When sugar 'X' is boiled with dilute H$$_2$$SO$$_4$$ in alcoholic solution, two isomers 'A' and 'B' are formed. 'A' on oxidation with HNO$$_3$$ yields saccharic acid where as 'B' is laevorotatory. The compound 'X' is

We need to identify sugar 'X' which on acid hydrolysis gives two isomers 'A' and 'B', where 'A' gives saccharic acid on oxidation with HNO$$_3$$ and 'B' is laevorotatory. When a monosaccharide is oxidized with HNO$$_3$$, both the aldehyde group and the terminal -CH$$_2$$OH group are oxidized to give a dicarboxylic acid (aldaric acid). Saccharic acid (glucaric acid) is the aldaric acid obtained from glucose, so 'A' is D-glucose.

'B' is laevorotatory (rotates plane-polarized light to the left). Fructose is laevorotatory (specific rotation = −92°), which is why it is also called levulose. Therefore, 'B' is D-fructose.

A disaccharide that on acid hydrolysis gives one molecule of glucose and one molecule of fructose is sucrose: $$\text{Sucrose} \xrightarrow{H_2SO_4} \text{Glucose} + \text{Fructose}$$. This process is called inversion of sugar. Note: Maltose gives two glucose units, lactose gives glucose + galactose, and starch gives only glucose units. None of these match.

Hence, the correct answer is Option B.