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Question #44151

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The Question

Compound \textbf{X}, 1-methoxy-4-(trimethylsiloxy)benzene, is treated with 1 equivalent of anhydrous \ce{HI} in toluene under reflux (383 K). Two competing cleavage pathways are theoretically possible, leading to either 4-(trimethylsiloxy)phenol + \ce{CH3I} or 4-methoxyphenol + \ce{(CH3)3SiI}. Identify the major products formed and the primary chemical principle governing this selectivity.

Options

A.
\ce{4-methoxyphenol} and \ce{(CH3)3SiI}; the cleavage is kinetically controlled by a highly stabilized pentacoordinate transition state at the electrophilic silicon atom.
B.
\ce{4-(trimethylsiloxy)phenol} and \ce{CH3I}; the \ce{Si-O} bond is exceptionally strong due to $p\pi-d\pi$ back-bonding, making the formation of the weak \ce{Si-I} bond thermodynamically disfavored.
C.
\ce{4-methoxyphenol} and \ce{(CH3)3SiI}; the methoxy oxygen is more basic, but the $+I$ effect of the silyl group thermodynamically drives the cleavage of the \ce{Si-O} bond.
D.
\ce{4-(trimethylsiloxy)phenol} and \ce{CH3I}; the bulky \ce{-Si(CH3)3} group sterically hinders the attack of \ce{I^-}, making the \ce{S_N2} cleavage of the methoxy group kinetically faster.

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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

Back Bonding Strength|Reaction Progress Coordinate|Ether Cleavage Patterns

Why This Question Is Tricky

Common Mistakes

The question tests the 'Hidden Variable' of the nucleophile's role in the thermodynamic feasibility of ether cleavage.

Students memorize that 'silyl ethers are highly acid-sensitive' and often assume they will cleave before alkyl ethers under any acidic conditions.

However, this sensitivity is specific to hydrolysis (aqueous acid), where a strong Si-O bond is traded for another Si-O bond.

Under anhydrous conditions with HI, the nucleophile is I^-.

$The formation of the weak Si-I bond cannot compensate for the cleavage of the exceptionally strong Si-O bond (stabilized by p\pi-d\pi back-bonding), making the reaction endothermic.$

Thus, the standard alkyl ether undergoes exothermic cleavage instead, bypassing the kinetic trap of silicon's pentacoordinate transition state.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Back Bonding Strength
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free