A disaccharide X cannot be oxidized by bromine water. The acid hydrolysis of X leads to a laevorotatory solution. The disaccharide X is:

Correct Answer:

Step-by-Step Explanation:

1. Understanding the given properties of disaccharide X:

• Inability to oxidize with bromine water: Bromine water (${\text{Br}}_2/{\text{H}}_2\text{O}$) is a mild oxidizing agent that oxidizes reducing sugars containing free hemiacetal or hemiketal groups (free anomeric carbon). Since disaccharide X cannot be oxidized by bromine water, it is a non-reducing sugar. This means both anomeric carbons of the constituent monosaccharide units are involved in the glycosidic linkage (i.e., C1 of α-D-glucopyranose and C2 of β-D-fructofuranose).

• Inversion of optical rotation upon acid hydrolysis: Sucrose is dextrorotatory, but upon acid hydrolysis, it breaks down into an equimolar mixture of D-(+)-glucose ([α]_D = +52.7°) and D-(-)-fructose ([α]_D = -92.4°). The net specific rotation of the resulting mixture is negative (-19.85°), making the hydrolyzed solution laevorotatory. This phenomenon is known as the inversion of cane sugar.

2. Identification of Disaccharide X:

Based on these characteristics, disaccharide X is sucrose.

Sucrose is composed of an α-D-glucopyranose unit and a β-D-fructofuranose unit linked via an $\text{α-1, β-2-glycosidic linkage}$.

3. Structural Analysis of the Correct Option:

Looking at the correct image structure:

• The ring on the left is a six-membered pyranose ring of α-D-glucose showing the standard configuration with the glycosidic oxygen attached to C-1.

• The ring on the right is a five-membered furanose ring of β-D-fructose with the glycosidic bond formed at C-2.

• Since both anomeric carbons (C-1 of glucose and C-2 of fructose) are involved in the linkage through an oxygen atom, there is no free hemiacetal OH group remaining, confirming that it represents sucrose.