E2 Bulky Base Least Substituted

[marymatthews]

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Chad's videos explains that the least substituted double bond is formed during an E2 reaction with a tertiary and bulky base.
However, all over the Chemistry DAT Destroyer, the least substituted double bond is formed during an E2 reaction with a SECONDARY and tertiary with bulky base.
So which is right? Does the least substituted occur with tertiary only, or secondary as well?

 

[Ibraiz]

In E2 mechanism, if its going to be least substituted is dependent on the tertiary base (nucleophile). It is not dependent on the tertiary or secondary halide (substrate).
So to answer your question, yes, the least substituted can be formed with secondary or tertiary halide given tertiary strong bulky base as a nucleophile.

 

[rnbaxter]

In E2 mechanism, if its going to be least substituted is dependent on the tertiary base (nucleophile). It is not dependent on the tertiary or secondary halide (substrate).
So to answer your question, yes, the least substituted can be formed with secondary or tertiary halide given tertiary strong bulky base as a nucleophile.

👍

 

[marymatthews]

I just watched the video with Chad again... he specifically said for the anti zaitsev product to form with E2, there needs to be a bulky base and a tertiary halide (electrophile)
he then gives an example with 2 bromo butane (a secondary halide) reacted with a bulky base, and the product formed is a zaitsev product (2 butene)
could you please clarify this?
thanks

 

[healthcaree]

Are you watching the HD quality or the original one?

If you look at "O2.6-Distinguishing-SN2-E2-SN1-E1" HD one, the example that he gives is

2-bromo-butane reacting with NaOCH3 (not bulky) in DMSO. This forms 2-butene like you said.

The reaction that he gives for bulky base is

1-bromo propane reacting with tertiary/bulky base which forms 1-propene.



good example he first gave was:

1-bromo-1-methyl cyclohexane reacting with tertiary/bulky base.

Here you have a tertiary halide; the double bond forms for the least substituted carbon (methylenecyclohexane) = anti-zaitsev product.

 

[marymatthews]

I am talking about the original non-HD video.... if you watch that it will show you what I am talking about....
1-bromo propane reacting with tertiary/bulky base will form 1-butene with or without a bulky base....

 

[healthcaree]

that is weird.. my destroyer problems only show anti-zaitsev product on tertiary halide + tertiary bulky base. possibly cuz i have 2008 edition.

most of my destroyer problems react with C2H5O-K+ which is not a bulky base. is it reacting with (CH3)3CO-K+ AND secondary halide to form anti-zaitsev product? if so, i would like to know the reasoning behind it too.. this is confusing now. b/c yes clearly chad mentions in the HD version also that tertiary halide/bulky base would form anti-zaitsev. If secondary halide, you don't always get the anti-zaitsev product (often get the zaitsev product).

 

[Ibraiz]

I am talking about the original non-HD video.... if you watch that it will show you what I am talking about....
1-bromo propane reacting with tertiary/bulky base will form 1-butene with or without a bulky base....

It could've been a typo but 1-bromo propane can not form 1-butene after reacting with a tertiary bulky base. If you were trying to write 1-bromobutane forms 1-butene then its only going to form 1 butene regardless of the nucleophile (its difficult to explain without images) because halide (Bromine) will leave from carbon 4 and it can only form a double bond with carbon 3 since we don't have rearrangement in E2 and this is not E1 (we have rearrangement in SN1 & E1 to form most substituted carbocation i.e. tertiary carbocation)