Question

(a) Name two water-soluble vitamins.

(b) How will you prove the absence of free \( - \mathrm{CHO} \) group in the cyclic structure of glucose molecule ?

Hint:

Name vitamin B and vitamin C. Glucose gives no Schiff test and no NaHSO3 addition product, and its pentaacetate does not react with hydroxylamine.

Answer 1

Step 1: Understand the question.
Part (a) asks for two vitamins that dissolve in water.
Part (b) asks for an experiment showing that ring-form glucose has no free aldehyde group.

Step 2: Answer part (a).
Vitamins are of two kinds: fat-soluble and water-soluble.
Vitamins A, D, E and K are fat-soluble.
The B group vitamins and vitamin C are water-soluble.
So two examples are vitamin B\(_1\) (thiamine) and vitamin C (ascorbic acid).
The body does not store them, so they are needed in the daily diet.

Step 3: Why free \( - \mathrm{CHO} \) is doubted in glucose.
Open-chain glucose has an aldehyde group at C-1.
In the ring form, the \( - \mathrm{OH} \) on C-5 adds to this C=O group.
This forms a cyclic hemiacetal, so no free \( - \mathrm{CHO} \) is left.
Glucose therefore behaves unlike a normal aldehyde in some tests.

Step 4: Tests that prove the absence of free \( - \mathrm{CHO} \).
(i) Glucose does not give the Schiff test. The pink colour of Schiff reagent is not restored.
(ii) Glucose does not form the hydrogensulphite addition product with \( \mathrm{NaHSO_3} \).
(iii) The pentaacetate of glucose does not react with hydroxylamine, so no oxime forms.
A free aldehyde would respond in all these tests.

Final Answer:
(a) Vitamin B and vitamin C. (b) Glucose does not give the Schiff test or the \( \mathrm{NaHSO_3} \) addition product, and its pentaacetate does not react with hydroxylamine.