JEE Main 2022 · 27 July · evening · Single correct · +4 / −1
Match List - I with Match List - II.
Choose the correct answer from the options given below:
| List - I | List - II | ||
|---|---|---|---|
| (A) | Glucose + HI | (I) | Gluconic acid |
| (B) | Glucose + Br₂ water | (II) | Glucose pentacetate |
| (C) | Glucose + acetic anhydride | (III) | Saccharic acid |
| (D) | Glucose + HNO₃ | (IV) | Hexane |
- (a) (A) - (IV), (B) - (I), (C) -(II), (D) - (III)
- (b) (A) - (IV), (B) - (III), (C) -(II), (D) - (I)
- (c) (A) - (III), (B) - (I), (C) -(IV), (D) - (II)
- (d) (A) - (I), (B) - (III), (C) -(IV), (D) - (II)
Answer and solution
Answer (a) (A) - (IV), (B) - (I), (C) -(II), (D) - (III)
Solution
Chemistry ma’am · 3 steps
Ask what each reagent touches: HI strips every oxygen, bromine water one end, nitric acid both ends.
- HI reduces fully. Prolonged heating with HI removes every oxygen from glucose and gives n-hexane. This shows that the six carbons sit in one straight chain. So A matches IV.
- Two oxidisers. Bromine water is a mild oxidiser. It oxidises only the -CHO group to -COOH, giving gluconic acid. Nitric acid is strong and also oxidises the terminal -CH₂OH, giving the dicarboxylic saccharic acid, so B matches I and D matches III.
Reaction What changes Product (A) Glucose + HI all oxygens removed (IV) hexane (B) Glucose + Br₂ water only -CHO becomes -COOH (I) gluconic acid (C) Glucose + acetic anhydride all five -OH groups acetylated (II) glucose pentaacetate (D) Glucose + HNO₃ -CHO and -CH₂OH both become -COOH (III) saccharic acid - So. The matching is A-IV, B-I, C-II, D-III, which is option (a).
The idea to keep. Bromine water oxidises only the aldehyde end and gives a monoacid. Nitric acid oxidises both ends and gives a diacid. HI goes the other way and reduces glucose all the way to hexane.
Answer from the NTA final answer key. Solution by PrepWiser.