JEE Chemistry 2025 Question with Solution

Given: We must identify which compounds give a yellow solid with NaOI/NaOH.

Find: Which labeled compounds undergo the iodoform reaction and hence form yellow $\mathrm{CHI_3}$.

This is the haloform reaction. A yellow precipitate is formed by compounds containing the methyl ketone group $\mathrm{CH_3COR}$, or by alcohols of the type $\mathrm{CH_3CH(OH)R}$ because they are oxidized to methyl ketones under these conditions.

From the given structures:

• (A) is $\mathrm{CH_3-CH(OH)-C_2H_5}$, a secondary alcohol of type $\mathrm{CH_3CH(OH)R}$, so it gives the iodoform test.
• (B) is $\mathrm{CH_3-CH_2-CH_2-OH}$, a primary alcohol, so it does not give the test.
• (C) is $\mathrm{CH_3COC_2H_5}$, a methyl ketone, so it gives the test.
• (D) is $\mathrm{CH_3COOH}$, a carboxylic acid, so it does not give the test.
• (E) is $\mathrm{CH_3CH_2CHO}$, an aldehyde other than ethanal, so it does not give the test.

Therefore, the compounds that give yellow solid are (A) and (C) only. Hence, the correct option is B.

Given: NaOI/NaOH gives a yellow solid in the iodoform test.

Find: Which of (A) to (E) satisfy the structural requirement.

The required structural conditions are:

1. A methyl ketone: $\mathrm{CH_3COR}$
2. Or a secondary alcohol oxidizable to a methyl ketone: $\mathrm{CH_3CH(OH)R}$

Checking each compound one by one:

• (A) fits $\mathrm{CH_3CH(OH)R}$ with $\mathrm{R=C_2H_5}$.
• (B) is $\mathrm{CH_3CH_2CH_2OH}$ and does not fit the required pattern.
• (C) directly fits $\mathrm{CH_3COR}$.
• (D) is a carboxylic acid, not a methyl ketone.
• (E) is propanal, not ethanal, so it does not respond positively.

Thus only (A) and (C) form the yellow precipitate of iodoform. The correct option is B.

The second provided approach contains inconsistent structural descriptions, but the conclusion supported by the valid structural analysis is B.