On oxidation with nitric acid, glucose as well as gluconic acid yield the same dicarboxylic acid, saccharic acid. What does this indicate about the structure of glucose?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A primary alcoholic (–OH) group is present
Gluconic acid already has a –COOH group at C1. Nitric acid turns it into a dicarboxylic acid, so a second –COOH must come from the other end of the chain, C6. Only a primary alcoholic group, –CH2OH, can be oxidised to –COOH without losing a carbon atom:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH} \xleftarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-CH2OH} \)
Glucose → Saccharic acid ← Gluconic acid
Why the other options are incorrect
Option (1) The carbonyl group is present as an aldehydic group — Incorrect. The aldehydic nature of the carbonyl group is shown by oxidation with a mild oxidising agent, bromine water, to gluconic acid.
Option (2) A carbonyl group (>C=O) is present — Incorrect. The carbonyl group is shown by the formation of an oxime with hydroxylamine and of a cyanohydrin with HCN; nitric acid acts on the other end of the chain.
Option (4) Five –OH groups are attached to different carbon atoms — Incorrect. Five –OH groups on different carbons are shown by the stable pentaacetate formed with acetic anhydride.
NCERT (2026-27 edition):
10.1.2.1 Glucose – “6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.”