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Question #44819

Difficulty: 8/10 | 🟢 Free

The Question

The mutarotation of D-glucose in aqueous solution at 298 K proceeds via an acyclic aldehydic intermediate (II). The mechanism is represented as:

α-D-glucopyranose\ce<=>[k1][k−1]I\ce<=>[k2][k−2]β-D-glucopyranose\alpha\text{-D-glucopyranose} \ce{<=>[k_1][k_{-1}]} I \ce{<=>[k_2][k_{-2}]} \beta\text{-D-glucopyranose}

The observed mutarotation rate constant is defined as kmut=kf,eff+kb,effk_{mut} = k_{f,eff} + k_{b,eff}, where kf,effk_{f,eff} and kb,effk_{b,eff} are the effective forward and backward rate constants for the direct α⇌β\alpha \rightleftharpoons \beta interconversion.

A student applies the Steady-State Approximation (SSA) to the intermediate II to derive kmutSSAk_{mut}^{SSA}. However, under specific acid-catalyzed conditions, the ring-opening of the α\alpha-anomer is rapid and reversible compared to the ring-closure to the β\beta-anomer (k−1≫k2k_{-1} \gg k_2), making the Pre-Equilibrium (PE) approximation more appropriate for finding the forward rate.

Which of the following correctly represents the expression for kmutSSAk_{mut}^{SSA} and the ratio of the effective forward rate constants (kf,effSSAkf,effPE)\left( \frac{k_{f,eff}^{SSA}}{k_{f,eff}^{PE}} \right)?

Options

  • A.

    kmutSSA=k1k2+k−1k−2k−1+k2k_{mut}^{SSA} = \frac{k_1 k_2 + k_{-1} k_{-2}}{k_{-1} + k_2} ; Ratio = k−1k−1+k2\frac{k_{-1}}{k_{-1} + k_2}

  • B.

    kmutSSA=k1k2−k−1k−2k−1+k2k_{mut}^{SSA} = \frac{k_1 k_2 - k_{-1} k_{-2}}{k_{-1} + k_2} ; Ratio = k2k−1+k2\frac{k_2}{k_{-1} + k_2}

  • C.

    kmutSSA=k1k−1+k2k−2k−1+k2k_{mut}^{SSA} = \frac{k_1 k_{-1} + k_2 k_{-2}}{k_{-1} + k_2} ; Ratio = k−1k−1+k2\frac{k_{-1}}{k_{-1} + k_2}

  • D.

    kmutSSA=k1k2+k−1k−2k−1+k2k_{mut}^{SSA} = \frac{k_1 k_2 + k_{-1} k_{-2}}{k_{-1} + k_2} ; Ratio = k−1+k2k−1\frac{k_{-1} + k_2}{k_{-1}}

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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

Steady State Approximation|D-Glucose Anomeric Equilibrium|Chemical Kinetics

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between the Steady-State Approximation (SSA) and the Pre-Equilibrium (PE) approximation in complex reaction mechanisms.

Students often blindly apply SSA to any low-concentration intermediate.

However, in mutarotation under specific conditions, the acyclic form is in rapid equilibrium with the anomers.

The trap lies in failing to recognize the exact kinetic form of reversible first-order reactions (kobs=kf+kbk_{obs} = k_f + k_b) and incorrectly assuming SSA and PE yield identical rate laws without evaluating the limit where k−1≫k2k_{-1} \gg k_2.

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Question Information

SubjectChemistry
ChapterChemical Kinetics - Class 12
ConceptSteady State Approximation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree