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ChemistryCBSE Class 12 · Board exam 2024, set 56/1/1 · 2 marks · Very short answer

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 ?

  1. (a)

    HI

    1 mark
  2. (b)

    Conc.

    1 mark

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. (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
  2. (b) Conc. oxidises both the –CHO group and the primary –CH₂OH group to –COOH, giving saccharic acid (glucaric acid):

    1 mark

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

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.