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Topic: Dehydrohalogenation of alkanes vs elimination reaction  (Read 10829 times)

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Offline vivekfan

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Dehydrohalogenation of alkanes vs elimination reaction
« on: June 13, 2009, 03:42:14 PM »
To me, the general form of the elimination reaction and the dehydrohalogenation reaction of an alkane to make an alkene look identical. What is the difference? Is the only difference the anti-periplaner geometry that is a requirement of the E2 reaction?

Offline azmanam

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Re: Dehydrohalogenation of alkanes vs elimination reaction
« Reply #1 on: June 13, 2009, 06:41:41 PM »
deydrohalogenation is an example of an elimination reaction.  So is dehydration.  Whether it's E1 or E2 will depend on reaction conditions.

try this page (especially the table at the bottom):
http://www.cem.msu.edu/~reusch/VirtTxtJml/alhalrx3.htm#hal6
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Offline kgerhard320

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Re: Dehydrohalogenation of alkanes vs elimination reaction
« Reply #2 on: June 14, 2009, 02:55:02 PM »
Dehydrohalogenations are always E2 (concerted)

Offline plankk

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Re: Dehydrohalogenation of alkanes vs elimination reaction
« Reply #3 on: June 14, 2009, 03:10:25 PM »
What about for example (Me)3CBr? I don't think that elimination for that compound will undergo by E2 mechanism.

Offline azmanam

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Re: Dehydrohalogenation of alkanes vs elimination reaction
« Reply #4 on: June 14, 2009, 03:16:54 PM »
yes, t-butyl bromide will undergo E2... again, depending on conditions.  KOtBu will eliminate by E2 mechanism.  However, maybe something like NaSMe in DMSO or acetonitrile will go by a unimolecular mechanism - and therefore give nontrivial amounts of elimination product (as a mixture with substitution).  But I agree with your sentiment.  I hesitate to say anything always proceeds by one particular mechanism.
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Offline orgopete

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Re: Dehydrohalogenation of alkanes vs elimination reaction
« Reply #5 on: June 15, 2009, 11:55:25 AM »
To me, the general form of the elimination reaction and the dehydrohalogenation reaction of an alkane to make an alkene look identical. What is the difference? Is the only difference the anti-periplaner geometry that is a requirement of the E2 reaction?

This question looks as though it is asking how an E1 reaction differs from an E2 elimination. In another question about elimination reactions, I suggested substitution v elimination is an often asked question. I raised that to highlight the difficulty one has in creating a clear picture for this and other chemical reactions.

Textbooks often have clear descriptions of SN1, SN2, E1, and E2 reactions. Then, when you begin to look at results, it becomes increasingly difficult to understand why or how these descriptions apply. I suggest that there are two extremes for substitution reactions. At one end, the leaving group departs to leave an intermediate that exists with sufficient life that no chirality information is retained before reacting with a nucleophile. At the other end, a nucleophile forms a bond with the carbon before the leaving group departs. In this reaction, chirality is reversed. In the reactions you do or read about, will be somewhere between these extremes.

Arguably, the leaving electrons of iodide have a weaker interaction with carbon than those of chloride. The degree to which bond formation must proceed in order for the carbon-iodide electron interaction to reach zero will be different than with chloride. That is, in these collision rate reactions with two pairs of electrons, the rates of the electron movements will simply differ. The expectation that they should be equal in an SN2 reaction should be doubtful.

If we compare an elimination reaction with a substitution reaction, we add an additional pair of electrons to this process. In my book, A Handbook of Organic Chemistry Mechanisms I discuss several reactions in light of these rate effects. I consider Zaitsev v Hoffmann eliminations to be a good exemplification of these effects.

Okay, back to the question. An E2 elimination is like a substitution reaction in which the bond forming electrons that push the leaving group out are from a neighboring pair of electrons and this best occurs from the back side. That is the reason an E2 elimination occurs via anti-periplanar geometry, but not necessarily Z v H elimination.

What if bond cleavage is much faster than displacement from neighboring electrons for elimination or intermolecular for substitution? What might slowdown bond formation? A similar question about amine basicity was discussed on this forum. The argument that carbon is an inductive electron donor was made. Rationally, electron donation by neighboring carbons will interfere with an attack by a nucleophile and will similarly enable greater bond cleavage (lengthen the carbon-leaving-group-electron-pair distance). Consequently, SN1 reaction occur and SN2 reactions do not occur with tertiary halides. It is the same rational for both effects.

If E1 and E2 elimination reactions are the intramolecular equivalents of substitution reactions, then bond cleavage should precede bond formation in an E1 elimination reaction. Hence, anti-periplanar geometry and concerted displacement does not need to be present. In E2 elimination reactions, bond formation is a kinetically important part of that reaction, hence anti-periplanar geometry is common. 

Quote from: azmanam
yes, t-butyl bromide will undergo E2... again, depending on conditions.  KOtBu will eliminate by E2 mechanism.  However, maybe something like NaSMe in DMSO or acetonitrile will go by a unimolecular mechanism - and therefore give nontrivial amounts of elimination product (as a mixture with substitution).  But I agree with your sentiment.  I hesitate to say anything always proceeds by one particular mechanism.

In light of my arguments, I am dubious of an SN1 reaction occurring as described. I argue the solvent change will retard carbon-halide bond cleavage and I believe it has the opposite effect. The solvent change will decrease bond cleavage. In order to undergo substitution, in a dipolar aprotic solvent, more bond formation must occur before bond cleavage (consistent with a description of SN2 reactions). The challenge of a substitution reaction with a tertiary halide is that electrostatic repulsion will interfere. The stronger electrostatic interaction will occur with the protons leading to an elimination. Because sulfide is a weaker base than butoxide, I would expect the elimination rate to be slower, but not to result in a different mechanism. I would predict that if an even weaker base were used (NaI), the elimination reaction rate will become very slow.
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