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Elimination reactions of cyclohexane derivatives

  • #1
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Homework Statement


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?
 

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Answers and Replies

  • #2
mjc123
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Is the mechanism the same in both cases?
Is there a hydride shift?
Is the leaving group -OH?
 
  • #3
TeethWhitener
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Obviously, this is an E1 mechanism
Are you sure about that?
Edit: ninja'd by mjc123.
 
  • #4
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Edit: I was actually wondering now that case 2 outcome should've been 2-methylcyclohexane instead (due to the more acidic hydrogen)?
 
  • #5
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Edit: I was actually wondering now that case 2 outcome should've been 2-methylcyclohexane instead (due to the more acidic hydrogen)?
Well, I think case 1 is definitely E1.
 
  • #6
TeethWhitener
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Well, I think case 1 is definitely E1.
Why? and do the same considerations apply for case 2?
 
  • #7
TeethWhitener
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2-methylcyclohexane
Side note: this molecule doesn't exist (why?)
 
  • #8
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Is the leaving group -OH?
The protonated hydroxyl - carbon bond breaks, that's for sure.
Is there a hydride shift?
Since you objected, I'm reconsidering whether the 2nd is E1.
 
  • #9
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Side note: this molecule doesn't exist (why?)
I'm sorry I meant 2-methylcyclohexene
 
  • #10
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Why? and do the same considerations apply for case 2?
Oh, no. The 2nd is an E2 mechanism
 
  • #11
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Ok, so I see that in case 2, the base will attack the 2° hydrogen in preference to the 3° one.
 
  • #12
TeethWhitener
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Oh, no. The 2nd is an E2 mechanism
Ok. (You still haven't said why, but) we can move on. What will the intermediate look like in both cases?
 
  • #13
TeethWhitener
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Ok, so I see that in case 2, the base will attack the 2° hydrogen in preference to the 3° one.
Of course this is true, given the product, but why?
 
  • #14
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Of course this is true, given the product, but why?
For E2, it's 1°>2°>3°
 
  • #15
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I get it. I mainly got confused because of thinking that both were E1. Also, I went into thinking inductive effects and all.
 
  • #16
TeethWhitener
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For E2, it's 1°>2°>3°
Why do you think this? It's not right.
 
  • #17
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  • #18
TeethWhitener
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Why? That's what we're taught.
What is the E2 product of 2-bromobutane?
 
  • #19
TeethWhitener
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Remember Zaitsev's rule.
 
  • #20
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2-Butene + 1-Butene
(major) (minor)
 
  • #21
TeethWhitener
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2-Butene
Right, so clearly
For E2, it's 1°>2°>3°
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?
 
  • #22
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Well, with Saytzeff's rule, even the case 2 product should've been same?
 
  • #23
TeethWhitener
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Well, with Saytzeff's rule, even the case 2 product should've been same?
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.
 
  • #24
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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
 

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  • #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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