The reaction is 2-methyl-5-phenyl-2-pentene with $\ce{H2SO4 }$

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I have tried multiple attempts at this question but am still getting it wrong. I have no idea why any of these answers are wrong.

I know that the chain off the aromatic ring is weak aromatic activator and an ortho-para director. I figured para would be better than ortho because of the size of the chain but obviously both answers were incorrect. I then figured that $\ce{H2SO4}$ is just a strong acid so its basically $\ce{H3O+}$ but adding $\ce{H}$ and $\ce{OH}$ across the double bond in the chain did not work either.

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    $\begingroup$ I think your top answer is the closest. If $\ce{H^+}$ from $\ce{H2SO4}$ adds first to the double bond, what adds to the other side? You have $\ce{HSO3^-}$ adding to the other side. Something is missing. Of course, it could be a cyclization due to Friedel-Crafts, the intramolecular verision of the top left reaction here. $\endgroup$ Commented Feb 29, 2016 at 3:22
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    $\begingroup$ Actually, the more I think about it, the more sure I am that you will have Friedel-Crafts-type cyclization occur. $\endgroup$ Commented Feb 29, 2016 at 3:29
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    $\begingroup$ Yes, you're right!!! I don't know how I didn't see the possibility of a Friedel- Crafts reaction, but thank you so much!! The final product was a benzene ring attached to a cyclohexane with two methyl groups coming off the bottom carbon. Once again thank you so much! I will update the question with the correct answer :) $\endgroup$
    – Kat S
    Commented Feb 29, 2016 at 4:07
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    $\begingroup$ No problem. By the way, for just an alkene, make sure you're adding $\ce{HSO4^-}$, not $\ce{HSO3^-}$. You would form an organosulfate, not an organosulfonate (or an organosulfuric acid, not an organosulfonic acid, really). $\endgroup$ Commented Feb 29, 2016 at 4:19

1 Answer 1


With the help of user SendersReagent in comments, the answer to the question was a simple intramolecular Friedel–Crafts reaction:

Mechanism for formation of 1,1-dimethyl-1,2,3,4-tetrahydronaphthalene


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