Identify the two compounds

The bottles contain:

• Iodobenzene: $C_6H_5I$
• Benzyl iodide: $C_6H_5CH_2I$

Reaction with aqueous $NaOH$

We test both separately by boiling with $NaOH$ solution.

(i) Benzyl iodide: $C_6H_5CH_2I$

This is an alkyl halide (benzylic halide), so it undergoes nucleophilic substitution readily: $C_6H_5CH_2I + NaOH \rightarrow C_6H_5CH_2OH + NaI$ Thus, iodide ions $I^-$ are produced in solution.

(ii) Iodobenzene: $C_6H_5I$

This is an aryl halide. Aryl halides do not undergo nucleophilic substitution easily with aqueous $NaOH$ under these conditions.

So, no appreciable $I^-$ is released: $C_6H_5I \;\text{does not react with aqueous } NaOH \text{ under ordinary boiling conditions}$

Why acidify with dilute $HNO_3$ before adding $AgNO_3$?

If excess $NaOH$ is present, then on adding $AgNO_3$, silver hydroxide / silver oxide may form and interfere.

So $HNO_3$ is added to neutralize $OH^-$: $NaOH + HNO_3 \rightarrow NaNO_3 + H_2O$

Hence, addition of $HNO_3$ is necessary. So option C is false.

Test with $AgNO_3$

If iodide ions are present, then: $Ag^+ + I^- \rightarrow AgI \downarrow$ A yellow precipitate of $AgI$ is obtained.

The question says substance B gave a yellow precipitate.

Therefore, B must be $C_6H_5CH_2I$ because only benzyl iodide produces $I^-$ on boiling with $NaOH$.

Identify A

Since B is $C_6H_5CH_2I$, the other bottle A must be: $A = C_6H_5I$

Check options

• A: A and $C_6H_5CH_2I$ — false
• B: B and $C_6H_5I$ — false
• C: Addition of $HNO_3$ was unnecessary — false
• D: A was $C_6H_5I$ — true

Final answer

The correct option is D.