Match List-I with List-II

List-I	List-II
A. Glucose + HI	I. Gluconic acid
B. Glucose + $$Br_2$$ water	II. Glucose pentacetate
C. Glucose + acetic anhydride	III. Saccharic acid
D. Glucose + $$HNO_3$$	IV. Hexane

We need to match the reactions of glucose with the corresponding products.

A. Glucose + HI

When glucose is heated with excess HI for a prolonged period, all the oxygen-containing groups are reduced and the product is hexane ($$C_6H_{14}$$). This proves that glucose has a straight chain of 6 carbon atoms.

$$\therefore$$ A $$\rightarrow$$ IV (Hexane)

B. Glucose + $$Br_2$$ water

Bromine water is a mild oxidizing agent. It selectively oxidizes the aldehyde group ($$-CHO$$) of glucose to a carboxylic acid group ($$-COOH$$), forming gluconic acid.

$$\therefore$$ B $$\rightarrow$$ I (Gluconic acid)

C. Glucose + Acetic anhydride

Glucose has five hydroxyl groups ($$-OH$$). Acetic anhydride acetylates all five $$-OH$$ groups to form glucose pentaacetate.

$$\therefore$$ C $$\rightarrow$$ II (Glucose pentacetate)

D. Glucose + $$HNO_3$$

Nitric acid is a strong oxidizing agent. It oxidizes both the aldehyde group and the terminal primary alcohol group of glucose, forming a dicarboxylic acid called saccharic acid (glucaric acid).

$$\therefore$$ D $$\rightarrow$$ III (Saccharic acid)

The correct matching is: A-IV, B-I, C-II, D-III

Hence, the correct answer is Option A.