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Question #44487

Difficulty: 8/10 | 🟢 Free

The Question

The acidity of 1,3-cyclopentadiene (\ce{C5H6}) is governed by the formation of the aromatic cyclopentadienyl anion (\ce{C5H5^-}). However, its pKapK_a is highly solvent-dependent due to solvation effects. At 298 K, the pKapK_a of \ce{C5H6} is 15.0 in water and 18.0 in dimethyl sulfoxide (DMSO).

Given the following standard Gibbs free energies of transfer (ΔGtr∘\Delta G^\circ_{tr}) from water to DMSO at 298 K: ΔGtr∘(\ceH+)=−19.0 kJ mol−1\Delta G^\circ_{tr}(\ce{H^+}) = -19.0 \text{ kJ mol}^{-1} ΔGtr∘(\ceC5H6)=−11.1 kJ mol−1\Delta G^\circ_{tr}(\ce{C5H6}) = -11.1 \text{ kJ mol}^{-1}

Calculate the standard Gibbs free energy of transfer of the cyclopentadienyl anion, ΔGtr∘(\ceC5H5−)\Delta G^\circ_{tr}(\ce{C5H5^-}), from water to DMSO at 298 K. (Take 2.303RT=5.7 kJ mol−12.303 RT = 5.7 \text{ kJ mol}^{-1})

Options

  • A.

    25.0 kJ mol−125.0 \text{ kJ mol}^{-1}

  • B.

    −13.0 kJ mol−1-13.0 \text{ kJ mol}^{-1}

  • C.

    47.2 kJ mol−147.2 \text{ kJ mol}^{-1}

  • D.

    9.2 kJ mol−19.2 \text{ kJ mol}^{-1}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Aromaticity (Huckel’s Rule)|Acidity and Basicity (pKa/pKb)|Basicity-Solvation Competition

Why This Question Is Tricky

Common Mistakes

The problem tests the thermodynamic cycle of acid dissociation across different solvents, highlighting the 'Basicity-Solvation Competition'.

While Huckel's rule explains the inherent stability of the cyclopentadienyl anion, its effective pKapK_a is heavily modulated by the solvent.

Students often assume polar aprotic solvents (like DMSO) uniformly destabilize all ions or fail to account for the fact that DMSO strongly solvates cations (like H+H^+) but poorly solvates anions due to the lack of hydrogen bonding.

The trap lies in correctly constructing the thermodynamic cycle and applying the correct signs for the transfer free energies, particularly recognizing that the transfer of the non-polar C5H6C_5H_6 is exergonic, while the transfer of the anion is highly endergonic.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterHydrocarbons - Class 11 Part 2
ConceptAromaticity (Huckel’s Rule)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree