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Preparation of Amines

🎓 Class 12 Chemistry CBSE Theory Ch 9 – Amines ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Preparation of Amines

આ મૂલ્યાંકન આના પર આધારિત હશે: Preparation of Amines

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Preparation of Amines — Six Routes to the C–N Bond

Section 9.4 gives six standard preparations. They are not six things to memorise separately: three of them preserve the carbon count, one raises it by one carbon, one lowers it by one carbon, and one is chosen specifically because it gives a pure primary amine. Sort them that way and conversion questions become almost mechanical.

The organising idea. Every route below either reduces a nitrogen already bonded to carbon (–NO₂, –CN, –CONH₂) or attaches nitrogen to carbon by nucleophilic substitution (ammonolysis, Gabriel). Ask first: does my target have more, fewer or the same number of carbons as my starting material?

9.4 Preparation of Amines

1. Reduction of nitro compounds

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 can be similarly reduced to the corresponding alkanamines.

Ar–NO₂  +  3H₂  —[Ni or Pd or Pt]→  Ar–NH₂  +  2H₂O
C₆H₅–NO₂  —[Sn + HCl, or Fe + HCl]→  C₆H₅–NH₂  (aniline)
Why iron scrap and HCl is the industrial choice. Reduction with iron scrap and hydrochloric acid is preferred because the FeCl₂ formed gets hydrolysed to release hydrochloric acid during the reaction. Thus only a small amount of hydrochloric acid is required to initiate the reaction — the acid is effectively regenerated, which makes the process cheap on the tonne scale.

2. Ammonolysis of alkyl halides

The carbon–halogen bond in alkyl or benzyl halides is easily cleaved by a nucleophile. An alkyl or benzyl halide reacting with an ethanolic solution of ammonia therefore undergoes nucleophilic substitution in which the halogen atom is replaced by an amino group. This cleavage of the C–X bond by an ammonia molecule is called ammonolysis. The reaction is carried out in a sealed tube at 373 K.

R–X  +  NH₃  →  R–NH₂·HX  —[NaOH]→  R–NH₂
R–NH₂  +  R–X  →  R₂NH  →  R₃N  →  R₄N⁺X⁻

The primary amine thus obtained behaves as a nucleophile itself and can react further with alkyl halide to form secondary and tertiary amines, and finally a quaternary ammonium salt. The free amine is obtained from the ammonium salt by treatment with a strong base:

R–NH₃⁺X⁻  +  NaOH  →  R–NH₂  +  NaX  +  H₂O
The drawback. Ammonolysis yields a mixture of primary, secondary and tertiary amines together with a quaternary ammonium salt. However, 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.
NH₃ R–NH₂ R₂NH R₃N R₄N⁺X⁻ quaternary salt +RX +RX +RX +RX Each amine formed is itself a nucleophile — hence the cascade. Large excess of NH₃ traps the reaction at the primary amine.
The ammonolysis cascade. Selectivity for the primary amine is achieved by flooding the reaction with ammonia.
Worked Example 9.1 — Writing ammonolysis equations

Question. Write chemical equations for: (i) reaction of ethanolic NH₃ with C₂H₅Cl; (ii) ammonolysis of benzyl chloride and reaction of the amine so formed with two moles of CH₃Cl.

(i) C₂H₅Cl + NH₃ (ethanolic, 373 K, sealed tube) → C₂H₅NH₂·HCl → (NaOH) → C₂H₅NH₂ (ethanamine). With excess ethyl chloride the cascade continues to (C₂H₅)₂NH, (C₂H₅)₃N and finally (C₂H₅)₄N⁺Cl⁻.

(ii) C₆H₅CH₂Cl + NH₃ → C₆H₅CH₂NH₂ (phenylmethanamine). Then with two moles of methyl chloride: C₆H₅CH₂NH₂ + CH₃Cl → C₆H₅CH₂NHCH₃ (N-methylphenylmethanamine), and a second mole gives C₆H₅CH₂N(CH₃)₂ (N,N-dimethylphenylmethanamine).

3. Reduction of nitriles

Nitriles on reduction with lithium aluminium hydride (LiAlH₄) or by catalytic hydrogenation produce primary amines. This reaction is used for ascent of the amine series — that is, for the preparation of amines containing one carbon atom more than the starting material.

R–CN  —[LiAlH₄ or H₂/Ni]→  R–CH₂–NH₂   (one extra carbon)
The two-step ascent. R–X —[KCN or NaCN]→ R–CN —[LiAlH₄]→ R–CH₂NH₂. Combining cyanide substitution with nitrile reduction is the standard way to add exactly one carbon and end with a primary amine.

4. Reduction of amides

Amides on reduction with lithium aluminium hydride yield amines. Here the carbon count is preserved — contrast this carefully with the Hoffmann degradation of the same amide, which loses a carbon.

R–CO–NH₂  —[LiAlH₄]→  R–CH₂–NH₂   (same number of carbons)

5. Gabriel phthalimide synthesis

Gabriel synthesis is used for the preparation of primary amines only. 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.

Phthalimide  —[KOH/ethanol]→  potassium phthalimide
  —[R–X, heat]→  N-alkylphthalimide  —[NaOH(aq), hydrolysis]→  R–NH₂  +  phthalic acid salt
Two things to remember about Gabriel synthesis. First, it gives an exclusively primary amine — free of the 2°/3°/quaternary contamination that plagues ammonolysis — because the nitrogen is blocked by the two acyl groups of the ring until the final hydrolysis. Second, aromatic primary amines cannot be prepared by this method, because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

6. Hoffmann bromamide degradation reaction

Hoffmann developed a method for preparing primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from the carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.

R–CO–NH₂  +  Br₂  +  4NaOH  →  R–NH₂  +  Na₂CO₃  +  2NaBr  +  2H₂O
(descent of series — one carbon lost as carbonate)
Carbon bookkeeping — the fastest way to choose a route SAME carbons Nitro reduction Ammonolysis of R–X Amide + LiAlH₄ Gabriel synthesis n → n ONE MORE carbon Nitrile reduction R–CN + LiAlH₄ ASCENT of series n → n+1 ONE FEWER carbon Hoffmann bromamide amide + Br₂/NaOH DESCENT of series n → n−1
Fig. 9.2: Sorting the six preparations by carbon bookkeeping. Note that the same amide gives R–CH₂NH₂ with LiAlH₄ but R–NH₂ with Br₂/NaOH.
Worked Example 9.2 — Conversions requiring one extra carbon

Question. Write chemical equations for: (i) CH₃–CH₂–Cl into CH₃–CH₂–CH₂–NH₂; (ii) C₆H₅–CH₂–Cl into C₆H₅–CH₂–CH₂–NH₂.

Read the carbon count first. In both cases the product has one carbon more than the starting halide. That immediately points to the cyanide/nitrile route.

(i) CH₃CH₂Cl + KCN → CH₃CH₂CN (propanenitrile) + KCl; then CH₃CH₂CN —[LiAlH₄]→ CH₃CH₂CH₂NH₂ (propan-1-amine).

(ii) C₆H₅CH₂Cl + KCN → C₆H₅CH₂CN (phenylacetonitrile) + KCl; then C₆H₅CH₂CN —[LiAlH₄]→ C₆H₅CH₂CH₂NH₂ (2-phenylethanamine).

Worked Example 9.3 — Working backwards from a Hoffmann product

Question. Write structures and IUPAC names of (i) the amide which gives propanamine by the Hoffmann bromamide reaction; (ii) the amine produced by the Hoffmann degradation of benzamide.

(i) Propanamine contains three carbons. Since Hoffmann degradation loses one carbon, the amide molecule must contain four carbon atoms. The starting amide is CH₃CH₂CH₂CONH₂, butanamide.

(ii) Benzamide (C₆H₅CONH₂) is an aromatic amide containing seven carbon atoms. Hence the amine formed is an aromatic primary amine containing six carbon atoms — C₆H₅NH₂, aniline (benzenamine).

🧪 Activity 9.2 — Two reagents, one amide, two different aminesL4 Analyse

This activity isolates the single most examinable contrast in Section 9.4: what happens to the same amide under two different reagents.

Predict: Butanamide, CH₃CH₂CH₂CONH₂, is treated in two separate flasks — flask A with LiAlH₄, flask B with Br₂ and aqueous NaOH. Will the two products have the same number of carbon atoms? Write your prediction first.
  1. Count the carbons in butanamide: four (three in the propyl chain, one in the C=O).
  2. Flask A (LiAlH₄): the reagent reduces the C=O to CH₂ but does not break any C–C bond. Count the carbons in the product.
  3. Flask B (Br₂/NaOH): the alkyl group migrates from the carbonyl carbon to nitrogen, and the carbonyl carbon leaves as carbonate. Count again.
  4. Name both products by IUPAC rules.

No — the two products differ by one carbon atom.

Flask A: CH₃CH₂CH₂CONH₂ + LiAlH₄ → CH₃CH₂CH₂CH₂NH₂, butan-1-amine (four carbons, retained). The hydride simply converts C=O into CH₂.

Flask B: CH₃CH₂CH₂CONH₂ + Br₂ + 4NaOH → CH₃CH₂CH₂NH₂, propan-1-amine (three carbons) + Na₂CO₃ + 2NaBr + 2H₂O. The propyl group migrates onto nitrogen and the old carbonyl carbon departs as carbonate.

The takeaway for exams: when a question gives you an amide and asks for a conversion, the reagent tells you the carbon count. LiAlH₄ keeps it, Br₂/NaOH drops it by one. Reading the carbon count of the target before choosing a reagent will solve most conversion problems in this chapter in one line.

Intext question

Intext 9.3 — How will you convert?

(i) Benzene into aniline. Nitrate, then reduce. C₆H₆ —[conc. HNO₃ + conc. H₂SO₄, 323–333 K]→ C₆H₅NO₂ —[Fe/HCl (or Sn/HCl)]→ C₆H₅NH₂.

(ii) Benzene into N,N-dimethylaniline. First make aniline as above, then exhaustively methylate with two moles of methyl iodide: C₆H₅NH₂ + 2CH₃I → C₆H₅N(CH₃)₂ + 2HI (the HI is mopped up by a base such as Na₂CO₃).

(iii) Cl–(CH₂)₄–Cl into hexane-1,6-diamine. The target has six carbons and the halide has four, so two carbons must be added — one at each end. Use the cyanide route at both ends: Cl(CH₂)₄Cl + 2KCN → NC–(CH₂)₄–CN (hexanedinitrile), then reduce both nitrile groups: NC–(CH₂)₄–CN —[LiAlH₄ or H₂/Ni]→ H₂N–(CH₂)₆–NH₂, hexane-1,6-diamine. (This is industrially important — it is one of the two monomers of nylon-6,6.)

Competency-Based Questions

A process chemist must supply 50 kg of pure butan-1-amine for a pharmaceutical intermediate. The specification is strict: the product must contain less than 0.5% of any secondary or tertiary amine. Three feedstocks are available in the plant — 1-chlorobutane, butanenitrile (CH₃CH₂CH₂CN) and pentanamide (CH₃CH₂CH₂CH₂CONH₂).

1. Explain why simple ammonolysis of 1-chlorobutane would fail the purity specification. L2 Understand

The primary amine formed first is itself a nucleophile, so it competes with ammonia for the remaining 1-chlorobutane. The product is therefore a mixture of butan-1-amine, dibutylamine, tributylamine and tetrabutylammonium chloride. Even with a large excess of ammonia the primary amine is only the major product, not the exclusive one, so the <0.5% limit on 2°/3° amines would not be met.

2. Show, with equations, how butan-1-amine can be obtained from 1-chlorobutane in a way that guarantees an exclusively primary product. L3 Apply

Use Gabriel phthalimide synthesis. (a) Phthalimide + KOH(ethanolic) → potassium phthalimide. (b) Potassium phthalimide + CH₃CH₂CH₂CH₂Cl, heat → N-butylphthalimide + KCl. (c) N-Butylphthalimide + NaOH(aq), hydrolysis → CH₃CH₂CH₂CH₂NH₂ + sodium phthalate. Because the nitrogen is blocked by the two ring carbonyl groups until the final hydrolysis, it can be alkylated only once — so no secondary or tertiary amine can form.

3. Both butanenitrile and pentanamide can give butan-1-amine. State the reagent needed in each case and account for the carbon count. L4 Analyse

Butanenitrile (4 C) → the target butan-1-amine also has 4 C, so we need a route that adds one carbon to the chain counting the nitrile carbon itself: CH₃CH₂CH₂CN —[LiAlH₄]→ CH₃CH₂CH₂CH₂NH₂. The nitrile carbon becomes the CH₂ bearing the –NH₂. Pentanamide (5 C) → the target has one carbon fewer, so use Hoffmann bromamide degradation: CH₃CH₂CH₂CH₂CONH₂ + Br₂ + 4NaOH → CH₃CH₂CH₂CH₂NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.

4. Fill in the blank: In the industrial reduction of nitrobenzene, iron scrap with a catalytic quantity of HCl is preferred to tin and HCl because the ______ formed is hydrolysed to regenerate hydrochloric acid. L1 Remember

FeCl₂ (ferrous chloride). Its hydrolysis releases HCl back into the reaction mixture, so only a small initiating quantity of acid is required — a decisive cost advantage on the industrial scale.

5. A junior chemist proposes making aniline by Gabriel phthalimide synthesis from chlorobenzene. Evaluate this proposal and suggest a correct route. L5 Evaluate

The proposal cannot work. Gabriel synthesis depends on the phthalimide anion displacing halide by nucleophilic substitution, but aryl halides do not undergo nucleophilic substitution with this anion — the C–Cl bond of chlorobenzene has partial double-bond character from resonance with the ring, and the ring carbon is not open to backside attack. Hence aromatic primary amines can never be made by Gabriel synthesis. Correct route: nitrate benzene to nitrobenzene, then reduce with Fe/HCl (or Sn/HCl, or H₂ over Ni/Pd/Pt) to give aniline. Alternatively, benzamide could be subjected to Hoffmann bromamide degradation.

Assertion–Reason Questions

For each pair choose: (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.

Assertion (A): Gabriel phthalimide synthesis is preferred over ammonolysis for preparing a pure primary amine.

Reason (R): In Gabriel synthesis the nitrogen atom is held in the phthalimide ring and can be alkylated only once, so no secondary or tertiary amine is produced.

Answer: A. Both statements are true and the reason is precisely why Gabriel synthesis gives an exclusively primary amine, whereas ammonolysis gives a mixture.

Assertion (A): The Hoffmann bromamide degradation of an amide gives an amine with one carbon atom fewer than the amide.

Reason (R): An alkyl or aryl group migrates from the carbonyl carbon of the amide to the nitrogen atom.

Answer: A. The migration described in R moves the carbon skeleton onto nitrogen and leaves the original carbonyl carbon to depart as carbonate, which is exactly why one carbon is lost.

Assertion (A): Reduction of a nitrile with LiAlH₄ is called ascent of the amine series.

Reason (R): LiAlH₄ is a stronger reducing agent than hydrogen gas over nickel.

Answer: C. The assertion is true — the product amine has one carbon more than the alkyl group of the starting halide from which the nitrile was made, so the series ascends. The reason, while a defensible remark about reagent strength, is irrelevant: the ascent arises from the extra carbon supplied by the cyanide ion, not from the reducing power of LiAlH₄. Catalytic hydrogenation of the same nitrile ascends the series equally well.
Coming next. Part 3 turns to Sections 9.5 and 9.6(1) — physical properties, intermolecular hydrogen bonding and the boiling-point order, followed by the basic character of amines and the full structure–basicity relationship that explains why the aqueous order is not the order you would predict from inductive effects alone.

Frequently Asked Questions

What are the six methods of preparing amines in NCERT Class 12 Chemistry Chapter 9?
The six methods are: (1) reduction of nitro compounds using H₂ with Ni/Pd/Pt or a metal in acid; (2) ammonolysis of alkyl halides with ethanolic ammonia in a sealed tube at 373 K; (3) reduction of nitriles with LiAlH₄ or by catalytic hydrogenation; (4) reduction of amides with LiAlH₄; (5) Gabriel phthalimide synthesis; and (6) Hoffmann bromamide degradation.
Why does ammonolysis of alkyl halides give a mixture of amines?
The primary amine formed in the first substitution still has a lone pair on nitrogen, so it is itself a nucleophile and attacks more alkyl halide. This produces a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. Using a large excess of ammonia makes the primary amine the major product by ensuring ammonia rather than the product amine captures most of the alkyl halide.
What is the difference between ascent and descent of the amine series?
Ascent means the amine obtained has one carbon atom more than the starting material — achieved by converting an alkyl halide to a nitrile with KCN and then reducing the nitrile with LiAlH₄. Descent means the amine has one carbon fewer — achieved by the Hoffmann bromamide degradation, in which an amide treated with Br₂ and NaOH loses its carbonyl carbon as carbonate.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Gabriel synthesis requires the phthalimide anion to displace a halide ion by nucleophilic substitution. Aryl halides do not undergo nucleophilic substitution with this anion, because the C–X bond in an aryl halide has partial double-bond character due to resonance with the ring and the ring carbon is not open to backside attack. Aniline must therefore be made by reducing nitrobenzene instead.
Why is iron scrap with HCl preferred for the industrial reduction of nitrobenzene?
The ferrous chloride (FeCl₂) produced during the reduction is hydrolysed and releases hydrochloric acid back into the reaction mixture. The acid is therefore effectively regenerated, and only a small initiating quantity of hydrochloric acid is needed. This makes the process markedly cheaper than using tin and HCl on an industrial scale.
Why does the same amide give different amines with LiAlH₄ and with Br₂/NaOH?
LiAlH₄ simply reduces the carbonyl group, converting R–CO–NH₂ into R–CH₂–NH₂ with no carbon–carbon bond broken, so the carbon count is preserved. Br₂ with NaOH causes the alkyl or aryl group to migrate from the carbonyl carbon to nitrogen, and the old carbonyl carbon leaves as carbonate, so the amine formed has one carbon fewer. Butanamide therefore gives butan-1-amine with LiAlH₄ but propan-1-amine with Br₂/NaOH.
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Chemistry Class 12 Part II – NCERT (2025-26)
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