JEE Challenger
More from Biomolecules
ChemMedAdv 2016 • Paper 2

Properties and Specific Rotation of Invert Sugar

Regarding 'invert sugar', select the correct statement(s): (Given: specific rotations of (+)-sucrose, (+)-maltose, L-(-)-glucose and L-(+)-fructose in aqueous solution are +66∘+66^\circ, +140∘+140^\circ, −52∘-52^\circ and +92∘+92^\circ, respectively)

Options

A

'invert sugar' is prepared by acid catalyzed hydrolysis of maltose

B

'invert sugar' is an equimolar mixture of D-(+)-glucose and D-(-)-fructose

Correct
C

specific rotation of 'invert sugar' is −20∘-20^\circ

Correct
D

on reaction with Br2\text{Br}_2 water, 'invert sugar' forms saccharic acid as one of the products

Topics & Concepts

BiomoleculesCarbohydrates

Step-by-Step Solution

To determine the correct statement(s) regarding 'invert sugar', let us analyze each option individually:

1. Hydrolysis and Composition of Invert Sugar (Options A and B):

  • Invert sugar is obtained by the acid-catalyzed or enzymatic hydrolysis of sucrose ((+)(+)-sucrose), which is dextrorotatory: Sucrose (Dextrorotatory)+H2O→H+D-(+)-glucose+D-(-)-fructose\text{Sucrose (Dextrorotatory)} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{D-(+)-glucose} + \text{D-(-)-fructose}
  • On the other hand, the hydrolysis of maltose yields two molecules of D−(+)\text{D}-(+)-glucose. Therefore, Option A is incorrect.
  • Since sucrose yields an equimolar mixture of D−(+)\text{D}-(+)-glucose and D−(−)\text{D}-(-)-fructose, Option B is correct.

2. Specific Rotation of Invert Sugar (Option C):

Enantiomers have equal magnitudes of specific rotation but opposite signs:

  • Given specific rotation of L−(−)\text{L}-(-)-glucose =−52∘= -52^\circ, the specific rotation of its enantiomer, D−(+)\text{D}-(+)-glucose, is: [α]D-glucose=+52∘[\alpha]_{\text{D-glucose}} = +52^\circ
  • Given specific rotation of L−(+)\text{L}-(+)-fructose =+92∘= +92^\circ, the specific rotation of its enantiomer, D−(−)\text{D}-(-)-fructose, is: [α]D-fructose=−92∘[\alpha]_{\text{D-fructose}} = -92^\circ

Since invert sugar is an equimolar mixture of D−(+)\text{D}-(+)-glucose and D−(−)\text{D}-(-)-fructose (both having identical molar mass M=180.16 g/molM = 180.16 \text{ g/mol}), the mass concentrations of both monosaccharides in the solution are equal. Therefore, the specific rotation of the mixture is the arithmetic mean of the specific rotations of its constituents: [α]invert sugar=[α]D-glucose+[α]D-fructose2[\alpha]_{\text{invert sugar}} = \frac{[\alpha]_{\text{D-glucose}} + [\alpha]_{\text{D-fructose}}}{2} [α]invert sugar=(+52∘)+(−92∘)2=−40∘2=−20∘[\alpha]_{\text{invert sugar}} = \frac{(+52^\circ) + (-92^\circ)}{2} = \frac{-40^\circ}{2} = -20^\circ

Hence, the specific rotation of invert sugar is −20∘-20^\circ. Therefore, Option C is correct.

3. Reaction with Bromine Water (Br2/H2O\text{Br}_2/\text{H}_2\text{O}) (Option D):

  • Br2\text{Br}_2 water is a mild oxidizing agent that selectively oxidizes the aldehyde group (−CHO-\text{CHO}) of aldoses (D\text{D}-glucose) to a carboxylic acid group (−COOH-\text{COOH}), forming gluconic acid: D-glucose→Br2/H2OGluconic acid\text{D-glucose} \xrightarrow{\text{Br}_2 / \text{H}_2\text{O}} \text{Gluconic acid}
  • It does not oxidize the primary alcohol group (−CH2OH-\text{CH}_2\text{OH}) at C-6 to forming a dicarboxylic acid (saccharic acid). Saccharic acid is formed only upon strong oxidation with concentrated HNO3\text{HNO}_3.
  • Ketoses like D\text{D}-fructose do not react with Br2\text{Br}_2 water.

Therefore, 'invert sugar' forms gluconic acid, not saccharic acid, on reaction with Br2\text{Br}_2 water. Hence, Option D is incorrect.

Conclusion:

The correct statements are B and C.

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