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Offline xstetsonx

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E2 reaction
« on: February 03, 2010, 07:57:26 PM »
cl-                      -I
benzene-  c-c   -H
H-                      -H

react with NaOH, methanol, and heat,
why is Iodine is more likely to be elminate?


sorry first time don't know how to draw the structure here.
plz forgive me
« Last Edit: February 03, 2010, 08:25:41 PM by xstetsonx »

Offline Schrödinger

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Re: E2 reaction
« Reply #1 on: February 03, 2010, 10:36:05 PM »
I hope you are referring to



If this is an E2 reaction :
Leaving group ability, as I have learned, is inversely proportional to the basicity of the leaving group. Here compared to I-, Cl- is a stronger base.


Why not E1? :
But then, why shouldn't the Chlorine leave? I mean... the subsequently formed carbocation will be stabilized by resonance right? I want someone to help us with this part please.
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Offline orgopete

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Re: E2 reaction
« Reply #2 on: February 04, 2010, 09:22:03 PM »
This is a good and difficult question to give a precise answer. From references in Org Syn of phenyl acetylene (Organic Syntheses, Coll. Vol. 1, p.438 (1941)), both alpha and beta-halostyrenes have been converted. I presume that the halostyrenes are prepared from the dihalides.

An analogous elimination of a 1,2-dibromo-1-phenylethane gave 2-bromo-1-phenylethene with KOtBu (Can J. Chem, 1977). A similar phase transfer elimination gave the alpha-bromo result (http://www.springerlink.com/content/wjl07u30033m4040/). In that paper they allude to all mechanisms possible. In the abstract, they suggest it is an E1cb-like reaction.

From that information, the iodide is the preferred leaving group because of the greater acidity of the benzylic hydrogen.

Why is it difficult to know what the product is and the mechanism? Even though I have considerable experience in organic chemistry, reactions can be constructed that are difficult to predict. This is such a case. In this example, a polar solvent is used and the base concentration is not specified. The difference in mechanism in such an instance can vary simply upon dilution. It would be possible to carry out the reaction is a manner more likely to give an E2 elimination (high concentrations, excess base, heat). However, this example allows the elimination to give a Zaitsev or Hoffmann product, different halogens, etc., so it is a difficult question.

Although I could not determine directly, I infer that beta-bromostyrene has been known for a longer time. I presume this is from the elimination from the dibromide. From that model, I would have expected the chloride to be eliminated. However, the leaving bromide has been replaced with a poorer chloride as leaving group and the group to be retained has been replaced with a better leaving iodide. It would be interesting to read a publication with this example.
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Offline Schrödinger

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Re: E2 reaction
« Reply #3 on: February 05, 2010, 03:48:03 AM »
In our case, we have
Let's consider more than one mechanism for the Elimination process, as has been pointed out be orgopete.

1. The formation of a benzylic anion via E1cb, and the subsequent elimination of .
2. The formation of benzylic carbocation via the elimination of , which would lead us to conclude an E1 reaction.
3. Two possible E2 mechanisms (Elimination of I- or elimination of Cl-)

Taking into consideration the polarity of the solvent, we can restrict our thoughts to 1 and 2, since we have stable ions being formed in these cases.

My question is thus:
Whether 1 or 2 is preferred depends on the concentration if the base. i.e., if the concentration of the base is high, E1cb would be favoured.

But this would also increase the rate of E2 reaction, which I had disregarded in the previous paragraph.

Can you please throw more light on the situation as to what would be expected to take place?
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