Here is the question:

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We have to find the order of their acidity.

Here is my thought process:

We check the acidity order by finding which conjugate base is more stable. Thus we remove $\ce{H+}$ from all of them. One question I have here is that in (IV), from which position should I remove $\ce{H+}$? Will it be the position from where resonance can take place or from the middlemost carbon?

In (I), (II), and (III), resonance can take place in all of them after removing $\ce{H+}$. Resonance will be more in (II) and (III) than in (I).


(II) and (III) are more stable than (I). I am uncertain about the stability between (II) and (III) and about (IV).

Please tell me if I have been wrong somewhere and how I could proceed further...

  • $\begingroup$ Consider the pKas of dinitromethane, malonitrile and malonate. Dintiromethane is by a long way the most acidic pKa 3.6, malonitrile is a bit more acidic than malonate 11 v 12.9. $\endgroup$
    – Waylander
    Mar 29, 2020 at 15:40

1 Answer 1


Okay, so if we talk about the acidic strength of a compound, consider adding a base to a solution of all the above. Practically, a reaction is bound to take place-if possible-due to thermodynamic reasons. Hence, reaction in IV will take place since the base will try to find the most acidic Hydrogen, which is why one should consider the removal of H atom from which resonance can take place.

Now for the order of acidic strength, I, II and III are individually more acidic than IV, due to extensive conjugation in all three but the last. This consequently makes the conjugate bases stable and acids stronger. Now for the order among these, notice that Cross-conjugation takes place in III, as we have COO- group. Hence, III will consequently be less stable and a weaker acid.

The order and reason stated by you for II and I are correct, and hence the order for acidic strength is:


i.e. option A.

  • $\begingroup$ After we remove H+, resonance can also take place in (IV). Moreover, more stable the compound is, less acidic it will be. Cross conjugation can also take place in NO2. $\endgroup$ Mar 29, 2020 at 15:15
  • $\begingroup$ Why does cross conjugation reduce stability? $\endgroup$ Mar 29, 2020 at 15:18
  • 1
    $\begingroup$ @aitchessbee I would suggest you check out the instability caused by cross conjugation yourself en.wikipedia.org/wiki/Cross-conjugation here. No, the more stable the compound after removing H atom, the less basic the conjugate base and hence the more acidic the compound. This is the basic principle which you yourself have used to solve some part of the question. And also, the bond in NO2 is a 'co-ordinate bond' and does not allow cross-conjugation. $\endgroup$ Mar 29, 2020 at 15:34

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