Chemistry · Practical Organic Chemistry
JEE Main 2025 — 8 April, Evening Shift — Question 15
Match the List-I with List-II
| List-I (Reagent) | List-II (Functional 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 | IV. Alcoholic - OH |
Choose the correct answer from the options given below:
- Option A:
A-III, B-II, C-IV, D-I
- Option B:
A-II, B-IV, C-III, D-I
- Option C:Correct
A-II, B-III, C-IV, D-I
- Option D:
A-II, B-III, C-I, D-IV
Answer: C
Step-by-step solution
| List-I (Reagent) | List-II (Functional detected) |
|---|---|
| A. Sodium bicarbonate solution | II. Carboxylic acid |
| B. Neutral ferric chloride | III. Phenolic - OH |
| C. Ceric ammonium nitrate | IV. Alcoholic - OH |
| D. Alkaline | I. Double bond/unsaturation |
(A) Carboxylic acid gives effervescence with sodium bicarbonate
(B) Phenolic-OH gives characteristic colour with neutral
(C) Alcoholic-OH gives red colour with ceric ammonium nitrate
(D) Purple colour of alkaline is discharged by multiple bond of alkenes and alkynes
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2025
- Subject
- Chemistry
- Chapter
- Practical Organic Chemistry
- Topic
- Qualitative analysis of Functional Groups