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Organic Reaction Question
Started by shane.
There's something else i'd like to clarify.
HCL in ROOR adds Markovnikov to the alkene.
HBr in ROOR, however, adds anti-Markovnikov to the alkene.
Correct?
You are correct. Thats actually in the Destroyer. If you don't have it, get it.
We're reacting 1-butene with Br2 and methanol and the product we get is 1-bromo, 2-butanol.
This reaction doesn't seem to make any sense to me, could someone possibly explain where each species is adding etc.
Thanks.
1-bromo2-butnol? Shouldnt it be 1-bromo 2-methoxybutane?
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There's something else i'd like to clarify.
HCL in ROOR adds Markovnikov to the alkene.
HBr in ROOR, however, adds anti-Markovnikov to the alkene.
Correct?
Yes ur right
We're reacting 1-butene with Br2 and methanol and the product we get is 1-bromo, 2-butanol.
This reaction doesn't seem to make any sense to me, could someone possibly explain where each species is adding etc.
Thanks.
methonal would go to more brached just like when reacting Br2 with H20
methonal would go to more brached just like when reacting Br2 with H20
I agree, When Br2 with H20 react, -OH goes to the more sub. side and that means -OCH3 will go to the more sub. side which in this case it the C-2
1-bromo2-butnol? Shouldnt it be 1-bromo 2-methoxybutane?
Indeed, it should.
My question lies in the mechanism I suppose. Is the Br2 first attacking the least substituted side of of the double bond, having a lone pair of electrons from the double bond move over onto C-2? This is where my confusion is.
Someone please correct me if I'm wrong...
First, Br will add across the double bond and you will form a propane like ring (the name escapes me at the moment) with bromine at the "tip". Then the methoxy (or alcohol) group will come in and add trans to the ring, breaking the ring and leaving the Br group on C1.
First, Br will add across the double bond and you will form a propane like ring (the name escapes me at the moment) with bromine at the "tip". Then the methoxy (or alcohol) group will come in and add trans to the ring, breaking the ring and leaving the Br group on C1.
Someone please correct me if I'm wrong...
First, Br will add across the double bond and you will form a propane like ring (the name escapes me at the moment) with bromine at the "tip".
bromonium ion
this type of triangular moiety is part of any halohydrin-type reaction such as this... a similar phenomenon occurs in oxymercuration where a mercurinium ion is formed
good luck 👍
*edit* sorry this is so brief. I would write a more extensive explanation... but I need to sleep so I can do decently on my ochem test tomorrow 😛
bromonium ion
this type of triangular moiety is part of any halohydrin-type reaction such as this... a similar phenomenon occurs in oxymercuration where a mercurinium ion is formed
good luck 👍
*edit* sorry this is so brief. I would write a more extensive explanation... but I need to sleep so I can do decently on my ochem test tomorrow 😛
Yeah the Br2 does come in to form whats called bridge bromonium ( a type of three membered ring with a Br at the tip (similar to epoxides only a Br instead of "O") than the methanol comes in and the bridge is broken forming an overall anti addation to the double bond with a methoxy group at the "C" that is more substituded.
Hope that helps
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