[Kate]

Hello. 

Got 2 questions:

1. Reaction of CH₃(CH₂)₄CHBr with CH₃CH₂O^- in CH₃CH₂OH gives mainly S_N2 or E2 ? And if it's E2 it follows Zaitsev's or Hofmann's product ?

I think it gives an E2 reaction with Hofmann's product.

2. Reaction of CH₃(CH₂)₄CHBr with (CH₃)₃O^- in (CH₃)₃OH gives mainly S_N2 or E2 ? And if it's E2 it follows Zaitsev's or Hofmann's product ?

I think it gives an E2 reaction with Hofmann's product.


Right or wrong ?

[fledarmus]

What leads you to those conclusions?

[discodermolide]

Hello. 

Got 2 questions:

1. Reaction of CH₃(CH₂)₄CHBr with CH₃CH₂O^- in CH₃CH₂OH gives mainly S_N2 or E2 ? And if it's E2 it follows Zaitsev's or Hofmann's product ?

I think it gives an E2 reaction with Hofmann's product.

2. Reaction of CH₃(CH₂)₄CHBr with (CH₃)₃O^- in (CH₃)₃OH gives mainly S_N2 or E2 ? And if it's E2 it follows Zaitsev's or Hofmann's product ?

I think it gives an E2 reaction with Hofmann's product.


Right or wrong ?

I don't think it's quite correct. Think about reaction 1 again. (should the formula not be CH3(CH2)4CH2Br?

[Dan]

Also, there is only one possible product form E2 - Hofmann/Zaitsev is irrelevant if it's a primary alkyl halide etc.

[Kate]

What leads you to those conclusions?

I think it's E2 for both.

Strong base, plus it's hindered.

I don't think it's quite correct. Think about reaction 1 again. (should the formula not be CH3(CH2)4CH2Br?

Yes, my mistake, it should be CH3(CH2)4CH2Br.

Anyway, SN2 or E2 ? I think it's E2 for both.

Also, there is only one possible product form E2 - Hofmann/Zaitsev is irrelevant if it's a primary alkyl halide etc.

Yeah, my mistake again. Thanks.

[fledarmus]

What is hindered? Your first reaction looks like a primary halide and a primary alkoxide.

(hint: http://en.wikipedia.org/wiki/Williamson_ether_synthesis)

[Kate]

What is hindered? Your first reaction looks like a primary halide and a primary alkoxide.

(hint: http://en.wikipedia.org/wiki/Williamson_ether_synthesis)

And it is a primary haloalkane with an alkoxide, with alkoxide acting as a strong base. In my organic class we havent talked about ether synthesis yet, so with what I only know so far (that being synthesis of alkenes, alkynes and a bit of alcohols), what does that reaction give ?!

Hindered base for you: http://www.chacha.com/question/what-is-the-definition-of-a-hindered-base-in-organic-chemistry

Also, I read in a organic chem book that an alkoxide in a solvent that is it's conjugated acid is more hindered than if it was in a different solvent. CH3CH2O- isnt a hindered base, but supposedly in CH3CH2OH it's harder for the alkoxide to shed those solvent molecules, thus becoming a "bit" hindered.

[fledarmus]

Yes, but ethoxide isn't sterically hindered. t-butoxide, your second base, is.

Seriously, read the article. Don't be put off by the name - it is a description of a reaction you will recognize.

[Kate]

Yes, but ethoxide isn't sterically hindered. t-butoxide, your second base, is.

Seriously, read the article. Don't be put off by the name - it is a description of a reaction you will recognize.

I edited the my previous post: "Also, I read in a organic chem book that an alkoxide in a solvent that is it's conjugated acid is more hindered than if it was in a different solvent. CH3CH2O- isnt a hindered base, but supposedly in CH3CH2OH it's harder for the alkoxide to shed those solvent molecules, thus becoming a "bit" hindered." True or not ?

OK, I saw the article, it's an SN2 reaction. Why would that be the major product of those 2 different reactions instead of an alkene via E2 ?

[fledarmus]

You will always get some competition between E2 and S_N2 reactions, if both reactions are possible. By using a really good nucleophile, you can get almost complete S_N2 reaction, and by using a strong base with no good nucleophile, you can get almost complete E₂ reaction.

The question you were given really represents the two extremes of steric hindrance and their effect on the reaction. Alkoxide ions in general are good nucleophiles, and bromide attached to an alkane is a good leaving group. Since your alkyl bromide is the same in both reactions, and both alkoxides are approximately the same strength as bases, the elimination reaction will take place at about the same rate under both reaction conditions. What you are looking at is how fast the S_N2 reactions will occur.

If the alkoxide is not sterically hindered, the S_N2 reaction will be quite a bit faster than the E₂ reaction. The carbon-bromine bond is strongly polarized, and attracts the negative charge of the alkoxide better than the proton on the next carbon. If the alkoxide is sterically hindered, however, it is too big to get close enough to the carbon, and has to be content with interacting with a proton instead. In general, the reaction of ethoxide or methoxide will be a rapid S_N2 substition, while the reaction with t-butoxide will be a slower E₂ elimination.

It is with the intermediate compounds that things like solvent effects become important. If you had a secondary alkyl bromide rather than a primary one, or a secondary alcohol rather than a primary one, or even something like t-amyl alcohol where it is primary but has a large group on the next carbon, you may slow the S_N2 reaction down to the point where you are getting a mixture of substitution and elimination products. In that case, you may need an even stronger nucleophile to drive the reaction to the substitution product. Although ethoxide is a strong nucleophile, as you say, in an ethanol solution there is still some ion-solvent interactions that need to be overcome for the reaction to occur. In this case, you can make the nucleophile even stronger by using a solvent which does not have such strong solvent-ion interactions. This is a very small steric effect, however, compared to an actual covalent bonded group on the alkoxide itself.

[Kate]

Thanks for the explanation.

What I get from all of this is that in the first reaction with ethoxide, the major product will be the S_N2 one, and with t-butoxide it will be the E2. I get that there's competition between the two by the way.