JEE Main 2025 · 8 April · evening · Single correct · +4 / −1
Match the LIST-I with LIST-II
Choose the correct answer from the options given below:
| LIST-I (Reagent) | LIST-II (Functional Group detected) |
|---|---|
| A. Sodium bicarbonate solution | I. double bond/unsaturation |
| B. Neutral ferric chloride | II. carboxylic acid |
| C. ceric ammonium nitrate | III. phenolic - OH |
| D. alkaline KMnO₄ | IV. alcoholic - OH |
- (a) A-II, B-III, C-I, D-IV
- (b) A-II, B-IV, C-III, D-I
- (c) A-III, B-II, C-IV, D-I
- (d) A-II, B-III, C-IV, D-I
Answer and solution
Answer (d) A-II, B-III, C-IV, D-I
Solution
Chemistry ma’am · 3 steps
Pair each reagent with the one visible change it gives: fizz, violet colour, red colour, pink colour fading.
- Look for the signal. Each reagent is a test that shows one clear sign with one functional group. Match each reagent to its sign and the group follows.
Reagent What you see Group detected A. NaHCO₃ solution Brisk effervescence of CO₂ II. carboxylic acid B. Neutral FeCl₃ Violet, blue or green colour III. phenolic -OH C. Ceric ammonium nitrate Yellow solution turns red IV. alcoholic -OH D. Alkaline KMnO₄ (Baeyer's test) Pink colour disappears I. unsaturation - Why bicarbonate. RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂ ↑. Phenols are weaker acids than carbonic acid, so they do not release CO₂. That makes this test specific to carboxylic acids.
- So. The match is A-II, B-III, C-IV, D-I, which is option (d).
The idea to keep. Learn functional group tests as reagent, sign and group together: NaHCO₃ fizz for acids, FeCl₃ colour for phenols, CAN red for alcohols, Baeyer's decolourising for C=C.
Answer from the NTA final answer key. Solution by PrepWiser.