Question:

Consider the reaction of bromoethane $$\left(CH_3CH_2Br\right)$$ with two different cyanide reagents in an aqueous ethanolic solution:

Reaction 1:

$$CH_3CH_2Br + KCN \rightarrow \text{Product A}$$

Reaction 2:

$$CH_3CH_2Br + AgCN \rightarrow \text{Product B}$$

Identify the major products A and B, respectively.

$$KCN$$ is predominantly ionic and dissociates completely in solution to give free $$CN^-$$ ions. Since the carbon end of the cyanide ion acts as the nucleophilic site, the major product formed is propanenitrile:

$$CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr$$

$$AgCN$$ is predominantly covalent and does not dissociate completely. The carbon atom remains bonded to silver, so the nitrogen atom attacks the alkyl halide, producing ethyl isocyanide:

$$CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr$$

Therefore,

• Product A = Propanenitrile (Ethyl cyanide)
• Product B = Ethyl isocyanide

Hence, the correct answer is Option (B).