I recently came across this reaction in my undergraduate labs. The experiment is taken from this article: J. Chem. Educ., 2008, 85, 413.

The condensation of 2-methoxynaphthalene with cinnamoyl chloride leads to the formation of 9-hydroxyphenalenone:

Reaction scheme

I think it's fairly clear that the first step is a Friedel-Crafts acylation. Subsequently to form the new C-C bond, there must be a Michael addition of some sort; however this would imply that the phenyl group leaves as "$\ce{Ph-}$". There are two things I'm not very sure about:

  1. How does the phenyl group even leave?

  2. How does the oxygen get demethylated? I am guessing that the mechanism should be similar to that with $\ce{BBr3}$, i.e. coordination of oxygen to the Lewis acid followed by nucleophilic attack of $\ce{Cl-}$ on the methyl carbon. Would that be right?

I looked up the original synthesis, which was reported quite a long time ago, in J. Org. Chem., 1941, 6, 558. The authors write that the reaction proceeds via the intermediacy of this compound:


which suggests that the initial nucleophile in the Michael addition is not the aromatic ring but rather the -OMe (or -OH) group. Unfortunately, that just makes things more confusing.


First, the methoxy- group can be demethylated by $\ce{AlCl3}$, beginning with $\ce{AlCl3}$ coordination to the ether oxygen, followed by an $\mathrm{S_N2}$ reaction:

$\hspace{3.6cm}$step 1

$\hspace{0cm}$enter image description here

The oxygen activates naphthalene's position 1, and the acid chloride will undergo a Friedel-Crafts acylation (as you noted):

$\hspace{.5cm}$step 3

Being a strong Lewis acid, $\ce{AlCl3}$ can coordinate with the carbonyl oxygen. This activates position 4 of the $\alpha$-$\beta$ unsaturated carbonyl towards electrophilic addition:

$\hspace{.75cm}$step 4

Electrophilic addition occurs at naphthalene's spatially adjacent position 8, the carbocation stabilized through the extensive conjugation of the molecule (as well as by the oxygen, as you noted):

$\hspace{1.75cm}$step 5

The spatially adjacent phenyl group accepts the proton and is ejected as benzene:

$\hspace{1cm}$step 6

Addition of water evolves $\ce{HCl}$ and precipitates $\ce{Al(OH)3}$, yielding the final product:

$\hspace{1.8cm}$step 7


I came across ringo's detailed mechanism of orthocresol's query about dephenylation during a Friedel-Crafts (FC) acylation. The penultimate step in ringo's mechanism involves the dephenylation. My thought is why a sigma bond of a phenyl ring is abstracting a proton that should be lost readily on its own to effect aromatization? The following modification to this step of an otherwise cogent mechanism is offered. The FC reaction liberates the strong acid H+ AlCl4-.

First, cation 1 loses a proton to generate species 2. Ipso protonation of the phenyl ring via the π-framework provides resonance stabilized cation 3 that collapses to afford aluminum salt 4 and benzene. Aqueous acid work-up gives 9-hydroxy-1H-phenalen-1-one 5.

enter image description here

  • $\begingroup$ In step 3, shouldn't the carbocation formed after protonation be present at the ipso position as at ipso position, it would become a three degree carbocation while at the current position(depicted in the mechanism) it is a two degree carbocation, noting that the extent of conjugation in both cases is similar? $\endgroup$ – YUSUF HASAN Dec 18 '18 at 10:47
  • $\begingroup$ I thought that the position shown for the carbocation might be preferred so that the negative inductive effect of oxygen is minimised, but then again, I am not sure how effective is that factor so far away in the molecule $\endgroup$ – YUSUF HASAN Dec 18 '18 at 10:49
  • 1
    $\begingroup$ @YUSUF HASAN: Protonation of structure 2 at the ortho-position does lead to a stable cation many times over BUT the cation does NOT lead to dephenylation. When ipso-protonation occurs as in 2 with the positive charge at the ortho-position and delocalized, dephenylation occurs. Notice in structure 2 the tail of the red arrow is at the ipso end of the double bond meaning that the double bond should be considered polarized ipso-negative, ortho-positive. $\endgroup$ – user55119 Dec 18 '18 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.