Question

In D-glucose and D-fructose, which OH participates in hemiatal/hemiketal formation?

• A. C–5 in both X and Y
• B. C–4 in X and C–5 in Y
• C. C–5 in X and C–6 in Y
• D. C–6 in both X and Y

Hint:

Ring Formation in Sugars.

• Aldehyde (glucose) + OH → hemiacetal.
• Ketone (fructose) + OH → hemiketal.
• In both glucose and fructose, C–5 OH group reacts to form a cyclic structure.

Answer 1

The Correct Option is A

• D-glucose (X): It has an aldehyde group at C–1. The ring (pyranose form) forms by reaction between the aldehyde at C–1 and the OH at C–5, forming a hemiacetal.
• D-fructose (Y): It has a ketone group at C–2. The ring (furanose form) forms by reaction between the ketone at C–2 and the OH at C–5, forming a hemiketal.

Hence, in both cases, the C–5 hydroxyl group participates in ring formation.

Answer 2

Step 1: Understand the concept of hemiacetal and hemiketal formation
Hemiacetal and hemiketal formations are intramolecular reactions in carbohydrates where a hydroxyl group reacts with a carbonyl group (aldehyde or ketone) within the same molecule. This leads to the formation of a cyclic structure — a key feature in the ring forms of sugars like glucose and fructose.

Step 2: Structure of D-glucose
D-glucose is an aldohexose, meaning it has an aldehyde group at carbon-1 (C-1). In aqueous solution, the –OH group on carbon-5 (C-5) attacks the aldehyde carbon (C-1), leading to the formation of a six-membered ring called pyranose. This results in the formation of a hemiacetal.

Step 3: Structure of D-fructose
D-fructose is a ketohexose, having a ketone group at carbon-2 (C-2). In its cyclic form, the hydroxyl group on carbon-5 (C-5) attacks the carbonyl carbon (C-2), resulting in a five-membered ring known as furanose. This leads to the formation of a hemiketal.

Step 4: Conclusion
In both D-glucose (X) and D-fructose (Y), the hydroxyl group on carbon-5 (C–5) is responsible for initiating the ring formation through hemiacetal or hemiketal linkage. Thus, C–5 OH participates in both cases.