When a benzene ring with a substituent undergoes an electrophilic aromatic substitution , the electrophile is installed on a specific position s depending on the substituent.
There are three relative positions for a disubstituted benzene ring: ortho, meta, and para. All the functional groups are divided into ortho -, para or meta -directors. To which one the group belongs, depends on how it stabilizes or destabilizes the transition state of the electrophilic substitution reaction. Activating groups the ones that stabilize the transition state and make the reaction faster turn out to direct the electrophile to the ortho, para positions , while the deactivating groups put them in the meta position, except for the halogens!
Halogens deactivate the aromatic ring, yet they direct the electrophile to the ortho and para positions. But why? How does that happen, right? Well, first a few words on the activators and deactivators. In short, the groups that donate electron density to the ring and make it electron-rich are activators. Measure content performance.
Develop and improve products. List of Partners vendors. Share Flipboard Email. Anne Marie Helmenstine, Ph. Chemistry Expert. Helmenstine holds a Ph. She has taught science courses at the high school, college, and graduate levels. Facebook Facebook Twitter Twitter. Updated October 02, Featured Video. 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. By analyzing the intermediate for the meta — product below, can you see why?
There are no particularly stable resonance forms. Since the positive charge is localized on C2, C4, and C6, a situation where the positive charge is directly adjacent to the electron-withdrawing CF 3 group is avoided. And the positive charge is thus delocalized nicely throughout the ring, unlike in the ortho- situation, above. Finally, the para- intermediate has the same problem as the ortho- ; an intermediate with a positive charge on C1, adjacent to the CF 3. This results in only two important resonance forms, which leads to a less delocalized and therefore less stable carbocation intermediate.
Bottom line: the meta carbocation intermediate is more stable than either the ortho — or the para- intermediates. Clearly, the carbocation with an adjacent oxygen bearing a lone pair is far more stable than a carbocation adjacent to an electron-withdrawing group like CF 3.
And that comprises the difference between an ortho-, para- director like OCH 3 and a meta- director like CF 3. Save my name, email, and website in this browser for the next time I comment.
Notify me via e-mail if anyone answers my comment. This site uses Akismet to reduce spam. Learn how your comment data is processed. Two important reaction patterns are observed. Great question. We said that Activating groups increase the rate of electrophilic aromatic substitution, relative to hydrogen. Deactivating groups decrease the rate of electrophilic aromatic substitution, relative to hydrogen. First: no activating groups are meta directors. Polar Aprotic? Are Acids! What Holds The Nucleus Together?
What are alkynes considered? OH is more activating than OR. It is even more activating when deprotonated to give O -. Draw the resonance forms of the conjugate base.
In which is the negative charge the most stable?
0コメント