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Question #44076

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The Question

A chemist investigates the kinetics of Schiff base formation between aniline ( $\ce{PhNH2}$ ) and benzaldehyde ( $\ce{PhCHO}$ ) at $298 \text{ K}$ . To maintain a constant acidic environment, the reaction is performed in an aqueous solution prepared by dissolving $0.10 \text{ mol}$ of anilinium chloride ( $\ce{PhNH3Cl}$ ) and $3.0 \times 10^{-3} \text{ mol}$ of $\ce{HCl}$ in $1.0 \text{ L}$ of water.

Given the acid dissociation constant of the anilinium ion is $K_a = 4.0 \times 10^{-5}$ at $298 \text{ K}$ .

Calculate the true degree of dissociation ( $\alpha$ ) of the anilinium ion (assume $1-\alpha \approx 1$ for the undissociated salt concentration), and identify the primary kinetic and electronic consequence of this specific pH environment on the aniline molecule.

Options

A.
$\alpha = 1.0 \times 10^{-2}$ ; The $+M$ effect of the $\ce{-NH2}$ group is quenched due to protonation, drastically reducing the active nucleophile concentration.
B.
$\alpha = 1.33 \times 10^{-2}$ ; The $+M$ effect of the $\ce{-NH2}$ group is quenched due to protonation, drastically reducing the active nucleophile concentration.
C.
$\alpha = 2.0 \times 10^{-2}$ ; The high $\ce{[H+]}$ optimally catalyzes the carbonyl activation, maximizing the nucleophilic attack rate.
D.
$\alpha = 1.0 \times 10^{-2}$ ; The anilinium ion acts as a stronger nucleophile than aniline due to the inductive pull of the positively charged nitrogen.

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Mechanism_Trap

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Concepts Behind This Question

Degree of Dissociation (α) approximation|Schiff Base pH Dependence|Aniline Resonance Pitfalls

Why This Question Is Tricky

Common Mistakes

This question tests the boundary conditions of the common-ion approximation in ionic equilibrium and its direct consequence on organic reaction mechanisms.

Students often blindly apply the approximation $C\alpha \ll C_{common}$ without verifying if the dissociation contribution is negligible.

Here, the $\ce{H+}$ from the weak acid dissociation ( $10^{-3} \text{ M}$ ) is $33\%$ of the strong acid concentration, invalidating the approximation.

Furthermore, it tests the 'Nucleophilic Paradox' of Schiff base formation: while acid is needed to activate the carbonyl, excessive acid protonates the amine, tying up the lone pair, quenching its $+M$ resonance stabilization, and rendering it non-nucleophilic.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Degree of Dissociation (α) approximation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free