(a) Evidence for a straight chain

Experiment : prolonged heating with HI

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

Glucose is reduced completely to n-hexane.

Reasoning :

This drastic reduction removes every oxygen-containing group but leaves the carbon skeleton untouched.

The product obtained is the straight-chain hexane, not a branched isomer such as 2-methylpentane or 2,3-dimethylbutane. Since the skeleton is unchanged by the reaction, the six carbon atoms of glucose must already have been in an unbranched chain.

(b) Evidence for five $-OH$ groups on different carbons

Experiment : acetylation with acetic anhydride

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

Two separate conclusions follow :

How many $-OH$ groups —

Exactly five acetyl groups are taken up, and acetylation converts one $-OH$ into one $-OCOCH_3$. Glucose must therefore contain five hydroxyl groups.

Why they must be on different carbons —

Two $-OH$ groups on the same carbon would constitute a gem-diol, which is unstable and immediately loses water to give a carbonyl group :

$$>C(OH)_2 \rightleftharpoons \ >C=O + H_2O$$

Since all five hydroxyl groups survive and are acetylated, each must sit on a separate carbon atom.

Putting the evidence together

Combined with the evidence for an aldehyde group (oxidation by bromine water to gluconic acid), these observations give the open-chain structure :

$$CHO-(CHOH)_4-CH_2OH$$

Six carbons in a straight chain, five $-OH$ groups on five different carbons, and one $-CHO$ group at the end.