E1

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Dencology

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Which of the following alkyl bromides exhibits SLOWEST rate of E1 elimination?
A. (CH3CH2)2CHBr B. (CH3)2CHBr C. (CH3)3CBr D. CH3CH2CH2Br E. CH3CH2Br
why is choice C wrong. isn't E1 is like SN1 in terms of carbocation. Tertiary carbocation is favored over primary? and isn't carbocation is the slowest step of the reaction.

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Correct but you are mixing up the fact that formation of the carbocation is the slow rate determining step and the fact that the question acts which i the slowest of these to do E1
But like you said c is tertiary so it would go SN1/E1 and while the formation of the carbocation at that tertiary site is rate determining of the choices it is the best suited to form this carbocation of the choices and would be the fastest.
That being sais I would say E is the slowest (actually wouldnt go E1 at all because the carbocation would be primary and no chance for rearrangement)
Hope that helps
 
Damn, second guessing is screwing me up.

When the Br leaves in E, you have a primary carbocation, and you can't change it to anything more stable.
 
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if D and E both dont go E1, would they even be the right answer? When the question is asking fo slowest E1, isn't that assuming that the rxn actually takes place? I was thinking in D, that E1 could occur because rearrangement is possible to form a secondary carbocation, thus would be the slow E1 rxn. Am I mistaken? What is the correct answer in your book?
 
if D and E both dont go E1, would they even be the right answer? When the question is asking fo slowest E1, isn't that assuming that the rxn actually takes place? I was thinking in D, that E1 could occur because rearrangement is possible to form a secondary carbocation, thus would be the slow E1 rxn. Am I mistaken? What is the correct answer in your book?

I thought you usually listed those as slowest anyways... because maybe it can occur under some circumstances??

Yeah I want to know the answer...
 
ah i found the answer. the answer is E and not D because E1 follows Sn1 in that there is a formation of a carbocation. In choice D, this carbocation can be stabilized by the inductive effects of the other alkyl groups whereas in choice E, the smaller chain leads to less of an inductive effect, thereby making it the least of all to undergo an E1 reaction.

based on that explanation (which is from achiever btw), if you had to put them in order from fastest reacting to slowest reacting, it would be C, A, B, D, E (A before B because more alkyl groups and thus higher inductive effect?) not positive on this, hoping someone could verify this for me. thanks =)
 
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