Passage
Preparation & Chemical reactions of amines
Preparation:
Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium, or platinum as catalyst. Nitriles on reduction with lithium aluminum hydride or catalytic hydrogenation produce primary amines. The Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide.
Chemical reactions:
Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides, and esters by nucleophilic substitution reaction. This reaction is known as acylation. Both aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide give corresponding isocyanide. Benzenesulphonyl chloride also known as Hinsberg’s reagent reacts with primary and secondary amines to form sulphonamide. Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. Aniline reacts with HNO3 in the presence of H2SO4 to give nitro derivatives. Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid or sulphanilic acid.
Which of the following reactions will not give primary amine?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → reduction of CH3CH2NC with LiAlH4 (product S)
Primary amines (\( \ce{R-NH2} \)) can be made by reducing nitriles or amides, or by the Hofmann bromamide degradation of amides. What decides the product is how many carbon atoms end up attached to the nitrogen:
In a nitrile (R–C≡N) and an amide (\( \ce{R-CO-NH2} \)), the nitrogen is attached to one carbon, so reduction gives \( \ce{R-CH2-NH2} \), a primary amine.
In an isocyanide (R–N≡C), the nitrogen is attached to two carbons, R and the terminal C. Reduction turns that C into a methyl group, giving \( \ce{R-NH-CH3} \), a secondary amine.
\( \displaystyle \ce{CH3CH2-N#C} \xrightarrow[\ce{H3O^+}]{\ce{LiAlH4}} \ce{CH3CH2-NH-CH3} \)
Option (1) CH3CH2C≡N with LiAlH4, H3O+ — Gives a primary amine. The nitrile is reduced to propan-1-amine.
\( \displaystyle \ce{CH3CH2C#N} \xrightarrow[\ce{H3O^+}]{\ce{LiAlH4}} \ce{CH3CH2CH2NH2} \)
Option (2) CH3CH2CONH2 with LiAlH4, H3O+ — Gives a primary amine. The C=O of the amide becomes \( \ce{CH2} \), giving propan-1-amine.
\( \displaystyle \ce{CH3CH2CONH2} \xrightarrow[\ce{H3O^+}]{\ce{LiAlH4}} \ce{CH3CH2CH2NH2} \)
Option (3) CH3CH2CONH2 with Br2/NaOH — Gives a primary amine. The Hofmann bromamide degradation gives ethanamine, with one carbon atom fewer than the amide.
\( \displaystyle \ce{CH3CH2CONH2 + Br2 + 4NaOH -> CH3CH2NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Option (4) CH3CH2NC with LiAlH4, H3O+ — Does not give a primary amine (gives the secondary amine N-methylethanamine). The isocyanide carbon becomes a methyl group on the nitrogen.