Open chain structure of D-glucose: $CH_2OH-(CHOH)_4-CHO$

(a) REACTION WITH HI (prolonged heating with excess HI / red P)

Step 1 - What HI does.
Hot HI is a powerful REDUCING agent. It replaces every $-OH$ group and the $-CHO$ group
by hydrogen atoms, reducing the whole molecule to the corresponding alkane.

Step 2 - Write the reaction.
$$CH_2OH-(CHOH)_4-CHO \xrightarrow[\Delta]{\text{HI}} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$$
$$\text{D-Glucose} \longrightarrow \textbf{n-Hexane}$$

Step 3 - Significance.
The formation of a STRAIGHT-CHAIN alkane proves that all SIX carbon atoms of glucose
are linked in an UNBRANCHED (straight) chain.

(b) REACTION WITH BROMINE WATER

Step 1 - What $Br_2$ water does.
Bromine water is a MILD OXIDISING agent. It oxidises only the aldehydic $-CHO$ group to
a carboxylic $-COOH$ group; the primary alcoholic group is untouched.

Step 2 - Write the reaction.
$$CH_2OH-(CHOH)_4-CHO \xrightarrow{Br_2/H_2O} CH_2OH-(CHOH)_4-COOH$$
$$\text{D-Glucose} \longrightarrow \textbf{D-Gluconic acid}$$

Step 3 - Significance.
This proves that the carbonyl group in glucose is an ALDEHYDE group $(-CHO)$ and not a
ketone group.

(c) REACTION WITH CONCENTRATED $HNO_3$

Step 1 - What conc. $HNO_3$ does.
Concentrated nitric acid is a STRONG OXIDISING agent. It oxidises BOTH the $-CHO$ group
AND the terminal primary alcoholic $-CH_2OH$ group to $-COOH$ groups.

Step 2 - Write the reaction.
$$CH_2OH-(CHOH)_4-CHO \xrightarrow{\text{conc. } HNO_3} COOH-(CHOH)_4-COOH$$
$$\text{D-Glucose} \longrightarrow \textbf{Saccharic acid (glucaric acid)}$$

Step 3 - Significance.
The formation of a DICARBOXYLIC acid proves the presence of a PRIMARY alcoholic
$(-CH_2OH)$ group in glucose.