Which of the following reactions show that glucose contains five \('OH'\) groups?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Acetylation of glucose with acetic anhydride gives the pentaacetate
The number of –OH groups in a compound is found by acetylation: every –OH group takes up one acetyl group (\( \ce{CH3CO} \)–). Glucose reacts with acetic anhydride to give glucose pentaacetate, so it has five –OH groups. Because the product is stable, the five –OH groups must be on different carbon atoms.
\( \ce{CHO(CHOH)4CH2OH + 5(CH3CO)2O -> CHO(CHOCOCH3)4CH2OCOCH3 + 5CH3COOH} \)
Why the other options are incorrect
Option (1) Glucose with HI on heating gives n-hexane — Incorrect. This shows that the six carbon atoms of glucose are joined in a straight chain; it says nothing about the number of –OH groups.
Option (3) Glucose with bromine water gives gluconic acid — Incorrect. A mild oxidising agent converts –CHO into –COOH, which shows the presence of an aldehyde group.
Option (4) Glucose with HNO3 gives saccharic acid — Incorrect. Both the –CHO and the terminal –\( \ce{CH2OH} \) are oxidised to –COOH, which shows that glucose has a primary alcohol group.
NCERT (2026-27 edition):
Structure of Glucose – …
