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susu6520
Joined: 04 Aug 2007 Posts: 3
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Posted: Sun Aug 05, 2007 10:19 pm Post subject: Question 52 |
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I read in several places that free radical halogenation would produce an anti-Markovnikov product with peroxide and a Markovnikov product without peroxide. How were we supposed to know in this question that the reaction included peroxide?
Thanks. |
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admin Site Admin
Joined: 08 Dec 2003 Posts: 2176
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Posted: Tue Aug 07, 2007 12:24 am Post subject: |
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You almost have it! It would be correct to say that halogenation of alkenes with Br produces the Markovnikov product (ionic medium) but if there is HOOH then the peroxide can induce free radicals. But once HBr becomes free radicals (how does not matter; ie HOOH or more commonly using heat/light/uv energy) then the consequence is the anti-Markovnikov product.
In other words, it's not necessary that you know if peroxide is involved. It is only important that you knew that it was a free radical addition of HBr.
ref:
www.chem.uky.edu/Courses/che230/RBG/lecnotes/J11.pdf
PS. mono bromination with free radical addition of Br2 produces a Markovnikov product. Now you can forget that because Markovnikov chemistry is not on the new MCAT! |
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susu6520
Joined: 04 Aug 2007 Posts: 3
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Posted: Tue Aug 07, 2007 5:27 pm Post subject: |
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| Wow, really? All Markovnikov chemistry is not on there? Including reactions like normal/Hoffman elimination? But we still need to know that if there's a positive charge, it should go to the more substituted carbon right? |
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admin Site Admin
Joined: 08 Dec 2003 Posts: 2176
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Posted: Wed Aug 08, 2007 1:23 am Post subject: |
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Nope, not even that! You might see a few such questions out of the 10 GS MCAT CBTs but we have been weaning them out and you will find none in
AAMC tests 8, 9 and 10 (the most recent released tests). |
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IrvPhill6636
Joined: 13 Nov 2009 Posts: 3
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Posted: Fri Apr 09, 2010 6:42 pm Post subject: |
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| Bromine is significantly more regioselective than Chlorine so shouldn't it be on the 2nd carbon? |
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admin Site Admin
Joined: 08 Dec 2003 Posts: 2176
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Posted: Tue Apr 20, 2010 10:47 am Post subject: |
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| I do not see why chlorine is brought into the question as there is no mention of chlorine in the question(?). As for the addition of HBr onto the alkene, the primary substituted product is correct (answer D) for the following: The addition of HBr to the alkene forms the product because the electrophile (H+) from the HBr follows the rule that applies to all electrophilic additions (Markovnikov’s rule) and adds therefore to the sp2 carbon bonded to the greater number of hydrogens. |
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