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Question #44406

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The Question

Consider the cleavage of diethyl ether ( $\ce{Et2O}$ ) at 298 K using sodium halide salts in pure carboxylic acid solvents. The cleavage proceeds via the initial protonation of the ether followed by a bimolecular nucleophilic attack ( $S_N2$ ) by the halide ion ( $\ce{X-}$ ):

Step 1: $\ce{Et2O + HA \rightleftharpoons Et2OH+ + A-}$ (fast equilibrium, constant $K_{eq}$ ) Step 2: $\ce{Et2OH+ + X- \rightarrow EtX + EtOH}$ (rate-determining step, rate constant $k_X$ )

Reaction 1: 1.0 M $\ce{NaI}$ and 1.0 M $\ce{Et2O}$ dissolved in pure Chloroacetic acid (CAA). Reaction 2: 1.0 M $\ce{NaBr}$ and 1.0 M $\ce{Et2O}$ dissolved in pure Trichloroacetic acid (TCA).

Given the following parameters:

$pK_a$ of $\ce{Et2OH+} = -3.5$
$pK_a$ of TCA $= 0.5$
$pK_a$ of CAA $= 2.5$
The nucleophilic rate constant ratio $k_I / k_{Br} = 40$ .

Assume the given $pK_a$ values accurately reflect the relative thermodynamic acidities in these solvent systems. Furthermore, assume the molar concentration of the solvent $[\ce{HA}]$ is identical in both reactions, autoionization of the solvents is negligible, complete dissociation of the sodium salts, and $\ce{Et2O}$ is the only significant base present (its concentration remains approximately 1.0 M).

What is the ratio of the initial rates of cleavage ( $r_1 / r_2$ ) for Reaction 1 compared to Reaction 2?

Options

A.
4.0
B.
0.4
C.
40
D.
400

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Ether Cleavage Patterns|Carboxylic Acid Strength|Inductive Effect Trends

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of chemical kinetics and thermodynamics in non-aqueous media.

The core trap lies in the equilibrium of ether protonation.

Students often mistakenly assume that the concentration of the protonated intermediate is directly proportional to the equilibrium constant ( $[\ce{Et2OH+}] \propto K_{eq}$ ).

However, because the solvent $\ce{HA}$ is the only source of protons and $\ce{Et2O}$ is the only base, the stoichiometry dictates that $[\ce{Et2OH+}] = [\ce{A-}]$ , leading to a square root dependence ( $[\ce{Et2OH+}] \propto \sqrt{K_{eq}}$ ).

Failing to account for this square root relationship leads to the incorrect answer 0.4.

Additionally, it tests the correct application of $pK_a$ values to find the equilibrium constant of an acid-base reaction.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Alcohols, Phenols, and Ethers - Class 12
Concept	Ether Cleavage Patterns
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free