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Chemistry Question 67 – JEE-MAIN 2026

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Match List - I with List - II. List - IList - IIMixture of CompoundsReagent used to distinguishA. Diethyl amine + Ethyl amineI. Bromine waterB. Acetaldehyde + AcetoneII. CHCl3+KOH,ΔC. Ethanol + PhenolIII. Neutral FeCl3D. Benzoic acid + Cinnamic acidIV. Ammonical silver nitrate Choose the correct answer from the options given below :

Identify the key functional groups in each pair of compounds in List-I to determine their chemical properties.

Step 1: Distinguishing Amines and Aldehydes/Ketones✦ Active

For A (Diethyl amine + Ethyl amine): Ethyl amine is a primary amine, which gives the Carbylamine reaction (foul-smelling isocyanide) with CHCl3+KOH,Δ. Diethyl amine is a secondary amine and does not give this test. Thus, A matches II.

For B (Acetaldehyde + Acetone): Acetaldehyde is an aldehyde, which reduces Tollens' reagent (ammonical silver nitrate) to form a silver mirror. Acetone is a ketone and does not give this test. Thus, B matches IV.

Step 2: Distinguishing Alcohols/Phenols and Saturated/Unsaturated Acids○ Expand

For C (Ethanol + Phenol): Phenol gives a characteristic coloration (violet, green, or blue) with neutral FeCl3 solution due to complex formation. Ethanol, being an alcohol, does not give this test. Thus, C matches III.

For D (Benzoic acid + Cinnamic acid): Cinnamic acid contains a carbon-carbon double bond, which will decolorize bromine water via an addition reaction. Benzoic acid is saturated and will not react with bromine water under these conditions. Thus, D matches I.

Step 3: Final Matching and Option Selection○ Expand

Combining the matches: A-II, B-IV, C-III, D-I. Comparing this with the given options, Option 4 is the correct choice.

💡 Teacher's Secret Hint

Ensure all individual matches are correct before selecting the final option.

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