The acidity of protons H in each of the following is:


  • a)
    I > II > III
  • b)
    II > III > I
  • c)
    II > I > III
  • d)
    III > I > II
Correct answer is option 'C'. Can you explain this answer?

Chemistry Question

4 Answers
Dinabandhu Das answered Jul 16, 2018
Though vacant d orbital in P is good idea but we could also thought of another way. I think in case of molecule 1 nitrogen capture the negative charge after protonation i.e. delocalization of the -ve charge in aromatic polycyclic ring will be less but in case of molecule 2 P have less electronegativity and hence delocalization of -ve charge can occur more easily with the aromatic rings. and in case of 3 the +I effect of CH3 disturbs the stability of -ve charge form after protonation.. If my explanation is helpful then please let me know.....

CHHATAR SINGH answered Feb 22, 2020
In case of molecule 1 nitrogen captures the negative charge after protonation i.e. delocalization of the -ve charge in aromatic polycyclic ring will be less but in case of molecule 2, P have less electronegativity and hence delocalization of -ve charge can occur more easily within the aromatic rings. and in case of 3 the +I effect of CH3 disturbs the stability of -ve charge form after protonation.

Gourab Banerjee answered Jun 26, 2018
In II the negative charge over conjugate base after deprotonation is stabised by vacant d orbital conjugation of P atom resulting higher stability. In I the conjugate base is stabilised by -I effect of positively charged N, whereas no such stabilizing factor is present in case of III rather destabilizing +I effect of -CH3 group is present . Higher stability of conjugate base results higher acidity.

Sharad Tyagi answered Jul 17, 2018
Why not option A???

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