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Allyl Bromide + HBr - Compounds formed during reaction.

  1. Aug 10, 2012 #1

    AGNuke

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    Allyl Bromide, during the addition of HBr gives

    [tex]CH_3-CHBr-CH_2Br[/tex]
    [tex]CH_2Br-CH_2-CH_2Br[/tex]
    [tex]CH_3-CH_2-CHBr_2[/tex]
    [tex]CH_3-CBr_2-CH_3[/tex]

    My take
    During the addition of HBr, electrophile will attack first, considering the shift of pi bond in allyl bromide as follows:

    [tex]BrCH_2-CH=CH_2 \longrightarrow BrCH_2-\overset{+}{C}H-\overset{-}{C}H_2[/tex]

    H+ will attack on negative C. Now, to add Br- on positive one, I get the first option.

    But the question is asking all possible products, it seems (possible intermediates/minor products). So, the possibilities of forming the other products is not theoretical impossible, I suppose.

    All I need is the mechanism, which can lead to the formation of other products. I am currently working on some mechanisms.
     
    Last edited: Aug 10, 2012
  2. jcsd
  3. Aug 10, 2012 #2
    This question is based on the electrophilic addition to alkenes.

    The reaction doesn't really proceed like this. Check the following link for the mechanism: http://www.chemguide.co.uk/mechanisms/eladd/symhbr.html#top
    It's obvious that the chief product is formed through Markovnikov addition but there is still a low yield of the Anti-Markovnikov product.
     
  4. Aug 10, 2012 #3

    AGNuke

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    The mechanism mentioned there omitted the intermediate step I mentioned.

    [tex]BrCH_2-CH=CH_2 \longrightarrow BrCH_2-\overset{+}{C}H-\overset{-}{C}H_2\xrightarrow[]{+H^+}BrCH_2-\overset{+}CH-CH_3[/tex]

    And I know there can be other potential minor products, that's why I asked the question.
     
  5. Aug 10, 2012 #4
    There's nothing omitted, the intermediate you mention doesn't form (at least in electrophilic addition). You can't break the pi-bond and then add H+. Refer a good book.
     
  6. Aug 10, 2012 #5

    AGNuke

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    Who's breaking pi bond? I am only shifting it. Check the arrows in the link.

    Moreover, here's an excerpt from Morrison and Boyd

    There's only one way to show the intermediate process, to shift the pi electrons.
     
    Last edited: Aug 11, 2012
  7. Aug 11, 2012 #6
    Note that you can't go about shifting pi electron clouds the way you did, and leave separated charges uncompensated for, and you definitely can't have something like that as the first step of a reaction mechanism.

    I suggest you consider conjugation in allyl bromide.
    Bromide has a lone pair of electrons it can donate, and the pi electron cloud can shift. Try using both of these facts to get a structure that would be in resonance with the structure of allyl bromide given.
     
  8. Aug 11, 2012 #7

    AGNuke

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    I only showed the cloud shift under the influence of the electrophile, I just show them, it makes it easier to make major product.
     
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