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Can anyone help with this o-chem question?

Discussion in 'Pre-Medical - MD' started by Harbindoc, Nov 26, 2002.

  1. Harbindoc

    Harbindoc Member
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    Hello, I am ready to throw my book across the room and I am wondering if anyone can help.

    The problem is as follows:

    Since I cant draw the molecule I wil name it.

    3,3 dimethyl-5-phenyl pentene. Gets H+/H2o added to it to give the folllowing product. 3,4-dimethyl-1-phenyl-2 pentanol.

    The question asks me to suggest a mechanisim for the reaction and explain why it occurs. I know that there is a methyl and an hydride shift, at least I think so... but I can't figure out why the OH group is being added to a lower substitued carbon.
    Can anyone help?

    Thanks!
     
  2. TroutBum

    TroutBum Senior Member
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    Hey Harbindoc, I'm a big O-chem dork so I'm working on this, but I have a question--as you've named it, the alkene is between the 1 and 2 carbon, not the 4 and 5 carbon (where the phenyl is), right?
     
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  3. TeinVI

    TeinVI Membership Revoked
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    damn....sorry i can't help you. it's been too long since i last took o chem
     
  4. Tweetie_bird

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    It's been long since i did O Chem, but I may have figured it out.

    Double check with somebody though, just to make sure.

    Ok, so you have come to the point where you have a secondary carbocation joined to a tert-butyl group that is also hooked up to a phenyl group attached to two CH2's.

    Your secondary carbocation wants to become tertiary carbocation to become stable. Therefore, you have a methyl shift. You now have a tert carbocation.

    Now, do you notice how if the 2nd CH2 (counting from the phenyl group) loses it's "H", then you get extension of conjugation? Basically, a benzyllic carbocation is more stable than a tert carbocation. Furthermore, if you have a hydride shift, you get extension of conjugation which is a favorable reaction.

    Thus, it the H moves and you get the second CH2 becoming CH which later gets the H20 added.

    Thus, it may seem like the H20 is going to the less stable group.

    I hope I'm right on the mechanism.
     
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  5. phil413ru

    phil413ru Senior Member
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    Like troutbum, I love O-Chem

    are you sure that phenyl group is carbon five, and not one? If it is on one, numbering would make molecule
    1-phenyl-3,3-dimethylpentene.
    If this is the molecule, the OH goes on carbon 2 because Carbon one can stabilize the positive charge due to presence of the benzene ring. Additionally, C2 would have a partial positive dipole-C1 with phenyl would slightly pull the negative charge more than simple hydrogens.

    If this is not above named molecule, I not sure because the double bond is btw c1 and c2, but hydoxyl group on c4.

    That is my hypothesis. Unfortunately, do not have my book, this is best shot. I can ask my prof (former) who Wrote book (Solomons and Fryhle) What book do you have?
     
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  6. TroutBum

    TroutBum Senior Member
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    Tweetie, I'm not sure I agree with you--I don't see anywhere in the mechanism where you would have a tert-butyl group.

    So far, all I've been able to come up with is this:

    The first step is addition of H from H30+ to the 1 carbon, resulting in a carbocation at the 2 carbon. Next, there is a methyl shift, where one of the methyls from the 3 carbon shifts to the 2 spot, so you now have 3,4 dimethyl-1-phenyl pentane, but with a carbocation on the 3 carbon instead of an H. Next, the H20 formed from the loss of the H in the first step reclaims a H from the (now) 2 carbon, which reforms the catalyst, H30+, with the electrons from this broken bond forming a double bond to eliminate the carbocation at the 3 carbon. (Sheesh, O-chem is tough when you can't use drawings!)

    So now you have 3,4 dimethyl-1-phenyl-pent-2-ene. And then you would just have addition of H to the 3 carbon and OH to the 2 carbon, yielding 3,4 dimethyl-1-phenyl-2-pentanol.

    So there is no hydride shift (which I'm not familiar with at all!), but this still raises the question of why the OH in the last step attaches to the 2 (less subsituted) carbon instead of the 3 carbon, where the carbocation is more favored to form (with the addition of H at the 2 carbon). Everything would be swell if there wasn't the extra carbon between the phenyl and the alcohol-bearing 2 carbon, because resonance from the phenyl would stabilize the cation on the 2 carbon.

    Tweetie mentioned that a benzyllic carbocation is very stable, which if true would work for the mechanism I've written out, because that's what you get in the last step, a benzyllic carbocation, but I'd need a refresher as to why that is. If you're talking about resonance, it wouldn't work because the double bonds from the phenyl wouldn't be able to "get past" the sp3 1 carbon. Tweetie, correct me if I'm wrong.

    So sorry, I don't know if that helps or not.
     
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  7. Gleevec

    Gleevec Peter, those are Cheerios
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    It could be that Im not understanding correctly, but I have no idea where your double bond is going since you didnt provide an IUPAC number before it. Can you just draw it in MSPaint and upload it as a pic? Or can you provide us with the location of the double bond?

     
  8. OP
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    Harbindoc

    Harbindoc Member
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    First, sorry it took me awhile to reply, I got busy and was unable to post a response. Turns out that there are both hydride and methyl shifts. In addition, the carbocation is resonance stablized allowing the OH group to attach to the least substituted group. I wish I could draw the reaction for everyone to see, but alas I am not that computer literate! Once again, everyone came through though! You guys are the best! Tweetie, when is your interview? I have been crossing my fingers for you for awhile now. Good Luck!!

    HAPPY THANKSGIVING SDN!!

    may the adcoms give you all something to be thankful for!


    :clap: :clap: :clap:

    Michele:love:
     
  9. LadyLuck

    LadyLuck Member
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    wow, it sounds like you guys are speaking another language! And here I am struggling with plain ol' gen chem.....
     
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