Question

(a) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ?

(b) How will you explain the presence of five -OH groups in glucose molecule which are attached to different carbon atoms ?

(c) Write the reaction of glucose with HCN.

Hint:

A DNA nucleotide splits into a base, 2-deoxyribose and phosphoric acid. Glucose pentaacetate proves five -OH groups, and HCN adds to the CHO group.

Answer 1

Step 1: Understanding the Concept:
A nucleotide is made of three parts: a nitrogenous base, a pentose sugar and a phosphate group. In DNA the sugar is 2-deoxy-D-ribose.
Hydrolysis splits a nucleotide back into these three parts.

Step 2: Products of hydrolysis, part (a).
A DNA nucleotide that contains thymine is made of thymine, 2-deoxy-D-ribose and phosphoric acid.
On hydrolysis it gives:
1. Thymine (the pyrimidine base).
2. 2-Deoxy-D-ribose (the sugar).
3. Phosphoric acid, \(\mathrm{H_3PO_4}\).

Step 3: Five -OH groups in glucose, part (b).
When glucose reacts with acetic anhydride, it forms glucose pentaacetate. This shows that glucose has five -OH groups.
\[ \mathrm{C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow C_{16}H_{22}O_{11} + 5CH_3COOH} \]
A carbon with two -OH groups on it is unstable. Such a compound loses water easily.
Glucose is a stable compound, so no two -OH groups can be on the same carbon. Hence the five -OH groups are on five different carbon atoms.

Step 4: Reaction with HCN, part (c).
Glucose has an aldehyde group (-CHO). HCN adds to it and forms a cyanohydrin.
\[ \mathrm{CH_2OH(CHOH)_4CHO + HCN \rightarrow CH_2OH(CHOH)_4CH(OH)CN} \]
This reaction shows the presence of the carbonyl group in glucose.

Final Answer:
(a) Thymine, 2-deoxy-D-ribose and phosphoric acid. (b) Acetylation gives glucose pentaacetate, so there are five -OH groups, and each sits on a different carbon since glucose is stable. (c) Glucose cyanohydrin is formed.