Step 1: Break the question into three parts.
Part (a) is about the structure of a nucleotide. Part (b) asks for the proof of five hydroxyl groups. Part (c) is a simple addition reaction.
Step 2: Hydrolysis of the thymine nucleotide.
Think of a nucleotide as base + sugar + phosphate. In DNA, the sugar has no -OH at carbon 2, so it is called deoxyribose.
The ester bond to phosphate and the glycosidic bond to the base are both broken by hydrolysis.
So the products are thymine, 2-deoxy-D-ribose and phosphoric acid. That is three products, as the three marks suggest.
Step 3: Explain the five -OH groups.
Treat glucose with excess acetic anhydride. All the -OH groups are acetylated and we get a pentaacetate. $\mathrm{C_6H_{12}O_6}$ gains five acetyl groups, so five -OH groups were present.
Now see why they are on different carbons. Two -OH groups on one carbon (a gem diol) would lose water and form a carbonyl group. Glucose is stable, so it has no such arrangement.
Glucose has six carbons. One is the aldehyde carbon, and the other five each carry one -OH group.
Step 4: Glucose with HCN.
The aldehyde carbon is attacked by cyanide, and the oxygen picks up hydrogen. A cyanohydrin results.
\[ \mathrm{CH_2OH(CHOH)_4CHO \xrightarrow{HCN} CH_2OH(CHOH)_4CH(OH)CN} \]
Final Answer:
(a) Thymine, 2-deoxy-D-ribose, phosphoric acid. (b) Glucose gives a pentaacetate with acetic anhydride, so it has five -OH groups on five different carbons. (c) Glucose cyanohydrin is formed.