Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: saN on November 18, 2007, 03:25:34 PM
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Isn't the left molecule cis, and the right molecule trans? Wouldn't the cis molecule react slower than the trans because of the beta hydrogen on the methyl is on the same side, yet needs to be anti inorder to react under E2? The trans will have the anti position in the boat form, and will undergo E2 easier.
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Absolutely correct.
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Isnt the right one cis and the left one trans?
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Yes, I missed that. However, the reasoning about trans-bromomethylcyclohexane reacting faster than the cis-cyclohexane is correct. Upon a second reading the sentence
The trans will have the anti position in the boat form, and will undergo E2 easier.
is wrong. The trans form will have the leaving group and beta hydrogen antiperiplanar in the chair form.
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Excuse me, but wouldnt the cis molecule react the quickest because it can for the more stable(more substituted) alkene when the C-Br and the C-H bond(from the C that also has the methyl) are both axial?
Then the second product from the trans molecule would be the less substituted alkene because the bonds could never be axial to give the more substituted alkene, it woul eliminate the proton from the C that does not have a methyl.
Another question, is there a more systematic name to label the Cs in the ring so that I dont have to keep saying "the C that has the methyl" and "the C that doesnt have the methyl"?
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Excuse me, but wouldnt the cis molecule react the quickest because it can for the more stable(more substituted) alkene when the C-Br and the C-H bond(from the C that also has the methyl) are both axial?
Then the second product from the trans molecule would be the less substituted alkene because the bonds could never be axial to give the more substituted alkene, it woul eliminate the proton from the C that does not have a methyl.
Another question, is there a more systematic name to label the Cs in the ring so that I dont have to keep saying "the C that has the methyl" and "the C that doesnt have the methyl"?
The trans isomer would be faster as Yggdrasil stated the abstracted proton must be anitperiplanar to the leaving substituent Br. The cis isomer would have to rotate conformation to make the C-2 (with methyl) available for proton abstraction. In the chair conformation the C-2 methyl would be axial and the bromine would be axial (elimination would not occur) For the cisisomer it would be more likely to abstract a proton from the C-6 position and form the less substituted cycloalkene.
You can address the isomers systematically by referring to the carbon with the methyl as C-2 in both instances. So we have cis C-2 and trans C-2.
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If the molecule is cis, than the leaving group is axial and the methyl group is equatorial. If the molecule is trans, than the leaving group and methyl group at both axial. Is this correct?
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In the chair conformation the C-2 methyl would be axial and the bromine would be axial (elimination would not occur)
Isn't this for the trans isomer?
Aren't E2 reactions with the beta hydrogen always anti to the leaving group? Why must it be antiperiplanar?
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Alright, I think I messed up my previous explanations. :-[ Here is an explanation to clear things up.
The molecule on the left is the trans isomer and the molecule on the right is the cis-isomer. In the trans-isomer, the bromine and methyl will both be equatorial in the most favorable conformation of the cyclohexane ring. In this conformation, there are no antiperiplanar hydrogens, so an E2 reaction cannot happen. For an E2 reaction to occur, the ring will have to flip to the conformation where both the methyl and bromine are axial. Since only a small fraction of the cyclohexenes will be in this conformation, your rate of reaction will be slow.
On the other hand, in the cis-cyclohexene, you have two conformations. In one conformation, the methyl is axial an the bromine is equatorial. In this conformation, the bromine is not antiperiplanar to any hydrogens. In the second conformation, the methyl is equatorial and the bromine is axial. This conformation places the bromine antiperiplanar to two hydrogen atoms. The difference here is that the equilibrium between the two conformations is not biased toward either form so near to half of the cyclohexanes will have the proper conformation for reaction.
For an illustration of this, see the picture below.
Antiperiplanar beta hydrogens are necessary because, to form a pi bond, the bonding orbitals for the leaving group and beta hydrogen must be coplanar so that they can overlap and form a pi bond.
Also, re: ultrashogun, the thermodynamic stability of the products is not relevant to the kinetics of a reaction (unless you're considering the backward rate of reaction). Key issues for the kinetics here are the availability of the correct conformation of your reactant and the stability of the transition state.
I hope this clears up any confusion I may have created previously.
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The trans isomer would be faster as Yggdrasil stated the abstracted proton must be anitperiplanar to the leaving substituent Br. The cis isomer would have to rotate conformation to make the C-2 (with methyl) available for proton abstraction. In the chair conformation the C-2 methyl would be axial and the bromine would be axial (elimination would not occur) For the cisisomer it would be more likely to abstract a proton from the C-6 position and form the less substituted cycloalkene.
Am I doing something wrong? I see it completely opposite the way you do. Replace ther terms "cis" and "trans" in your post and thats how I see it.
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Also, re: ultrashogun, the thermodynamic stability of the products is not relevant to the kinetics of a reaction (unless you're considering the backward rate of reaction). Key issues for the kinetics here are the availability of the correct conformation of your reactant and the stability of the transition state.
But doesnt the lower energy product also have the lower energy transition state?
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Not necessarily. If you can make a reaction irreversible, you can sometimes use this fact to isolate the kinetically favored product instead of the thermodynamically favored product.
Unfortunately, I can't think of any examples of this now? Anyone else know of any?
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By choosing the proper reaction conditions (usually low temperatures) the 1,2 addition product of HX to a diene can be produced as the major product in preference to the generally more stable 1,4 addition product.
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This helps a lot. I was trying to build it out of my molecules, and boy do they help so much! Thanks for all your help.
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The difference here is that the equilibrium between the two conformations is not biased toward either form so near to half of the cyclohexanes will have the proper conformation for reaction.
Why is this statement true? Wouldn't the preferred isomer be where the methyl group is equatorial and bromine axial since the methyl won't be interacting with any of the hydrogens?
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Well, you get interactions between the bromine and hydrogens when the methyl is equatorial. I'm not sure how the size of a methyl compares with the size of a bromine.
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According to published "A values" (a measure of 1-3 diaxial interactions), the bromine atom (A = 0.38-0.49) is considerably less sterically demanding than the methyl group (A= 1.70). A list of these values can be found at the following link:
http://www.chem.wisc.edu/areas/organic/index-chem.HTM
You will need to select "A values" from the list at the left.
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Sweet. Thanks RBF. So, saN, you are correct. The preferred conformer will be the one with the methyl group equatorial and the bromine axial.