will this undergo E1,E2,SN1, or SN2?

[joonkimdds]

I was wondering which reaction will this undergo.

It doesn't have aprotic but I think that doesn't mean it's impossible to undergo SN2.

it has small unhindered base/nucleophile so I think E2 is not an answer.

It has strong base/nucleophile that prefers SN2 and E2.

it has protic that prefers SN1 and E1.

I don't know which one it is.

---

Members don't see this ad.

 

[sciencegod]

e2. it has a very strong nucleophile so on a secondary halide e2 will be prefered.

 

[chessxwizard]

E2 since it's a strong base which creates a pi bond that is conjugated with the ring system.

 

[joonkimdds]

e2. it has a very strong nucleophile so on a secondary halide e2 will be prefered.

But shouldn't it be hindered base to make it E2?
Ethoxide is an unhindered base/nucleophile.

I thought that
strong base/nucleophile = both SN2 and E2
but if it's hindered = E2
if not hindered = SN2

 

[chessxwizard]

Read my response and think about it.

 

[MaskItOrCasket]

Read my response and think about it.

yup agreed.... that ring is gonna trump all of them

 

[doc3232]

I think E1/E2 because it is in the benzylic position. The carbocation is VERY stable because of the aromatic ring. I don't think you can say E2 and not E1 despite the strong base.

 

[sciencegod]

E2 since it's a strong base which creates a pi bond that is conjugated with the ring system.

good point

 

[sciencegod]

as it happens, the reaction would be e2 on a cyclohexane also.

 

[5HTburb]

http://www.mhhe.com/physsci/chemist...raphics/carey04oc/ref/ch11benzylicsystem.html

Go towards the bottom of the page (E1 vs E2)

Combination of good LG with SB and benzylic halide, E2. Check out the site though.

 

[sciencegod]

it has protic that prefers SN1 and E1.

I don't know which one it is.

its still e2 if the protic solvent is the conjugagte acid of the strong nucleopihile.

 

[Sublimation]

If you enter any chemistry laboratory, and ask the TA or researchers to define which mechanism a reaction will undergo or describe the different mechanisms. You will get an array of answers. yes this is an predominantly E2 reaction, however because there is a lack of heat, you can just as easy argue an sn1 due to stabilization of the carbocation by the benzene ring (resonance stabilized). However, in a lab you will end up with both mixtures and obviously due to the fact that we have a secondary carbon the major product will be a zeitsev << I think i speeled that wrong, Alkene. Here this chart helped me out alot in orgoI. Hope this helps.