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elimination vs substitution

Discussion in 'MCAT Discussions' started by drlove87, 08.17.08.

  1. drlove87

    drlove87

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    How do u distinguish between nucleophilic substitution and elimination? For example, if you are given the following question: What is the major product formed from the reaction of 2-bromo-2-methylpentane with sodium ethoxide?
    I know that sodium ethoxide is a base, however, isn't a nucleophile, in a way, similar to a base? please clarify
     
  2. landofoo

    landofoo

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    anything with a lone pair of e- can be a nucleophile..so technically this can be a nucleophile, acutally a decent nucleophile.

    The difference between Substitution and Elimination is that substitution needs a small unhindered base to be able to squeeze itself into the molecule to substitutue/ backside attack.

    Big bulky bases cannot do this and can only pluck off a hydrogen when a leaving group is present. bullkier bases will give the hoffman product while smaller will give the zaitsev product.

    This specific question looks like it will be a substitution. you will probably get 2-ethoxy-2methyl-pentane + NaBr.
     
  3. drlove87

    drlove87

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    This question was taken from one of the old Kaplan exams and i thought that the answer should be 2-ethoxy-2methyl-pentane + NaBr as well. However, Kaplan says it should be an elimination reaction, E2 to be more specific, and the right answer should be 2 methyl 2 pentene
     
  4. GoldShadow

    GoldShadow

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    In this case, a substitution would be unlikely because the alkyl halide is tertiary; it is too hindered for a substitution, so elimination will predominate.
     
  5. landofoo

    landofoo

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    you guys are right...i didn't take degree of substituents into account. So, yeah, I guess E2 would be the victor here. sorry about the mix up
     
  6. Lola84

    Lola84

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    Elimination dominates when you have you have a big bulky nucleophile, because of the presence of steric hindrance. In addition, you'd have to look at the strength of your nucleophile and the state of the carbon being attacked (i.e. if it's primary, secondary, or tertiary) in order to determine if it will undergo a bimolecular (SN2 or E2) or unimolecular (SN1 or E1) reaction.

    Sodium ethoxide is a very good nucleophile and it is relatively small. 2-bromo-2-methylpentane has a tertiary carbon being attacked; therefore, it is a bulky electrophile and will undergo a frist-order reaction. Because of these two conditions, you will have an SN1 reaction. If your nucleophile was, let's say, sodium tertbutoxide, it would undergo E1.

    So the major product is an SN1 product.
     
  7. Lola84

    Lola84

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    It would not be an E2 reaction because you are working with a tertiary carbon electrophile. It would not be elimation because you do not have a bulky nucleophile.
     
  8. Lola84

    Lola84

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    Substitution first-order (SN1) is likely because a tertiary carbocation is formed. The transition state has an sp2 hybridized carbon and ethoxide can attack from either side.

    Elimination will not predominate because ethoxide is relatively small and it is, therefore, unnecessary to take away a beta hydrogen to form an alkene.
     
  9. landofoo

    landofoo

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    Lola84 = FAIL O-chem
     
  10. w8ting2exhale

    w8ting2exhale what a journey...

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    Ok, what I think is it is less likely to be SN1 or E1 because even though it's a highly substituted alkane, ethoxy is a pretty strong nucleophile. E1/SN1 rxns are particularly favored by weak nucleophiles, like ethanol instead of ethoxy. So stronger nucleophiles like methoxy, ethoxy, etc. and especially small ones should make you think sn2/E2. Now it will not be SN2 because of steric hindrance. Therefore it must be E2. I hope this helps. Also, did they mention heat? heat is usually a clue that elimination will occur.
     
  11. Adamska

    Adamska

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    um..Lola, strong base tends to favor E2 over anything else, and ethoxide reaction with tertiary halides is exclusive to E2...
     
  12. Lola84

    Lola84

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    Well, my bad... looks like I'm incorrect. Sorry if I have caused any confusion.

    I actually ran these experiments this year. I isolated my major product and my NMR spectra showed a substitution product (an ether). When I increased the amount of my base, the rate of reaction didn't change at all. So, it lead me to those conclusions. It's possbile that I could have done something wrong.

    I'm pretty sure a mixture of products would be formed, though, but I guess the textbook answer for the major product would be E2.

    As for landofoo's comment... geez... I'm honestly just trying to help.
     
  13. Adamska

    Adamska

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    ^If it were a secondary halide then yes a mixture of substitution and elimination would occur. For tertiary it is quantitatively and qualitatively only elimination. I think it's just that you didn't use a strong enough base for your experiment.
     
  14. Raigon

    Raigon This is an emergency.

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    Landofoo was probably joking. Don't take the comment personally. And yeah, you guys are right - elimination predominates with a strong base.
     
  15. GoldShadow

    GoldShadow

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    By the way, your avatar is awesome.
     

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