(A) (a) Reducing sugar

Definition : A carbohydrate that contains a free aldehyde or free ketone group (or a potential one at a free anomeric carbon) and is therefore able to reduce Tollens' reagent and Fehling's solution is called a reducing sugar.

The sugar itself is oxidised while it reduces $Ag^+$ to metallic silver (a silver mirror) or $Cu^{2+}$ to $Cu_2O$ (a red-brown precipitate).

Which sugars qualify : all monosaccharides are reducing. Among disaccharides, maltose and lactose are reducing, but sucrose is not — both its anomeric carbons are locked in the glycosidic linkage.

(A) (b) (i) Fibrous versus globular proteins

(A) (b) (ii) Nucleotide versus nucleoside

$$\text{Base} + \text{Sugar} \rightarrow \text{Nucleoside} \xrightarrow{H_3PO_4} \text{Nucleotide}$$

OR

(B) (a) Glucose with hydroxylamine

The aldehyde group condenses with $H_2N-OH$, eliminating water to give an oxime :

$$\underset{\text{glucose}}{CHO-(CHOH)_4-CH_2OH} + H_2N-OH \rightarrow \underset{\text{glucose oxime}}{CH=N-OH-(CHOH)_4-CH_2OH} + H_2O$$

What it proves : glucose contains a carbonyl group.

(B) (b) Glucose with acetic anhydride

All five hydroxyl groups are acetylated :

$$C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow \underset{\text{glucose pentaacetate}}{C_6H_7O(OCOCH_3)_5} + 5CH_3COOH$$

What it proves : glucose has five $-OH$ groups, each on a different carbon — two on the same carbon would form an unstable gem-diol.

(B) (c) Glucose with concentrated $HNO_3$

This strong oxidising agent attacks both terminal groups :

$$\underset{\text{glucose}}{CHO-(CHOH)_4-CH_2OH} \xrightarrow{\text{conc. }HNO_3} \underset{\text{saccharic acid}}{COOH-(CHOH)_4-COOH}$$

What it proves : glucose has one aldehyde group and one primary alcoholic group, at the two ends of the chain.