Glucose on reaction with bromine water gives.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → gluconic acid
Bromine water is a mild oxidising agent. It oxidises only the aldehyde group of glucose to a carboxylic acid; the –\( \ce{CH2OH} \) and –CHOH groups are not affected. The product is gluconic acid:
\( \ce{CHO-(CHOH)4-CH2OH} \) \( \xrightarrow{\text{Br}_2\ \text{water}} \) \( \ce{COOH-(CHOH)4-CH2OH} \)
This also shows that the carbonyl group of glucose is an aldehydic group.
Why the other options are incorrect
Option (2) — Incorrect. COOH at both ends is saccharic acid, formed with a strong oxidising agent, nitric acid, which also oxidises the –\( \ce{CH2OH} \) group.
Option (3) — Incorrect. It keeps the –CHO and oxidises the –\( \ce{CH2OH} \) end; this is not formed, because bromine water oxidises only the –CHO group.
Option (4) — Incorrect. A dialdehyde; it would need oxidation of –\( \ce{CH2OH} \) to –CHO with the aldehyde group untouched, which bromine water does not do.
NCERT (2026-27 edition):
Glucose – “… 4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.”