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What rule is this (O-Chem)?

Discussion in 'DAT Discussions' started by FROGGBUSTER, Aug 1, 2011.

  1. FROGGBUSTER

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    [​IMG]

    Why is the second product more favored than the other?
     
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  3. rmm30

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    That's a good question. I'm curious to know too. Furthermore to get decent yields via E1 elimination with an alcohol normally you have to use a big fat acid like Phosporic or Sulfuric. Why is H30+ being used here instead?
     
  4. indigenoustw

    indigenoustw Am I picking my nose or showing my shaka?
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    To be honest, I won't even know which will be the major or minor product if I see this question on my test coz both are tri substituted alkenes. But since we know which is major, there's one explanation for it.

    As tertiary carbocation forms, the more lengthy carbon chain stabilizes the positive charge more so than the other substituents because it has more electrons to donate to the electrophilic carbon (just like EAS). Since that's likely the case, the second product will predominate because the energy barrier to the formation of this product is lower, thus the major product.

    I can't think of any other way to explain this....and I don't think there's a rule for it lol
     
  5. quocstazz

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    hey guys, The H+/H2o in here is use to protonate the OH group on the reactant. after protonated, the OH becomes H2O, which is a good leaving group. the substrate will leave, leaving back a carbocation, the reason why the bottom picture is the major product is because the carbocation that i was left behind is most stable because of markovnikov rule i think

    its either markonikov or zaitsev
     
  6. Double Bonded

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    I miss ochem now that my DAT is done. :(


    It's probably something like the heat makes the kinetic product form faster than the thermodynamic one. Ask chad on his forum if you're using his videos, he's really helpful, even on stuff not from the website and he gets back to you in 1-2 days (usually).

    Markovinkov is for addition reactions.

    Zaitsev rule says that the most substituted pi bond gets formed but both are tri-substituted. This is a weird question. It might have something to do with inductive effects :confused: but I'd just ask Chad or some other expert.



    Where is this question from btw?
     
    #5 Double Bonded, Aug 2, 2011
    Last edited: Aug 2, 2011
  7. FROGGBUSTER

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  8. yappy

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    Anyone else wish they could borrow chad's brain for the DAT? I wish I could send him as a stand in lol.
     
  9. Maasfoor

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    the propyl group on the right side of the minor product is less favored because it is bulky and it's interfering more with ethyl group below it. While in the major product you have equal groups leading to less hindrance. I think! :)
     
    #8 Maasfoor, Aug 2, 2011
    Last edited: Aug 2, 2011
  10. LetsGo2DSchool

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    Nah, this doesn't make sense. If you look at where the double bond is formed in elimination, the major product is formed from this bulkier propyl chain while the minor product is formed from a smaller ethyl chain. So there are more sterics involved in forming the major product.
     
  11. LaughingGas

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    I thought thermodynamic/kinetic product was more relevant when it comes to an addition reaction. Plus if heat is applied, you should get more thermodynamic product than kinetic, cz usually thermodynamic products have lower energy as product but higher transition statw energy.
     
    #10 LaughingGas, Aug 2, 2011
    Last edited: Aug 2, 2011
  12. Double Bonded

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    Right you are LaughingGas, my mistake.
     
  13. Maasfoor

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    In the mechanism of either product the hydrogen would be removed from a secondary carbon to form the double bond so I don't think hindrance in the mechanism plays a big role in the outcome for this particular problem. Therefore the stability of the end product is what determines the major product.
     
    #12 Maasfoor, Aug 2, 2011
    Last edited: Aug 2, 2011
  14. egan

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    The more stable product is always the more substituted alkene. Take a look at #229 on 2011 destroyer and it should clear it up a bit. Hope this helps! Let me know if I'm mistaken.:)
     
  15. HannibalLecter

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    I think this question has nothing to do with Zaistev, Markovinokovs or whatever because the two products all obey those rules, its just that if the double bond was introduced into the alkane chain where it would likely be placed:

    Right smack dab in the middle or towards the end of one side of the chain and I don't know where I read it but there was rule on which is more stable (maybe boiling point or melting point) that the more interior a double bond is meaning the more centered the double bond is in the alkane chain the more stable it is. Since product B, is right smack dab in the middle, I suspect thats the reason why it is more stable than product A.
     

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