Which of these carbocations are more stable?

Question

The line of text that follows this figure in my work-book reads:

Cation (A) is more stable than cation (B), since the former is resonance stabilized but the latter isn't (due to steric effects).

Don't both carbocations show resonance stabilization? More importantly, isn't cation (B) more stable than (A)?

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Based on what I've learnt, the following should occur:

1) Cation (A)

The sp² carbon (in the isopropyl group) acts as a -M group and causes the shift of a pair of π electrons from the benzene ring towards the $\ce{C-C^{+}}$ bond, thereby creating a double bond in its place. So the resulting structure should look like,

The mesomeric effect would also ensure that the positive charge "travels" across the ring, and decreases the electron density at ortho- and para- positions. The delocalization of the positive charge across the benzene ring should confer some degree of stability to the carbocation.

2) Cation (B)

Once again, a pair of π electrons from the benzene ring would shift towards the $\mathrm{C-C^{+}}$ bond, leaving behind this:

And again, the mesomeric effect would delocalize the positive charge over the benzene ring and create electron deficient centers at the ortho- and para- positions.

However, the presence of two isopropyl groups at ortho- positions (with respect to the orignal charge-carrying isopropyl group) will result in the "supression" of the positive charge via inductive and esomeric effects, hence, conferring additional stability to this molecule.

Hence, cation (B) should be more stable than (A).

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Yet, my conclusion is different from that provided by the book. Well, I guess that the presence of three adjacent isopropyl groups (in cation-B) does seem sterically unfavorable, but how would that prevent resonance effects in the molecule (as the book seems to imply)?

So, is my conclusion correct? Have I arrived at it correctly? If the book's right, then why is it right?

Answer 1

As we know (to some extent), resonance in a molecule occurs due to sideways overlapping of p-π orbitals.

Now in the molecule B, there is high steric hinderance near the carbocation due to which the plane of the "carbon with carbocation" changes, so that it can feel minimum repulsion from the other groups. This forces the carbocation to remain in the mutually perpendicular plane, which inhibits the resonance.

Thus, this makes A (the less hindered carbocation) more stable than B.

Answer 2

The carbon atom with the positive charge can only participate in resonance with the ring if they lie in the same plane. Due to steric effects/inhibition that particular C atom will not be coplanar with the rest of the ring anymore. Lesser delocalisation of the positive charge makes structure B less stable.