in my intro organic class, the instructor mentioned that stable radicals tend to form and determine the main product.
thus, when brominating say ethyl benzene, the bromine will latch on the benzylic carbon because the benzylic radical has the most resonance
structures.
my question is, if in drawing the resonance structures, the unpaired electron moves around the ring, why does the bromine latch onto the benzylic
carbon and not somewhere on the ring?
and if this is true, then why in the bromination of 2,5-dimethoxy phenethylamine does the bromine primarily bond to the 4 position?Arrhenius - 12-10-2010 at 23:40
The benzyl radical is the predominant resonance structure. While certainly, we should expect side products arising from reaction at all reasonable
radical resonance positions, in reality we only see reaction of the benzylic radical. Why? Because moving the radical onto one of the ring carbons
requires that we break aromaticity. This is an unfavorable process, and while it does happen frequently, the benzyl radical is highly favored
thermodynamically.
Now, as for the bromination of 2,5-dimethoxyphenylethylamine to 2C-B, you've already answered your own question. Check the experimental conditions.
Is this a radical bromination, or simply an electrophilic aromatic substitution?
[Edited on 13-10-2010 by Arrhenius]DDTea - 13-10-2010 at 01:55
my question is, if in drawing the resonance structures, the unpaired electron moves around the ring, why does the bromine latch onto the benzylic
carbon and not somewhere on the ring?
To be really anal here, the electron isn't actually jumping around the ring! The actual structure is a quantum superposition of all of the resonance
structures. Accurate descriptions are good for you in O. Chem spirocycle - 13-10-2010 at 04:58
^ i realized that, i meant that when you draw the structures the unpaired electron jumps around, not that it physically jumps around
Quote:
Now, as for the bromination of 2,5-dimethoxyphenylethylamine to 2C-B, you've already answered your own question. Check the experimental conditions. Is
this a radical bromination, or simply an electrophilic aromatic substitution?
So even though the 2c-H is mixed directly with bromine, I'm guessing the lack of UV light means that its not a radical bromination?jwarr - 13-10-2010 at 17:17
Now I'll add another question here: The methoxy groups on 2C-H are ortho/para directing (3,4, and 6 positions in this case). It seems reasonable that
bromination at the 6 position does not occur due to steric crowding. But why is bromination at the 3 position not a major product?spirocycle - 13-10-2010 at 17:26
why are they ortho/para directing?
how does the ethylamine group direct?
note: i realized how much I suck at chemistry when the thought occured to me that drug synths are more readily available than innocuous lab preps
[Edited on 14-10-2010 by spirocycle]jwarr - 13-10-2010 at 17:57
spirocycle: look at the resonance structures. placing a positive charge on the carbon adjacent to the alkoxy group allows an additional resonance
structure with a pi bond between the carbon and the oxygen.
You just answered my question: I forgot to think of the alkyl group, which directs ortho/para as well due to inductive stabilization of the tertiary
carbocation.
Can anyone here perhaps comment on the hypothetical ratios that would be obtained? I'm simply curious how strong the affect of alkyl group would be.
I wouldn't think it would be all that significant but perhaps I'm wrong. Would you think the 6-bromo substitution would be found in any significant
portions (>10%?)spirocycle - 14-10-2010 at 13:13
if the methoxy groups are para to each other, how can they direct para?
and I have heard that if the temperature is too high, a 6-bromo impurity will form
jwarr - 14-10-2010 at 19:33
In general alkoxy groups are ortho/para directing. Obviously they will only be ortho directing in this case.
