[iSherry]

Hi! I'm currently studying basic level organic chemistry on my own and I'm stuck on a problem regarding the following compounds:


The question is asking us to combine our knowledge of alkene addition reactions and Electrophilic aromatic substitution. It asks which of above compounds would react quicker with HBr. Now, my initial guess was that since the methoxy group is activating and the nitro group is deactivating, the nitrostyrene would react slower because of its fewer resonance structures. However, I am fairly certain that I am missing something, since I don't exactly know how the compounds would react with HBr. I thought that the methoxy-compound would react like this:



The proton would bond with the carbon with the least substituents (Markovnikov's rule) and the positive charge would then be stabilized with the pi electrons in the ring. My problem is though that I do not understand what happens with the Bromide ion. It can't bond with the carbon with the most substituents because there is no charge on it. EAS is a substitution reaction but what am I even substituting here?

This all feels like it is very obvious but I simply can't detangle the mess that has formed in my head. I would really appreciate some help.

EDIT: The deactivating group would react slower, of course. Not the activating group.

[orthoformate]

My problem is though that I do not understand what happens with the Bromide ion. It can't bond with the carbon with the most substituents because there is no charge on it.

What do you mean here?

[iSherry]

My problem is though that I do not understand what happens with the Bromide ion. It can't bond with the carbon with the most substituents because there is no charge on it.

What do you mean here?

I mean that the carbon with the most substitients is the third one (counting from the right) and according to Markovnikov's rule the halogen bonds to the carbon with the most substituents. However, in my case the positive charge is on the second carbon which does not have the most substituents. So I don't really know what to to with the bromide ion.

[orthoformate]

In this picture, you have donated the proton to the double bond correctly. The carbocation is at the benzylic position and you have observed Markovnikov's rule, good work.

Now: You have a formal positive charge drawn, and I think you know that the negative charge is on the bromide ion. Where could that bromide ion land?

what would happen if it landed at the benzylic position? What would happen if it landed at either the ortho or para positions?

Draw all the resonance structures. show me what you think would/could happen at each point.

[iSherry]

I am in the middle of a move and cannot sit down to do this today, but I will tomorrow and get back to you then. Thank you for taking the time to help me!

[iSherry]

I tried to draw the possible resonance structures but I am just so confused. I don't know how to place a positive formal charge in either the para or ortho position in relation to the MeO group. It seems that only meta positions are available. Also, if the Br- were to bond in the benzylic position then wouldn't the compound be neutral? Thus finishing the mechanism?

I'm sorry for the blurry photo. All charges on the resonance structures (marked with a circle around them) are positive.

[orthoformate]

Hi,

I'm sorry, I should have made it clear that i meant ortho and para relative to the vinylic group that your attacking with the HBr. You have put the positive charges where they are supposed to go, good work.

You forgot one resonance form where the methoxy oxygen octet resonance stabilizes the carbocation at the para position.

You are correct in thinking that if you put the bromide down on the benzylic position the compound would be neutral and the mechanism would be complete. In my opinion, this is the correct answer.

You can try to put the bromide ion down at the other points on the ring, but you will find that you cannot re-establish aromaticity if you do that.


I think when your teacher said "use what you know about electrophilic aromatic substitution" your teacher meant for you to to draw the resonance structures. If you draw them the same way with the nitro group present, you will see that the cation locates itself geminal to the nitro...this will slow your HBr attack on the double bond down, possibly even lead to no reaction on the nitro containing compound.

Is this making sense? You said you were confused, but you actually seem to have a really good grasp of this stuff.