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Question #44583

Difficulty: 8/10 | 🟢 Free

The Question

The chiral aldehyde $(S)$ -2-methoxypropanal is treated with \ce{PhMgBr} in the presence of 1 equivalent of \ce{TiCl4} at $-78^\circ\text{C}$ to yield $\textbf{P}$ as the major diastereomer. The reaction proceeds via a rigid chelated intermediate complex $\textbf{C}$ . Assume \ce{Ti} maintains a coordination number of 6 in the intermediate $\textbf{C}$ . According to Valence Bond Theory, what is the hybridization of the Titanium center in $\textbf{C}$ , and what is the absolute configuration of $\textbf{P}$ ?

Options

A.
$d^2sp^3$ and $(1S, 2S)$
B.
$sp^3d^2$ and $(1S, 2S)$
C.
$d^2sp^3$ and $(1R, 2S)$
D.
$sp^3$ and $(1R, 2S)$

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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

Valence Bond Theory Hybridization|Grignard Reagent Sensitivity|Coordination Compounds - Class 12

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Coordination Chemistry (VBT) and Advanced Stereochemistry (Cram-Chelate model).

The first trap is assuming an octahedral \ce{Ti^{4+}} complex uses outer $d$ orbitals ( $sp^3d^2$ ) instead of the available inner $3d$ orbitals ( $d^2sp^3$ ).

The second trap is the stereochemical outcome: students must recognize that the presence of a chelating Lewis acid (\ce{TiCl4}) overrides the standard Felkin-Anh model (which would yield the $(1R, 2S)$ diastereomer) and instead enforces a rigid Cram-chelate intermediate, directing Grignard attack from the less hindered face to yield the $(1S, 2S)$ product.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Coordination Compounds - Class 12
Concept	Valence Bond Theory Hybridization
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free