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Question #44858

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The Question

Consider the reaction of 2-iodo-2-methyl-1,1-diphenylpropan-1-ol with 1.0 equivalent of anhydrous \ce{FeCl3} in \ce{CH2Cl2} at 298 K. The reaction proceeds rapidly to yield a major organic product \textbf{Y} and a transition metal complex \textbf{Z}. Given the standard reduction potentials ( $E^\circ_{\ce{Fe^{3+}/Fe^{2+}}} = +0.77$ V, $E^\circ_{\ce{I2/I^-}} = +0.54$ V), which of the following correctly identifies \textbf{Y}, the final oxidation state and d-electron configuration of the metal in \textbf{Z}, and the mechanistic rationale?

Options

A.
2-methyl-1,2-diphenylpropan-1-one; \ce{Fe^{2+}} ( $d^6$ ); Reduced Lewis acidity allows OH lone pair to drive 1,2-phenyl shift
B.
3,3-diphenylbutan-2-one; \ce{Fe^{3+}} ( $d^5$ ); Strong OH coordination forces OH departure and 1,2-methyl shift
C.
2-methyl-1,2-diphenylpropan-1-one; \ce{Fe^{3+}} ( $d^5$ ); Redox-inert Lewis acid abstracts iodide, maintaining $d^5$ state
D.
3,3-dimethyl-2,2-diphenyloxirane; \ce{Fe^{2+}} ( $d^6$ ); Redox generates base for intramolecular \ce{S_N2} displacement

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Variable Oxidation State Stability|Oxidation State Traps|Pinacol-Pinacolone Rearrangement|SN1 Carbocation Rearrangement

Why This Question Is Tricky

Common Mistakes

This question tests the concept of 'Redox-Switchable Lewis Acidity' and the integration of inorganic redox potentials with organic reaction mechanisms.

The trap is assuming \ce{FeCl3} acts purely as a redox-inert Lewis acid (like \ce{AlCl3}).

If \ce{Fe^{3+}} did not undergo reduction, its strong Lewis acidity would lead to tight coordination with the hydroxyl group, tying up the oxygen lone pair and potentially altering the migratory aptitude or reaction pathway (e.g., favoring epoxide formation or stalling the rearrangement).

Recognizing the spontaneous reduction to \ce{Fe^{2+}} ( $d^6$ ) is crucial to understanding why the hydroxyl group remains free to drive the classic Pinacol rearrangement.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Variable Oxidation State Stability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free