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Question #44969

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution of L-Lysine (pKa1=2.20pK_{a1} = 2.20 for −\ceCOOH-\ce{COOH}, pKa2=8.90pK_{a2} = 8.90 for α−\ceNH3+\alpha-\ce{NH3+}, pKa3=10.50pK_{a3} = 10.50 for ϵ−\ceNH3+\epsilon-\ce{NH3+}) is reacted with an excess of D-glucose at 298 K. A selective Maillard reaction occurs, forming a Schiff base at the less sterically hindered and intrinsically more basic amino group of L-Lysine. The pKapK_a of the conjugate acid of the newly formed imine nitrogen is 3.50. Assuming the pKapK_a values of the unreacted groups remain unchanged and no other side reactions occur, what is the isoelectric point (pI) of the isolated Schiff base conjugate?

Options

  • A.

    6.20

  • B.

    7.00

  • C.

    5.55

  • D.

    9.70

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Amino Acid Zwitterion pH Sensitivity|Reducing Sugar Criteria|Osazone Formation Logic

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of biomolecular reactivity (Maillard reaction) and acid-base equilibria.

The primary trap is regioselectivity: students must deduce that the ϵ\epsilon-amino group reacts due to lower steric hindrance and higher basicity.

The secondary trap is recalculating the isoelectric point for the modified polyprotic system.

Students who mechanically calculate the pI of basic amino acids as the average of the two highest pKapK_a values will incorrectly answer 9.70, failing to account for the chemical transformation that consumes the side-chain basicity and fundamentally shifts the zwitterionic boundaries.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptAmino Acid Zwitterion pH Sensitivity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree