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Question #44268
The Question
Consider the thermodynamic basicity of the following tertiary amines in aqueous solution at 298 K: (I) 1-Azabicyclo[2.2.2]octane (Quinuclidine) (II) Triethylamine (III) Triisopropylamine (IV) 1-Azabicyclo[1.1.1]pentane
Which of the following correctly represents the decreasing order of their basic strength () and accurately identifies the primary structural phenomena governing their relative basicities?
Options
- A.
Order: I > II > III > IV. (I) is the strongest as its rigid framework ties back the alkyl groups, exposing the lone pair. (III) is weaker than (II) because Pauli exclusion-driven steric repulsion (F-strain) between the bulky isopropyl groups and the hydration sphere hinders \ce{N-H+} formation, overriding the +I effect. (IV) is the weakest because extreme angle strain drastically increases the s-character of the lone pair orbital.
- B.
Order: III > II > I > IV. (III) is the strongest due to the maximum +I (inductive) effect of three isopropyl groups. (I) is weaker than (II) because the rigid bicyclic framework prevents structural relaxation upon protonation. (IV) is the weakest because the bridgehead nitrogen is forced into a planar geometry, delocalizing the lone pair.
- C.
Order: I > IV > II > III. (I) is the strongest due to the combined +I effect of the three ethylene bridges. (IV) is highly basic because the constrained geometry forces the lone pair into a highly directional pure p-orbital. (III) is the weakest because severe B-strain (back-strain) forces the nitrogen to adopt a non-basic planar configuration.
- D.
Order: II > I > III > IV. (II) is the strongest because flexible alkyl chains allow maximum solvation of the conjugate acid. (I) is weaker due to the absence of hyperconjugative stabilization from the bridgehead carbons. (IV) is the weakest because Pauli repulsion between the bridgehead \ce{C-H} bonds and the nitrogen lone pair completely prevents protonation.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
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Concepts Behind This Question
Pauli Exclusion Principle|Acidity and Basicity (pKa/pKb)|Inductive Effect Trends
Why This Question Is Tricky
Common Mistakes
Students often rely on a simplified heuristic: 'More alkyl groups = more +I effect = stronger base.' This leads them to incorrectly predict Triisopropylamine as the strongest base (Option B).
The trap lies in ignoring the Pauli Exclusion Principle-driven steric repulsion (F-strain) where the bulky electron clouds of the isopropyl groups physically block the incoming proton and hydration sphere, overriding the Inductive Effect.
Furthermore, students may fail to account for the hybridization shift in highly strained bridgehead nitrogens like 1-Azabicyclo[1.1.1]pentane.
By Bent's rule, extreme angle strain forces the lone pair into an orbital with high s-character, rendering it exceptionally non-basic.
This problem demands a synthesis of Inductive Effects, Steric Hindrance, and Hybridization theory.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Structure of Atom - Class 11 Part 1 |
| Concept | Pauli Exclusion Principle |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |