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Topic: cis/trans dehydrohalogenation products  (Read 16475 times)

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ektabh

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cis/trans dehydrohalogenation products
« on: December 16, 2004, 06:52:51 PM »
Hi,
 I was wondering how come we have 2 products when we dehydrohalogenate cis-1-bromo-2-methylcyclohexane and we have only one product if we dehydrohalogenate trans-1-bromo-2-methylcyclohexane? ???

I thought that it wouldn't matter if the reactant is cis or trans... I am confused :-\, please *delete me*! Thank you!

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Re:cis/trans dehydrohalogenation products
« Reply #1 on: December 16, 2004, 07:46:47 PM »
I assume that this dehalogenation takes place with base (i.e. an E2 elimination).  Think about the requirements of the E2 mechanism and you will find the answer.

dexangeles

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Re:cis/trans dehydrohalogenation products
« Reply #2 on: December 17, 2004, 02:21:42 AM »
I'd like to know the answer to this question too :)

The only thing I could think of that causes the difference is structure:

CIS having Br in equatorial and methyl as axial; thus forming a more stable confirmation which I believe exists more in solution
with this is mind, there will be two possibilities: a major and minor product
Zaitsev's would predict the major to be the tri-substituted cyclohexene (1-methylcyclohexene), the minor is the di-substituted 3-methyl-1-cyclohexene
Is this correct?

Trans having one product due to steric hindrance, right?  Or am I wrong?  The stable confirmation having both methyl and Br being equatorial; thus the methyl blocks....

WAIT A MINUTE!!!!  everything I've said pertains to Substitution, now I'm lost

so, what's the answer? lol
I assume that saying this is a dehydrohalogenation reaction, the "requirements" for Elimination reaction is being used...so that contradicts the other thought I have which pertains to an unhindered base possibly going through a possible Substitution reaction; thus giving cis 2 products (or maybe 3), and trans 1 product (or maybe 2)

Is there such a thing as a 100 percent reaction at all?


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Re:cis/trans dehydrohalogenation products
« Reply #3 on: December 17, 2004, 02:21:59 PM »
Well, you got off to the right start.  The important factor is the conformation of the ring.

Let's start at the start; here is a generic elimination reaction to form ethylene from iodoethane:



The leaving group and the proton that is pulled off by the base must be anti-periplanar to one another, meaning they are in the same plane, but pointing in opposite directions.  See the Newman projection below:



I have also shown a scheme with the orbitals of the starting material and the product shown.  These alignment of these orbitals is the reason that the anti-periplanar geometry is necessary for reaction by the E2 mechanism.  The sp3 hybrid orbitals will "un-hybridize" to eventually form the pi bond in the product.  As you know, for pi bonding, the two p-orbitals must be in the same plane.  Forming the alkene directly is a much lower energy reaction pathway than the alternative (deprotonation of a non-anti-periplanar hydrogen followed by rotation to place the resulting anion anti-periplanar to the leaving group followed by elimination).  Specifically, the electrons from the C-H bond are released from the C-H bond and then donated into the adjacent sigma anti-bonding orbital (not shown) of the C-I bond.

So, let's apply these principles to the cyclic case at hand.



In both cases there are two possible sites for deprotonation, the two carbons adjacent to the carbon bearing the bromine (labeled 1 and 2).  In order to achieve the anti-periplanar geometry that is necessary, the bromine must be axial (draw out all the possible Newman projections if you don't believe me).  Let's look at carbon 2, since it is the interesting one.  In this case, when we place the bromine axial we see that the proton at carbon 2 is anti-periplanar and elimination can proceed as we would expect.  If you drew out the Newman projection for carbon 1, you would find that elimination can occur from that side as well.  So, we could get two products from the cis starting material.

Now for the trans isomer:



In this case when we apply the same model we find that the bromine and the hydrogen at carbon 2 cannot achieve an anti-periplanar arrangement!  So elimination cannot occur from carbon 2.  Were you to draw out the Newman projection for carbon 1, you would find that a hydrogen can be anti-periplanar to the bromine so elimination can occur from carbon 1.  Only one product would be expected from this starting material.

Cool, huh?

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Re:cis/trans dehydrohalogenation products
« Reply #4 on: December 17, 2004, 03:00:50 PM »
15 scooby snacks given, geesh that was detailed.
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Re:cis/trans dehydrohalogenation products
« Reply #5 on: December 17, 2004, 04:45:19 PM »
Heh.

We aim to please.  :)

dexangeles

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Re:cis/trans dehydrohalogenation products
« Reply #6 on: December 17, 2004, 05:44:03 PM »
Awesome!!!  I can't believe I forgot about the anti coplanar structure is best.....I need to read again...

that's prolly why I bomed my finals  :'(

I'm down to 92 percent in Organic  :-\

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