Sciencemadness Discussion Board

Resonance stabilisation

maniacscientist - 1-1-2008 at 11:25

Hi

and a great new year to all!

I was reading and experimenting with o, p and meta directing groups
as well as predicting and experimentally confirming inductive effects on reactivity of certain reactions.

Anyway, I still wonder why some substituents on the benzene ring direct to ortho and para, so substituents with an electronpair, which is donated to the system (+M- effect) and not meta -the point is, I don´t understand the resonance structure and how this works to justify the activation of the ortho and para carbon on the benzene, by donating electron pairs to these postition exclusively and why is the meta position within the deactivated ring favorable?

Could anyone help me out, or is this just another it´s like it is-scenario?

maniac

vulture - 1-1-2008 at 13:39

Quote:

I don´t understand the resonance structure and how this works to justify the activation of the ortho and para carbon on the benzene, by donating electron pairs to these postition exclusively and why is the meta position within the deactivated ring favorable?


First of all, it's important that you realize that this counts for electrophilic aromatic substitution only. Make sure you understand that mechanism and why it is favorable in that case to have a higher electron density on the reactive aromatic position.

Now as for resonance structures, if you draw all resonance structures for benzene with a electrondonating substituent it should become obvious, that no matter how hard you try, you'll always end up at the same positions with your negative charge.

As for electron withdrawing substituents, you'll need to do exactly the same, but with a positive charge. It is also important to realize that the meta position isn't activated in this case, but _less_ deactivated than the others.

maniacscientist - 4-1-2008 at 12:25

Hi,

I know about SE-reactions.

In an enceclopaedia it´s explained with acetamide as an example, which is logical to me, as are all conjugate alkenes, but how can it be explained with the benzene ring?

Here it´s the situation, that the resoance means, all electrons are evenly distributed among the whole ring, right? It´s a psi electroncloud with a distance of somewhere inbetween a double bond and a single bond.
Then there´s some H atoms within each carbone with a sigma bond, right? Do these interfere with the resonace structure?

So I don´t see, how one could draw a resonance structure, which would explain subsequent ortho and para substitutions? The pics I´ve seen, just draw an electronpair donated to the ring at the o and p, but I don´t see, why it couldn´t be drawn to the meta postition? -to me there´s no obvious rule which seemsd applicable not to draw the electronpair to the meta position.

Nicodem - 4-1-2008 at 12:56

Quote:
Originally posted by maniacscientist
The pics I´ve seen, just draw an electronpair donated to the ring at the o and p, but I don´t see, why it couldn´t be drawn to the meta postition? -to me there´s no obvious rule which seemsd applicable not to draw the electronpair to the meta position.

If you would have at least tried to draw the resonance structures, you would have known it is not possible to draw a resonance with negative charge in meta positions.

Try reading http://www.cem.msu.edu/~reusch/VirtTxtJml/benzrx1.htm
And also do some exercises in drawing resonance structures for groups with differing M and I effects, including the ones with greater M than I effect and vice versa. Its all in making exercise, quite simple otherwise.

chemrox - 4-1-2008 at 21:07

Nice link!

maniacscientist - 15-1-2008 at 13:39

I tried, there´s one thing I don´t unmderstand in the explanation: Br as an example has three free electron pairs, of which it donates one (electron or pair? wit help of a catalyst FeBr3?) to the benzene-ring and then it says there´d be ne electronpair which can be donated to the aromatic ring?
Still I prefer to stick to the statistic, as a chem. student told me it´s quantum mechanics to describe the ortho and para directing of a second ring substitutent.

To the q: Where´re the other two electron pairs or should it be four to reach the nobel gas structure?

chemrox - 15-1-2008 at 21:09

Br- as it approaches the benzene molecule has four pairs. Don't assume any hybridization or assume sp as it was in the Br-Br bond. Only one pair will be oriented correctly to bond with the sp2 orbitals available for bonding on benzene. You get a sigma bond between the px or pz orbital on Br and one of the sp2 orbitals in the plane of the rings not already part of the sigma framework.

vulture - 17-1-2008 at 10:10

Quote:

I tried, there´s one thing I don´t unmderstand in the explanation: Br as an example has three free electron pairs, of which it donates one (electron or pair? wit help of a catalyst FeBr3?) to the benzene-ring and then it says there´d be ne electronpair which can be donated to the aromatic ring?


I hope that chemistry student flunks because what he told you is absolute BS. Look up the mechanism! FeBr4- and Br+ is generated which is attacked by the aromatic, not the other way round. And it can probably calculated with QM, but you'll need several days of computing power to calculate something, which has a large statistical error, which you can do in 5 minutes by drawing resonance structures for crying out loud! Unless you really want to be part of the chemistry chairforce, ofcourse.

maniacscientist - 21-1-2008 at 06:47

thnaks, that´s some good help, which came along, so now I´m just having a problem with the definition mesomerism and, what is called in my language: boarder structure.

So for 1,3-butadiene, the structures are: delta+C-C=C-C[delta-] and C=C-C=C (H-atoms left out ) for example a Nitrobond it´s one negative charge on both oxygen of the nitro, a double bond at the 1 to the 2 C-atom of benzene and a doluble bond at the 1 to the 6th carbone of benzene, which leaves me with two additiona double bonds of the same positions, just it´s the para position?

O- O-
\ /
N
// \\
| |
\\ //
It has to be conjugate, I see, so is this above the resonance structure of a Nitrobenzene? Or, for conjugation´s sake, has one of the pi-bond to become a negative charge on the "outside" of the ring at the ortho or para position and is the reason, that within each of this position the maximum of other structures being possible, that these are therefore favoured?

[Edited on 21-1-2008 by maniacscientist]