The given reaction is 2-methylbutanamide reac | vedclass

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(B) The reaction is the $Hofmann$ $Bromamide$ degradation,which converts an amide into a primary amine with one less carbon atom.Reactant: $CH_3-CH(C_

Vedclass · Accepted Answer

It is optically active and has one less carbon than the reactant

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(B) The reaction is the $Hofmann$ $Bromamide$ degradation,which converts an amide into a primary amine with one less carbon atom.Reactant: $CH_3-CH(C_2H_5)-CONH_2$ ($2$-methylbutanamide).Product: $CH_3-CH(C_2H_5)-NH_2$ ($sec$-butylamine).$1$. The product $sec$-butylamine has a chiral center at the carbon attached to the $-NH_2$ group,making it optically active.$2$. The product has $4$ carbon atoms,while the reactant has $5$ carbon atoms,so it has one less carbon.$3$. Primary amines give the isocyanide test.Therefore,option $B$ is correct.

The given reaction is $2$-methylbutanamide reacting with $KOBr$ and $\Delta$. What is the correct statement about the product?

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