Question

Among the following compounds I-IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH (ii) dil. $HNO_3$ (iii) $AgNO_3$? 

• A. I
• B. II
• C. III
• D. IV

Hint:

Halides attached to $sp^2$ carbons (aryl/vinyl) are inert to this test, while those on $sp^3$ carbons (alkyl/benzyl) react readily. Silver iodide is the only one giving a strong yellow precipitate.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This sequence of reagents is a test for halide ions in organic compounds. NaOH induces nucleophilic substitution or elimination to release halide ions ($X^-$). Addition of dil. $HNO_3$ neutralizes the base, and $AgNO_3$ reacts with $X^-$ to form a silver halide precipitate.
Step 2: Detailed Explanation:
1. Analysis of Compounds:
Compound I and II are aryl halides (chlorobenzenes). Chlorines attached directly to the benzene ring do not undergo nucleophilic substitution easily due to partial double bond character.
Compound III is an aryl bromide. Similarly, the Br is inert to NaOH at room temperature.
Compound IV contains an iodo-methyl group (benzyl-type iodide) and a chloro group on the ring. The $-CH_2I$ group is highly reactive towards nucleophilic substitution.
2. Reaction of IV:
Reacting IV with NaOH releases iodide ions ($I^-$).
$Ar-CH_2I + OH^- \rightarrow Ar-CH_2OH + I^-$
3. Precipitation:
After neutralizing with $HNO_3$, the addition of $AgNO_3$ results in:
$Ag^+ + I^- \rightarrow AgI \downarrow$ (Yellow precipitate)
Silver chloride ($AgCl$) is white, and silver bromide ($AgBr$) is pale yellow, but silver iodide ($AgI$) is distinctively yellow.
Step 3: Final Answer:
Compound IV forms the yellow precipitate.