Why must the hydrogen in E2 reactions be anti- and planar to the leaving group?

  • Thread starter Ahmed Abdullah
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In summary, the proton is withdrawn from the alkyl halide and the carbon begins to re-hybridize, forming an alkenes.
  • #1
Ahmed Abdullah
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In the presence of polar, non-protic solvents (i.e., DMSO or DMF), alkyl halides undergo reaction with bases to generate alkenes by an E2 mechanism. In the E2 reaction, the proton removed by the base must be anti- to the leaving group, and they must also be periplanar (the hydrogen, the leaving group and the two carbons must all be in the same plane). In the E2 reaction, both the alkyl halide and the base are present in the rate-limiting transition state, making the reaction bimolecular and concerted.

Now my question is why hydrogen which is removed must be anti-(trans-) and planar relative to the leaving halogen??

I have found in some books it is about electron coming from backside and forming C-C double bond. HERE what is meant by electron coming from backside.:grumpy:

Your sensible response will broad my understanding. Thnx in anticipation.
 
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  • #2
The base withdraws the proton in the E2 mechanism. This causes the carbon to begin to rehybridize in the rate determining step from sp3 to sp2. For there to be simultaneous loss of halogen (as X+) they must be interacting by the newly rehybridizing carbon's nascent p orbital. This is the same p orbital that will eventually form the double bond. You can see from a geometrical analysis that the departing proton, nascent p-p orbital formation and halogen leaving group must be coplanar. Aligning the two p "dumbell" shaped orbitals at right angles precludes any interaction, they must be aligned head to head. The only way for this to happen is for the halogen to be either trans or eclipsed cis to the incoming base. For steric reasons the cis orientation is disfavored.
 
  • #3
Isn't something about orientation of molecular orbital?, due to the overlap required beween the sigma* C-X and sigma C-H orbitals to form the double bond (which is the combination of two coplanar p orbitals)?
 
  • #4
The leaving group, halogen, will produce a p orbital for overlap with the adjacent carbon's newly forming p orbital. For there to be simultaneity, these two orbitals must be aligned in a coplanar fashion.
 
  • #5
Another way to think about it is that the leaving group will run off and leave a carbocation at the other end, which is then promptly taken up by the electron coming in 'the back side'- imagine people in two rooms, someone left the front room and another person replaces that person from the back room. This means there is an empty 'space' for a new person to come into the back room - this electron is the new person! That's an easy way to think about it at any rate- you can go from there! The overlapping orbitals will have an empty space which is filled up by that electron...
 

1. What is an E2 mechanism?

An E2 mechanism is a type of elimination reaction in organic chemistry in which a molecule loses a leaving group and a proton simultaneously, resulting in the formation of a double bond. It is a one-step process and follows the bimolecular rate law.

2. How does an E2 mechanism differ from an E1 mechanism?

An E2 mechanism differs from an E1 mechanism in that it does not involve a carbocation intermediate. In E1 reactions, the leaving group first dissociates to form a carbocation, which then undergoes deprotonation to form a double bond. E2 reactions occur in a single step, with the leaving group and proton being eliminated at the same time.

3. What are the requirements for an E2 reaction to occur?

An E2 reaction requires a strong base, a good leaving group, and the presence of a hydrogen atom on the carbon adjacent to the leaving group. The base must be strong enough to remove the proton and form the double bond, while the leaving group must be able to leave easily.

4. What factors affect the rate of an E2 reaction?

The rate of an E2 reaction is affected by the strength of the base, the strength of the leaving group, and the steric hindrance around the carbon with the leaving group and the adjacent carbon. A more bulky base or a more hindered carbon will result in a slower reaction rate.

5. What are some examples of E2 reactions?

E2 reactions are commonly seen in the synthesis of alkenes from halogenated compounds, such as the dehydration of alcohols or the dehydrohalogenation of alkyl halides. They are also involved in the elimination of hydrogen halides from cycloalkanes to form alkenes.

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