I need to compare the acidic strength in these molecules:

enter image description here

My Approach:

In compound I: Removing a proton would lead to conjugation between -ve charge, pi bond and lone pair of O. Moreover the inductive effect due to 'O' will stabilize the charge.

In compound II: Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation, but relatively less than one. Moreover 'O' is far away, so inductive effect does not play a role here.

From this I concluded that compound 'I' must be more acidic. But the answer says that the second compound is more acidic.

Where am I going wrong?

  • $\begingroup$ Btw, In the second compound O is as far away as in first compound. $\endgroup$
    – King Tut
    Mar 19, 2018 at 2:45

1 Answer 1


This has a simpler answer than you expect. First of all, your reasoning is wrong because inductive effect is much less dominant than conjugation of electron pairs. So, if the latter is unequal in both cases, we should first prioritize it over inductive effects.

Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation

It seems you already know that the beta hydrogen of carbonyls is most acidic. Both molecules have carbonyl groups, and both have beta acidic hydrogens. So, you're on the right track!

Whichever molecule is more acidic, it will have a more stable conjugate base.

Note that in the second anion, the excess negative charge is completely delocalised over to the carbonyl group. Compare that with the first anion, where both the excess negative charge and the oxygen's lone pair are delocalising with the same carbonyl group. Thus, the excess negative charge is unable to completely delocalise with the carbonyl group.

Can you now identify why the first molecule is less acidic?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.