Question

For ‘invert sugar’, the correct statement(s) is (are).
(Given: specific rotations of (+)-sucrose, (+)-maltose, L-(-)-glucose and L-(+)-fructose in aqueous solutions are $+ 66^\circ$, $+ 140^\circ$, $+ 52^\circ$ and $- 92^\circ$ respectively)
This question has multiple correct options.
A. ‘Invert sugar’ is prepared by acid catalysed hydrolysis of maltose.
B. ‘Invert sugar’ is an equimolar mixture of D-(+)-glucose and D-(-)fructose.
C. Specific rotation of the ‘invert sugar’ is $- 20^\circ$
D. On reaction with water, ‘invert sugar’ forms saccharic acid as one of the products.

Answer 1

Sucrose in dextrorotatory in nature ($+ 66^\circ$) and on hydrolysis, a change occurs in the rotation and the products formed are equimolar mixtures of dextro and levo forms of its substituents.

Complete step by step answer:
On hydrolysis or on reaction with an enzyme known as sucrase, sucrose gives its constituent sugars i.e. glucose and fructose. The formula of sucrose is ${{\text{C}}_{12}}{{\text{H}}_{22}}{{\text{O}}_{11}}$ and its specific rotation is given as ${\text{ + 66}}^\circ$. Naturally, sucrose occurs in dextro form only.
On hydrolysis, sucrose gives an equimolar mixture of D-(+)-Glucose and D(-)-Fructose which have specific rotations as $+ 52^\circ$ and $- 92^\circ$ respectively.
Hydrolysis reaction is given as –
${{\text{C}}_{12}}{{\text{H}}_{22}}{{\text{O}}_{11}}{\text{ }}$ ${\text{ + }}$ ${{\text{H}}_2}{\text{O}}$ $\xrightarrow{{{{\text{H}}^ + }}}$ ${{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}$ ${\text{ + }}$ ${{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}$
D-(+)-Glucose ($+ 52^\circ$) D-(-)Fructose ($- 92^\circ$)

Net specific rotation is given as-
(Average is taken considering that both monomers are present 1 mole each)
\alpha {\text{ = }}$$$$\dfrac{{ + 52^\circ {\text{ - 92}}^\circ }}{2}$$$${\text{ = }}\dfrac{{ - 40^\circ }}{2}$$$${\text{ = - 20}}^\circ
So, statement B and C are correct.

Note:
Sucrose is a disaccharide made up of monosaccharide glucose and fructose and maltose is a disaccharide made up of two glucose units. So, option A is incorrect.
Bromine water is a weak oxidising agent. So, it will oxidise the aldehyde group to carboxylic acid and itself get reduced. In order to get sachharic acid we need a strong oxidising agent like concentrated sulphuric acid . So, option D is also incorrect.