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Question #45102

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The Question

The \ce{S_N2} reaction of neopentyl bromide (\ce{(CH3)3CCH2Br}) with \ce{NaN3} in DMF at $300\text{ K}$ is notoriously slow compared to bromoethane (\ce{CH3CH2Br}). Kinetic data reveals that the rate constant for the neopentyl system ( $k_{neo}$ ) is $10^{-6}$ times that of the ethyl system ( $k_{ethyl}$ ). Thermochemical analysis using Transition State Theory (TST) shows that the activation enthalpy ( $\Delta H^\ddagger$ ) for the neopentyl system is $17.235\text{ kJ mol}^{-1}$ higher than that of the ethyl system. Assuming the transmission coefficient $\kappa = 1$ for both reactions, calculate the difference in their activation entropies ( $\Delta S^\ddagger_{neo} - \Delta S^\ddagger_{ethyl}$ ) caused by the severe steric crowding (B-strain) restricting the orientation factor ( $p$ ). [Given: At $300\text{ K}$ , $2.303 RT = 5.745\text{ kJ mol}^{-1}$ and $2.303 R = 19.15\text{ J K}^{-1}\text{ mol}^{-1}$ ]

Options

A.
$-57.45 \text{ J K}^{-1} \text{ mol}^{-1}$
B.
$-38.30 \text{ J K}^{-1} \text{ mol}^{-1}$
C.
$0 \text{ J K}^{-1} \text{ mol}^{-1}$
D.
$+57.45 \text{ J K}^{-1} \text{ mol}^{-1}$

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

Steric Hindrance (B-Strain)|Collision Frequency vs. Effective Collisions|Units of Rate Constant (k)

Why This Question Is Tricky

Common Mistakes

This question targets the 'Lethal Collision' trap in chemical kinetics.

Students often erroneously assume that the Arrhenius pre-exponential factor ( $A$ ) is a constant geometric collision frequency, ignoring the 'Steric Factor' ( $p$ ).

By transitioning to Transition State Theory (TST), the problem exposes that severe steric hindrance (B-strain) in neopentyl-like \ce{S_N2} transition states doesn't just increase activation enthalpy ( $\Delta H^\ddagger$ ), but drastically reduces the probability of the correct orientation.

This manifests as a highly negative entropy of activation ( $\Delta S^\ddagger$ ).

Failing to account for this leads to overestimating the rate constant by orders of magnitude (Option C assumes $\Delta \Delta S^\ddagger = 0$ , while Option B is a common math error).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Steric Hindrance (B-Strain)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free