a will become aromatic after 1st step which is very stable
b will become antiaromatic (least stable)
I got the same but answer is a>b>c
Maybe it's wrong
Nope a won't become aromatic
For aromaticity we need all sp2 hybridizd carbon and resonance all over the ring.
But yeah a will be more stable than c so order must be a>c>b
I doubt the answer is incorrect
a will be more stable due to allylic carbonation right?