Why meta directing

If you look through the list of ortho- , para- directors, you might recognize that many of them are also activating groups. Yes indeed. What is up with the halogens, and how is it that they can be deactivating i. However, the first place to start is that it has to do with the stability of the carbocation intermediate in electrophilic aromatic substitution reactions.

More specifically, how does each substituent affect the stability of that intermediate? It might be worth going back and revisiting some of the factors that affect the stability of carbocations.

And also, if you prefer to look at it from the opposite side of the coin, here are some of the factors which make carbocations more unstable. Great article! But, what if you have a Ph group as an substituent?

Is it considered as a alkyl group as far a being a o,p, director? Excellent question. It would be considered an electron donating group, yes — an ortho,para director. Hi James, thank you for the brilliant article?

What if we have a trityl as a substituent? I believe it should be a meta director since the carbon would be quite electrophilic due to the -I effect of the phenyl group. Why would you choose to have trityl as a substituent? You have three phenyl groups which could interfere with your desired EAS. This article is quite detailed. But how does this ortho para affect the difference in boiling points and melting points when it comes to isomerism.

Regerence: B. Moodie, J. Penton, and K. Copy to clipboard. United States. Anionic systems are always electron donors. Other availability. Find in Google Scholar. Search WorldCat to find libraries that may hold this journal. Similar Records. In the previous post we introduced ortho- ,para- and meta- directors in electrophilic aromatic substitution. Previous to that we covered the mechanism of electrophilic aromatic substitution , and showed that the mechanism proceeds through a carbocation intermediate.

Carbocations are electron-poor species with six electrons in their valence shell. So, any substituent which can donate electron density toward the carbocation will be stabilizing.

This includes:. Given everything written above, which of the two carbocations below will be more stable? If you can answer this question well, you are most of the way towards understanding the difference between ortho-, para- and meta- directors.

Electrophilic aromatic substitution of methoxybenzene tends to give ortho- and para- products with very little meta-. We break a C—C pi bond and form a new C—E bond where E is intended to represent a generic electrophile , generating a resonance-stabilized carbocation:. If you look closely, you should note that we can form four resonance forms.

In this resonance form, all of the carbon atoms have a full octet of electrons. This is an example of pi donation. Note the difference! Attack of the electrophile at C-3 results in a carbocation which can be delocalized via resonance to C2, C4, and C6. This makes the meta- carbocation intermediate much less stable than the ortho- carbocation intermediate. Here we can again draw resonance forms with carbocations on C1, C3, and C5, as well as a fourth resonance form where the attached oxygen atom donates an electron pair to the carbocation on C1, resulting in a full octet at carbon.

This is a situation essentially the same as the ortho- intermediate. So, by analysis of the resonance forms and mark my word, you may well be asked to do the same on a test in the near future we can see that the intermediate carbocations resulting from ortho- and para — addition are considerably more stable than the intermediate from meta — addition. This explains is why ortho- and para- products dominate, and the meta- product is minor. The transition state leading to the ortho- and para- addition products is much lower in energy than the meta —.

One question — why do you think that para- might be favored e. So what about CF 3? Why does it give meta- products? As with the ortho- intermediate above, we can draw three resonance forms placing the carbocation on C1, C3, and C5, respectively. Therefore we would expect this to be a very minor resonance contributor, with the result that the positive charge is only delocalized over C3 and C5.