ChemistryCBSE Class 12 · Amines

Preparation of Amines: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Amines are prepared by reducing nitro compounds (Fe/HCl or $H_2$/Pd), nitriles and amides ($LiAlH_4$); by ammonolysis of alkyl halides (which gives a mixture); by Gabriel phthalimide synthesis (pure primary aliphatic amines); and by Hofmann bromamide degradation, which gives a primary amine with one carbon fewer than the amide.

Understand Preparation of Amines

NCERT lists six methods. For each, learn the reagent, what kind of amine forms, and how the carbon count changes.

1. Reduction of nitro compounds. Nitroarenes are reduced to aromatic amines by hydrogen with a finely divided catalyst (Ni, Pt or Pd), or by a metal with acid (Sn/HCl or Fe/HCl): $C_{6} H_{5} N O_{2} + 3 H_{2} P d C_{6} H_{5} N H_{2} + 2 H_{2} O$ . Iron scrap with hydrochloric acid is preferred industrially, because the $F e C l_{2}$ formed is hydrolysed and releases HCl again, so only a small amount of acid is needed to start the reaction. Nitroalkanes are reduced to alkanamines the same way.

2. Ammonolysis of alkyl halides. Ammonia, a nucleophile, displaces the halogen: $R - X + N H_{3} \to R - N H_{3}^{+} X^{-}$ . Treating the ammonium salt with a strong base liberates the free amine: $R N H_{3}^{+} X^{-} + N a O H \to R N H_{2} + N a X + H_{2} O$ . The drawback is that the primary amine formed is itself a nucleophile and reacts further, giving secondary and tertiary amines and finally a quaternary ammonium salt. A large excess of ammonia favours the primary amine. The order of reactivity of halides is $R I > R B r > R C l$ .

3. Reduction of nitriles. $R - C \equiv N + 2 H_{2} N i or L i A l H_{4} R - C H_{2} N H_{2}$ . The amine has one carbon more than the alkyl halide used to make the nitrile, so this sequence (R–X → R–CN → R–CH₂NH₂) is used to ascend the series.

4. Reduction of amides. $L i A l H_{4}$ followed by water reduces an amide to an amine with the same number of carbons: $R C O N H_{2} \to R C H_{2} N H_{2}$ .

5. Gabriel phthalimide synthesis. Phthalimide is treated with ethanolic KOH to form potassium phthalimide. Its nitrogen anion is a good nucleophile and reacts with an alkyl halide to give an N-alkylphthalimide, which is hydrolysed by aqueous NaOH to the primary amine and the sodium salt of phthalic acid. Because nitrogen can be alkylated only once, the method gives pure primary amines with no secondary or tertiary contamination. It cannot make aromatic primary amines, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

6. Hofmann bromamide degradation. An amide heated with bromine in aqueous or ethanolic NaOH gives a primary amine: $R C O N H_{2} + B r_{2} + 4 N a O H \to R N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$ . The alkyl or aryl group moves from the carbonyl carbon to the nitrogen, and the carbonyl carbon leaves as carbonate, so the amine has one carbon fewer than the amide (descent of the series). Benzamide gives aniline; propanamide gives ethanamine.

Summary of carbon count: nitrile reduction +1 (from R–X); amide reduction 0; Hofmann −1.

Preparation of Amines: formulas and key facts

Reduction of nitrobenzene: $C_{6} H_{5} N O_{2} F e / H C l or H_{2} / P d C_{6} H_{5} N H_{2}$; Fe scrap + HCl preferred: $F e C l_{2}$ hydrolysis regenerates HCl.
Ammonolysis: $R X + N H_{3} \to R N H_{3}^{+} X^{-} N a O H R N H_{2}; R N H_{2} \to R_{2} N H \to R_{3} N \to R_{4} N^{+} X^{-}$; Mixture; excess $N H_{3}$ favours 1°. Reactivity $R I > R B r > R C l$ .
Nitrile reduction (ascent): $R C \equiv N H_{2} / N i or L i A l H_{4} R C H_{2} N H_{2}$; One carbon more than R–X.
Amide reduction: $R C O N H_{2} L i A l H_{4}, H_{2} O R C H_{2} N H_{2}$; Same number of carbons.
Gabriel phthalimide synthesis: $phthalimide K O H (a l c) K-phthalimide R X N-alkylphthalimide N a O H (a q) R N H_{2}$; Pure 1° aliphatic amines; not for aryl amines.
Hofmann bromamide degradation: $R C O N H_{2} + B r_{2} + 4 N a O H \to R N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$; Amine has one carbon fewer (descent).

How to solve Preparation of Amines questions

Questions ask for products, reagents for a conversion, reasons (why Gabriel fails for aniline, why ammonolysis gives a mixture), or step-up/step-down of carbon chains.

1Compare carbon counts of starting material and target amine: same, one more, or one less.
2One more: alkyl halide → nitrile (KCN) → reduce ( $L i A l H_{4}$ or $H_{2}$ /Ni).
3One less: make the amide and use Hofmann bromamide ( $B r_{2}$ /NaOH).
4Same: reduce the amide ( $L i A l H_{4}$ ), reduce a nitro compound, or use Gabriel synthesis (from an alkyl halide) for a pure primary amine.
5For aromatic amines, start from the nitro compound (Fe/HCl) or from a benzamide (Hofmann); never Gabriel.
6Write each step with its reagent and condition; name intermediates.
7For mass questions, use mole ratios from the balanced equation (1 amide → 1 amine in Hofmann).

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.

Reduction of nitrobenzene to aniline

Suppose a small dye-making unit reduces 24.6 g of nitrobenzene to aniline using iron scrap and a little hydrochloric acid. The chemist records an 80% yield.

Write the overall change, explain why only a little HCl is needed, and calculate the mass of aniline obtained. (Nitrobenzene 123 g mol⁻¹, aniline 93 g mol⁻¹)

Try it first, then show the solutionSolution

$C_{6} H_{5} N O_{2} F e / H C l C_{6} H_{5} N H_{2}$ (overall, $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ ).
The $F e C l_{2}$ formed is hydrolysed, releasing HCl, so the acid is regenerated during the reaction.
Moles of nitrobenzene $= 24.6/123 = 0.20$ mol; theoretical aniline $= 0.20 \times 93 = 18.6$ g.
Actual yield $= 0.80 \times 18.6 = 14.9$ g.

Answer: About 14.9 g of aniline; HCl is regenerated by hydrolysis of $F e C l_{2}$ .

Memory hook: Iron and a pinch of acid turn nitrobenzene into aniline because the acid keeps coming back.

Tricks to remember Preparation of Amines

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

Nitrile adds, Hofmann subtracts
R–X → R–CN → R–CH₂NH₂: plus one carbon. RCONH₂ → RNH₂ (Hofmann): minus one carbon.
Use it when: Ascent and descent questions.
Gabriel gives one, and only aliphatic
Phthalimide N can take only one alkyl group, so only 1° amines; aryl halides refuse to join.
Use it when: Gabriel reasoning questions.
Hofmann's 1-1-4 → 1-1-2-2
1 amide + 1 $B r_{2}$ + 4 NaOH → 1 amine + 1 $N a_{2} C O_{3}$ + 2 NaBr + 2 $H_{2} O$ .
Use it when: Writing the balanced Hofmann equation.

Preparation of Amines: solved examples

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

Example 1: Hofmann product · 1 mark · MCQBoard level

Propanamide on heating with $B r_{2}$ and aqueous NaOH gives:
(a) propan-1-amine (b) ethanamine (c) butan-1-amine (d) propanenitrile

Try it first, then show the solutionSolution

Hofmann bromamide degradation removes the carbonyl carbon: $C_{2} H_{5} C O N H_{2} \to C_{2} H_{5} N H_{2}$ .

Answer: (b) Ethanamine.

Example 2: Gabriel synthesis · 2 marks · Very short answerBoard level

Write the steps of the Gabriel phthalimide synthesis of ethanamine. Why can aniline not be prepared by this method?

Try it first, then show the solutionSolution

Phthalimide + ethanolic KOH → potassium phthalimide; + $C_{2} H_{5} C l$ → N-ethylphthalimide; hydrolysis with aqueous NaOH → $C_{2} H_{5} N H_{2}$ + sodium phthalate (1 mark).
Aniline would need chlorobenzene as the halide; aryl halides do not undergo nucleophilic substitution with the phthalimide anion (1 mark).

Answer: Phthalimide → K-phthalimide → N-ethylphthalimide → ethanamine; aryl halides do not react.

Example 3: Three conversions · 3 marks · Short answerHarder board level

How will you convert: (a) chloroethane into propan-1-amine (b) benzamide into aniline (c) nitrobenzene into aniline? Give reagents and equations.

Try it first, then show the solutionSolution

(a) $C_{2} H_{5} C l K C N C_{2} H_{5} C N L i A l H_{4} or H_{2} / N i C H_{3} C H_{2} C H_{2} N H_{2}$ (one carbon added) (1 mark).
(b) Hofmann: $C_{6} H_{5} C O N H_{2} + B r_{2} + 4 N a O H \to C_{6} H_{5} N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$ (1 mark).
(c) $C_{6} H_{5} N O_{2} F e / H C l (or Sn/HCl, or H_{2} / P d) C_{6} H_{5} N H_{2}$ (1 mark).

Answer: (a) KCN then $L i A l H_{4}$ . (b) $B r_{2}$ /NaOH. (c) Fe/HCl.

Example 4: Problems with ammonolysis · 3 marks · Short answerHarder board level

(a) Write the reaction of bromoethane with ammonia and explain why a mixture of amines is obtained. (b) How can the formation of the primary amine be favoured? (c) Arrange RCl, RBr and RI in order of reactivity in this reaction.

Try it first, then show the solutionSolution

(a) $C_{2} H_{5} B r + N H_{3} \to C_{2} H_{5} N H_{3}^{+} B r^{-}$ ; with a base, $C_{2} H_{5} N H_{2}$ is freed. The primary amine is also a nucleophile and attacks more bromoethane, giving $(C_{2} H_{5})_{2} N H$ , $(C_{2} H_{5})_{3} N$ and $(C_{2} H_{5})_{4} N^{+} B r^{-}$ (1½ marks).
(b) Use a large excess of ammonia, so the alkyl halide more often meets $N H_{3}$ than the amine (½ mark).
(c) $R I > R B r > R C l$ (weaker C–X bond reacts faster) (1 mark).

Answer: (a) Further alkylation gives 1°, 2°, 3° amines and a quaternary salt. (b) Excess $N H_{3}$ . (c) $R I > R B r > R C l$ .

Example 5: One amide, two amines · 5 marks · Long answerHOTS / competency

An amide A ( $C_{3} H_{7} N O$ ) gives amine B ( $C_{2} H_{7} N$ ) when heated with $B r_{2}$ and NaOH, and amine C ( $C_{3} H_{9} N$ ) when reduced with $L i A l H_{4}$ . (a) Identify A, B and C. (b) Write both equations. (c) Calculate the mass of B formed from 3.65 g of A, assuming complete reaction. (A = 73, B = 45 g mol⁻¹) (d) Suggest a way to prepare C from bromoethane.

Try it first, then show the solutionSolution

(a) A: propanamide, $C H_{3} C H_{2} C O N H_{2}$ . B: ethanamine, $C H_{3} C H_{2} N H_{2}$ . C: propan-1-amine, $C H_{3} C H_{2} C H_{2} N H_{2}$ (1½ marks).
(b) $C H_{3} C H_{2} C O N H_{2} + B r_{2} + 4 N a O H \to C H_{3} C H_{2} N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$ ; $C H_{3} C H_{2} C O N H_{2} L i A l H_{4}, H_{2} O C H_{3} C H_{2} C H_{2} N H_{2}$ (1½ marks).
(c) Moles of A $= 3.65/73 = 0.050$ ; B formed $= 0.050$ mol; mass $= 0.050 \times 45 = 2.25$ g (1 mark).
(d) $C H_{3} C H_{2} B r K C N C H_{3} C H_{2} C N L i A l H_{4} C H_{3} C H_{2} C H_{2} N H_{2}$ (1 mark).

Answer: (a) A propanamide, B ethanamine, C propan-1-amine. (c) 2.25 g. (d) KCN, then $L i A l H_{4}$ .

Where students lose marks in Preparation of Amines

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

Mistake: Keeping the carbon count the same in Hofmann degradation
Wrong working
$C H_{3} C H_{2} C O N H_{2} \to C H_{3} C H_{2} C H_{2} N H_{2}$ with $B r_{2}$ /NaOH
Right working
$C H_{3} C H_{2} C O N H_{2} \to C H_{3} C H_{2} N H_{2}$
Do this: Hofmann removes the C=O carbon as carbonate.
Usually costs: 1 mark.
Mistake: Using Gabriel synthesis for aromatic amines
Do this: Use reduction of nitroarenes (or Hofmann from benzamide) for aniline.
Usually costs: 1 mark.
Mistake: Unbalanced Hofmann equation
Wrong working
$R C O N H_{2} + B r_{2} + N a O H \to R N H_{2} + N a B r$
Right working
$R C O N H_{2} + B r_{2} + 4 N a O H \to R N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$
Do this: Learn the full stoichiometry.
Usually costs: ½ to 1 mark.
Mistake: Expecting a pure primary amine from ammonolysis
Do this: Say that a mixture forms and that excess ammonia favours the primary amine.
Usually costs: 1 mark in reasoning questions.

Exam tips

In conversion questions, write the carbon count of start and finish before choosing reagents.
Name each named reaction (Gabriel, Hofmann bromamide) explicitly; it carries credit.
Mention 'ethanolic KOH' for phthalimide and 'aqueous NaOH' for the final hydrolysis.
For nitro reduction, quote both catalytic hydrogenation and metal/acid methods.
Write $L i A l H_{4}$ followed by $H_{2} O$ for amide and nitrile reductions.

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
Iron scrap and hydrochloric acid are preferred for reducing nitrobenzene because:
(a) iron is a catalyst
(b) $F e C l_{2}$ formed is hydrolysed, releasing HCl, so little acid is needed
(c) it gives secondary amines
(d) HCl is not needed at all
Hint
Think about the acid being regenerated.
Answer
(b).
Q2 · 2 marks · Very short answerBoard level
Write the general equation for the Hofmann bromamide degradation and state how the number of carbon atoms in the product compares with the amide.
Hint
Carbonate removes one carbon.
Answer
$R C O N H_{2} + B r_{2} + 4 N a O H \to R N H_{2} + N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O$ . The amine has one carbon atom fewer than the amide.
Q3 · 2 marks · Very short answerHarder board level
Write the products: (a) $C H_{3} C H_{2} C N H_{2} / N i$ (b) $C_{6} H_{5} C O N H_{2} L i A l H_{4}, H_{2} O$
Hint
Nitriles and amides both give RCH₂NH₂.
Answer
(a) $C H_{3} C H_{2} C H_{2} N H_{2}$ (propan-1-amine). (b) $C_{6} H_{5} C H_{2} N H_{2}$ (benzylamine, phenylmethanamine).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Hofmann bromamide degradation of benzamide gives aniline.
Reason (R): In this reaction the aryl group migrates from the carbonyl carbon to the nitrogen atom.
(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
Which carbon leaves as carbonate?
Answer
(a). The phenyl group moves to N and the carbonyl carbon leaves as carbonate, so $C_{6} H_{5} C O N H_{2}$ gives $C_{6} H_{5} N H_{2}$ .
Q5 · 2 marks · Very short answerHarder board level
Calculate the mass of methanamine obtained from 2.95 g of ethanamide by Hofmann bromamide degradation, assuming complete conversion. (Ethanamide 59 g mol⁻¹, methanamine 31 g mol⁻¹)
Hint
1 : 1 mole ratio.
Answer
Moles $= 2.95/59 = 0.050$ ; mass of $C H_{3} N H_{2} = 0.050 \times 31 = 1.55$ g.
Q6 · 3 marks · Short answerHarder board level
Starting from ethanenitrile, ethanamide and propanamide respectively, which method gives ethanamine in each case? Name the reagent and say whether the carbon chain grows, stays or shrinks.
Hint
Count carbons.
Answer
Ethanenitrile ( $C H_{3} C N$ ) + $L i A l H_{4}$ or $H_{2}$ /Ni → ethanamine (chain grows by one relative to $C H_{3}$ –X; same carbon count as the nitrile). Ethanamide + $L i A l H_{4}$ → ethanamine (same). Propanamide + $B r_{2}$ /NaOH (Hofmann) → ethanamine (shrinks by one).
Q7 · 3 marks · Short answerHOTS / competency
Suggest three different methods to prepare benzylamine ( $C_{6} H_{5} C H_{2} N H_{2}$ ), naming a starting material and reagent for each. Why does Hofmann degradation of benzamide not give benzylamine?
Hint
Benzyl chloride, benzonitrile, benzamide.
Answer
(1) Benzyl chloride by Gabriel synthesis (K-phthalimide, then NaOH(aq)). (2) Benzonitrile ( $C_{6} H_{5} C N$ ) reduced with $L i A l H_{4}$ or $H_{2}$ /Ni. (3) Benzamide ( $C_{6} H_{5} C O N H_{2}$ ) reduced with $L i A l H_{4}$ . Hofmann degradation removes the carbonyl carbon, so benzamide gives aniline ( $C_{6} H_{5} N H_{2}$ ), not benzylamine.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student wants pure butan-1-amine. She considers (P) heating 1-bromobutane with alcoholic ammonia, and (Q) the Gabriel phthalimide synthesis starting from 1-bromobutane.
(a) Which route gives a mixture, and what does the mixture contain? (1 mark)
(b) Write the steps of route Q. (2 marks)
(c) If she needed pentan-1-amine from 1-bromobutane instead, what two-step route would add the extra carbon? (1 mark)
Hint
Ammonolysis over-alkylates; nitriles add a carbon.
Answer
(a) Route P: butan-1-amine, dibutylamine, tributylamine and tetrabutylammonium bromide. (b) Phthalimide + ethanolic KOH → potassium phthalimide; + 1-bromobutane → N-butylphthalimide; + aqueous NaOH (hydrolysis) → butan-1-amine + sodium phthalate. (c) 1-Bromobutane + KCN → pentanenitrile; reduce with $L i A l H_{4}$ (or $H_{2}$ /Ni) → pentan-1-amine.

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.Reagent for Hofmann bromamide degradation?Show answer
$B r_{2}$ with aqueous or ethanolic NaOH.
2.Which method gives only primary aliphatic amines?Show answer
Gabriel phthalimide synthesis.
3.Reduction of nitriles changes the carbon count how?Show answer
It keeps the nitrile's count, which is one more than the alkyl halide used.
4.Reagent to reduce nitrobenzene industrially?Show answer
Iron scrap and HCl.
5.Order of halide reactivity in ammonolysis?Show answer
$R I > R B r > R C l$ .
6.Product of Hofmann reaction on benzamide?Show answer
Aniline.

Preparation of Amines: frequently asked questions

What are the methods of preparation of amines in Class 12?

Reduction of nitro compounds, ammonolysis of alkyl halides, reduction of nitriles, reduction of amides, Gabriel phthalimide synthesis, and Hofmann bromamide degradation.

Why does Gabriel synthesis give only primary amines?

The phthalimide nitrogen carries only one replaceable hydrogen, so it can be alkylated only once; hydrolysis then releases a single primary amine.

What is the Hofmann bromamide degradation?

Heating an amide with bromine and sodium hydroxide gives a primary amine with one carbon atom fewer than the amide.

Why does ammonolysis give a mixture of amines?

Each amine formed is also a nucleophile and can react with more alkyl halide, giving secondary and tertiary amines and a quaternary ammonium salt.

How can I increase the carbon chain by one while making an amine?

Convert the alkyl halide to a nitrile with KCN, then reduce the nitrile with LiAlH_4 or H_2/Ni.