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2 June, 04:16

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

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  1. 2 June, 04:40
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    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.
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