Question

Maltose on hydrolysis gives two monosaccharide units. The incorrect statement about the monosaccharides formed is

• A. Both are $\alpha$-D-glucose units only
• B. One is $\alpha$-D-glucose and second one is $\beta$-D-fructose
• C. Both are reducing sugars
• D. In maltose, they are joined through 1,4-glycosidic linkage

Hint:

Memorize the composition of common disaccharides: - Sucrose (table sugar) $\rightarrow$ Glucose + Fructose - Lactose (milk sugar) $\rightarrow$ Glucose + Galactose - Maltose (malt sugar) $\rightarrow$ Glucose + Glucose Also, remember that all monosaccharides and most disaccharides (except sucrose) are reducing sugars.

Answer 1

The Correct Option is B

Let's analyze the properties of maltose and its hydrolysis products.
Maltose is a disaccharide, also known as malt sugar.
The hydrolysis of maltose yields two molecules of glucose. Specifically, it yields two molecules of $\alpha$-D-glucose.
Maltose $\xrightarrow{\text{Hydrolysis}}$ $\alpha$-D-glucose + $\alpha$-D-glucose
Now let's evaluate the given statements based on this information. The question asks for the incorrect statement about the products (which are two glucose units).
(A) Both are $\alpha$-D-glucose units only: This is the correct product of maltose hydrolysis. This statement is correct.
(B) One is $\alpha$-D-glucose and second one is $\beta$-D-fructose: This describes the hydrolysis products of sucrose, not maltose. Sucrose hydrolysis gives one molecule of glucose and one molecule of fructose. This statement is incorrect.
(C) Both are reducing sugars: The product is glucose. Glucose is a monosaccharide with a free hemiacetal group, which can open to form an aldehyde group. Aldehyde groups can be oxidized, making glucose a reducing sugar. Since both units are glucose, this statement is correct.
(D) In maltose, they are joined through 1,4-glycosidic linkage: This statement is about the structure of maltose itself, not the products. In maltose, the two $\alpha$-D-glucose units are linked by a glycosidic bond between the C1 of the first glucose unit and the C4 of the second glucose unit. This is called an $\alpha$-1,4-glycosidic linkage. This statement is correct.
The question asks for the incorrect statement. Statement (B) is factually incorrect for the hydrolysis of maltose.