# Resonance stabilized carbanion

I am sure that the answer is (b) because I am familiar with the other rings and I think they are showing resonance (and it is a single choice question so it was easy to guess)

I wish to know why is (b) not showing resonance (because I think it can as far as I know) and if I am wrong about the answer then please correct me.

(b) does not show resonance because the second resonance structure (with negative charge at oxygen and double bond in the ring) violates Bredt's rule. The carbon $\ce{sp^3}$ orbital with the negative charge and the $\ce{p_z}$ orbital of the adjacent carbonyl C cannot overlap because they are not in plane, so the charge cannot be delocalized into the carbonyl group.

You are right about compound B. Try to build a model of the resonance form for B where charge is delocalized to the adjacent carbonyl. Here is a picture of the situation.

Note what is called a "bridgehead double bond" in the resonance form on the right. Such a structure would be a violation of Bredt's Rule, which states that double bonds can't be placed in bridgehead positions if there would be a lot of ring strain as a result. Again, try to build a model of the molecule on the right and see how strained that bridgehead double bond makes the system.

For any cyclic compound to exhibit resonance, the following rules have to be obeyed:

1) $(4n+2)\pi$ electrons.

2) Conjugation. Conjugation is not possible for either b) or d).

3) Planarity. Compound b) is not planar.

And of course, compound b) violates Bredt's rule as well.

Hence, neither b) nor d) can be resonance stabilised.

• These are rules for aromaticity not conjugation. – Mithoron Aug 13 '15 at 23:18