Intramolecular Friedel-Crafts Reactions

by James

in Aromatic Reactions, Organic Chemistry 2

Intramolecular Reactions

Here is an instant formula for an organic chemistry exam question.

Start with a straightforward reaction that students understand fairly well,  like the Williamson Ether synthesis….

(drawn weirdly, for a good purpose)

…now, just make a simple modification by adding one bond. Voila. Instant stumper!

The second question above is an example of an intramolecular reaction, where the nucleophile and electrophile are on the same molecule, and the result of their reaction is that a ring is formed. In past posts on this subject, I’ve used the analogy of a belt. It still works.

Why are intramolecular reactions good exam questions? Because they sort out the students who learn the reactions by memorizing a table of simple examples, and those who actually know (and most importantly, can apply!) the pattern of bonds formed and bonds broken. Furthermore, it involves no new concepts, which makes it totally fair game.

[You think I am joking about the “instant stumper” comment? I am not. As one student memorably put it, “Intramolecular Reactions Make Me Crumple Into The Fetal Position Crying For My Mommy” ]

Intramolecular variants exist for a lot of different reactions, and it comes up so often that it’s worth mentioning separately. The Friedel-Crafts alkylation and Friedel-Crafts acylation reactions are no exception.

Intramolecular Friedel-Crafts Alkylation

Here’s an example of an intermolecular Friedel-Crafts alkylation. The nucleophile is the aromatic ring, and the electrophile is the alkyl chloride. Add a little catalyst (AlCl3) and boom!  electrophilic aromatic substitution. 

Now let’s change things up just a little bit. We’ll attach the alkyl halide to the ring via a new carbon-carbon bond, and then add the catalyst. What’s the product?

No new concepts! Same pattern of bonds that form and break. But if you haven’t seen an example like this before, it might throw you completely off your memorized notes. That’s the plan!  [Cue recording of  “evil laugh”].

via GIPHY

OK. Here’s the mechanism, and the final product is drawn out below left.

I want you to note that the overall pattern of bonds that form and bonds that break is exactly the same in the intramolecular case as it is in the intermolecular case, namely: form C-C and H-Cl,  break C-H and C-Cl.

Timeless advice for drawing out the mechanism for a reaction like this:

  • number the carbons! it’s really easy “drop” a carbon in your drawings,  and that will lose you points.
  • draw the “ugly version” first, and THEN re-draw to make it look nice. Don’t worry about making it look pretty until you have drawn in the bonds that form and break.

In the case of the Friedel-Crafts, the intramolecular version works best for making 6-membered rings, but it’s also possible to use the Friedel-Crafts to make 5- and 7- membered rings as well (not shown). 

Here’s a slightly more advanced practice problem that starts with benzene. Can you draw the final product? (answer below)

Intramolecular Friedel-Crafts Acylation

Once you’ve seen the intramolecular Friedel-Crafts alkylation, the intramolecular Friedel-Crafts acylation is not exactly going to come as a surprise.

Again, I want you to verify that the bonds being formed and broken are exactly the same in each case. The only difference is that in the second case, the nucleophile and the electrophile are attached to each other through a tether.

One little wrinkle that you probably won’t see, but what the heck. We’re used to seeing acyl halides (and anhydrides) in the Friedel-Crafts, but one interesting thing to note about the intramolecular Friedel-Crafts acylation is that carboxylic acids can participate too. For example, treating this carboxylic acid with a strong acid such as H2SO4 results in an intramolecular Friedel-Crafts acylation:

One reason why this works well for the intra- versus the intermolecular case is that the nucleophile is held so closely to the electrophile. This has the same effect as if the concentration of the electrophile was increased dramatically. You might see other examples where you can “get away with” using a poor nucleophile (or electrophile) in a given reaction if it’s done in an intramolecular fashion.

Here’s a challenge question for you. Can you draw the product of this sequence of Friedel-Crafts reactions?

(answer below)

Conclusion

No really groundbreaking new concepts here, but it’s always helpful to keep alert for intramolecular examples of reactions. This concept never goes away. You will see it again!

In our next post in this series we’ll cover a completely different class of substitution reactions on aromatic compounds that works well with electron-poor aromatic groups. It’s called Nucleophilic Aromatic Substitution.


notes:

Rearrangement vs. ring closure. We’ve seen that primary alkyl halides can rearrange via hydride (and alkyl) shifts to give the “more stable” carbocation intermediates. So what happens when a primary alkyl halide is involved in an intramolecular Friedel-Crafts alkylation reaction? Does rearrangement happen first, or is ring closure faster?

This is not the kind of question you can answer simply by thinking about it. When there are competing reaction rates, the only way to know for sure is through experiment.

This was studied, and the full paper on the study of rates of rearrangement versus ring closure is here. [J. Org. Chem, 1966, 31, 89]. Would indicate that closure to the 6-membered ring is faster than closure to the 5 membered ring in this case.

Answer #1

Answer #2. 

 

Reference

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{ 3 comments… read them below or add one }

Victor

I wonder if there would be any takers to propose the mechanism for the tetralone formation via the above mentioned Friedel-Crafts :) I used to put that one on an exam for like 3 or 4 years. Students didn’t appreciate the challenge…

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Kirk Meester

In practice problem #1 it appears that you changed the reactant from a seven carbon chain in the practice problem to a six carbon chain in the answer key.

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James

Shoot. Thank you Kirk, will fix.

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