Question

Which of the following compound(s) are stable in their ionic structure?

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (D)

The correct options are
B
D

Order of stability:
Aromatic > Non-aromatic > Antiaromatic
Option (a):
It is planar, has conjugation and the number of $\pi$ electrons in ring system are 4, which follow $\left(4n\right)\pi e^{-}$ rule, thus making it antiaromatic and the least stable.
Option (c) :
The ring that has negative charge on carbon is planar , has conjugation has $6\pi e^{-}$ i.e. $(4n+2)\pi e^{-}$ , making it aromatic.
The ring that has positive charge on carbon is planar, has conjuagtion and $4\pi e^{-}$ i.e. it has $\left(4n\right)\pi e^{-}$ . So, it is anti-aromatic and thus formation of charges on that ring is highly unlikely due to its unstability.

So, ionic form of that compound is unstable due to 1 anti-aromatic ring.

In option (b) it forms a tropylium cation which is planar , has conjugation and has 6$\pi$ electron. So, ionic form is highly stable due to aromaticity.
Structure in option (d) forms a two ring system which are both planar, have conjugation and have 6 and 2 $\pi$ electrons respectively, thus both are obeying the $(4n+2)\pi e^{-}$ Huckel's rule. Hence, both rings are becoming aromatic on charge separation.
Hence, ionic form is stable.
So, structures in option (b) and (d) are stable in their ionic form.