Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert

Match List-I with List-II

  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

Choose the correct answer from the options given below

ChemistryBiomoleculesJEE MainJEE Main 2022 (27 Jul Shift 2)
Options:
  • 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
Solution:
1847 Upvotes Verified Answer
The correct answer is: A-IV,B-I,C-II,D-III

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.