Question

Glucose prefers an open-chain conformation in aqueous solution(T/F)

Answer 1

Answer:

False

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.