Question

The correct statements about the following sugars X and Y are:

(A) X is a reducing sugar, Y is a non-reducing sugar.
(B) X is a non-reducing sugar, Y is a reducing sugar.
(C) The glycosidic linkages in X and Y are $\alpha $ and $\beta $ respectively.
(D) The glycosidic linkages in X and Y are $\beta $ and $\alpha $ respectively.

Answer 1

Hint: Think about the topic of carbohydrates. Two disaccharides are given in the question and four options are given. We need to determine whether they are reducing sugars or not. Also, we need to find out the type of glycosidic linkages present in both the sugars and then choose the correct option.

Complete answer:
- Let’s start by taking a look at the given compounds, X and Y.
- A carbohydrate is said to be a reducing sugar if the carbon at first position contains a free hydroxyl group (-OH) attached to it. If at C-1 position, a free hydroxyl group is absent then the sugar is a non-reducing sugar.
- To find out the type of linkage present in the disaccharide, we need to check the monosaccharide of which C-1 position is forming the linkage. If the monosaccharide is $\alpha $ then the linkage is said to be $\alpha $-linkage and if the monosaccharide is $\beta $ then the linkage is said to be $\beta $-linkage.
- Now, let’s see the compounds X and Y one by one.

- In X, there are no free hydroxyl groups present and so, this is a non-reducing sugar. The monosaccharide forming the linkage is glucose which is present as $\alpha $-glucose. Therefore, an $\alpha $-linkage is formed.

- In Y, there is one free hydroxyl group present , so this is a reducing sugar. The monosaccharide forming the linkage is glucose which is present as $\beta $-glucose. Therefore, an $\beta $-linkage is formed.
- Therefore, X is a non-reducing sugar, Y is a reducing sugar and the glycosidic linkages in X and Y are $\alpha $ and $\beta $ respectively.

Therefore, the correct statements are option (B) and (C).

Note:
Remember for a sugar to be classified as a reducing sugar, at least one free hydroxyl group must be present at C-1 position. If no free hydroxyl groups are present then the sugar is a non-reducing sugar. If the C-1 atom of the monosaccharide forming the linkage is an $\alpha $-anomer then the linkage will be $\alpha $-linkage and if the C-1 atom of the monosaccharide forming the linkage is a $\beta $-anomer then the linkage will be $\beta $-linkage.