entify the products (1) and (2) in the following reactions:C2H5Cl + KCN (1) and C2H5Cl + KNO2

To identify the products (1) and (2) in the given reactions, we need to analyze the reactants and the reaction conditions.

Reaction 1: C2H5Cl + KCN In this reaction, C2H5Cl (ethyl chloride) reacts with KCN (potassium cyanide). The reaction between an alkyl halide and a cyanide ion is known as a nucleophilic substitution reaction.

The cyanide ion (CN-) acts as a nucleophile and replaces the chloride ion (Cl-) in ethyl chloride. The reaction proceeds as follows:

C2H5Cl + KCN → C2H5CN + KCl

Therefore, the product (1) in this reaction is C2H5CN (ethyl cyanide), and the byproduct is KCl (potassium chloride).

Reaction 2: C2H5Cl + KNO2 In this reaction, C2H5Cl (ethyl chloride) reacts with KNO2 (potassium nitrite). The reaction between an alkyl halide and a nitrite ion is also a nucleophilic substitution reaction.

The nitrite ion (NO2-) acts as a nucleophile and replaces the chloride ion (Cl-) in ethyl chloride. The reaction proceeds as follows:

C2H5Cl + KNO2 → C2H5NO2 + KCl

Therefore, the product (2) in this reaction is C2H5NO2 (ethyl nitrite), and the byproduct is KCl (potassium chloride).

In summary: Reaction 1: C2H5Cl + KCN → (1) C2H5CN + KCl Reaction 2: C2H5Cl + KNO2 → (2) C2H5NO2 + KCl