Question

Which one of the following compounds will give a yellow precipitate when reacted with \( I_2/\text{NaOH} \)?

• A. \( C_6H_5OH \)
• B. \( C_6H_5CH_2OH \)
• C. \( (CH_3)_3C - OH \)
• D. \( C_6H_5CHOH - CH_3 \)

Hint:

The iodform reaction gives a yellow precipitate only for alcohols that have the structure \( -CHOH-CH_3 \), where the hydroxyl group is adjacent to a methyl group.

Answer 1

The Correct Option is D

Step 1: Understand the reaction with \( I_2/\text{NaOH} \).
When an alcohol reacts with iodine in the presence of sodium hydroxide, it undergoes a reaction known as the iodform reaction. The reaction produces a yellow precipitate of \( CHI_3 \) (iodoform) if the alcohol has the structure \( -CHOH-CH_3 \), where the hydrogen atom is attached to the carbon adjacent to the hydroxyl group.

Step 2: Analyze the compounds.
- \( A \) (C6H5OH): This is phenol. The iodform test does not give a positive result for phenols.
- \( B \) (C6H5CH2OH): This is a benzyl alcohol. It does not undergo the iodform reaction.
- \( C \) ((CH3)3C-OH): This is a tertiary alcohol, which does not undergo the iodform reaction.
- \( D \) (C6H5CHOH-CH3): This is an alcohol with a structure \( -CHOH-CH_3 \), which gives a yellow precipitate with iodine in the presence of sodium hydroxide.

Step 3: Conclusion.
The correct compound that gives a yellow precipitate with \( I_2/\text{NaOH} \) is option (D).