Electron Withdrawing vs Electron Donating groups

Question

Hello, could someone please explain how to determine if an R group is more electron withdrawing or donating in a benzene ring (e.g. phenols etc.) (especially explain if oxygen is electrowithdraw or donating?)

I have never taken O-chem.

How does e-withdrawing vs electron donating affect the pKas?

Do the positions of the R group affect the pKas?

Answer 1

-The way that this can be done is by drawing resonance structures. If the resonance structure ends up taking the electrons out of the benzene ring (for instance, a ketone bonded to the benzene ring), then the group is an electron withdrawing group. If it ends up with the electrons going into the benzene ring, then it is an electron donating group.

-Oxygen would initially be thought of as electron withdrawing since it is very electronegative, but because it can participate in resonance stability, it is actually electron donating in this context; essentially, if the electronegativity outweighs the resonance effect, it would be electron withdrawing. The opposite would give us an electron donating group (as with oxygen).

-In general, alkyl groups (carbon groups, with only carbon and hydrogen), will donate electron density, and this will raise pKa (and decrease acidity).

-An electron withdrawing group will generally lower the pKa of the molecule (making it more acidic), since the conjugate base that results will be more stable since the electrons can be spread out over more atoms. The more spread out the electrons are, the more stable it is (and the weaker the conjugate base, and the lower the pKa).

-Electron donating groups will increase the pKa since they will tend to de-stabilize the conjugate base, which would make it a stronger conjugate base (and in turn, a weaker acid).

-The R group position can affect pKa, since if we had a phenol and the electron withdrawing group such as the carbonyl or nitro group were meta (one carbon in between, not next to each other as in ortho or 180 degrees apart as in para), the electrons could not participate in resonance with the nitro group and become delocalized. With the phenol (OH group) and nitro either ortho and para with respect to one another, they could be delocalized onto the nitro group and this would decrease the pKa since the conjugate base is more stable (and the acid is more acidic with a lower pKa).

Answer 2

If you haven't taken O-Chem, this might be challenging for you. But here we go:

In the case of phenol, the -OH group has both and inductive electron-withdrawing effect and a mesomeric electron donating effect.

This mesomeric effect can happen because the lone pair of electrons in oxygen can interact with the pi-bonds in the benzene ring. For reactions involving the benzene ring, this effect is very important.

You probably already know that an acid has a conjugated base. One way to compare the strength of acids is to compare the stability of their conjugated bases. The more stable they are, stronger the acid is (and lower is its pKa).

Now let's compare the acidity of phenol and phenylmethanol. Due to mesomeric effects, the phenoxi base is more stable than the phenylmethoxi base. Therefore phenol has a lower pKa than phenylmethanol. Note that the -OH group in phenol has a electron-donating mesomeric effect, therefore helping stabilize the conjugated-base.

If you have just one R group, the position doesn't affect. If you have more groups, then you have to consider always the stability of the conjugated base. Both inductive ans mesomeric effects.

Wikipedia has a great introduction to this topic:https://en.wikipedia.org/wiki/Mesomeric_effect

Hope I have helped!