[Justin]

Hi, I was hoping someone could tell me how to determine where the NO2 group will be located in an electrophilic nitration reaction on armoatic rings. I can't seem to find a principal at work here other than the further from any other bonds the better, which seems to only be true some of the time.
Does anyone understand how to predict where the NO2 group will locate?

[russellm72]

OK,

First point read a textbook. The nitration of aromatic compounds depends on other groups on the aromatic nucleus and can be activating (o,p directing) or deactivating (m directing). I strongly suggest u read a text book as its way too long to explain here.

In addition, stereochemistry is not involved if we are talking about the nitration of aromatic rings I am assuming you meant regiochemistry??

Enjoy your read.

R.

[Justin]

I understand the mechanism of forming an electrophile and generating a sigma complex then eliminating a H, its just the location on the ring that I'm strugling with.
well now I understand what I'm looking to learn to understand this. That's all I need. Thank you.

[Will]

It depends on the effects of the other substituents on the aromatic ring. Some have a electron withdrawing effect on the aromatic ring (-NO₂, ammonium, phosphonium, carbonyls, sulfonates and cyanide) and then electrophilic substitution will normally occur in the meta position. If the substituent has an electron donating effect (like -OH, -NH₂, and alkyl groups) then electrophilic substitution will normally occur in the ortho and para positions. The halogens have a deactivating effect because of their electronegativity, but because of overlap of their orbitals with the aromatic system (not so much for the larger halogens) they are ortho and para directing.
Steric hinderance is also a factor, even adding a small Ac group to the N in phenylamine hinders electrophilic substitution at the ortho position.

I strongly suggest u read a text book as its way too long to explain here.

ressellm is right, the topic is quite long and a textbook would help a lot as you need diagrams etc.