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Topic: HBr and Br2  (Read 6715 times)

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Offline Lithium

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HBr and Br2
« on: July 17, 2008, 07:30:43 PM »
Okay, I have only two questions.  One has to do with 1-butene and HBr and the other with 1-butene Br2.  The question said to assume a racemic mixture for chiral products.  I wasnt sure what exactly what that meant!  I know racemic means both R and S and chiral means 4 different products.  My ? is should I show sterochemistry or not?
  I have options a-e A.) 1-bromobutene no sterochem B.) 2-bromobutene with the front wedge c.) 2-bromobutan-1-0l
d.) 1,2-Bromobutane with only the 2 bromo showing the front wedge and the other bromine is in stick fashion or e.) not a-d    For the first question Im sure its either B or E I just dont know if I need to show the front wedge only or the stick.  for the second question i dont know if its okay to show the second bromine in the stick or it also needs to be shown in th eback wedge.

Offline macman104

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Re: HBr and Br2
« Reply #1 on: July 17, 2008, 08:04:33 PM »
Sorry, you really need to post the exact question.  Or explain better, because right now it doesn't make sense.

Offline Lithium

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Re: HBr and Br2
« Reply #2 on: July 17, 2008, 08:22:20 PM »
okay you have a alkene with the double bond on the end.  its a butene.  The butene is then reacted with HBr now am i suppose to show sterochemistry?  As in a front wedge or no?  The directions said to assume a racemic mexture for chiral products!  The second reaction was Br2.  I know intial it forms a ring which breaks and puts to bromines.  My question is if one of the bromine is shown on the front wedge on the second carbon and just the line on the first carbon for bromine is that fine as an answer or would onl the back wedge be the acceptable answer? 

Offline macman104

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Re: HBr and Br2
« Reply #3 on: July 17, 2008, 08:56:56 PM »
I believe that what they mean by assume a racemic mixture, is that you don't draw any wedges or dashes, but instead just lines, as you cannot say what the stereochemistry is.

Offline Lithium

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Re: HBr and Br2
« Reply #4 on: July 17, 2008, 09:08:42 PM »
okay i uploaded the question

Offline macman104

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Re: HBr and Br2
« Reply #5 on: July 17, 2008, 11:14:48 PM »
I'm honestly not completely sure what is meant for that.  But honestly, just ignore that part, it doesn't affect the correct answers at all.  You'll have to ask your teacher what they mean for that.  It may be one of those defaut statements professors might put on that.

Offline Lithium

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Re: HBr and Br2
« Reply #6 on: July 18, 2008, 01:27:13 AM »
So the answers for those two should be E it seems like alot of Es? Can anyone else provide some input.

Offline nj_bartel

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Re: HBr and Br2
« Reply #7 on: July 18, 2008, 01:42:45 AM »
It's to make the question very clear, so you don't mark E based on stereoisomerism.

Offline Lithium

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Re: HBr and Br2
« Reply #8 on: July 18, 2008, 12:13:40 PM »
Anyone else have any ideas or thoughts to share on those two problems

Offline azmanam

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Re: HBr and Br2
« Reply #9 on: July 18, 2008, 12:20:27 PM »
The question instructions are saying that for any chiral compounds the professor drew in the possible answers, you are to assume that the product is in equal amount with its enantiomer - thus, the chiral compound is a racemic mixture.

Your answers are correct.  However, your intermediate in problem 2 is not correct.  There is no carbocation in the peroxide mechanism.  Your intermediate for 3 is also incorrect.  In the first step of the mechanism, one atom of bromine adds syn across a double bond.  A wedge and a dash is incorrect.  Additionally, the bromonium intermediate is charged.

(edited for clarity)
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline Lithium

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Re: HBr and Br2
« Reply #10 on: July 18, 2008, 12:46:24 PM »
Thanks for your *delete me*!

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