JEE Chemistry 2026 Question with Solution

Given: Structures of four disaccharides are given and we have to identify the non-reducing sugar.

Find: Which among maltose, lactose, sucrose, and cellobiose is non-reducing.

Concept: Reducing sugars have a free anomeric carbon atom with a free $-OH$ group on the hemiacetal carbon, so they can open to form an aldehyde or ketone group. Non-reducing sugars do not have a free anomeric carbon because both anomeric carbons are involved in glycosidic linkage.

Step 1: Analyze each disaccharide.

• Maltose: formed from two glucose units with $\alpha(1 \to 4)$ linkage. One anomeric carbon is free, so it is a reducing sugar.
• Lactose: formed from glucose and galactose with $\beta(1 \to 4)$ linkage. One anomeric carbon is free, so it is a reducing sugar.
• Sucrose: formed from glucose and fructose with $\alpha(1 \to 2)$ and $\beta(2 \to 1)$ linkage. Both anomeric carbons are involved in the glycosidic bond, so it is non-reducing.
• Cellobiose: formed from two glucose units with $\beta(1 \to 4)$ linkage. One anomeric carbon is free, so it is a reducing sugar.

Step 2: Conclusion. Sucrose is the only non-reducing sugar among the given disaccharides.

Therefore, the correct option is C.

Given: A set of disaccharides is provided.

Find: The non-reducing sugar.

A quick test is to check whether both anomeric carbons are tied up in the glycosidic bond. If both are involved, the sugar is non-reducing. If at least one anomeric carbon remains free, the sugar is reducing.

Among the listed sugars, only sucrose has both anomeric carbons involved in bond formation.

Therefore, the correct option is C.