MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Alcoholic silver cyanide converts alkyl halides into alkyl isocyanides.Explanation:Ethyl bromide reacts with alcoholic silver cyanide ($\mathrm{AgCN}$) to give ethyl isocyanide as the main product.The chemical equation is:$$\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Br} + \mathrm{AgCN}_{\text{(alc)}} \rightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{NC} + \mathrm{AgBr}$$The cyanide ion ($\mathrm{CN}^-$) is an ambident nucleophile because it can attack through either the carbon atom or the nitrogen atom.When $\mathrm{AgCN}$ is used, it is largely covalent in nature.The silver ion binds with the carbon end, making that end unavailable for attack.Hence, the nitrogen end attacks the alkyl halide, forming an isocyanide.In contrast, alcoholic $\mathrm{KCN}$ gives a nitrile because the carbon end attacks.Therefore, the product $\mathrm{P}$ is ethyl isocyanide, $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{NC}$.Answer:Option A: $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{NC}$