Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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.

Last updated date: 13th Jun 2024
Total views: 51.9k
Views today: 1.51k
Verified
51.9k+ views
Hint: 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.

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.