Statement I: A solution of sucrose in water is dextrorotatory, but on hydrolysis in the presence of a little HCl it becomes laevorotatory.
Statement II: Sucrose on hydrolysis gives unequal amounts of glucose and fructose, which causes the change in the sign of rotation.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Statement I is correct but Statement II is incorrect.
Acid hydrolysis of sucrose gives one molecule of glucose and one of fructose:
\( \displaystyle \underset{\text{Sucrose}}{\ce{C12H22O11}} + \ce{H2O} \xrightarrow{\ce{H^+}} \underset{\text{D-(+)-Glucose}}{\ce{C6H12O6}} + \underset{\text{D-(-)-Fructose}}{\ce{C6H12O6}} \)
Statement I: A solution of sucrose in water is dextrorotatory, but on hydrolysis in the presence of a little HCl it becomes laevorotatory. — True. Sucrose is dextrorotatory, but the glucose–fructose mixture formed is laevorotatory; this change in the sign of rotation is why the product is called invert sugar.
Statement II: Sucrose on hydrolysis gives unequal amounts of glucose and fructose, which causes the change in the sign of rotation. — False. The mixture is equimolar. The sign changes because the laevorotation of fructose (−92.4°) is larger than the dextrorotation of glucose (+52.5°), so equal amounts of the two give a net laevorotation.
NCERT (2026-27 edition):
10.1.3 Disaccharides – “(i) Sucrose: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(–) fructose. … Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–) and the product is named as invert sugar.”