DESTROYER OCHEM question #34

[potbelly]

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Hi all,

I am little confused about Sn2 and E2 products...
In #34, if you look at the first step, a double alpha proton is removed by NaOC2H5 and strong and bulky base is formed.

This will react with CH3CH2Br via E2 and CH2=CH2 will be formed.
This was how I thought.

However, destroyer says after deprotonation, it will undergo SN2.

I know that it is a primary alkyl bromide.
If it is strong and bulky base, doesn't the reaction favor E2 rather than SN2? Am I thinking wrong?? What do you guys think?

 

[xplayer2885]

Hi all,

I am little confused about Sn2 and E2 products...
In #34, if you look at the first step, a double alpha proton is removed by NaOC2H5 and strong and bulky base is formed.

This will react with CH3CH2Br via E2 and CH2=CH2 will be formed.
This was how I thought.

However, destroyer says after deprotonation, it will undergo SN2.

I know that it is a primary alkyl bromide.
If it is strong and bulky base, doesn't the reaction favor E2 rather than SN2? Am I thinking wrong?? What do you guys think?

Im confused myself by this question..An im looking at destroyer as we speak.NaoC2H5, is a good weakly basic nucleophile....Im a lil dumfounded as to why theres not a good leaving group?????

 

[Sublimation]

Hi all,

I am little confused about Sn2 and E2 products...
In #34, if you look at the first step, a double alpha proton is removed by NaOC2H5 and strong and bulky base is formed.

This will react with CH3CH2Br via E2 and CH2=CH2 will be formed.
This was how I thought.

However, destroyer says after deprotonation, it will undergo SN2.

I know that it is a primary alkyl bromide.
If it is strong and bulky base, doesn't the reaction favor E2 rather than SN2? Am I thinking wrong?? What do you guys think?

A base will undergo Elimination and nucleophile will undergo substitution. from wat i see if u deprotonate....ur gonna have the lone pair come down and form a pibond and kick the bromine out.......i dont see how u can have a Sn2.....unless the base reacts with the ethylbromide prior to being deprotonated....it its deprotonated....then its too late to have an sn2.