JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
Which among the above compound/s does/do not form Silver mirror when treated with Tollen's reagent?
- (a) (I), (III) and (IV) only
- (b) Only (IV)
- (c) Only (II)
- (d) (III) and (IV) only
Answer and solution
Answer (c) Only (II)
Solution
Chemistry ma’am · 3 steps
Do not judge by the drawing alone. Ask what each compound becomes in solution, an aldehyde or a ketone?
- What Tollens needs. Tollens' reagent, [Ag(NH₃)₂]⁺, is a mild oxidant that attacks a -CHO group and deposits a silver mirror. Any compound that is an aldehyde, or turns into one in the reagent, gives a positive test. Ketones do not.
- Enols and hemiacetals. An enol, with -OH on a C=C carbon, quickly tautomerises to its carbonyl form. A cyclic hemiacetal, with -OH on a carbon next to a ring O, opens up to a free aldehyde in solution.
Compound What it is Form in solution Silver mirror? (I) 2-methylpropanal (CH₃)₂CH-CHO, an aldehyde Yes (II) enol, CH₂=C(OH)CH₃ tautomerises to acetone, CH₃COCH₃, a ketone No (III) enol, HO-CH= on a cyclohexane ring tautomerises to cyclohexanecarbaldehyde, C₆H₁₁-CHO Yes (IV) cyclic hemiacetal ring opens to 5-hydroxypentanal, HO(CH₂)₄CHO Yes - So. Only (II) ends up as a ketone, so only (II) fails to give a silver mirror. The answer is (c).
The idea to keep. For Tollens, hunt for hidden aldehydes. An enol HO-CH=C tautomerises to -CHO, and a ring O-CH-OH hemiacetal opens to one. An enol that gives a ketone stays negative.
Answer from the NTA final answer key. Solution by PrepWiser.