Preparation of amines

How are amines prepared?

In plain words: there are three kinds of route. Reductions turn –NO₂, –C≡N or –CONH₂ into an amine and keep every carbon; substitutions put nitrogen where a halogen was; Hofmann degradation cuts out the carbonyl carbon. So the carbon count changes in only two cases: Hofmann degradation removes one carbon, and the nitrile route adds one when it starts from an alkyl halide, because cyanide brings that carbon in (NCERT Example 9.2, page 264).

Reduction of nitro compounds

• Nitro compounds are reduced to amines by hydrogen gas over finely divided nickel, palladium or platinum, or by metals in acid; nitroalkanes are reduced in the same way to alkanamines (NCERT page 262).
• Nitrobenzene gives aniline with H₂/Pd in ethanol, or with Sn + HCl or Fe + HCl (NCERT page 262); in strongly acidic solution the product is an anilinium salt, and base work-up releases free aniline.
• Iron scrap and hydrochloric acid is preferred, because the FeCl₂ formed is hydrolysed and releases hydrochloric acid during the reaction, so only a little acid is needed to start it (NCERT page 262).

Ammonolysis of alkyl halides

The heated sealed-tube description is a theoretical reaction condition. Practical work requires approved pressure-rated apparatus and supervision; an ordinary closed test tube is not suitable.

• An alkyl or benzyl halide heated with ethanolic ammonia under NCERT’s sealed-tube conditions at 373 K undergoes nucleophilic substitution: –NH₂ replaces the halogen. This cleavage of the C–X bond by ammonia is ammonolysis (NCERT page 262).
• The primary amine is itself a nucleophile, so it reacts further with the alkyl halide: RNH₂ → R₂NH → R₃N → R₄N⁺X⁻, the quaternary ammonium salt (NCERT pages 262–263).
• The amine is set free from its ammonium salt by a strong base: R–NH₃⁺X⁻ + NaOH → R–NH₂ + H₂O + NaX (NCERT page 263).
• The disadvantage is a mixture of primary, secondary and tertiary amines and the quaternary salt; a large excess of ammonia makes the primary amine the major product. The reactivity of halides with amines is RI > RBr > RCl (NCERT page 263).
• NCERT slip (page 263): in Example 9.1 (i), the last step adds C₂H₅Cl to N,N-diethylethanamine, (C₂H₅)₃N, yet the quaternary salt is printed as (C₂H₅)₃N⁺Cl⁻. The product carries four ethyl groups: (C₂H₅)₄N⁺Cl⁻, tetraethylammonium chloride (reasoned from NCERT’s own scheme, R₃N → R₄N⁺X⁻, page 263). In an exam, write (C₂H₅)₄N⁺Cl⁻.

Reduction of nitriles and amides

• Nitriles are reduced to primary amines by lithium aluminium hydride or by catalytic hydrogenation: R–C≡N → R–CH₂–NH₂ (H₂/Ni, or Na(Hg)/C₂H₅OH, in NCERT’s scheme) (NCERT page 263).
• This is used for the ascent of the amine series, making an amine with one carbon atom more than the alkyl halide used to make the nitrile (NCERT Example 9.2). The reduction itself keeps every carbon: the extra one is the nitrile carbon, brought in when cyanide replaces a halogen (NCERT Example 9.2, page 264), and it becomes the new CH₂ (reasoned).
• Primary amides RCONH_₂ are reduced by LiAlH₄, then water, to amines with the same number of carbons: R–CONH₂ → R–CH₂–NH₂ (NCERT page 263).

Gabriel phthalimide synthesis

• Phthalimide with ethanolic KOH gives the potassium salt of phthalimide; heating this with an alkyl halide gives an N-alkylphthalimide, whose alkaline hydrolysis (aqueous NaOH) gives the primary amine and sodium phthalate (NCERT page 264).
• Aromatic primary amines cannot be prepared by Gabriel synthesis, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion (NCERT page 264); so chlorobenzene does not give aniline this way. Strictly, this holds for ordinary, unactivated aryl halides (reasoned).

Hofmann bromamide degradation

• Treating a primary amide RCONH_₂ (R alkyl or aryl) with bromine in aqueous or ethanolic NaOH gives a primary amine with one carbon less: the alkyl or aryl group migrates from the carbonyl carbon to the nitrogen (NCERT page 264).

• NCERT’s Example 9.3 (page 265): butanamide, CH₃CH₂CH₂CONH₂, gives the three-carbon amine, and benzamide gives aniline.
• NCERT slip (pages 264–265), and a naming note: NCERT writes “Hoffmann” on pages 264–265 but “Hofmann’s bromamide reaction” in its exercises (page 279); beyond the NCERT text, the reaction is named after its discoverer, August Wilhelm von Hofmann, with one f (Wikipedia, “Hofmann rearrangement”). Expect either spelling. In Example 9.3, NCERT calls the product “propanamine”; its own rules give propan-1-amine, since propan-2-amine also exists (NCERT Table 9.1, page 261), so write the locant in an exam.

Solved examples

Common mistakes to avoid

Quick revision summary

• Nitro compounds: H₂ with Ni, Pd or Pt, or Sn/Fe and HCl (Fe and HCl preferred).
• Ammonolysis: RX + ethanolic NH₃, pressure-rated apparatus, 373 K (textbook conditions); gives a mixture up to R₄N⁺X⁻; excess NH₃ favours RNH₂; RI > RBr > RCl.
• Nitriles (LiAlH₄ or H₂/Ni): one carbon more than the alkyl halide that gave the nitrile; amides (LiAlH₄): same carbons.
• Gabriel (potassium phthalimide, RX, then NaOH): primary aliphatic amines only.
• Hofmann (primary amide, Br₂, NaOH): one carbon less; formamide is the special case giving ammonia.

Questions students ask

What is the mechanism of the Hofmann bromamide reaction, and why does it need a primary amide?

In the Hofmann bromamide reaction, beyond the NCERT text, bromine and NaOH form hypobromite, which turns the amide, RCONH₂, into an N-bromoamide; the base takes the second N–H proton, and R shifts from carbon to nitrogen as Br⁻ departs, forming an isocyanate, R–N=C=O (Wikipedia, “Hofmann rearrangement”). Water then converts the isocyanate into a carbamic acid, which falls apart into CO₂ (Na₂CO₃ in NCERT’s equation, page 264) and the amine with one carbon less (Wikipedia). The nitrogen loses both its hydrogens on the way to the isocyanate, so only a primary amide reacts; an N-substituted amide such as acetanilide cannot (reasoned).

Is the Gabriel phthalimide synthesis SN1 or SN2, and which alkyl halides does it suit?

The Gabriel phthalimide synthesis rests on an S_N2 step, a point beyond the NCERT text: the phthalimide anion, formed when ethanolic KOH takes the N–H proton of phthalimide (NCERT page 264), attacks the alkyl halide’s carbon and displaces the halide ion in one step (S_N2; Wikipedia, “Gabriel synthesis”). So it suits primary alkyl halides but seldom secondary ones, unless unusually reactive (Wikipedia), since the bulky anion reaches a crowded carbon only with difficulty (reasoned). The amine can also be freed with hydrazine instead of NCERT’s alkaline hydrolysis, the Ing–Manske procedure (Wikipedia).

What is the difference between the Hofmann bromamide reaction and the Hofmann elimination?

The Hofmann bromamide reaction turns a primary amide into a primary amine with one carbon less, using bromine and NaOH (NCERT page 264), while the Hofmann elimination, beyond the NCERT text, makes an alkene from an amine: the amine is turned into a quaternary ammonium hydroxide, which on heating gives the alkene, a tertiary amine and water (Wikipedia, “Hofmann elimination”; reasoned). The first keeps nitrogen in the product; the second strips it from the carbon chain, giving mainly the least substituted alkene, the Hofmann product behind Hofmann’s rule (Wikipedia). Both honour August Wilhelm von Hofmann (Wikipedia).

How is aniline made on an industrial scale?

Catalytic hydrogenation of nitrobenzene is an industrial route to aniline, applying the same reduction principle as NCERT’s first preparation method. Catalyst, temperature and pressure depend on the process; a single temperature range or annual production total should not be treated as a universal chemical fact.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why is reduction with iron scrap and hydrochloric acid preferred for nitro compounds?

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   The FeCl₂ formed is hydrolysed and releases hydrochloric acid during the reaction, so only a small amount of acid is needed to start it (NCERT page 262).

2. Q2. How can ammonolysis be made to give mainly the primary amine?

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   By using a large excess of ammonia; otherwise the primary amine reacts further with the alkyl halide and a mixture of amines and the quaternary salt results (NCERT page 263).