Question:
Match List I with List II.
Match List I with List II.
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When identifying functional groups, use specific tests that target hydroxyl, carbonyl, and aldehyde groups. Each test is unique to the functional group it detects.
Updated On: Apr 4, 2026
- I-Q, II-S, III-P, IV-R
- I-P, II-Q, III-R, IV-S
- I-P, II-Q, III-S, IV-R
- I-P, II-Q, III-S, IV-S
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The Correct Option is A
Solution and Explanation
Step 1: Analyzing ethanol (I).
Ethanol is an alcohol and gives a positive test with Ceric ammonium nitrate, which is a confirmatory test for alcohols. Hence, ethanol → Q.
Step 2: Analyzing phenol (II).
Phenol gives a characteristic violet color with Neutral FeCl₃, which is a specific test for phenolic compounds. Hence, phenol → P.
Step 3: Analyzing ethanoic acid (III).
Ethanoic acid reacts with NaHCO₃ to produce brisk effervescence due to the release of CO₂ gas. This is a confirmatory test for carboxylic acids. Hence, ethanoic acid → S.
Step 4: Analyzing acetaldehyde (IV).
Acetaldehyde is an aldehyde and gives a positive test with Schiff reagent, producing a pink color. Hence, acetaldehyde → R.
Step 5: Conclusion.
The correct matching is:
\[ \text{I–Q, II–P, III–S, IV–R} \]
Final Answer: I–Q, II–P, III–S, IV–R.
Ethanol is an alcohol and gives a positive test with Ceric ammonium nitrate, which is a confirmatory test for alcohols. Hence, ethanol → Q.
Step 2: Analyzing phenol (II).
Phenol gives a characteristic violet color with Neutral FeCl₃, which is a specific test for phenolic compounds. Hence, phenol → P.
Step 3: Analyzing ethanoic acid (III).
Ethanoic acid reacts with NaHCO₃ to produce brisk effervescence due to the release of CO₂ gas. This is a confirmatory test for carboxylic acids. Hence, ethanoic acid → S.
Step 4: Analyzing acetaldehyde (IV).
Acetaldehyde is an aldehyde and gives a positive test with Schiff reagent, producing a pink color. Hence, acetaldehyde → R.
Step 5: Conclusion.
The correct matching is:
\[ \text{I–Q, II–P, III–S, IV–R} \]
Final Answer: I–Q, II–P, III–S, IV–R.
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