What will be the product for the following reactions?

\(CH_3-X \overset{KCN}{\longrightarrow} A\)

\(CH_3-X \overset{AgNO_2}{\longrightarrow}B\)

\(A: CH_3CN\); \(B:CH_3NO_2\)

The correct answer is option 1. \(A: CH_3CN\); \(B:CH_3NO_2\).

Let us break down the reactions in detail:

1. Reaction with Potassium Cyanide (KCN)

Reaction: \(CH_3-X \overset{KCN}{\longrightarrow} CH_3CN \)

Potassium cyanide (\(KCN\)) dissociates in solution to give potassium ions (\(K^+\)) and cyanide ions (\(CN^-\)).  The cyanide ion (\(CN^-\)) is a good nucleophile. In a nucleophilic substitution reaction (\(S_N2\)), the nucleophile attacks the carbon atom bonded to the leaving group (X) in the alkyl halide (\(CH_3-X\)). This attack displaces the leaving group (X), forming a new bond between the carbon and the cyanide ion. The product of this reaction is methyl cyanide (\(CH_3CN\)), also known as acetonitrile. The cyanide group is attached to the carbon where the halogen (X) was originally bonded.

Reaction with Silver Nitrite (AgNO_2)

Reaction: \(CH_3-X \overset{AgNO_2}{\longrightarrow} CH_3NO_2\)

Silver nitrite (\(AgNO_2\)) dissociates in solution to give silver ions (\(Ag^+\)) and nitrite ions (\(NO_2^-\)). The nitrite ion (\(NO_2^-\)) is ambident, meaning it can bind through either nitrogen or oxygen. In this reaction, it typically binds through nitrogen to form a nitro compound. In the nucleophilic substitution reaction with an alkyl halide (\(CH_3-X\)), the nitrite ion attacks the carbon atom bonded to the leaving group (X), leading to the formation of nitromethane (\(CH_3NO_2\)). The product of this reaction is nitromethane (\(CH_3NO_2\)). The nitro group (\(-NO_2\)) is attached to the carbon where the halogen (X) was originally bonded.

Summary

For the reaction with \(KCN\): Methyl cyanide (\(CH_3CN\)) is formed.

For the reaction with \(AgNO_2\): Nitromethane (\(CH_3NO_2\)) is formed.

Thus, the correct answer is option 1: \(A: CH_3CN\); \(B: CH_3NO_2\)