Step 1: Proposed open-chain structure of glucose.
Glucose ($C_6H_{12}O_6$) is proposed to have an unbranched 6-carbon chain with one aldehyde at C1 and five hydroxyl groups on different carbon atoms (C2 to C6). Both the straight chain and the five $-OH$ groups need chemical proof.
Step 2: Proof of straight chain in glucose.
When glucose is reduced vigorously with red phosphorus and hydroiodic acid (HI/P), all oxygen functions are removed. The product is $n$-hexane: \[ C_6H_{12}O_6 \xrightarrow{HI/P} CH_3CH_2CH_2CH_2CH_2CH_3 \;(n\text{-hexane}) \]
Step 3: Significance of $n$-hexane formation.
$n$-Hexane is the straight-chain (unbranched) isomer of hexane. Its formation conclusively proves that all 6 carbons in glucose are in a continuous straight chain. If there were any branching, a branched hexane isomer would have been produced instead.
Step 4: Proof of five $-OH$ groups on different carbon atoms.
Glucose reacts with acetic anhydride ($(CH_3CO)_2O$) in the presence of pyridine. It forms glucose pentaacetate (five acetyl groups incorporated): \[ \text{Glucose} + 5(CH_3CO)_2O \xrightarrow{\text{pyridine}} \text{Glucose pentaacetate} + 5CH_3COOH \]
Step 5: Significance of pentaacetate formation.
Five acetyl groups are incorporated, one per $-OH$ group. This proves that glucose has exactly five hydroxyl ($-OH$) groups. Since the aldehyde group does not react with acetic anhydride under these conditions, and the 6-carbon chain can have at most one OH per carbon (alongside the CHO at C1), the five $-OH$ groups must be on five different carbon atoms (C2 through C6).
Step 6: Summarise both proofs clearly.
(a) Straight chain: Reduction of glucose with HI/P gives $n$-hexane, proving an unbranched 6-carbon chain. (b) Five $-OH$ on different carbons: Glucose forms a pentaacetate with $(CH_3CO)_2O$, proving exactly five hydroxyl groups each on a separate carbon. \[ \boxed{\text{(a) }n\text{-hexane from HI/P;\; (b) pentaacetate from }(CH_3CO)_2O} \]