Match the carbonyl compounds in List-I…

Match the carbonyl compounds in List-I with the correct set of observations in List-II. Ethanal | 1. Reduces Tollens' reagent; no yellow precipitate…

Single correct option

Match the carbonyl compounds in List-I with the correct set of observations in List-II.

| List-I | List-II |
|---|---|
| P. Ethanal | 1. Reduces Tollens' reagent; no yellow precipitate with

\mathrm{I_2/NaOH}

|
| Q. Butan-2-one | 2. Yellow precipitate with

\mathrm{I_2/NaOH}

; does not reduce Tollens' reagent |
| R. 2-Hydroxyacetophenone | 3. Reduces Tollens' reagent and also gives a yellow precipitate with

\mathrm{I_2/NaOH}

|
| S. Benzaldehyde | 4. Violet colour with neutral

\mathrm{FeCl_3}

and a yellow precipitate with

\mathrm{I_2/NaOH}

|
| | 5. Violet colour with neutral

\mathrm{FeCl_3}

; no yellow precipitate with

\mathrm{I_2/NaOH}

A. P→3, Q→2, R→4, S→1
B. P→1, Q→2, R→4, S→3
C. P→3, Q→2, R→5, S→1
D. P→3, Q→4, R→2, S→1

Show answer and solution

Answer: (A) P→3, Q→2, R→4, S→1

P: ethanal is an aldehyde (reduces Tollens' reagent) and has the

\mathrm{CH_3CO{-}}

group attached to H, so it gives the iodoform test: 3. Q: butan-2-one is a methyl ketone, positive iodoform, but a ketone does not reduce Tollens' reagent: 2. R: 2-hydroxyacetophenone has a phenolic OH (violet colour with neutral

\mathrm{FeCl_3}

) and the

\mathrm{COCH_3}

methyl-ketone group (iodoform positive); it is not an aldehyde, so Tollens' is negative: 4. S: benzaldehyde reduces Tollens' reagent but has no

\mathrm{CH_3CO}

\mathrm{CH_3CH(OH)}

unit, so no iodoform: 1. Entry 5 describes a phenol without a methyl ketone and matches none. Correct code: P→3, Q→2, R→4, S→1.

Report a mistake

← Previous Qualitative and Practical Chemistry question Next Qualitative and Practical Chemistry question →

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake