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Updated Q:

Inquiring to understand the efficacy of intramolecular Friedel-Craft like reaction on stability on the ring structures with =o, c=o bonds.

Keen to know if ring structures are challenged

  1. solely based on specific species of Lewis acid used in reaction acting as electrophiles or
  2. based on the changes in the pH (by any acid) that corresponds to pKa dependent stability of the core-structure (as shown in the rough sketch below).
  3. or as a combination of both? (with the reasoning that in FC, as species like Al3+ may not just remain spectators... but catalysts)

Previous:

AlCl3 is a Lewis acid. In Friedel-Craft reaction, purpose of AlCl3 is to produce electrophile, which later adds to benzene nucleus. This electrophilic aromatic substitution allows the synthesis of monoacylated products from the reaction between arenes and acyl chlorides or anhydrides. The products are deactivated and do not undergo a second substitution. Normally, a stoichiometric amount of the Lewis acid catalyst is required for both the substrate and the product form complexes. AlCl3 (and other Lewis acids like it) will coordinate to halogens, and facilitate the breaking of these bonds. In doing so, it increases the electrophilicity of its binding partner, making it much more reactive.

LaCl3 and FeCl3 are mild Lewis Acid. According Lewis Acid, the more deficient of electron to complete octate, the more acidic in strength. AlCl3 have vacant orbital to fulfil octate, where as FeCl3 have already octate form so AlCl3 is more acidic than FeCl3. What about LaCl3?

  1. Can the Lewis dot structure be shown for LaCl3 as an electrophile in this reaction? (it's important in this context)

  2. Can FeCl3, LaCl3 substitute AlCl3 in above Friedel-Craft reaction?

  3. What impact can Friedel-Craft have on the following reversible reaction? If FC doesn't apply to it, then on what account AlCl3(aq) Lewis acid (electrophile) can result in it?

enter image description here ​

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    $\begingroup$ FeCl3 has been used with anisole: pubs.acs.org/doi/pdf/10.1021/ed073p272 $\endgroup$
    – user55119
    Commented Oct 14, 2021 at 2:11
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    $\begingroup$ In its initial form (October 16th), your question was about «Can $\ce{FeCl3}$, $\ce{LaCl3}$ substitute $\ce{AlCl3}$ in above Friedel-Craft reaction?» After multiple edits and answers already received, at present (October 19th), the question now is about efficiency and selectivity of catalysts and conditions of Friedel-Crafts reactions. In my perception, this is no edit to clarify a question. Instead, it is moving the target a question addresses. Please refrain from this. Instead, rise a separate question, including your lines of thought. $\endgroup$
    – Buttonwood
    Commented Oct 19, 2021 at 8:06
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    $\begingroup$ @bonCodigo - Please refrain from multiple edits that turn a query into a moving target. If you have other questions please ask them separately. $\endgroup$
    – Todd Minehardt
    Commented Oct 19, 2021 at 14:52

2 Answers 2

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Yes, $\ce{FeCl3}$ is a well known Lewis acid that is used in place of $\ce{AlCl3}$ for FC reaction. See the paper linked by user55119 and also this question.

Other $\ce{MX_n}$ types of reagents used in FC reactions are $\ce{BF3}$, $\ce{BeCl2}$, $\ce{TiCl4}$, $\ce{SbCl5}$, $\ce{SnCl4}$, $\ce{SeCl4}$, $\ce{TeCl4}$, $\ce{InCl3}$, $\ce{NbCl5}$, $\ce{IrCl3}$ or $\ce{RhCl3}$.

Regarding lanthanum, they are mostly used in the form of triflates ($\ce{La(OTf)3)}$). Other lanthanoids used as triflates are neodymium, ytterbium, europium and samarium along with yttrium.

Reference

  1. Rueping M, Nachtsheim BJ. A review of new developments in the Friedel-Crafts alkylation - From green chemistry to asymmetric catalysis. Beilstein J Org Chem. 2010 ;6:6. doi:10.3762/bjoc.6.6 (There is a periodic table which mentions which metal salts can as be used as FC reagents)
  2. Striving towards improved Friedel-Crafts acylation catalysts, International Nuclear Information System (INIS), Scott, N.M.; Deacon, G.B, 1998 (link)
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    $\begingroup$ @bonCodigo I agree on the point that Friedel-Crafts reactions require anhydrous conditions. Still, many proceed in solution for easier transfer of heat and substance, and because some of the Lewis acids mentioned either are a liquid (e.g., $\ce{TiCl4}$, $\ce{SnCl4}$), or are easier to work with/to purchase as a reagent solution (e.g., $\ce{BF3}$ as adduct with diethyl ether, $\ce{BF3\cdot OEt2}$). $\endgroup$
    – Buttonwood
    Commented Oct 14, 2021 at 13:20
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    $\begingroup$ $\ce{AlCl3}$ and $\ce{FeCl3}$ are hydrolyzed in contact with water, and are not usable for Friedel-Crafts synthesis $\endgroup$
    – Maurice
    Commented Oct 16, 2021 at 9:22
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    $\begingroup$ Your question 3 has nothing to do with a Friedel-Crafts reaction... $\endgroup$
    – Maurice
    Commented Oct 16, 2021 at 14:13
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    $\begingroup$ @bonCodigo In same tune like @Maurice, $\ce{AlCl}$ decomposes by action of water (to generate corrosive $\ce{HCl}$). First intentional encounter in the elementary lab classes org chem, to initiate the work up. But $\ce{TiCl4}$ and $\ce{SnCl4}$ used for e.g., other FC reactions, these decompose by water, too. 3) What is this for a drawing added? If you use the arrow for mesomers, which may alter bond order but not create new / delete old bonds (which were a reaction) between atoms, nor alter the atom count. But the formula on the right has one C and two H more than the on the left. $\endgroup$
    – Buttonwood
    Commented Oct 16, 2021 at 16:47
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    $\begingroup$ I wish I could accepted two answers or more. I upvoted yours as it serves previous part of the question. Let's keep yours? I am thinking perhaps, I need to post a new question with the actual title of the question above. $\endgroup$
    – bonCodigo
    Commented Oct 20, 2021 at 9:46
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Your question stays fuzzy, and the addition of the scheme does not add much clarification. The arrow between mesomers is not the one about a reaction, nor about a chemical equilibrium. (Typically, R and R' designate substituents at the periphery of a chain, or a cyclic core structure; but not about what different spaces within a chain. This just as a side note.)

Assuming you would like to submit alcohol 1 to conditions of a Friedel-Crafts reaction with $\ce{AlCl3}$

enter image description here

the alcohol would react with the Lewis acid to yield $\ce{HCl}$ (like a hydrolysis) and alkyl chloride 2. In principle this might lead to an intramolecular Friedel-Crafts alkylation 3. However it is more likely that Lewis acid $\ce{AlCl3}$ yields a complex with the Lewis basic carbonyl oxygen of the ester 4 (and, similarly, on 5, too).

Or, do you intend to use a large excess of $\ce{AlCl3}$ to proceed further? If you intend to open the ester (similar to the Harworth reaction between maleic anhydride and benzene), the then simultaneous presence of an acyl chloride and now two alkyl chlorides will render the reaction messy.

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  • $\begingroup$ The R and R' represents extended rings with polar groups. The particular species tend to have few different pKa values. When environment becomes acidic, they expect to have the ring open at c-o-c and becomes coo- (as above). This is reversible relative to the pH of the solution (corresponding pKa of the species). I just wanted to confirm from an organic chemist point of view, that this can actually happen and the closest reaction I could think of was FC (intra-molecular). So I posted the question. $\endgroup$
    – bonCodigo
    Commented Oct 18, 2021 at 0:00
  • $\begingroup$ what reaction (even if it's not FC) can result in opening and closing of that particular position of the ring? $\endgroup$
    – bonCodigo
    Commented Oct 18, 2021 at 0:10
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    $\begingroup$ The cyclization 2 to 3 is likely, because it would be an intramolecular reaction and the ring formed includes five carbon atoms. If 4 opens, I speculate the alcohol moiety of the ester equally is replaced installing a second alkylhalogen potentially engaging in FC type reactions. Formation of a cycle of now 4 carbon atoms is more costly in energy, but some competition with the other group is possible. It then becomes more likely that the FC reaction favours an intermolecular pathway, especially with the then present acyl chloride (cf. FC acylation). $\endgroup$
    – Buttonwood
    Commented Oct 18, 2021 at 15:57

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