| Hydrohalogenation of Butene with Hydrogen Bromide |
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3dObject Member
Joined: 24 May 2006 Posts: 11
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Posted: Sat Jun 24, 2006 6:30 pm Post subject: Hydrohalogenation of Butene with Hydrogen Bromide |
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The answer to the following reaction is given as choice E. I thought it should be D because, if the two carbon atoms at the double bond are linked to a different number of hydrogen atoms, the halogen is found preferentially at the carbon with less hydrogen substituents (Markovnikov's rule). Would this not be the second carbon of the butene? Please advise.
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opuntia Regular
Joined: 15 Jun 2005 Posts: 82 Location: Maldives(the chain of islands)
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Posted: Sun Jun 25, 2006 6:02 pm Post subject: |
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| It'll be the second carbon of the butene because the major product (2-bromobutane) is always formed through the most stable carbocation.
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niks Junior Member
Joined: 29 Mar 2006 Posts: 4
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Posted: Sat Jul 01, 2006 4:04 pm Post subject: |
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| I too think you are correct. I too reviewed the Markovnikov's rule. The structure should be CH3---CHBr---CH2---CH3
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adrian Regular
Joined: 20 Apr 2006 Posts: 44 Location: Bucharest, Romania
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Posted: Sat Jul 01, 2006 9:21 pm Post subject: the addition has not an ionic mechanism |
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| There is a little detail (UV light) than means we cannot use the Markovnikov rule. The rule is OK for ionic addition, but here we have a radicalic addition. So, the most probably shall be 1-Bromoderivative (the intermediate is a free radical)
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victor Junior Member
Joined: 09 Jul 2006 Posts: 4 Location: Yogyakarta, Indonesia
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Posted: Sun Jul 09, 2006 2:19 am Post subject: |
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I don't think it'll be different, because HBr is in the liquid form and will be dissosiated into H+ and Br-. So, Markovnikov will run the reaction..
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bw800402 Junior Member
Joined: 09 Aug 2006 Posts: 5
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Posted: Wed Aug 09, 2006 11:59 pm Post subject: |
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Markovnikov's Rule will not apply!!
The use of HBr and UV light will not give H+ and Br-.
The UV light photochemically intitiates a radical reaction. Through this reaction 1 electron from bromine forms a bond with 1 electron from the double bond with the Br going to the less substituted side of the double bond and leaving 1 free electron radical on the more substituted carbon because a secondary radical is more stable than a primary radical. This secondary radical then reacts with the hydrogen, giving the anti-Markovnikov addition of HBr.
Hope this helps,
bw800402
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kristoph Junior Member
Joined: 06 Apr 2007 Posts: 8
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Posted: Sat Apr 14, 2007 2:22 pm Post subject: |
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| So, what exactly about this addition identifies it as a free radical addition. Is it only the UV light? Nothing about regioselectivity, is this structure cis or trans? If it is just the UV light, then what about the UV light causes free radical addition? I have searched the internet high and low, no luck on any solid info about UV addition.
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RobJim Senior Member
Joined: 13 Feb 2005 Posts: 320 Location: Los Angeles, CA
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Posted: Thu May 17, 2007 6:19 am Post subject: |
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| Yes, it's the UV light. I guess the photons of UV light are just the right energy to break chemical bonds, making radicals. It's a method for making haloalkanes from alkanes; the UV light makes radicals, allowing the reaction.
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