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Question #44431

Difficulty: 8/10 | 🟢 Free

The Question

One mole of Maltose is treated with an excess of periodic acid ( $\ce{HIO4}$ ) in an aqueous medium at 298 K. Assuming the reducing end of Maltose undergoes oxidation exclusively via its open-chain form, let $X$ be the total number of moles of $\ce{HIO4}$ consumed, $Y$ be the total number of moles of formic acid ( $\ce{HCOOH}$ ) produced, and $Z$ be the total number of moles of formaldehyde ( $\ce{HCHO}$ ) produced. What are the values of $X, Y,$ and $Z$ respectively?

Options

A.
$X=5,\ Y=3,\ Z=1$
B.
$X=7,\ Y=6,\ Z=1$
C.
$X=6,\ Y=5,\ Z=0$
D.
$X=4,\ Y=2,\ Z=1$

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

Glycol Periodic Acid Cleavage|Disaccharide Hydrolysis Products

Why This Question Is Tricky

Common Mistakes

This problem tests the regioselectivity of Periodic Acid Cleavage on complex carbohydrates.

The primary trap is the 'Hidden Variable' of the glycosidic linkage acting as an ether block.

Students often memorize that free aldohexoses consume 5 equivalents of HIO4 and blindly apply this to the reducing end of a disaccharide.

However, the 1->4 glycosidic bond at C4 lacks a free hydroxyl group, preventing cleavage at the C3-C4 and C4-C5 bonds.

Failing to account for this ether block leads to an overestimation of HIO4 consumed and HCOOH produced (Option B).

Additionally, students must correctly identify that the non-reducing acetal ring does not open, limiting its cleavage to the C2-C3-C4 contiguous diols.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Alcohols, Phenols, and Ethers - Class 12
Concept	Glycol Periodic Acid Cleavage
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free