Comprehension:
Physical and Chemical properties of Amines
Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp3 hybridised. Each of the three sp3 hybridised orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines. The fourth orbital of nitrogen in all amines contains an unshared pair of electrons. Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base as well as act as a nucleophile. Lower aliphatic amines are soluble in water because they form hydrogen bond with water molecules and also form strong intermolecular hydrogen bonding. Solvation, +I and +R groups affect the electron density over nitrogen atom. Their pKb value lies in the range of 3 to 4.22. The most widely used method for preparing amines involves the reduction of nitro group, nitriles, amides, etc. using different reagents like H2/Pd, Sn+HCl, H2/Ni, LiAlH4+H2O, Br2+NaOH etc. Primary amines can be easily detected by the carbylamine test and Hinsberg's reagent. Diazonium salt can be synthesized from amines easily for synthesis of variety of new compounds.
Which of the following compounds on reduction with H2 + Ni / Na (Hg) / C2H5OH yield a primary amine?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Methyl cyanide
Primary amines can be made by reducing compounds in which nitrogen is joined to a carbon that becomes a –CH2–NH2 group. Each starting compound behaves differently:
Nitriles (R–C≡N): reduced by \( \ce{H2} \)/Ni or by sodium amalgam and ethanol (Mendius reduction) to primary amines with one extra carbon compared with the alkyl group R.
\( \displaystyle \ce{R-C#N} \xrightarrow{\ce{H2}/\ce{Ni}\text{ or }\ce{Na(Hg)}/\ce{C2H5OH}} \ce{R-CH2-NH2} \)
Isocyanides (R–N≡C): the carbon joined to nitrogen becomes a methyl group on nitrogen, so reduction gives a secondary amine.
\( \displaystyle \ce{R-NC + 4[H] -> R-NH-CH3} \)
Amides (R–CONH2): reduced to primary amines by the strong reducing agent LiAlH4, not by these milder reagents.
\( \displaystyle \ce{R-CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2-NH2} \)
Phthalimide: used in the Gabriel phthalimide synthesis: it is converted to its potassium salt with ethanolic KOH, heated with an alkyl halide, and then hydrolysed to give a primary amine. It is not reduced.
\( \displaystyle \text{phthalimide} \xrightarrow{\text{KOH (alc.)}} \text{K-phthalimide} \xrightarrow{\ce{R-X}} \text{N-alkylphthalimide} \xrightarrow{\ce{NaOH}\text{ (aq)}} \ce{R-NH2} \)
Methyl cyanide (ethanenitrile) is a nitrile, so its reduction gives the primary amine ethylamine:
\( \displaystyle \ce{CH3-C#N + 4[H] -> CH3-CH2-NH2} \)
Why the other options are incorrect
Option (2) Ethylisocyanide — Incorrect. \( \ce{C2H5NC} \) is reduced to N-methylethanamine, \( \ce{C2H5NHCH3} \), a secondary amine.
Option (3) Acetamide — Incorrect. Amides need LiAlH4; \( \ce{H2} \)/Ni and Na(Hg)/\( \ce{C2H5OH} \) do not reduce them readily.
Option (4) Phthalimide — Incorrect. It gives a primary amine through the Gabriel synthesis (alkylation and hydrolysis), not by reduction with these reagents.
NCERT (2026-27 edition):
Reduction of Nitriles
