What is the mechanism by which benzene, cyclohexene and $\ce{AlCl3}$ form cyclohexylbenzene?

I believe that cyclohexene will form a bond with Al with its double bond. However, how do we get rid of the $\ce{AlCl3}$ now attached to the previous cyclohexene after the benzene attaches to the module?


The Friedel-Crafts alkylation is another example of an electrophilic aromatic substitution. I've drawn the mechanism below showing the formation of cyclohexylbenzene. The reaction is typically run in the presence of an acid and the aluminum chloride serves as a catalyst to enhance the electrophilic properties of the proton, at the end of the reaction the aluminum chloride has been regenerated.

enter image description here

In the insert box I've shown that when the reaction is carried out with n-propylchloride isopropyl benzene is produced, not n-propylbenzene. A carbocation intermediate is generated in thr Friedel-Crafts alkylation and will rearrange to a more stable carbocation if possible. In the cyclohexene case, rearrangement of the intermediate carbocation just regenerates the same cyclohexyl carbocation, so the rearrangement is "invisible" in this case (unless we label the starting cyclohexene).

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  • $\begingroup$ Hi ron, I was particularly asking for the case where there is no acid. My textbook says the reaction can be done with only AlCl3, which is what was confusing me. Does this make sense? $\endgroup$ – yolo123 Dec 18 '14 at 22:06
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    $\begingroup$ If you mix aluminum chloride and an alkene a carbocation will still be generated and the product will be a mixture of the various possible alkene isomers weighted according to their thermodynamic stability. In this Friedel-Crafts example (without a proton source) the cyclohexyl carbocation is weakly bound with the aluminum chloride (ion pair). After this carbocation attacks benzene, the resultant cyclohexylbenzenonium carbocation will just transfer what was a benzene proton up to the cyclohexyl ring releasing (and regenerating ) the aluminum chloride. $\endgroup$ – ron Dec 18 '14 at 22:16
  • $\begingroup$ Why would the cyclohexyl carbocation be weakly bound? Would it not covalently share its double bond with the AlCl3? $\endgroup$ – yolo123 Dec 18 '14 at 22:21
  • $\begingroup$ Could you show the transfer of the proton too? Thanks in advance. $\endgroup$ – yolo123 Dec 18 '14 at 22:22
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    $\begingroup$ I bet that this reaction has never been done with rigorous exclusion of water. Water will hydrolyze AlCl3 to give the catalytic HCl necessary for Ron's top mechanism. $\endgroup$ – jerepierre Dec 18 '14 at 22:57

MECHANISM OF ALKYLATION: At first, electrophile methyl carbonium ion ($\ce{CH3}$) is formed from $\ce{CH3Cl3}$ and $\ce{AlCl3}$, which, at the second step, attracted by $\pi$-electron of the benzene ring to form sigma complex. At last step, the approach of nucleophile like, $\ce{AlCl4-}$, one proton is given out from sigma complex to form stable product toluene

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