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JEE Chemistry 2026 Question with Solution

From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are :

Four cyclic carbohydrate structures labeled A, B, C and D are shown, with substituents such as CH2OH, OCH3, OH and ring oxygen marked for comparison of anomeric carbon functionality.
  • A

    A and D

  • B

    B and C

  • C

    A and B

  • D

    B and D

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Four cyclic structures A to D are to be tested with Tollen's reagent, [Ag(NH3)2]OH[Ag(NH_3)_2]OH.

Find: Which structures will not react with Tollen's reagent.

Principle: Tollen's reagent gives a positive test with compounds that can generate an aldehyde form. In cyclic carbohydrate-type structures, hemiacetals can open to the aldehyde form, whereas acetals/glycosides cannot open under these conditions.

So, check the anomeric carbon — the carbon bonded to two oxygens.

  • If the anomeric carbon has an -OH group, the structure is a hemiacetal and is reducing, so it reacts with Tollen's reagent.
  • If the anomeric carbon has an -OR group such as -OCH3, the structure is an acetal/glycoside and is non-reducing, so it does not react.

From the given structures:

  • A and C are hemiacetals, so they react with Tollen's reagent.
  • B and D are acetals or glycosides, so they do not react with Tollen's reagent.

Therefore, the structures that will not react are B and D.

The correct option is D.

Common mistakes

  • Mistake: Looking for only an aldehyde group drawn explicitly in the cyclic form. Why it is wrong: cyclic hemiacetals may not show a free aldehyde directly, yet they can open to the aldehyde form. What to do instead: identify whether the ring form is a hemiacetal or an acetal by examining the anomeric carbon.

  • Mistake: Treating every cyclic sugar structure as reducing. Why it is wrong: if the anomeric -OH is replaced by -OR, the compound becomes a glycoside/acetal and does not give Tollen's test. What to do instead: check whether the anomeric carbon bears -OH or -OR.

  • Mistake: Confusing any methoxy substituent in the molecule with non-reducing behavior. Why it is wrong: the key issue is not the presence of -OCH3 anywhere, but whether it is attached at the anomeric carbon. What to do instead: first locate the carbon bonded to two oxygens, then inspect its substituent.

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