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What will be the major product of the following reaction?

Please also help me with the mechanism. I'm not sure whether C- acylation or O- acylation will occur here. And if O- acylation does occur, I think the heat will cause a Fries' rearrangement. What will be the major product of this rearrangement?

Also, if simple Friedel Crafts acylation occurs, will AlCl3 affect the reaction due to formation of -O-AlCl3 (Lewis acid-base interaction), something like what happens in the case of Friedel Crafts reactions with aniline? (I know aniline cannot undergo FC reactions due to deactivating effect of -NH-AlCl3 interactions)

I'm unable to upload a pic of the question due to some technical problem, but what I mean is reaction of ortho cresol and butyrolactone in AlCl3, along with heat.

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    $\begingroup$ When you write "a cyclic ester" do you mean a lactone? $\endgroup$ – Waylander Jan 15 at 20:19
  • $\begingroup$ Waylander it is a lactone.Recently a question on product formation between butyro lactone and ortho cresol in presence of aluminium chloride was asked in national exam conducted nation wide.How ever phenols may not undergo freidal crafts reaction. $\endgroup$ – Chakravarthy Kalyan Jan 16 at 3:27
  • $\begingroup$ @Chakravarthy Kalyan, Yes that is the question I am referring to. The official answer given has the carbonyl acylation at the position para with respect to methyl group. Is that right? I was thinking the acylation should be done at para position with respect to hydroxy group. $\endgroup$ – Ankit Kumar Misra Jan 16 at 5:53
  • $\begingroup$ It's possible that different conditions provide different products. $\endgroup$ – Zhe Jan 17 at 14:21
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I can take a guess since no one else has provided an accepted answer.

First, I don't think lactones will take part in FC-acylation on an arene, but rather prefer alkylation as mentioned in this paper. In the case of o-cresol and heat, this could be a possible explanation for your question why the O-acylation takes place instead of the C-acylation.

Second, when the Fries rearrangement occurs, and the acylium ion is released in the intermediate, the phenolic oxygen is complexed with the aluminum chloride (second to last step here). If you imagine an ortho-methyl group, as well as the complexed aluminum chloride, it's plausible substitution to the remaining ortho-site is not sterically favorable. This would force it to substitute elsewhere on the ring.

As for why "elsewhere" prefers para to the methyl group and not the oxygen—going by what you wrote in your comment—I'm not sure. The para alumino-oxy substitution should produce the more stabilizing resonance structure for the arenium intermediate. At least, in one other reaction of electrophilic aromatic substitution with ortho-cresol, namely its nitration, the 3/5-nitro products are not found. Maybe it's a solvent matter. Wikipedia states a non-polar solvent prefers ortho in the Fries rearrangement; perhaps if the ortho site is blocked, it will settle for para to a methyl substituent.

Edit

In a summary of the comments, the m-hydroxy-ketone was called into question as the correct product. I mentioned none of the papers I found of similar reactions furnished anything but the para or ortho product. Here are the papers I looked through:

paper 1 - first paragraph: o-tolyl acetates yield predominantly p-hydroxy-ketones.

paper 2 - tables 2 and 3

paper 3 - table 1 (page 378)

paper 4 - 5th paragraph

A few others: book p 223, paper, paper

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  • $\begingroup$ Thanks for your answer. But I still don't understand this... does the complexation of phenolic oxygen with AlCl3 convert it to a deactivating group? Why then do Friedel Crafts reactions and Fries rearrangements occur in phenols? And if it doesn't convert it to a deactivating group, does it just lower the directive power of the -OH group such that -Me becomes more dominant? $\endgroup$ – Ankit Kumar Misra Jan 16 at 18:17
  • $\begingroup$ To add to the confusion, this book states that the phenolic -OH group does not interact with Lewis acids like AlCl3 to the extent of affecting reactions. books.google.co.in/… $\endgroup$ – Ankit Kumar Misra Jan 16 at 18:22
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    $\begingroup$ I don't expect the phenol to initiate with aluminum chloride. O-acylation isn't a FC mechanism. O-acylation is catalyzed by an acid or base, and is kinetically favored under heat. I don't have literature on it, but my assumption would be that aluminum chloride acts as a lewis acid to coordinate to the carbonyl group of the lactone. The phenol attacks the labilized carbonyl to form an orthoester. The phenol hydrogen shifts to the neighboring oxygen, which is subsequently released, and the ester is formed. From there, a Fries rearrangement takes place, as described here: bit.ly/2VXs2yq $\endgroup$ – Blaise Jan 17 at 11:16
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    $\begingroup$ As for why methyl takes precedence in being para directing, I do not know. Are you sure the para-methyl product is correct? I would have also expected it to be para to the hydroxyl group. For example, this paper ( pubs.rsc.org/en/content/articlepdf/1958/jr/jr9580002926) describes o-tolyl acetates with AlCl$_3$ as resulting in a para-hydroxy arrangement, and other papers of similar reactions have the same result. $\endgroup$ – Blaise Jan 17 at 12:13
  • $\begingroup$ As the mechanism in the link shows, does Fries rearrangement always shift the attached acyl group to the ortho and para positions (with respect to phenolic hydroxy group)? Or does the product also depend upon other substituents on the aromatic ring? Are you aware of any such examples where Fries rearrangement shifts the acyl group to meta with respect to -OH due to other substituents of the phenyl ring? $\endgroup$ – Ankit Kumar Misra Jan 17 at 12:18
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It will follow the same mechanism as friedel crafts acylation.Just break the C-O bond of that cyclic Ester and add it to para (with respect to OH group) and the H removed will be added to the last O- and form OH group. enter image description here

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  • $\begingroup$ My doubt is, will the presence of AlCl3 as a medium here, affect the reaction, by interacting with the hydroxy group to form -O-AlCl3? My teachers are giving this explanation and saying that the -O-AlCl3 group will have lesser directive influence than the methyl group on the benzene ring, thus acylation will be done at the para position with respect to methyl group. $\endgroup$ – Ankit Kumar Misra Jan 16 at 6:01
  • $\begingroup$ @Ankit Kumar Misra phenol is an ambidentate nucleophile.Nucleophillic attack can take place can take place at C C -ACYLATION and O - Acylation.O acylation takes place via nucleophillic acyl substitution. IF 1 equivalent of AlCl3 is used then it will not function as a catalyst.In the paper given excess AlCl3 is not given.Product may not be formed. $\endgroup$ – Chakravarthy Kalyan Jan 16 at 6:25
  • $\begingroup$ @Chakravarthy Kalyan, so O-acylation product will be formed. And if we then assume excess AlCl3 and heat, what will happen next? I think a Fries rearrangement. What will be the product then? Which position of the benzene ring will the acyl group migrate to? $\endgroup$ – Ankit Kumar Misra Jan 16 at 9:58
  • $\begingroup$ @Ankit Kumar Misra heat or photo has not been given , thetefore fries rearrangement or photo fries rearrangement would not take place most likely.If these conditions ate given then acyl cation will participate in eas para to O assuming that one lone pair is in coordinate covalent bond with Al and other is not.This makes it to have greater directing then methyl group $\endgroup$ – Chakravarthy Kalyan Jan 16 at 10:33
  • $\begingroup$ @Chakravarthy Kalyan: Okay, thanks for the clarification. By the way, heat was given in the question, so I think Fries rearrangement should be considered here too. I think that means the product with acylation at the position para to -OH should be the major product. $\endgroup$ – Ankit Kumar Misra Jan 16 at 17:16

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