Chemistry · Ch 13 — Amines
Reduction of alkyl cyanide (alkanenitriles)
Reduction of alkyl cyanide (alkanenitriles)
An alkyl cyanide (also called an alkanenitrile), reduced with sodium metal dissolved in ethanol, gives a primary amine -- this specific combination of reagents is a named reaction, the Mendius reduction. The same overall reduction can also be brought about instead using lithium aluminium hydride. Because the nitrile's carbon (the one triple-bonded to nitrogen in -C#N) becomes, after reduction, the -CH2- carbon directly bonded to the new -NH2 group, the product primary amine has EXACTLY the same total number of carbon atoms as the starting nitrile: R-C#N + 4 [H], with Na/C2H5OH (or with LiAlH4 in ether, followed by aqueous acid workup, H3O+), gives R-CH2-NH2, a primary amine. Since an alkyl cyanide is itself very commonly made in a separate, prior step from an alkyl halide plus KCN (R-X + KCN gives R-CN + KX, a substitution that itself ADDS one carbon atom, the nitrile carbon, onto the original alkyl group), the full two-step sequence starting from an alkyl halide -- first to the nitrile, then reduced here to the amine -- delivers an overall primary amine with ONE MORE carbon atom than the original alkyl halide. This whole two-step, carb …
Worked out. Worked problem: convert methyl bromide into ethylamine, and comment on the carbon count of starting material versus product. Solution, in two stages: (1) CH3-Br + KCN gives CH3-CN (methyl cyanide) + KBr -- this is the alkyl-halide-to-nitrile substitution that adds one carbon (the nitrile carbon) to the chain. (2) CH3-CN, reduced with Na/C2H5OH (Mendius reduction), gives CH3-CH2-NH2 (ethylamine). Methyl bromide has one carbon, the final product ethylamine has two; since the number of carbons has increased across the overall two-step sequence, the text explicitly names this a 'step-up' conversion (contrasted later, in section 13.3.6, with Hofmann degradation, a 'step-down' c …