solve Preparation of Amines questions

Preparation questions in ISC are conversions and A, B, C chains. Count carbons first; that tells you which method fits.

1Compare the carbon count of the starting material and the amine required.
2Same count: reduce a nitro compound, a nitrile made from the same chain, or an amide with LiAlH₄; or use Gabriel for a primary amine from RX.
3One more carbon: RX + KCN → RCN, then reduce (LiAlH₄ or Na/C₂H₅OH).
4One fewer carbon: make the amide (RCOOH → RCOCl or ammonium salt → RCONH₂) and do Hofmann degradation.
5If a pure primary amine is needed, avoid ammonolysis (mixture); use Gabriel or reduction.
6Write every step with reagents, by-products and balanced equations; use [H] for reducing agents.

Preparation of Amines in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Choosing a method that gives a pure primary amine

Suppose a pharmaceutical laboratory needs pure benzylamine, C₆H₅CH₂NH₂, as a starting material for a drug. A trainee proposes heating benzyl chloride with ethanolic ammonia; a senior chemist suggests Gabriel synthesis.

Explain why the trainee's method gives an impure product, write the first equation, and outline the Gabriel route.

Try it first, then show the solutionSolution

Ammonolysis: $C_{6} H_{5} C H_{2} C l + N H_{3} \to C_{6} H_{5} C H_{2} N H_{2} + H C l$ .
Benzylamine is itself a nucleophile and reacts with more benzyl chloride, giving dibenzylamine, tribenzylamine and the quaternary salt: a mixture.
Gabriel: phthalimide + ethanolic KOH → potassium phthalimide; with benzyl chloride → N-benzylphthalimide (SN2); alkaline hydrolysis → benzylamine + phthalate.
The nitrogen in N-benzylphthalimide has no H left to be replaced, so only the primary amine forms.

Answer: Ammonolysis gives a mixture of 1°, 2°, 3° amines and a quaternary salt; Gabriel synthesis gives pure benzylamine.

Memory hook: Gabriel guards the nitrogen: one alkyl group in, one primary amine out.

Tricks to remember Preparation of Amines

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hofmann Hacks one carbon off
Hofmann bromamide: amide minus the C=O carbon = amine with one carbon fewer.
Use it when: Descending a series.
Cyanide climbs
RX + CN⁻ then reduce: the chain climbs by one carbon.
Use it when: Ascending a series.
Gabriel is great but not for aryls
Gabriel gives pure primary aliphatic amines; aryl halides don't react, so no aniline by Gabriel.
Use it when: Reason questions.

Preparation of Amines: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Hofmann degradation of propanamide · 1 mark · MCQBoard level

Propanamide is warmed with bromine and aqueous NaOH. The product is:
(a) propan-1-amine
(b) ethanamine
(c) butan-1-amine
(d) N-methylethanamine

Try it first, then show the solutionSolution

Hofmann bromamide degradation removes the carbonyl carbon as carbonate.
CH₃CH₂CONH₂ (3 C) → CH₃CH₂NH₂ (2 C).

Answer: (b) ethanamine.

Example 2: Aniline from benzamide · 2 marks · Short answerBoard level

Write the balanced equation for the preparation of aniline from benzamide and name the reaction.

Try it first, then show the solutionSolution

$C_{6} H_{5} C O N H_{2} + B r_{2} + 4 N a O H \to C_{6} H_{5} N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$ (1½ marks).
Hofmann bromamide degradation (½ mark).

Answer: Hofmann bromamide degradation, as shown.

Example 3: Ammonolysis of 1-chlorobutane · 3 marks · Short answerHarder board level

1-Chlorobutane is heated with ethanolic ammonia in a sealed tube. (a) Write the equation for formation of the primary amine. (b) Explain why secondary and tertiary amines also form. (c) How can the yield of the primary amine be increased?

Try it first, then show the solutionSolution

(a) $C H_{3} C H_{2} C H_{2} C H_{2} C l + N H_{3} \to C H_{3} C H_{2} C H_{2} C H_{2} N H_{2} + H C l$ (1 mark).
(b) Butan-1-amine has a lone pair on N and is a better nucleophile than NH₃; it attacks more 1-chlorobutane to give dibutylamine, then tributylamine and the quaternary salt (1 mark).
(c) Use a large excess of ammonia, so that alkyl halide molecules are much more likely to meet NH₃ than the amine (1 mark).

Answer: (a) Butan-1-amine + HCl. (b) Amines are nucleophiles and alkylate further. (c) Excess NH₃.

Example 4: Three amines from one acid · 3 marks · Short answerHarder board level

Starting from propanoic acid, show how you would prepare (a) ethanamine, (b) propan-1-amine and (c) butan-1-amine. Give reagents for each step.

Try it first, then show the solutionSolution

(a) CH₃CH₂COOH → (NH₃, then heat) CH₃CH₂CONH₂ → (Br₂/NaOH, Hofmann) CH₃CH₂NH₂ (1 mark).
(b) CH₃CH₂CONH₂ → (LiAlH₄) CH₃CH₂CH₂NH₂ (amide reduction keeps all three carbons) (1 mark).
(c) CH₃CH₂COOH → (LiAlH₄) CH₃CH₂CH₂OH → (PBr₃) CH₃CH₂CH₂Br → (KCN) CH₃CH₂CH₂CN → (LiAlH₄) CH₃CH₂CH₂CH₂NH₂ (1 mark).

Answer: (a) Amide + Hofmann. (b) Amide + LiAlH₄. (c) Acid → alcohol → bromide → nitrile → reduce.

Example 5: Gabriel synthesis with a yield · 5 marks · Long answerHOTS / competency

Propan-1-amine is prepared by Gabriel synthesis from 7.35 g of phthalimide (M = 147) and excess 1-bromopropane.
(a) Write the three steps with equations. [3]
(b) If 2.36 g of propan-1-amine (M = 59) is obtained, find the percentage yield. [1]
(c) Why can't aniline be prepared by this method? [1]

Try it first, then show the solutionSolution

(a) $C_{6} H_{4} (C O)_{2} N H + K O H \to C_{6} H_{4} (C O)_{2} N K + H_{2} O$ (1 mark).
$C_{6} H_{4} (C O)_{2} N K + C_{3} H_{7} B r \to C_{6} H_{4} (C O)_{2} N C_{3} H_{7} + K B r$ (1 mark).
$C_{6} H_{4} (C O)_{2} N C_{3} H_{7} + 2 N a O H \to C_{6} H_{4} (C O O N a)_{2} + C_{3} H_{7} N H_{2}$ (1 mark).
(b) Phthalimide $= \frac{7.35}{147} = 0.050$ mol; theoretical amine $= 0.050 \times 59 = 2.95$ g; yield $= \frac{2.36}{2.95} \times 100 = 80%$ (1 mark).
(c) Aryl halides do not undergo nucleophilic substitution with the phthalimide anion (C–X bond strengthened by resonance), so the N-aryl phthalimide cannot form (1 mark).

Answer: (a) Three steps as shown. (b) 80%. (c) Aryl halides do not react with potassium phthalimide.

Where students lose marks in Preparation of Amines

The slips examiners see most often, and what to write instead.

Mistake: Forgetting that Hofmann degradation removes a carbon
ISC examiners reported 'ethylamine' chosen as the product from ethanamide (2024).
Do this: Count carbons: ethanamide (2 C) gives methanamine (1 C).
Usually costs: 1 mark.
Mistake: Writing molecular hydrogen for LiAlH₄ reductions
Examiners noted H₂ written instead of nascent hydrogen [H] for LiAlH₄ (2025).
Wrong working
$C_{2} H_{5} C N + 2 H_{2} \to C_{3} H_{7} N H_{2}$ (LiAlH₄)
Right working
$C_{2} H_{5} C N + 4 [H] L i A l H_{4} C H_{3} C H_{2} C H_{2} N H_{2}$
Do this: Show LiAlH₄ over the arrow and 4[H] in the equation.
Usually costs: ½ to 1 mark.
Mistake: Losing a carbon after a nitrile
Examiners reported C₂H₅NH₂ and C₂H₅OH written instead of propylamine and propan-1-ol after C₂H₅Br → C₂H₅CN → reduction (2025).
Do this: The CN carbon stays: C₂H₅CN reduces to CH₃CH₂CH₂NH₂ (three carbons).
Usually costs: 2 marks in a chain.
Mistake: Saying ammonolysis gives only secondary amines
In a 2024 assertion-reason item many candidates misjudged the statement about ammonolysis.
Do this: Ammonolysis uses alcoholic ammonia and gives a mixture of primary, secondary and tertiary amines and a quaternary salt.
Usually costs: 1 mark.

Exam tips

Write a carbon count next to each compound in a chain; it catches most mistakes.
Name the reaction where one exists: Hofmann bromamide degradation, Gabriel phthalimide synthesis, Mendius reduction.
Balance Hofmann equations with 4NaOH, Na₂CO₃ and 2NaBr.
Mention that aryl amines are made by reducing nitroarenes, not by Gabriel synthesis or ammonolysis.
Use IUPAC names (ethanamine, propan-1-amine, N-methylethanamine); examiners asked for structures and IUPAC names in 2025.

Practice questions on Preparation of Amines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which method gives a primary amine containing one carbon atom fewer than the starting compound?
(a) Gabriel synthesis
(b) reduction of a nitrile
(c) Hofmann bromamide degradation
(d) reduction of an amide with LiAlH₄
Hint
The carbonyl carbon leaves as carbonate.
Answer
(c).
Q2 · 2 marks · Short answerBoard level
Write the balanced equation for the reduction of ethanenitrile with LiAlH₄ and name the product.
Hint
Show the reducing agent as 4[H].
Answer
$C H_{3} C N + 4 [H] L i A l H_{4} C H_{3} C H_{2} N H_{2}$ ; ethanamine.
Q3 · 2 marks · Short answerHarder board level
Why does ammonolysis of an alkyl halide not give a pure primary amine? What conditions are used for the reaction?
Hint
The product is also a nucleophile.
Answer
The primary amine formed attacks more alkyl halide, so secondary and tertiary amines and a quaternary salt also form. The halide is heated with ethanolic ammonia in a sealed tube at about 373 K.
Q4 · 3 marks · Short answerHarder board level
How would you prepare 4-methylaniline from toluene? Give equations and conditions.
Hint
Nitrate first, then reduce.
Answer
Nitrate toluene with conc. HNO₃ and conc. H₂SO₄ (–CH₃ is o/p directing) and separate 4-nitrotoluene. Reduce it with Sn (or Fe) and HCl, then add alkali: $C H_{3} C_{6} H_{4} N O_{2} + 6 [H] \to C H_{3} C_{6} H_{4} N H_{2} + 2 H_{2} O$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Hofmann bromamide degradation of ethanamide gives methanamine.
Reason (R): In this reaction the carbonyl carbon of the amide is removed as carbonate.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Count carbons.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Convert ethanamine into propan-1-amine, giving reagents and equations for each step.
Hint
Remove N, add a carbon, put N back.
Answer
$C_{2} H_{5} N H_{2} + H N O_{2} \to C_{2} H_{5} O H + N_{2} + H_{2} O$ (NaNO₂/HCl, cold); C₂H₅OH → (PBr₃) C₂H₅Br; C₂H₅Br + KCN → C₂H₅CN + KBr; $C_{2} H_{5} C N + 4 [H] L i A l H_{4} C_{3} H_{7} N H_{2}$ .
Q7 · 3 marks · Short answerHOTS / competency
Ethyl cyanide and ethyl isocyanide are each reduced with LiAlH₄. Name the products, classify each amine, and suggest a test to tell them apart.
Hint
Which carbon is attached to nitrogen in each?
Answer
$C_{2} H_{5} C N + 4 [H] \to C H_{3} C H_{2} C H_{2} N H_{2}$ : propan-1-amine (primary). $C_{2} H_{5} N C + 4 [H] \to C_{2} H_{5} N H C H_{3}$ : N-methylethanamine (secondary). Carbylamine test: warm with CHCl₃ and alcoholic KOH; only the primary amine gives the offensive smell of an isocyanide.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a dye works makes aniline by reducing nitrobenzene with iron scrap and a small amount of hydrochloric acid, and then liberates the free amine with alkali.
(i) Write the overall equation for the reduction, using [H]. [1]
(ii) Why is iron and HCl preferred to tin and HCl industrially, and why is only a little acid needed? [1]
(iii) What mass of aniline (M = 93) can be obtained from 12.3 g of nitrobenzene (M = 123) if conversion is complete? [1]
Hint
1 mol nitrobenzene gives 1 mol aniline.
Answer
(i) $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ . (ii) Iron scrap is cheap, and the FeCl₂ formed is hydrolysed, releasing HCl during the reaction, so only a little acid is needed to start it. (iii) $\frac{12.3}{123} = 0.10$ mol; $0.10 \times 93 = 9.3$ g.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which method gives pure primary amines?Show answer
Gabriel phthalimide synthesis (aliphatic only).
2.Reagents for Hofmann bromamide degradation?Show answer
Bromine and NaOH (aqueous or ethanolic).
3.Product of reducing ethyl isocyanide?Show answer
N-Methylethanamine (a secondary amine).
4.Industrial reducing agent for nitrobenzene?Show answer
Iron scrap and hydrochloric acid.
5.Why can't aniline be made by Gabriel synthesis?Show answer
Aryl halides do not react with potassium phthalimide.
6.How many carbons does the amine from Hofmann degradation of butanamide have?Show answer
Three (propan-1-amine).

Preparation of Amines: frequently asked questions

What is Hofmann bromamide degradation?

An amide warmed with bromine and NaOH gives a primary amine with one carbon fewer, the carbonyl carbon leaving as sodium carbonate.

What is Gabriel phthalimide synthesis?

Potassium phthalimide reacts with an alkyl halide to give an N-alkylphthalimide, which is hydrolysed to give a pure primary amine.

Why does ammonolysis give a mixture of amines?

Each amine formed is a nucleophile and reacts further with the alkyl halide, giving more highly substituted amines.

How do you increase the carbon chain by one while making an amine?

Convert the alkyl halide to a nitrile with KCN and reduce the nitrile with LiAlH₄ or Na/ethanol.

How is aniline prepared?

By reducing nitrobenzene with Sn or Fe and HCl (or catalytic hydrogenation), then liberating the amine with alkali.