When sugar 'X' is boiled with dilute `H_2SO_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

Options

A. Maltose
B. Sucrose
C. Lactose
D. Starch

Explanation

To solve the problem, we need to identify the sugar 'X' based on the provided information about the isomers 'A' and 'B' formed when boiled with dilute `H2SO4` in an alcoholic solution.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
When sugar 'X' is boiled with dilute `H2SO4`, it undergoes hydrolysis. This means that the sugar is broken down into simpler sugars (monosaccharides).

2. **Identifying the Sugar**:
The question mentions that two isomers 'A' and 'B' are formed. The most common disaccharide that fits this description is sucrose (C12H22O11). When sucrose is hydrolyzed, it produces glucose and fructose.

3. **Identifying the Isomers**:
- **Isomer A**: This is likely glucose, which is a dextrorotatory sugar (it rotates plane-polarized light to the right).
- **Isomer B**: This is likely fructose, which is a levorotatory sugar (it rotates plane-polarized light to the left).

4. **Oxidation of Isomer A**:
The problem states that isomer 'A' (glucose) on oxidation with `HNO3` yields saccharic acid. This is consistent with glucose, as it can be oxidized to form saccharic acid.

5. **Conclusion**:
Since we have identified that the sugar 'X' is sucrose, which hydrolyzes to form glucose (isomer A) and fructose (isomer B), we conclude that the compound 'X' is sucrose.

### Final Answer:
The compound 'X' is sucrose (C12H22O11).