CBSE Class 12 Chemistry 2024: question 19
Set in the CBSE Class 12 board exam of 2024, paper 56/1/1, section B, question 19, for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q19 · 2 marks · Very short answerBiomolecules
What happens when D-glucose is treated with the following reagents ?
- (a)1 mark
HI
- (b)1 mark
Conc.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
HI is a strong reducing agent; nitric acid is a strong oxidising agent.
Hint 2
Nitric acid oxidises both terminal carbons of glucose — which groups are at the two ends?
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(a) On prolonged heating with HI, glucose is reduced to n-hexane, showing that its six carbon atoms are in a straight chain:
- 1 mark
(b) Conc. oxidises both the –CHO group and the primary –CH₂OH group to –COOH, giving saccharic acid (glucaric acid):
Answer: (a) n-Hexane is formed; (b) Saccharic (glucaric) acid is formed.
Where the marks are
- HI gives n-hexane (1 mark)
- Conc. HNO₃ gives saccharic acid (1 mark)
Revise the chapter
- Chapter: Biomolecules (notes, formulas, where marks are lost and every past question)
- Topic: Glucose: Structure and Reactions
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