Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rutherford on December 30, 2012, 04:30:39 AM
-
When the attached compound reacts with bromine under light, theoretically, two monobromine derivates could be obtained, but in reality only one is produced. Why?
-
How about the radical stability? The benzylic radical is the most stable, why?
-
Don't know. I know for carbocations and anions. What affects the stability?
-
Draw out the resonance forms for the stabilisation of the benzyl radical.
-
I always forget to think about resonance in these situations. If the hydrogen from the CH group is substituted, the electron on the carbon becomes delocalized under the influence of the ring, so the radical is stabilized, meaning that only the hydrogens from the methyl groups can be substituted with bromine.
I suppose that it is always this case, when an alkyl group is attached to benzene ring. The H atoms on the C atom attached directly to the ring can't be substituted. I tried to draw the resonance structures. One is attached. Hope that it is good.
-
I always forget to think about resonance in these situations. If the hydrogen from the CH group is substituted, the electron on the carbon becomes delocalized under the influence of the ring, so the radical is stabilized, meaning that only the hydrogens from the methyl groups can be substituted with bromine.
I suppose that it is always this case, when an alkyl group is attached to benzene ring. The H atoms on the C atom attached directly to the ring can't be substituted. I tried to draw the resonance structures. One is attached. Hope that it is good.
I don't quite agree with the text in red. If you generate a radical from the methyl groups then it is primary and not very stable.
In your molecule under these conditions the benzyllic position is the most reactive. So I would expect bromination there.
-
If the primary is unstable, it should react fast with bromine because of that.
Why is the benzylic position more reactive when the electron gets delocalized? The radical should be stabilized and therefore less reactive.
-
The very fact that it is stabilised makes it the most reactive position.
If the radical is not stabilised it can readily undergo an H radical transfer to give a more stabilised radical or other side reactions. Similar stability order to carbocations.
-
I don't get it. Stabilized means stable, not very reactive. The noble gasses have stable configurations so they aren't so reactive.
I tried to reformulate it this way: The benzyl radical is more stable, so it is energetically favored to be produced in one step of substitution (this would be similar to the carbocations i.e. when Markovnikof's rule is used). I agree with this:''If the radical is not stabilised it can readily undergo an H radical transfer to give a more stabilised radical or other side reactions.''
One last question: Is it really true that not even a single molecule is made with a primary hydrogen replaced by bromine? This wouldn't seem reasonable, as in addition of e.g. HCl to pentene, both product are going to be made, on is the major product, the other is present in traces.
-
I don't get it. Stabilized means stable, not very reactive. The noble gasses have stable configurations so they aren't so reactive.
You are referring to ground state stabilization. In that case yes, stable means less reactive. Disco is saying that, of the two intermediate radicals that could form (the primary vs. the secondary+benzylic), one of them is more stable than the other. So the energy hill that we have to climb is smaller if we climb to the more stable intermediate. The smaller energy hill = more reactive.
-
I can't see the logic here.
If something is stable, it is easier made e.g. Cl2.
If something is unstable, it is harder made (the production is endothermic) and it tends to get stabilized, so it reacts fast with something e.g. Cl*.
-
I can't see the logic here.
If something is stable, it is easier made e.g. Cl2.
If something is unstable, it is harder made (the production is endothermic) and it tends to get stabilized, so it reacts fast with something e.g. Cl*.
Is Int1 more stable or Int2?
Would you expect Product C or D to dominate?
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2Fdkcrs.gif&hash=3cff7220342876b38d8cdccec16aa5069601adfc)
-
Int2 more stable-D dominates. Good?
-
Int2 more stable-D dominates. Good?
Yes. So that answers your question doesn't it? Are you still in a conundrum about disco's statement? "The very fact that it is stabilised makes it the most reactive position."
Do you get the logic now?
-
I knew that, but I don't understand why do you call it reactivity. I thought that reactivity is "the rate at which a chemical substance tends to undergo a chemical reaction".
The primary radical reacts faster meaning that it is more reactive, but it won't be produced here, as it is energetically favored that the secondary hydrogen leaves benzene.
Overall, the secondary radical is produced and then reacted with bromine, so I suppose that you were naming this as bigger reactivity.
Never thought about this in that way. Thanks to everyone for the help.
-
The benzyllic position, i.e. Ph-CH2- is usually very reactive as the radical, or the cation, and even the anion can be stabilised by resonance interaction with the pi system of the aromatic ring.
Typically in the mass spec. of compounds containing a benzyllic substituent you often see a mass of 91 which is the benzyl radical cation (in the EI mass spec). This is really quite stable. In fact it can form the tropylium cation. See
http://en.wikipedia.org/wiki/Tropylium_tetrafluoroborate (http://en.wikipedia.org/wiki/Tropylium_tetrafluoroborate)
Now that is digging in the deep recesses of my addled brain cell.
-
Thanks for the extra info.
I just wanted to return to the question I posted: "Is it really true that not even a single molecule is made with a primary hydrogen replaced by bromine?"
-
"Is it really true that not even a single molecule is made with a primary hydrogen replaced by bromine?"
Not true.
-
What happens with toluene + bromine + light?
-
The primary hydrogen will be replaced as the C6H5CH2· is stabilized now.
"Is it really true that not even a single molecule is made with a primary hydrogen replaced by bromine?"
Not true.
Then the question in my first post is wrong. Very rigorously, two derivates are made in reality.
-
What happens with hexane + bromine + light?
-
No resonance now, but I remember that I read somewhere that primary, secondary and tertiary hydrogens in alkanes have different reactivities (I think that the tertiary are the most reactive). Not sure what affects this.
-
No resonance now, but I remember that I read somewhere that primary, secondary and tertiary hydrogens in alkanes have different reactivities (I think that the tertiary are the most reactive). Not sure what affects this.
This is the stability of the radical that is formed. This stability parallels that of carbocations.
So a primary radical is not very stable it is likely to rearrange through the abstraction of a H radical to a secondary radical, this can also abstract another H radical or you can get alkyl shifts to generate tertiary radicals.
So to answer my question, hexane + bromine + light will give a mixture of brominated products.
-
The only thing I forgot to wrote in my post was the answer :P.
-
Happy new year ;D
-
Happy new year :D!
-
The primary hydrogen will be replaced as the C6H5CH2· is stabilized now.
"Is it really true that not even a single molecule is made with a primary hydrogen replaced by bromine?"
Not true.
Then the question in my first post is wrong. Very rigorously, two derivates are made in reality.
Very rigorously, all possible derivatives are made in almost all reactions you can think of. If you frame the question the way you did: "even a single molecule".
When you have ~1023 molecules flying around even the most probablistically unlikely event is very sure to sometimes happen (at Room Temp. ). Even a Markovniov addition has almost surely at least one molecule of the anti-Markovnikov product.
I'd love to see cases where a product is strictly forbidden at a molecular level; that'd have to be some electronic or quantum mechanical restriction. But based on energetics or activation energies alone I think molecular level absolute selectivity is very difficult.
I don't think this makes the question wrong. You are only mis-interpreting it.
-
Okay, when it is said that only one product is obtained, it is actually meant that there is a major product and a very little abundant product. I think that everything is clear here now. Thanks again.