ChewyDrop

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Is this consider the right answer? I am working problems from my old ochem hw and the answer key is off.
 

aj30

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why isnt the Br in the tertiary carbon? thats an allylic position.
That's the other right answer, maybe even the dominant one. However once you form the radical there it can shift and give you the answer you wrote above.
For the DATs purposes you should probably just focus on the Br going to the tertiary position.
 
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ChewyDrop

ChewyDrop

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That's the other right answer, maybe even the dominant one. However once you form the radical there it can shift and give you the answer you wrote above.
For the DATs purposes you should probably just focus on the Br going to the tertiary position.
wouldn't the double bond be more stable?
 

aj30

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wouldn't the double bond be more stable?
The double bond would be more stable in the first picture you showed but the radical is more stable leading up to the alternative product. Don't worry about that though; that's part of why both products form, perhaps in different percentages.
 

orgoman22

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Is this consider the right answer? I am working problems from my old ochem hw and the answer key is off.
NOT A CHANCE !!!!! The bromine would remove the tertiary allylic hydrogen.....which would hands down have the SMALLEST bond dissociation energy ! The final product would be in HIGH yield as well....perhaps 95%......It would be ( E ) - 4-bromo-5 methyl -2- hexene.

Hope this helps !!!

Dr. Romano
 

fit2

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So this is the stuff about thermodynamic vs kinetic control. What you have is the thermodynamic product (more stable alkene less stable intermediate secondary) This usually happens at higher temperatures....room temp and higher .if you had negative temperatures you can control the rxn to form the other product...the kinetic product which would be the opposite product from the picture (less stable alkene more stable intermediate tertiary)
 

aj30

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NOT A CHANCE !!!!! The bromine would remove the tertiary allylic hydrogen.....which would hands down have the SMALLEST bond dissociation energy ! The final product would be in HIGH yield as well....perhaps 95%......It would be ( E ) - 4-bromo-5 methyl -2- hexene.

Hope this helps !!!

Dr. Romano
See McMurry ch. 11 where he always discusses all the different products that can form in response to questions on allylic bromination. That being said, I did mention that for the DATs purposes the tertiary allylic will be the answer.
 

aj30

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NOT A CHANCE !!!!! The bromine would remove the tertiary allylic hydrogen.....which would hands down have the SMALLEST bond dissociation energy ! The final product would be in HIGH yield as well....perhaps 95%......It would be ( E ) - 4-bromo-5 methyl -2- hexene.

Hope this helps !!!

Dr. Romano
As another point, in McMurry's problems bond dissociation of the tertiary H isn't the deciding factor because obviously, no matter the product, that H will be pulled off. It's a question of what happens next-does the Br come into the tertiary allylic radical or does the molecule rearrange through resonance before the Br comes in.
 
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ChewyDrop

ChewyDrop

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NOT A CHANCE !!!!! The bromine would remove the tertiary allylic hydrogen.....which would hands down have the SMALLEST bond dissociation energy ! The final product would be in HIGH yield as well....perhaps 95%......It would be ( E ) - 4-bromo-5 methyl -2- hexene.

Hope this helps !!!

Dr. Romano
upload_2016-8-21_20-38-40.pngLike this? Just a tad bit confuse on why no radical rearrangement occurs.:(
 

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ChewyDrop

ChewyDrop

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Yes that's the answer for the DAT.
And a rearrangement can occur but won't necessarily be the major product.
ah ok! When I was taking ochem I was always being hammered for remember rearrangements. Thanks!
 

fit2

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Right! Think about the intermediate in the second case it's tertiary radical vs secondary radical when you draw out the resonance.
I thought thermodynamic vs kinetic depends on the conditions. but thanks Dr. Romano for your post. I'll keep in mind that if nothing is specified the tertiary radical product will be the major one
 
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ChewyDrop

ChewyDrop

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Right! Think about the intermediate in the second case it's tertiary radical vs secondary radical when you draw out the resonance.
I thought thermodynamic vs kinetic depends on the conditions. but thanks Dr. Romano for your post. I'll keep in mind that if nothing is specified the tertiary radical product will be the major one
So if you have a internal double bond already, don't worry about rearrangement?
 

aj30

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So if you have a internal double bond already, don't worry about rearrangement?
For some reason for the DAT nobody focuses on rearrangements that can occur during allylic bromination (although yes the books talk about it at length) so just discount that.
Keep things simple I suppose.
 

fit2

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So if you have a internal double bond already, don't worry about rearrangement?
Well think about which resonance form is more stable tertiary>secondary>primary. So here because the tertiary radical is more stable than the secondary...the Br is added where the radical is more stable. Does that make sense? you can't discount the possibility of rearrangement...you just gotta think about which intermediate is more stable. It's not too bad you will most likely have to choose from 2 in a case like this
 

aj30

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Well think about which resonance form is more stable tertiary>secondary>primary. So here because the tertiary radical is more stable than the secondary...the Br is added where the radical is more stable. Does that make sense? you can't discount the possibility of rearrangement...you just gotta think about which intermediate is more stable. It's not too bad you will most likely have to choose from 2 in a case like this
The fact that the pre-rearrangement radical is more stable than the post will probably cause the pre to dominate. However, the final product (with its double bond) resulting from the post is actually more stable than the product of the pre (tri sub. double bond). So maybe this is a kinetic vs. thermodynamic issue-but for that DAT, just go with the tertiary. There's no way they'd get this deep into it.