Product of Friedel-Crafts cyclisation of (3-methylpent-4-en-1-yl)benzene

Question

My proposed mechanism and final product is:

But the book says the answer is:

I know that a 6 membered ring is more stable than a 5 membered ring so I made the product with the 6 membered ring (however, the carbocation was a secondary carbocation).

However, the book claims the answer to be the one with the 5 membered ring (the carbocation here was a tertiary carbocation).

Is the answer provided in the book wrong?

Then I suppose the carbocation thus formed rearranges to 3° and then is being attacked by benzene. Later through electrophilic substitution the product given in the book is obtained. — Suraj S, Apr 30 '17 at 16:33
Related: Why are 7-membered rings less likely to form than 5- and 6- membered rings? — Yashas, May 06 '17 at 04:23
The comments seem to underestimate the fact that a tertiary carbocation is more favorable due to inductive effect. So, then it will form pentacarbon ring on the second step. And I don't think there is much strain in a pentacarbon ring. — Mockingbird, May 06 '17 at 07:28
I think the carbocation will rearrange to a more stable one. — Mockingbird, May 06 '17 at 07:38

score 4 · Accepted Answer · answered May 27 '19 at 23:08

Though the question is old and pretty much answered in the comments, I ran into this question and draw the mechanism, so I thought about summarizing all that was brought up in an answer.

The carbocation is first formed on the secondary carbon. However, molecules are more stable when the charges are delocalized (this page talks about the stability of carbocations). When the positive charge is in the secondary carbon, it can only "share the burden" of having this charge with 2 neighbor carbons. So the tertiary carbon donates an hydrogen through a sigma to atomic orbital interaction (the sp3 hybdrization allows a π-type interaction) resulting in the more stable tertiary carbocation seen in the mechanism below:

Then the reaction follows analogously to the pathway you deduced for the 6 membered ring: First a π to atomic orbital interaction between the aromatic ring and the tertiary carbocation, which I believe to be the Rate Determining Step since it breaks the ring aromacity (bigger energy gap). Then finally the sigma to atomic interaction between the hydrogen and the ring carbocation, that restores aromaticity and releases the hydrogen.

Product of Friedel-Crafts cyclisation of (3-methylpent-4-en-1-yl)benzene

1 Answers1