For the Organic Chem-heads out there...

Discussion in 'Community' started by themadchemist, Jun 30, 2004.

  1. themadchemist macrumors 68030

    themadchemist

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    #1
    I've got a burning orgo question for which I can't seem to find the answer. Consider the addition of a hydrogen halide to an alkyne.

    The addition of one equivalent merely involves the pi electrons of the triple bond attacking the hydrogen in H-X, thereby displacing the halide. Thus, one forms a vinyl cation on the more substituted carbon (Markovnikov addition). X- then adds to the carbocation, giving both E and Z isomers.

    So now you have an alkene with a halogen substituent on one of the double-bond carbons.

    When you have an alkyne whose triple-bond carbons are equally substituted, you get a mixture of products.

    This is all pretty simple. The problem comes when you add another equivalent of H-X.

    So, let's say you started with but-2-yne, a symmetrical alkyne.

    Addition of, say, hydrobromic acid (HBr) yields 2-bromo-but-2-ene:

    H3CBrC=CHCH3

    When you add another equivalent of HBr, experiments demonstrate that one obtains only the geminal dihalide:

    H3CBr2C-CH2CH3

    In other words, both bromides add to the same carbon. This suggests that the carbocation in the second addition formed on the carbon that already has the bromine. The bromine is electron-withdrawing and it should, therefore, destabilize the carbocation. If the alkyne were unsymmetrical, then maybe the argument could be made that the electron-donating effect of an additional alkyl group would outweigh the electron-withdrawing effect of the bromine, although I am even skeptical of this. However, in the case of the symmetrical alkyne, carbocation formation on the non-halogenated carbon would still provide the electron-donating benefits of an alkyl while avoiding the electron-withdrawing effect of the halogen.

    Why, then, does the geminal dihalide form? I'd really appreciate any insight anyone could provide.
     
  2. eaalthof@hotmai macrumors newbie

    Joined:
    Jun 30, 2004
    #2
    Bromine isn't particularly electron withdrawing

    However, the true answer lies with Br's lone pairs and a little bit with Br's size. The lone pairs stabilize the carbocation by donating electron density to it (imagine it as a kind of "resonance" the term for this is actually hyperconjugation [resonance is actually a subunit of hyperconjugation]). This stabilization greatly decreases the energy of the carbocation next to the bromine such that the second Br- attack creates the (geminal) 2,2 disubstituted product. This stabilization is great enough that if you start with 1-Bromo-1-butene and use HBr, the 1,1-dibromobutane product.
     
  3. themadchemist thread starter macrumors 68030

    themadchemist

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    #3
    Ah, that makes sense. Thank you. I didn't think about bromine's high polarizability. Moreover, I suppose that a carbocation's electrophilicity would be far greater than that of any halogen covalently bonded to form an octet. Then, the lone pairs would prove particularly useful in stabilizing the carbocation.

    I imagine this effect would be much weaker with, say fluorine, whose atomic radius is small and polarizability is low. Then, would you see a mixture of products or would the electrophilicity of the carbocation still exceed that of the highly electronegative fluorine sufficiently to draw some electron density toward it despite fluorine's low polarizability?

    edit: Oh, and I'm glad that my query encouraged you to join and post...Or, so it seems.
     
  4. RandomDeadHead macrumors 6502

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    #4
    Was I the only person who thought this thread was about something totally different? :confused:
     
  5. themadchemist thread starter macrumors 68030

    themadchemist

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    #5
    what did you think? I can't find any sexual double entendre in the thread title, which is usually what causes confusion.
     
  6. G5orbust macrumors 65816

    G5orbust

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    Jun 14, 2002
    #6
    Im thinking he saw a drug connotation but I could be wrong...
     
  7. themadchemist thread starter macrumors 68030

    themadchemist

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    #7
    Hmm, I SUPPOSE. So as not to disappoint, I could post the synthetic pathway (though not the lab instructions, since I don't have them) to synthesize meth amphetamine from common laboratory ingredients. ;)

    What?! It was a problem in my textbook!
     
  8. G5orbust macrumors 65816

    G5orbust

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    #8

    I think that qualifies for "subliminal messenging in literature"
     
  9. themadchemist thread starter macrumors 68030

    themadchemist

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    #9
    Yeah...But whoever said textbooks couldn't be practical!

    If we asked for real world problems, we got 'em.
     
  10. MongoTheGeek macrumors 68040

    MongoTheGeek

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    Its not so much where you are as when you are.
    #10
    One of my books in college described to how to manufacture RDX. It also had formulas to determine the k of a spring without measuring it.

    Mark's Standard Handbook.
     

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