In the 2C-x family of drugs, the x part is whatever halogen has replaced the hydrogen atom which would be found at the 4th position.

What I'm wondering is, why is it the 4th position?

Looking at the molecule, I don't see why it favours position 4 explicitly. Would the 3rd position not be equally as appealing?

I'm obviously missing something, or just haven't covered that yet in my learning, or maybe even somewhere in between...

http://img143.imageshack.us/img143/3364/2cxtc9.th.jpg

The only thing I can think is that the methyl part of the methoxy group at position 2 is bent as the picture suggests, which would maybe make the 4 position a more favourable bonding location, due to electron repulsion?

This has been bugging me for the last few hours, so if someone could clear it up I'd be grateful.

Thanks

Well, the first of many answers is "because that's how the molecule automatically self-assembles," and since your longer question seems to revolve around 'why does it automatically do this' (as opposed to speculative bioactivity advantages), that's what we'll adress...

It's simple : like charges repel.

As a side-result of this, and probably a bunch of other complex factors, the substitution of benzene with just about any guiding group that is not a nitro group will try desperately to get as far away from the other substitution as possible. That just happens to be the 4-position, since numbering is, well... relativistic 'n all, and the 4-position is "the one directly opposite whatever you call the 1-position" 'n such...

'course, it's true that charge-localization, electron-withdrawing, and resonance get a little bit more downright wierd when the substituent is nonpolar, like an alkyl chain... but generally, the rule of thumb is that if the guiding substituent is not a nitro group, it's going in the para position...

Wikipedia -> 2C -> Answer.

-F☺rd

So it is due to the repulsion experienced from the other electrons?

I thought it would be something like that.

Ford: I already checked that page, all it says is that the 2C chemicals have a substituent in the 4 position, and that it is lipophilic. It doesn't go into the details of why, which is more what I wanted to make sure I had the right idea about.

______________

This leads to another question:

How hard would it be to create a 2C substance in which the substituent is on, say, position 3? And how major of a difference would be observed?

I'm thinking that by starting with 2C-I for example, the 4 position would already be occupied, so upon further reaction with another substance (Br or something) the Br would be forced to react on either points 3 or 6.

You would then need to remove the Iodine somehow...

Doesn't sound easy, and quite possibly wouldn't be worth it, or wouldn't work at all, but I'm curious...

You actually sparked some intererst on me with that, i will do some searching and let you know but i am sure that if you where to brominate 2C-I you would get a bunch of crap, some 2C-I and maybe some 2C-B. I mean off the top of my head i can see the Br pushing out the iodine, but dunno i will do some searching as i suck at predicting chemical changes

I am still searching and so far no refs or mention on that, won't give up tho'

I appreciate the help man!

I've been looking for some things myself, with no success.

Just wondering, though, as I've never done anything like this, but how would one predict what would happen if a compound (2C-i in this case) was to be reacted in some way (again, in this case, the reaction would be a bromination).

What sort of things should I be looking for and trying to find out?

Would be great to have a nudge in the right direction from any members who know about working out things like this...

I can definitely answer this for you.


I don't know how much organic you've had, but reactions with benzene rings have two types of substituents: activating and deactivating. Activating compounds contribute to the resonance structure of the ring, like NH2 and methoxy substituents. Deactivating substituents are like a carboxyl group; they withdraw electrons from the ring, because the carbon directly attached to the ring has a delta + charge. Activating substituents direct added groups [like the halogen] ortho or para. The deactivating compounds usually direct the arriving group to the meta position. The only exception to this rule is that halogens or C-X groups direct the arriving groups to the ortho/para positions, even though it's deactivating.

So when you are adding a substituent, you always add it according to the position of the MOST activating substituent. In this case, that would be NH2, which is going to direct the substituent to the para position to itself. This would be in the position that the -X group is in the picture.

It adds para as opposed to ortho because there is less steric hindrance to the NH2 group there. If it was a 2:1 stoich ratio, the second one would add ortho. This would also happen if the para position was blocked; the first one would go ortho.

Does that help?

Excellent. Love your user name too. Tiny death god eh?

I presume from SN commments that this is correct Thanks a lot!