Question

Which carbon atoms of glucose, numbered from 1 to 6 forms a hemiacetal structure between -CHO and -OH groups.?

• A. C-2
• B. C-3
• C. C-4
• D. C-5

Hint:

The aldehyde carbon C-1 reacts with the hydroxyl on C-5 to form the six-membered pyranose ring.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Glucose has an aldehyde at C-1 and hydroxyl groups on C-2 to C-6. In solution it forms a ring because the aldehyde reacts with one of its own -OH groups.

Step 2: Detailed Explanation:
The -OH on C-5 attacks the -CHO group on C-1. This forms a hemiacetal (an -O- linking C-1 and C-5). The ring has five carbon atoms and one oxygen, so it is a six-membered ring called pyranose.

Step 3: Why the other options are wrong.
C-2, C-3 or C-4 hydroxyls would give rings of 3, 4 or 5 members, which are strained or not formed in glucose.

Final Answer:
The hemiacetal forms with the C-5 hydroxyl, option (D). \[ \boxed{\text{C-5}} \]