JEE Main 2022 — Biomolecules Question with Solution
JEE Main 2022 (27 Jul Shift 2)
Question
Match List-I with List-II
| List-I | List-II | ||
| A | Glucose | I | Gluconic acid |
| B | Glucose water | II | Glucose pentacetate |
| C | Glucose acetic anhydride | III | Saccharic acid |
| D | Glucose | IV | Hexane |
Choose the correct answer from the options given below
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Show full solutionCorrect option: A
Step-by-step explanation
A) On prolonged heating with HI, Glucose forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.
B) Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.
C) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.
D) On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose
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This is a previous-year question from JEE Main 2022, covering the Biomolecules chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.