Undergoes E2 reaction, what factor determines the side of double formation?

[vDDmaniaC]

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I always keep it mind that formation follow the stability and tend to form more stable bond. In this reaction, why Cl goes to Hc and Hd side, not Ha side? H2 side there are two carbons attached so it seems like having more stability.

 

[PinkDDS]

For E2 reactions you want to have the halogen and the H anti. Even if it does not seem very stable you still need to eliminate the H anti to the halogen.

 

[harrygt]

Cl will leave with Hd [the only anti-coplannar H], giving you a double bond there. Ha is not anti-coplanar with Cl, so E2 can't eliminate Ha.

 

[vDDmaniaC]

What a basic......... why didn't I know this................
But they usually don't give a question like visualizing picture where H is co planar with Cl stuff. I hope it won't be on my exam.

 

[doc3232]

I am confused here, I hate E2 E1 SN2 SN1...
So when you do E2, you must have the hydrogen that leaves be anti to the leaving group?

 

[doc3232]

What a basic......... why didn't I know this................
But they usually don't give a question like visualizing picture where H is co planar with Cl stuff. I hope it won't be on my exam.

expect it...probably would be one of tougher ones...I hope.

 

[yakuza]

Also be careful because sometimes you'll have to draw the chair confirmation to see which H is anti

 

[DROCKINDAHOUSE]

Helpful hint: E2 always corresponds to SN2 conditions and E1 to SN1 conditions (seriously). So with that in mind....what happens during an SN2 rxn? Well the nuc/base molecule approaches BACKSIDE right??? Well if it was to approach backside, and the rxn conditions were controlled such that elimination was favored over substitution (i.e. high heat), then the base molecule would abstract the proton that is ANTI to the halogen. At least thats the way I remember it.

 

[DROCKINDAHOUSE]

I always keep it mind that formation follow the stability and tend to form more stable bond. In this reaction, why Cl goes to Hc and Hd side, not Ha side? H2 side there are two carbons attached so it seems like having more stability.

Even though you may want to form the more substituted alkene, there is too much steric hindrance due to the presence of the bulky alkyl group for the nuc/base molecule to abstract that proton in an e2 reaction.