Question

When sucrose is hydrolysed the optical rotation values are measured using a polarimeter and are given in the following table:

S. No.	Time (hours)	Specific Rotation
1	0	+ 66.5°
2	∞	- 39.9°

1. Account for the two specific rotation values.
2. What is the specific name given to sucrose based on the above observation?
3. One of the products formed during the hydrolysis of sucrose is a glucose, that reacts with hydroxylamine to give compound A. Identify compound A.

Answer 1

1. The reactant Sucrose is dextrorotatory. On hydrolysis it give glucose dextrorotatory and fructose which is leavoroatatory. The specific rotation of fructose is higher than glucose. 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.
2. Invert sugar, The hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (-) and the product is named as invert sugar.
3. Glucose
   \[\begin{array}{cc}
   \ce{CH = N - OH}\\
   |\phantom{.............}\\
   \ce{(CHOH)4}\phantom{.......}\\
   |\phantom{.............}\\
   \ce{CH2OH}\phantom{.......}\end{array}\]