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In the following reaction there are three possible aryl groups that can migrate: an unsubstituted phenyl group, a p-methoxyphenyl group, and a p-nitrophenyl group:

Preferential migration of p-methoxyphenyl group

Why does the electron-rich p-methoxyphenyl group migrate instead of the phenyl or p-nitrophenyl group, especially when it leads to a carbocation that is less stabilised (since it will no longer be conjugated with the stabilising p-methoxyphenyl group, cf. the crossed-out arrow)?

orthocresol
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Katia
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  • Probably due to an increase in tendency to stablize the positive charge. But this assumption is not true always as there are reactions where the contrary occurs. – Solid - NMR Aug 01 '20 at 09:53
  • This is a shift which follows a migratory aptitude which differs slightly depending on the reaction you are talking about. In your case, the migration will take place with the most donating ring to stabilize the cation by a sort of "neighboring group participation" .See this for more details – Yusuf Hasan Aug 01 '20 at 09:57
  • Also see ron's answer here for a general discussion of migratory aptitude – Yusuf Hasan Aug 01 '20 at 09:58
  • But why not think of it in another way, if the nitrobenzyl migrated, we will get a more stable carbocation with the NO2 being further away and (OCH3Bn) being closer to the carbocation, which means closer electron donating group and further electron withdrawing group = more stable carbocation. – Katia Aug 01 '20 at 10:05
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    @Katia Did you check this link that I gave above? Your carbocation will be stabilized by a similar "neighbouring group participation" by the formation of a phenoium cation ispecies. While this migration is occuring, the positive charge of the carbocation will be delocalized over the migrating ring. Now try the migrating ring was nitrobenzyl, the -NO2 would in fact highly destabilize the transition state by it's electron withdrawing effect – Yusuf Hasan Aug 01 '20 at 12:04
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    @YusufHasan indeed, this would certainly be the kinetic product. Any thoughts on thermodynamic product? Presumably, the other structure Katia proposed would be the thermodynamically favored product (but it would have a hard time getting there!) – jezzo Aug 01 '20 at 12:42
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    @jezzo Yes, I believe that Katia's proposed product might fit the bill of thermodynamically controlled product, as I can see that the product #2 shown in the diagram has the positive charge delocalized over the anisole ring. But again, the activation energy barrier to get to this product would be quite high due to the destabilized transition state, hence under normal conditions, I would propose the kinetically controlled product(that is, product #1 in the diagram) unless condition are specified for thermodynamic control – Yusuf Hasan Aug 01 '20 at 13:17
  • That is because migratory aptitude depends on electron density.Here, -m groups withdraw electrons from benzene ring making it slightly less electron dense. – Ruchi s. Aug 26 '22 at 13:09

1 Answers1

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The answer varies slightly depending on whether the migration is one-step or two-step via an intermediate.

In either case, the respective transition state or intermediate looks like this:

mechanism

You can see that the middle structure would be stabilized most by a group that is able to stabilize the ring via resonance, for example, an alkoxy substituent.

If the mechanism is single-step, then that substitution would stabilize the transition state and make this migration the fastest.

If the mechanism is two-step, then we can invoke the Hammond postulate to say that the transition state leading to the higher energy intermediate will resemble it in structure and thus also be stabilized by the same kind of substitution, leading to the same conclusion.

We therefore suggest that the propensity of migration is X=OMe > X=H > X=NO₂.

Zhe
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  • Thank you so much! any reference? where did you learn this? – Katia Aug 01 '20 at 21:14
  • I have no idea where or even if I learned this. You gave an observation, and I used what I know about principles of organic chemistry to provide a rationale. This is a core skill for a chemist. – Zhe Aug 01 '20 at 22:04
  • This should be covered in a first- or second-year organic chemistry course, @Katia. Likely not this specific reaction, but the general skill of drawing resonance structures for intermediate products and assessing their relative stabilities in order to predict the most likely products of the reaction. Any college-level organic chemistry textbook should have a treatment of this in general terms, although it likely won't be a specific page or even chapter. As Zhe said, it's more of a general or core skill. – Cody Gray - on strike Aug 02 '20 at 03:10
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    @Katia if the answer has helped you, please consider accepting it (click the green check mark near the upvote/downvote buttons). See: What should I do when someone answers my question? for more information. – orthocresol Aug 02 '20 at 07:31