(a) Reaction with HI (on prolonged heating)

$$C_6H_{12}O_6 \xrightarrow[\Delta,\ \text{long time}]{HI} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$$

Glucose is reduced completely to n-hexane.

What it proves : all six carbon atoms of glucose lie in a straight, unbranched chain. Every oxygen-containing group is stripped away by this drastic reduction.

(b) Reaction with bromine water

Bromine water is a mild oxidising agent. It attacks only the $-CHO$ group, converting it into $-COOH$, and leaves the primary alcoholic $-CH_2OH$ untouched :

$$\underset{\text{glucose}}{CHO-(CHOH)_4-CH_2OH} \xrightarrow{Br_2\ \text{water}} \underset{\text{gluconic acid}}{COOH-(CHOH)_4-CH_2OH}$$

What it proves : the carbonyl group in glucose is specifically an aldehyde. A ketone would not be oxidised by so mild a reagent — which is why this reaction distinguishes glucose from fructose in neutral medium.

(c) Reaction with concentrated $HNO_3$

Concentrated nitric acid is a strong oxidising agent. It oxidises both terminal groups — the $-CHO$ at C-1 and the $-CH_2OH$ at C-6 — into $-COOH$ groups :

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

The product is saccharic acid (glucaric acid), a dicarboxylic acid.

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

Summary of what each reagent reveals