Verified Solution • 2025–26 Session
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QJEEM24Ch84b8: JEE Main 2024 Chemistry - Class 11 JEEM24Ch84b8 Solution: Match List I with List II .tg…
Match List I with List II
| LIST I (Test) | LIST II (Observation) | ||
|---|---|---|---|
| A. | I. | Yellow orange or orange red precipitate formed | |
| B. | Ceric ammonium nitrate test | II. | Reddish orange colour disappears |
| C. | Ferric chloride test | III. | Red colour appears |
| D. | 2, 4 - DNP test | IV. | Blue, Green, Violet or Red colour appear |
Choose the correct answer from the options given below:
Multiple Choice Options
- A A-II, B-III, C-IV, D-I
- B A-III, B-IV, C-I, D-II
- C A-I, B-II, C-III, D-IV
- D A-IV, B-I, C-II, D-III
Step-by-Step Verified Solution
Let's match the given tests in List I with the correct observations in List II based on known chemical reactions and their typical outcomes:
List I with List II
A. water test- This test is used to detect unsaturation in organic compounds. If the compound has a double bond (alkenes) or a triple bond (alkynes), the in water (bromine water) will react and decolorize.
- Correct Observation: Reddish orange color of bromine water disappears (II).
B. Ceric ammonium nitrate test
- This test is used to detect alcohols. Ceric ammonium nitrate forms a red-colored complex with alcohols.
- Correct Observation: Red color appears (III).
C. Ferric chloride test
- Commonly used to detect phenols, the ferric chloride test leads to the formation of a colored complex with phenols, typically varying shades like purple, green, blue, or even red, depending on the specific phenol.
- Correct Observation: Blue, Green, Violet, or Red color appear (IV).
D. 2,4-DNP test
- This test is used for the detection of carbonyl compounds (aldehydes and ketones). 2,4-Dinitrophenylhydrazine (2,4-DNP) reacts with carbonyl groups to form a yellow, orange, or red precipitate.
- Correct Observation: Yellow orange or orange red precipitate formed (I).
Conclusion:
Based on these observations, the correct matching of tests with outcomes should be:
- A with II
- B with III
- C with IV
- D with I
This makes the correct choice:
Option A: A-II, B-III, C-IV, D-I