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Homework Help: Elimination reactions of cyclohexane derivatives

  1. Jun 14, 2018 #1
    1. The problem statement, all variables and given/known data
    When trans-2-methylcyclohexanol is subjected to acid-catalyzed dehydration, the major product is 1-methylcyclohexene.
    IMG_20180614_192151.JPG
    However, when trans-1-bromo-2-methylcyclohexane is subjected to dehydrohalogenation, the major product is 3-methylcyclohexene. IMG_20180614_192708.JPG
    The attempt at a solution
    Obviously, this is an E1 mechanism and the hydride shifts occur from either of the adjacent (to the carbocation) carbons in each case.
    Also, -Br is a much better leaving group than -OH (but I don't know if it's relevant here).
    The 3° hydrogen in case 2 is more acidic due to the greater -I effect of -Br. Right?
    How will this affect the outcome of the reaction?
    How can we account for the different outcomes anyway?
     
  2. jcsd
  3. Jun 14, 2018 #2

    mjc123

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    Is the mechanism the same in both cases?
    Is there a hydride shift?
    Is the leaving group -OH?
     
  4. Jun 14, 2018 #3

    TeethWhitener

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    Are you sure about that?
    Edit: ninja'd by mjc123.
     
  5. Jun 14, 2018 #4
    Edit: I was actually wondering now that case 2 outcome should've been 2-methylcyclohexane instead (due to the more acidic hydrogen)?
     
  6. Jun 14, 2018 #5
    Well, I think case 1 is definitely E1.
     
  7. Jun 14, 2018 #6

    TeethWhitener

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    Why? and do the same considerations apply for case 2?
     
  8. Jun 14, 2018 #7

    TeethWhitener

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    Side note: this molecule doesn't exist (why?)
     
  9. Jun 14, 2018 #8
    The protonated hydroxyl - carbon bond breaks, that's for sure.
    Since you objected, I'm reconsidering whether the 2nd is E1.
     
  10. Jun 14, 2018 #9
    I'm sorry I meant 2-methylcyclohexene
     
  11. Jun 14, 2018 #10
    Oh, no. The 2nd is an E2 mechanism
     
  12. Jun 14, 2018 #11
    Ok, so I see that in case 2, the base will attack the 2° hydrogen in preference to the 3° one.
     
  13. Jun 14, 2018 #12

    TeethWhitener

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    Ok. (You still haven't said why, but) we can move on. What will the intermediate look like in both cases?
     
  14. Jun 14, 2018 #13

    TeethWhitener

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    Of course this is true, given the product, but why?
     
  15. Jun 14, 2018 #14
    For E2, it's 1°>2°>3°
     
  16. Jun 14, 2018 #15
    I get it. I mainly got confused because of thinking that both were E1. Also, I went into thinking inductive effects and all.
     
  17. Jun 14, 2018 #16

    TeethWhitener

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    Why do you think this? It's not right.
     
  18. Jun 14, 2018 #17
    Why? That's what we're taught.
     
  19. Jun 14, 2018 #18

    TeethWhitener

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    What is the E2 product of 2-bromobutane?
     
  20. Jun 14, 2018 #19

    TeethWhitener

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    Remember Zaitsev's rule.
     
  21. Jun 14, 2018 #20
    2-Butene + 1-Butene
    (major) (minor)
     
  22. Jun 14, 2018 #21

    TeethWhitener

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    Right, so clearly
    Is incorrect. If it were correct, you'd expect the product to be 1-butene.

    Moving on. The important thing to think about is the intermediate. What does it look like in the 1-bromo-2-methylcyclohexane E2 reaction?
     
  23. Jun 14, 2018 #22
    Well, with Saytzeff's rule, even the case 2 product should've been same?
     
  24. Jun 14, 2018 #23

    TeethWhitener

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    Right. Which tells you that there's something else going on. Draw out the intermediate. It might even help to build a 3D model of the molecule.
     
  25. Jun 15, 2018 #24
    I got it. It has to do with the anti coplanar state preferred during E2 eliminations. In cyclohexane derivatives, the anti coplanar state is achieved only when the leaving group and the hydrogen adjacent to it are axial. Here, we've got trans-1-bromo-2-methylcyclohexane, whose chair confirmation would be as follows, and clearly the hydrogen on C3 is the only axial hydrogen adjacent to the -Br. Hence the product.
    Of course, the more stable confirmation of the reactant will be with the -Br being equatorial & -CH3 being axial, but it has to interconvert to this less stable intermediate to react.
    IMG_20180615_163911.JPG
     
  26. Jun 15, 2018 #25

    TeethWhitener

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    This is the right idea. The picture is almost correct, but on the carbon labeled 3, the axial H that is abstracted by the base will be pointed downward. The main point is that the leaving group and the proton have to be antiperiplanar in E2 reactions.

    Good work.
     
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