Question

How do you explain the following? \[ (a)\;\text{Presence of a straight chain in glucose} \] \[ (b)\;\text{Presence of five }-OH\text{ groups in glucose which are attached to different carbon atoms} \]

Hint:

Glucose gives \(n\)-hexane with HI, proving straight chain; glucose pentaacetate proves five \(-OH\) groups.

Answer 1

Concept:
Structural features of glucose are confirmed by characteristic chemical reactions. A straight carbon chain is proved by reduction of glucose to \(n\)-hexane. The presence of hydroxyl groups is proved by acetylation.

Step 1: Presence of straight chain in glucose.
When glucose is heated with hydrogen iodide for a long time, it gets completely reduced to \(n\)-hexane. \[ \text{Glucose} \xrightarrow[\Delta]{HI} n\text{-hexane} \] Since \(n\)-hexane is a straight-chain hydrocarbon, this proves that the six carbon atoms in glucose are arranged in a straight chain.

Step 2: Presence of five hydroxyl groups.
Glucose reacts with acetic anhydride to form glucose pentaacetate. \[ \text{Glucose}+5(CH_3CO)_2O \rightarrow \text{Glucose pentaacetate} \] Formation of pentaacetate proves that glucose contains five hydroxyl groups.

Step 3: Hydroxyl groups are attached to different carbon atoms.
Glucose gives a stable pentaacetate. This indicates that the five hydroxyl groups are present on five different carbon atoms. If two \(-OH\) groups were attached to the same carbon atom, the structure would be unstable. Hence: \[ \boxed{\text{Straight chain is proved by reduction of glucose to }n\text{-hexane.}} \] \[ \boxed{\text{Five }-OH\text{ groups are proved by formation of glucose pentaacetate.}} \]