Question: The treatment of ethyl bromide with alcoholic silver nitrite gives:
Step 1 (Governing Chemical Principle): [0.5 Mark]
The reaction of haloalkanes with silver nitrite ($AgNO_2$) is a nucleophilic substitution reaction. Since $AgNO_2$ is primarily covalent in nature, the nitrogen atom acts as the nucleophile rather than the oxygen atom, leading to the formation of a nitroalkane.
The general reaction equation is:
$$R-X + AgNO_2 \xrightarrow{\text{alc.}} R-NO_2 + AgX$$Step 2 (Step-by-Step Chemical Working): [0.25 Mark]
Substituting ethyl bromide ($C_2H_5Br$) into the reaction:
- Reactant: Ethyl bromide ($CH_3CH_2Br$)
- Reagent: Alcoholic silver nitrite ($AgNO_2$)
- Reaction: $CH_3CH_2Br + AgNO_2 \rightarrow CH_3CH_2NO_2 + AgBr$
The product formed is nitroethane ($C_2H_5NO_2$).
Step 3 (Final Concluded Answer): [0.25 Mark]
The correct option is (B) nitroethane.
💡 EL10 Concept Insight: Think of $AgNO_2$ like a person holding a door handle with their nitrogen hand; because the bond between silver and nitrogen is strong, the nitrogen is the one that grabs onto the ethyl group, turning it into a nitro compound instead of a nitrite.