Identify the final product (C) in the reaction given below; CH_3CH_2I…

The correct answer is Option (2) → \(CH_3CH_2NH_2\)

The sequence first adds one carbon (through a cyanide) and then removes one carbon (through Hofmann bromamide degradation), so the amine at the end has the same number of carbon atoms as the starting halide.

First, ethyl iodide reacts with sodium cyanide by nucleophilic substitution; the cyanide ion replaces iodide and the chain grows by one carbon:

    \( \displaystyle \ce{CH3CH2I + NaCN -> CH3CH2CN\,(A) + NaI} \)

Partial hydrolysis of the nitrile with mild alkali stops at the amide:

    \( \displaystyle \ce{CH3CH2CN + H2O} \xrightarrow{\ce{OH-}\text{, partial}} \ce{CH3CH2CONH2\,(B)} \)

Propanamide is then treated with bromine and sodium hydroxide. In this Hofmann degradation the ethyl group migrates from the carbonyl carbon to nitrogen, and the carbonyl carbon is lost as carbonate, giving an amine with one carbon less than the amide:

    \( \displaystyle \ce{CH3CH2CONH2 + Br2 + 4NaOH -> CH3CH2NH2\,(C) + Na2CO3 + 2NaBr + 2H2O} \)

So C is ethanamine, \( \ce{CH3CH2NH2} \).

Shortcut: Count the carbons: 2 (ethyl iodide) + 1 (cyanide) − 1 (Hofmann) = 2. Only Option (2) is an amine with two carbon atoms.