Answer:
False
Step-by-step explanation:
Glucose is a monosachharide carbohydrate, with the molecular formula C₆H₁₂O₆.
Glucose molecule can exist in two forms-
1. Open chain form
2. cyclic form
The open chain form of the glucose is an unbranched 6 carbon atom chain. The carbon 1 of the molecule is an aldehyde group and the rest of the five carbon atoms have one hydroxyl group each.
The cyclic form of the glucose can be-
a. Pyranose: The pyranose form is a 6-membered cyclic ring, which consists of 5 carbon atoms and 1 oxygen atom in the ring.
b. Furanose: The furanose form is a 5- membered cyclic ring, which consists of 4 carbon atoms and 1 oxygen atom in the ring.
In an aqueous solution, 99% glucose molecule exists in the cyclic pyranose form as it is energetically more stable.
Therefore, in aqueous solution, the glucose molecule does not prefer the open-chain conformation.
Therefore, the statement is false.