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Question #44998
The Question
One mole of a reducing disaccharide \ce{A} (), which yields only D-glucose upon complete acidic hydrolysis, is subjected to such hydrolysis. The resulting solution is neutralized and then treated with dilute \ce{Ba(OH)2} (aq) at 298 K until the Lobry de Bruyn-van Ekenstein equilibrium is established. The equilibrium mixture is found to contain exactly 30% D-fructose by mole fraction. The solution is then neutralized again and subjected to complete oxidative cleavage using excess \ce{HIO4} (aq). What is the total number of moles of formic acid (\ce{HCOOH}) and formaldehyde (\ce{HCHO}) produced, respectively?
Options
- A.
8.8 mol, 2.6 mol
- B.
10.0 mol, 2.0 mol
- C.
4.4 mol, 1.3 mol
- D.
9.4 mol, 2.6 mol
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Disaccharide Hydrolysis Products|Epimerization vs Anomerization|Alpha-Hydrogen Acidity
Why This Question Is Tricky
Common Mistakes
This question tests the student's ability to integrate disaccharide hydrolysis with the base-catalyzed Lobry de Bruyn-van Ekenstein transformation and subsequent periodate cleavage stoichiometry.
The primary 'Hidden Variable' trap is treating the hydrolysis product statically (assuming 2 moles of glucose remain) and ignoring the basic conditions that convert a portion of the aldohexose into a ketohexose.
Because aldohexoses and ketohexoses yield different molar ratios of formic acid and formaldehyde upon periodate cleavage (ketohexoses yield CO2 from the C2 carbonyl instead of HCOOH), failing to account for this isomerization leads to an incorrect stoichiometric calculation.
A secondary trap is forgetting that 1 mole of disaccharide yields 2 moles of monosaccharides.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Biomolecules - Class 12 |
| Concept | Disaccharide Hydrolysis Products |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |