Which carbon atom has a different configuration of -OH group in two cyclic hemiacetal forms of glucose ?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → C - 1
The open-chain structure of glucose cannot explain some of its reactions. For example, it does not give Schiff’s test and does not react with \( \ce{NaHSO3} \), even though it has an aldehyde group.
These facts are explained by a cyclic structure. The –OH group on C-5 adds to the –CHO group at C-1, forming a six-membered ring (a cyclic hemiacetal). In this process, C-1 becomes a new chiral centre, so two cyclic forms are possible, α-D-glucose and β-D-glucose. They differ only in the orientation of the –OH group on C-1.
Why the other options are incorrect
Option (2) C - 5 — Incorrect. The –OH group on C-5 is used up in forming the ring, and its oxygen becomes the ring oxygen. C-5 has the same configuration in both forms, and it is this configuration that makes the sugar a D-sugar.
Option (3) C- 4 — Incorrect. The configuration at C-4 is the same in both cyclic forms; only C-1 changes.
Option (4) C - 3 — Incorrect. The configuration at C-3 is the same in both cyclic forms; only C-1 changes.
NCERT (2026-27 edition):
Glucose – …
