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Question #44393

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from Phenol ( $\ce{P}$ ):

$\ce{P ->[ (i) NaOH, CO2, 400 K, 5 atm ][ (ii) H3O^+ ] Q (Major)}$

$\ce{Q + CH3OH (excess) ->[ H2SO4 (cat.) ][ \Delta ] R}$

Let $v_R$ be the initial rate of formation of $\ce{R}$ . An isomer of $\ce{Q}$ , 4-hydroxybenzoic acid ( $\ce{S}$ ), is subjected to identical Fischer esterification conditions to form product $\ce{T}$ with an initial rate $v_T$ . Which of the following correctly compares $v_R$ and $v_T$ , and identifies the primary mechanistic rationale?

Options

A.
$v_R < v_T$ ; Intramolecular H-bonding in Q stabilizes the ground state, decreasing carbonyl electrophilicity, and the ortho-OH sterically hinders the tetrahedral intermediate.
B.
$v_R > v_T$ ; The ortho-OH group in Q acts as an internal acid catalyst, facilitating carbonyl protonation and accelerating nucleophilic attack.
C.
$v_R < v_T$ ; The ortho-OH group exerts a stronger -I effect than the para-OH group, which strongly destabilizes the protonated carbonyl intermediate.
D.
$v_R > v_T$ ; Intramolecular H-bonding in Q stabilizes the $sp^3$ tetrahedral intermediate, lowering the activation energy compared to S.

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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

Kolbe-Schmitt Reaction Mechanism|Esterification Steric Hindrance|Esterification Reversibility

Why This Question Is Tricky

Common Mistakes

The question tests the 'Ortho Effect' in Fischer esterification, specifically targeting the misconception that an ortho-OH might accelerate the reaction via internal catalysis or inductive effects.

Students often apply simple Linear Free Energy Relationships (LFER) or electronic effects (-I, +M) without accounting for 3D spatial constraints.

The hidden variable is the dual role of the ortho-hydroxyl group: it thermodynamically stabilizes the ground state via intramolecular H-bonding (reducing electrophilicity) and kinetically hinders the reaction via steric shielding of the tetrahedral intermediate, causing a significant deviation from expected para-substituted reaction rates.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Alcohols, Phenols, and Ethers - Class 12
Concept	Kolbe-Schmitt Reaction Mechanism
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free