Write the reactions of D-Glucose with the following:
(a) HI
(b) \(Br_2\) water
(c) Conc. \(HNO_3\)
(a) HI
(b) \(Br_2\) water
(c) Conc. \(HNO_3\)
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Solution and Explanation
Glucose has an open-chain aldohexose structure with one aldehyde group, five carbons bearing hydroxyl groups (one primary, four secondary) and a straight carbon chain. Different reagents attack different parts: HI reduces the whole chain, mild oxidising agents oxidise only the aldehyde, and strong oxidising agents oxidise both ends.
Step 1 (a) with HI:
On prolonged heating with HI, glucose is completely reduced; all C\(=\)O and C\(-\)OH groups are removed and a straight chain alkane is formed. Glucose gives n-hexane.
\(C_6H_{12}O_6 \xrightarrow{HI,\ \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3\ (n\text{-hexane})\). This shows glucose has a straight chain of six carbons.
Step 2 (b) with \(Br_2\) water:
Bromine water is a mild oxidising agent and oxidises only the aldehyde (\(-CHO\)) group to a carboxylic acid (\(-COOH\)), giving gluconic acid (a monocarboxylic acid).
\(CHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2/H_2O} COOH-(CHOH)_4-CH_2OH\ (gluconic\ acid)\). This confirms the presence of an aldehyde group.
Step 3 (c) with conc. \(HNO_3\):
Concentrated nitric acid is a strong oxidising agent and oxidises both the terminal aldehyde group and the terminal primary alcohol group to \(-COOH\), giving a dicarboxylic acid, saccharic acid (glucaric acid).
\(CHO-(CHOH)_4-CH_2OH \xrightarrow{conc.\ HNO_3} COOH-(CHOH)_4-COOH\ (saccharic\ acid)\).
Answer: (a) n-Hexane; (b) Gluconic acid; (c) Saccharic acid (glucaric acid).
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