JEE Main202227 Jul 2022Evening ShiftChemistryBiomoleculesActual
Match List-I with List-II List-I List-II A Glucose + HI I Gluconic acid B Glucose + Br 2 water II Glucose pentacetate C Glucose + acetic anhydride III Saccharic acid D Glucose + HNO 3 IV Hexane Choose the correct answer from the options given below
Options
- AA - IV , B - I , C - II , D - III
- BA - IV , B - III , C - II , D - I
- CA - III , B - I , C - IV , D - II
- DA - I , B - III , C - IV , D - II
Correct answer
A. A - IV , B - I , C - II , D - III
Step-by-step solution
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 grou