12 min readPathway sheet1-mark MCQ or assertion-reason on a product or reagent; 2 or 3-mark Gabriel phthalimide synthesis or Hoffmann bromamide degradation with equations; also a NEET favourite

Preparation of amines

Amines are made by reducing nitro compounds, nitriles and amides, by ammonolysis, by the Gabriel synthesis, and by Hoffmann bromamide degradation (one carbon fewer).

Reducing nitrobenzeneC₆H₅NO₂ → (Sn + HCl, Fe + HCl, or H₂/Pd) C₆H₅NH₂

Gabriel synthesisPhthalimide → (ethanolic KOH) → (RX, heat) → (aqueous NaOH) → pure primary amine RNH₂

Hoffmann bromamide degradationRCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Carbon countRX → RCN → RCH₂NH₂ adds one carbon; Hoffmann degradation removes one

Reduction methods

1★
Nitro compounds: C₆H₅NO₂ + 3H₂ → (Pd, ethanol) C₆H₅NH₂ + 2H₂O
H₂ over finely divided Ni, Pd or Pt works, and so does a metal in acid: Sn + HCl, or Fe + HCl
2
C₆H₅NO₂ → (scrap iron + HCl) C₆H₅NH₂
Preferred: the FeCl₂ formed is hydrolysed and gives back HCl, so only a little HCl is needed to start the reaction
3★
Nitriles: R-C≡N → (H₂/Ni, or Na(Hg)/C₂H₅OH, or LiAlH₄) R-CH₂-NH₂
A primary amine with one carbon more than RX used to make the nitrile: an ascent of the amine series
4
Amides: R-CO-NH₂ → (i) LiAlH₄ (ii) H₂O → R-CH₂-NH₂
Same number of carbons as the amide

★ key step

Ammonolysis of alkyl halides

1
R-X + NH₃ → (ethanolic NH₃, sealed tube, 373 K) R-NH₃⁺X⁻
Nucleophilic substitution: the lone pair of NH₃ attacks the carbon of C-X. This cleavage of the C-X bond by ammonia is called ammonolysis
2
R-NH₃⁺X⁻ + NaOH → R-NH₂ + H₂O + NaX
A strong base frees the amine from its salt
3★
RNH₂ → (RX) R₂NH → (RX) R₃N → (RX) R₄N⁺X⁻
The amine formed is itself a nucleophile, so a mixture of 1°, 2° and 3° amines and the quaternary salt forms. A large excess of NH₃ gives mainly the primary amine. Reactivity: RI > RBr > RCl

★ key step

Gabriel phthalimide synthesis

1
C₆H₄(CO)₂NH + KOH (ethanolic) → C₆H₄(CO)₂N⁻K⁺ + H₂O
The N-H of phthalimide is acidic because two C=O groups pull electrons, so KOH removes the H⁺: potassium phthalimide
2★
C₆H₄(CO)₂N⁻K⁺ + R-X → (heat) C₆H₄(CO)₂N-R + KX
N-Alkylphthalimide
3
C₆H₄(CO)₂N-R + 2NaOH (aq) → (heat) R-NH₂ + C₆H₄(COONa)₂
Alkaline hydrolysis gives the pure primary amine and sodium phthalate
4
Limitation: aryl halides cannot be used
Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, so aniline and other aromatic primary amines cannot be made this way

★ key step

Hoffmann bromamide degradation

1★
R-CO-NH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Heat the amide with bromine in aqueous or ethanolic NaOH
2
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Ethanamide gives methanamine. Benzamide, C₆H₅CONH₂, gives aniline in the same way
3
Why one carbon is lost
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: a descent of the series

★ key step

Gabriel vs Hoffmann

Gabriel phthalimide synthesisStarts from an alkyl halide. The amine has the same number of carbons as RX. Gives pure primary aliphatic amines only; no aromatic amines.

Hoffmann bromamide degradationStarts from an amide. The amine has one carbon fewer. Gives primary aliphatic and aromatic amines (benzamide gives aniline).

Both give primary amines free of secondary and tertiary amines, which ammonolysis cannot do.

Which method for which job

You need	Use
Aniline from nitrobenzene	Sn + HCl, Fe + HCl, or H₂/Pd
A pure primary aliphatic amine from RX, same carbons	Gabriel phthalimide synthesis
An amine with one carbon more than RX	RX + KCN → RCN, then LiAlH₄ or H₂/Ni
An amine with one carbon fewer than the amide	Hoffmann bromamide degradation
An amine with the same carbons as the amide	LiAlH₄, then H₂O
Cheap amines where a mixture is acceptable	Ammonolysis of an alkyl halide

In real life

Aniline is made on a large scale by reducing nitrobenzene. It is the starting point for azo dyes used on cotton and silk, for sulpha drugs and for rubber chemicals.

Exam angle

3 marks: “Write the Gabriel phthalimide synthesis of ethanamine. Why can aniline not be prepared by this method?” Phthalimide + ethanolic KOH → potassium phthalimide; + C₂H₅Br, heat → N-ethylphthalimide; + NaOH(aq), heat → C₂H₅NH₂ + sodium phthalate. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

2 marks: “Write the Hoffmann bromamide degradation of benzamide. How does the number of carbon atoms change?” C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The amine has one carbon fewer: the carbonyl carbon leaves as carbonate.

2 marks: “Why is ammonolysis of alkyl halides not a good method for a pure primary amine? How can the yield of primary amine be raised?” The primary amine formed is a nucleophile and reacts with more RX, giving 2° and 3° amines and a quaternary salt. Using a large excess of ammonia favours the primary amine.

Remember it

Gabriel keeps the carbons, Hoffmann loses one, the nitrile route gains one. Ammonolysis gives a mixture.