I am studying Friedel–Crafts alkylation reactions and my textbook states the following proposition [1, p. 1090]:

Another very good means of introducing $n$-propyl group is to use cyclopropane as the alkylating agent.

benzene + cyclopropane to propylbenzene

I do understand the mechanism of Friedel–Crafts alkylation, but I am not able to write the mechanism for this particular reaction. Any hints are appreciated.


  1. Shahi, R. Essential Organic Chemistry for JEE Main & Advanced; Arihant Prakashan: New Delhi. ISBN: 978-93-13192-10-7.
  • 1
    $\begingroup$ Please -- not only for ChemSE, but as a general applicable technique -- not only quote, but reference the source of a quote. Cyclopropane has considerable ring tension (building it once with a molecular kit, I literally broke once one of the bonds ...) but so far I thought alkanes always had to be halogenated, then permitting $\ce{AlCl3}$ to yield an electrophilic reagent. $\endgroup$
    – Buttonwood
    Dec 19, 2020 at 9:31
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    $\begingroup$ This paper uses cyclopropanes to alkylate electron rich benzenes with Lewis acid catalysis but as it is behind a paywall I can only read the abstract. It may discuss the mechanism chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/… $\endgroup$
    – Waylander
    Dec 19, 2020 at 9:59
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    $\begingroup$ @Waylander From the SI of the publication you indicate (annotated NMR spectra), accessible without subscription, it looks like cyclopropane always bears an acceptor decoration (1,1-dicarboxylate) and an electron donor (e.g., 2-phenyl group). By the title of the papers citing this work (equally accessible), this donor-acceptor pattern seems to repeat, too. It is not so obvious that it is about cyclopropane itself (boiling at $\pu{-33 ^\circ{}C}$). $\endgroup$
    – Buttonwood
    Dec 19, 2020 at 10:34

1 Answer 1


This was a very interesting question, and I found the answer (or the reference thereto) in Thomas's "Anhydrous Aluminum Chloride in Organic Chemistry" (1941). On page 501 is referenced the very thorough work of Ipatieff et al. on the very subject in question, "alkylation by naphthenes".

First, we can say that because there is no isomerization of the chain (n-propyl benzene is formed, and not isopropyl) that neither chloropropane nor propylene are intermediates. Friedel-Crafts alkylation with either of those leads to at least some isopropylbenzene formation, because of the better stability of the secondary carbocation.

Second, and most convincing, Ipatieff et al. state that "The presence of hydrogen chloride is essential for the alkylation of aromatics with olefins or naphthenes when aluminum chloride is used as a catalyst.". This leads them to propose a hydrogen chloride-aluminum chloride complex, "hydrogen aluminum tetrachloride" or "hydrogen tetrachloroaluminate", $\ce{HAlCl_4}$, which then reacts with cyclopropane to form n-propyl aluminum tetrachloride. This intermediate is the active alkylation agent, and goes on to produce only n-propyl benzenes.

$\ce{HCl + AlCl_3 -> AlCl_3\cdot HCl or HAlCl_4}$

$\ce{(CH_2)_3 + HAlCl_4 -> CH_3CH_2CH_2ClAlCl_3}$

$\ce{CH_3CH_2CH_2ClAlCl_3 + C_6H_6 -> C_6H_5CH_2CH_2CH_3 + HAlCl_4}$

So far as I can tell, this theory has never been refuted and stands as the current state of the art.

Ipatieff, Pines, and Schmerling, J. Org. Chem., 5, 253 (1940)

  • $\begingroup$ The book also states that the reaction is highly favorable. So much that it could occur even below 0°C. $\endgroup$ May 19 at 18:07

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