E2 major product

[ioud]

I am really confused about the E2 mechanism

From what I know about E2 reactions,

* When a small nucleophile attacks a hydrogen, the major product is the most substituted alkene

* When a big nucleophile attacks a hydrogen (antiperiplanar in both cases), the major product is the less substituted alkene

But I don't understand the fact that sometimes only 'E' or 'Z' alkene forms and sometimes a mixture of them forms.
I don't seem to get the right product every time I attempt to solve E2 reaction problems.

For example, when 1-bromo-2 methyl butane with a deuterium ion attached to carbon 1 (same carbon as bromine is attached) reacts with tert butoxide, why is it that only a Z product forms? If its b/c of the antiperiplanar reason, how can I rotate the atoms the right way so I can get the Z product?

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[PiBond]

I am really confused about the E2 mechanism

From what I know about E2 reactions,

* When a small nucleophile attacks a hydrogen, the major product is the most substituted alkene

* When a big nucleophile attacks a hydrogen (antiperiplanar in both cases), the major product is the less substituted alkene

But I don't understand the fact that sometimes only 'E' or 'Z' alkene forms and sometimes a mixture of them forms.
I don't seem to get the right product every time I attempt to solve E2 reaction problems.

For example, when 1-bromo-2 methyl butane with a deuterium ion attached to carbon 1 (same carbon as bromine is attached) reacts with tert butoxide, why is it that only a Z product forms? If its b/c of the antiperiplanar reason, how can I rotate the atoms the right way so I can get the Z product?

The E/Z mixture arises from differet beta-hydrogens that are possible to abstract that are in the antiperiplanar geometry. If there's only one choice in hydrogens to abstract then you only get one isomer.

For your second question I would need to see the stereochemistry of the molecule to tell you which way to rotate it. However, draw your newman projection looking down the bond between your leaving group (Bromine) and the beta-hydrogen (on carbon 2) and then rotate it until you find conformation in which they are anti. At that point, look at the other groups on the molecule and you'll observe the Z isomer.

 

[CaptainSSO]

Pretty sure that this is more detail than necessary for the MCAT, but with E2 the base will attack the antibonding orbital of the carbon with the hydrogen (C2 in this case)-- the leaving group needs to be antiperiplanar to the hydrogen in order for rearrangement to not occur.

And as ^ alluded to, pictures would be very, very helpful here.