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Difficulty: Very hardSynthetic Polymers, Carbohydrates, Amino Acids, and Proteins

An unknown disaccharide XX with the molecular formula C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11} does not reduce Fehling's solution. Upon acid-catalyzed hydrolysis, XX yields an equimolar mixture of two isomeric hexoses, YY and ZZ. Compound YY rapidly produces a deep red color when heated with Seliwanoff's reagent, whereas compound ZZ is oxidized by bromine water to form a monocarboxylic acid. Which of the following correctly identifies disaccharide XX and explains why it fails to react with Fehling's solution prior to hydrolysis?

  1. Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.Answer
  2. B
    Maltose; because the glycosidic bond links carbon-1 of one glucose unit to carbon-4 of another, masking both reducing ends.
  3. C
    Lactose; because both monosaccharide units exist strictly in locked cyclic ring forms that cannot open to form aldehydes.
  4. D
    Sucrose; because fructose contains a ketone group which must first be hydrolyzed into an aldehyde group before Fehling's reduction can occur.

Answer

Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.
Seliwanoff's test specifically identifies ketohexoses such as fructose through rapid dehydration to hydroxymethylfurfural and reaction with resorcinol. Bromine water selectively oxidizes aldoses like glucose to aldonic acids without oxidizing ketoses. A disaccharide yielding glucose and fructose upon hydrolysis is sucrose. Sucrose is a non-reducing sugar because its glycosidic bond connects C-1 of glucose and C-2 of fructose, locking both anomeric carbon atoms and preventing ring opening to form reactive carbonyl groups.

Step-by-Step Solution

1
Analyze the chemical test results of the hydrolysis products YY and ZZ.
Compound YY gives a positive Seliwanoff's test (rapid red color), which is characteristic of a ketose (fructose). Compound ZZ is oxidized by bromine water (a mild oxidizing agent), which selectively oxidizes aldoses (glucose) to aldonic acids.
Seliwanoff's reagent differentiates ketoses from aldoses, while bromine water differentiates aldoses from ketoses.
2
Identify the disaccharide XX based on its hydrolysis products.
Disaccharide XX hydrolyzes into glucose and fructose, identifying XX as sucrose.
Sucrose (C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11}) is composed of one glucose unit and one fructose unit.
3
Determine the structural basis for the non-reducing nature of sucrose.
In sucrose, the glycosidic bond connects C-1\text{C-1} (α\alpha-anomeric carbon of glucose) to C-2\text{C-2} (β\beta-anomeric carbon of fructose).
Because both potential reducing centers (anomeric carbons) are locked in the glycosidic bond, sucrose lacks a free hemiacetal or hemiketal group, rendering it a non-reducing sugar that does not reduce Fehling's reagent.

Key Concept

Glycosidic bond linkages and reducing vs. non-reducing carbohydrate behavior
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Topics

Synthetic Polymers, Carbohydrates, Amino Acids, and Proteins

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