Concept:
Optical rotation depends on the net effect of all optically active components present in a solution.
On hydrolysis, sucrose breaks into two monosaccharides—glucose and fructose—each having its own specific rotation.
Step 1: Nature of sucrose.
Sucrose is dextrorotatory, having a specific rotation of approximately:
\[
[\alpha]_D \approx +66.5^\circ
\]
Hence, Statement–I (first part) is correct.
Step 2: Effect of hydrolysis of sucrose.
On hydrolysis:
\[
\text{Sucrose} \xrightarrow{\text{hydrolysis}} \text{Glucose} + \text{Fructose}
\]
Specific rotations:
\[
\text{Glucose: } +52.5^\circ \quad (\text{dextrorotatory})
\]
\[
\text{Fructose: } -92.4^\circ \quad (\text{laevorotatory})
\]
The magnitude of laevorotation of fructose is greater than the dextrorotation of glucose.
Therefore, the resulting mixture (called invert sugar) becomes laevorotatory.
Thus, Statement–I (second part) is also correct.
Step 3: Evaluate Statement–II.
Statement–II claims:
\[
\text{Laevorotation of glucose}>\text{Dextrorotation of fructose}
\]
This is incorrect because:
Glucose is dextrorotatory, not laevorotatory
Fructose is laevorotatory, not dextrorotatory
Magnitude of rotation of fructose (\(\sim 92^\circ\)) is greater than that of glucose (\(\sim 52^\circ\))
Hence, Statement–II is incorrect.
Step 4: Final conclusion.
Statement–I is correct
Statement–II is incorrect
Therefore, the correct option is (3).