orgo confusion

[IWantH2O]

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(CH3)3CCHBrCH3 -----> (CH3)3CC=CH2 the base used was NaOH

but why isnt there a methyl shift....is it becaue its an E2 and there is no carbocation intermediate or can there still be a shift in E2

 

[jdent]

(CH3)3CHBrCH3 -----> (CH3)3C=CH2 the base used was NaOH

but why isnt there a methyl shift....is it becaue its an E2 and there is no carbocation intermediate or can there still be a shift in E2

did u mean to write (CH3)3CH + BrCH3 ?

 

[IWantH2O]

did u mean to write (CH3)3CH + BrCH3 ?

i edited it....do u no the answer to my question

 

[jdent]

there cant be a shift since the carbon is already a tertiary carbon and therefore cant get a double bond because that would make it carbon with 5 bonds- I dont see any other shift possible.

 

[funktion]

there is no shift because there is no carbo cation formed in this reaction. a proton would be abstracted from the methyl group by OH- [(CH3)3CCHBrCH3] resulting in a carbanion and a methyl shift would only increase instability so it will not happen in this situation. so the proton is abstracted leaving a carbanion then double bond forms and Br- leaves with excess charge.

 

[futuredent22]

There indeed would be a methyl shift if this was a E1 reaction. However, because we have a strong base, it goes E2. jdent, if your confused about the shift, try drawing the molecule out and removing the Br and try to do a shift to make a more stable carbocation.

 

[harrygt]

Let's keep it simple. You are using a strong base [NaOH], so that means the reactions is going to go through an E2 mechanism. There are no rearrangements in E2.
In order to have an E1 reaction, you have to use weak bases like an alcohol.

 

[doc3232]

(CH3)3CCHBrCH3 -----> (CH3)3CC=CH2 the base used was NaOH

but why isnt there a methyl shift....is it becaue its an E2 and there is no carbocation intermediate or can there still be a shift in E2

This question is unfair.
It is a secondary carbon.
A methyl shift could occur.
Finally, this reaction will go by SN1, SN2, E1, and E2. OH- is a great nucleophile.
We can only be sure of primary and teritiary carbons.
When you do secondary, all hell breaks loose and we can only find out what really happens experimentally.