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Question #44342

Difficulty: 8/10 | 🟢 Free

The Question

Consider the base dissociation constants (KbK_b) of the following aromatic amines in aqueous solution at 298 K: \begin{enumerate} \item[(I)] \ce{C6H5NH2} \item[(II)] \ce{2,6-(CH3)2C6H3NH2} \item[(III)] \ce{C6H5N(CH3)2} \item[(IV)] \ce{2,6-(CH3)2C6H3N(CH3)2} \end{enumerate} Which of the following represents the correct decreasing order of basic strength?

Options

  • A.

    (IV)>(III)>(I)>(II)(IV) > (III) > (I) > (II)

  • B.

    (II)>(I)>(IV)>(III)(II) > (I) > (IV) > (III)

  • C.

    (III)>(IV)>(I)>(II)(III) > (IV) > (I) > (II)

  • D.

    (IV)>(II)>(III)>(I)(IV) > (II) > (III) > (I)

Check Your Answer

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

VSEPR Lone Pair Positioning|Ortho-Effect in Anilines

Why This Question Is Tricky

Common Mistakes

The question tests the critical boundary between the Ortho-effect (Steric Inhibition of Solvation) and Steric Inhibition of Resonance (SIR).

Students often apply a single rule universally—either assuming ortho-groups always decrease basicity due to steric hindrance (choosing C), or assuming they always increase basicity via +I/SIR (choosing B).

The 'Hidden Variable' is the size of the N-substituent: primary amines suffer the thermodynamic solvation penalty without SIR, while tertiary amines undergo SIR, which electronically supercharges the basicity enough to overcome the solvation penalty.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Bonding and Molecular Structure - Class 11 Part 1
ConceptVSEPR Lone Pair Positioning
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree