- Exam
- JEE Main
- Year
- 2022
- Date
- 27 July
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- Biomolecules
JEE Main 2022 · 27 July · Evening Shift · Chemistry · Biomolecules
Match List-I with List-II List-I List-II A Glucose + HI I Gluconic...
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
Options
- A
- B
- C
- D
Correct answer
A.
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 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