JEE Main 2025 Apr 8 Shift 2, Chemistry Q17: Qualitative and Quantitative Analysis
Match the LIST-I with LIST-II
| 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 KMnO | IV. alcoholic -OH |
Choose the correct answer from the options given below:
- A
A-II, B-III, C-IV, D-I
- B
A-II, B-III, C-I, D-IV
- C
A-III, B-II, C-IV, D-I
- D
A-II, B-IV, C-III, D-I
Sodium bicarbonate solution releases CO gas (effervescence) on reaction with a carboxylic acid, so it detects the carboxylic acid group: A-II.
Neutral ferric chloride forms a characteristic violet-coloured complex with phenols, so it is used to detect a phenolic -OH group: B-III.
Ceric ammonium nitrate reacts with alcohols to give a red colouration through formation of a cerium-alkoxide complex, so it detects an alcoholic -OH group: C-IV.
Alkaline KMnO is decolourised in the presence of a carbon-carbon double or triple bond as the unsaturation is oxidatively cleaved or dihydroxylated, so it detects unsaturation: D-I.
Combining these gives A-II, B-III, C-IV, D-I, matching option (1).
Concept behind this question
Qualitative and Quantitative AnalysisNotes, formulas and examples →More previous year questions on Qualitative and Quantitative Analysis
Practice more Chemistry
Concept-wise practice with instant solutions on Fundamenthol.
Start practicing →