Help Me with O-Chem

[Harbindoc]

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Hey I was wondering if anyone could help with this stupid problem I can't figure it out!

The problem is:

1,4-dimethyl benzene is reacted with F2C=CFCH2Br in an AlCl3 catalyst. Explain why the preferred product has the F2C=CFCH2 group in the ortho position to the number 1 methyl group. Hint: the relative carbocation stabilities are important.


PLEASE HELP!!!!! I have been trying to figure this out for an hour!!! THANKS!

 

[VienneseWaltz]

I think your TA or classmates would be a more appropriate source of information.

 

[Doctor Octopus]

That's a weird question. All four positions that are avaliable for electrophilic aromatic substitution are equivalent. I don't see how one could be favored over another...

 

[dark_scientist]

It would be a lot easier for me if I was able to draw if for you, but I'll try to explain it the best I can:

The aluminum tri chloride catalyst makes the F2C=CFCH2 species an excellent electrophile, making it susceptible to attack from the 1,4 dimethyl benzene compound. From my understanding, the methyl groups activate the ring for electrophilic aromatic substitution (creating partial negative charges at the ortho positions to the methyl group). Thus, the double bond attacks the positive charge on the methyl group of F2C=CFCH2. This functionaly group will attach to the ortho position for two reasons: 1) steric hindrance from the methyl group and 2) a tertiary carbocation is more stable than a secondary one - ie a carbocation is formed at the location of the methyl group.

I hope that helps.

 

[dark_scientist]

Originally posted by Doctor Octopus
That's a weird question. All four positions that are avaliable for electrophilic aromatic substitution are equivalent. I don't see how one could be favored over another...

That's true, neither position is favored - you will still end up with a product that has been substituted at an ortho position. Only one addition occurs because the F2C=CH2 group is deactivating the ring for further addition.

 

[Doctor Octopus]

Originally posted by dark_scientist
That's true, neither position is favored - you will still end up with a product that has been substituted at an ortho position. Only one addition occurs because the F2C=CH2 group is deactivating the ring for further addition.

Right, and the question asks why the position ortho to the #1 Methyl group is favored, as if their really was some difference between the two groups, or between any of the avaliable positions which are both ortho and meta to one or the other methyl group. Poorly worded and pointless question, if you ask me.

 

[Harbindoc]

Thanks for all the help. I agree with all of your sentiments which is why the question has been driving me crazy! Since it is a ring with two methyl groups at the #1 and 4 position isn't it symmetrical? I mean if it is ortho to one methyl group it is meta to the other and vice versa. I am not sure what the difference is in position. It was assigned by my prof on a problem set that is due tomorrow. I went to ask her about it today, and she just reiterated that carbocation stability is important... that was about all she would tell me. I will post and let everyone know the outcome of the problem when I get it back. For now, I give up!! 🙄 Thanks again for all your help.

 

[UCLA2000]

Originally posted by Harbindoc
Hey I was wondering if anyone could help with this stupid problem I can't figure it out!

The problem is:

1,4-dimethyl benzene is reacted with F2C=CFCH2Br in an AlCl3 catalyst. Explain why the preferred product has the F2C=CFCH2 group in the ortho position to the number 1 methyl group. Hint: the relative carbocation stabilities are important.


PLEASE HELP!!!!! I have been trying to figure this out for an hour!!! THANKS!

Disclaimer #1 I have not done ochem in years.
Disclaimer #2 I didn't really think about the problem that much.

I took a quick glance, completely ignored structure and saw a Bromine. Then I read the words preferred ortho product.

First thing to pop into my head was Bromine is huge (and that's why it does a backside attack). If you have a preferred product it's gotta be a steric hindrance thing.

 

[Ramoray]

ortho to 1, and ortho to 4 is the same product its like there is no way it has to be ortho to 1, cause 1 is just a naming thing, like ortho to 4 would look identical, i think the structure might be 1, 3 dimethyl benzene, then ortho to 1 would also be para to the 3 methyl making it the favored product due to hiindrince ofr one and it being directed to ortho by the 1 methyl and para by the 3 mehtyl.. 1,4 methyl doenst have a favored product The problem is wrong

 

[pallhaco doce]

Ramoray is right. There will be no difference in which position is favored with 1,4 methyl substituents. The question does not make sense unless you have (1) different substituents, or (2) different positions for the same substituent (i.e., 1,3). Even in the 1,3 case, the ortho for one -CH3 will be para for the other and vice versa. Of course, your prof might be trying to ask a trick question of some sort, which is just lame.

Now, back to checking my mailbox.