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Bulky base -- Zaitsev/Hoffman question

Discussion in 'DAT Discussions' started by LetsGo2DSchool, Aug 16, 2011.

  1. LetsGo2DSchool

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    I understand that a bulky base will yield the less substituted product (Hoffman) in an elimination reaction. However, Chad mentions that's not always the case -- you can get the Zaitsev product if dealing with a bulky base and secondary halide.

    For purposes of the DAT, how should I treat this:
    bulky base = always Hoffman, or
    bulky base = Zaitsev/Hoffman depending on the bulkiness of halide?
     
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  3. Demps

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    really? that's interesting... I don't quite remember Chad saying that.

    but say both Zeitsev and Hoffmann elimination products are given as possible choices, I would definitely pick Hoffmann product for Terbutoxide + 2ndary halide combo.
     
  4. LetsGo2DSchool

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    Yeah, I'm watching his videos for the 2nd time around and I'm realizing there are a lot of stuff I missed the first time. He mentions it in video #2.5 and again in 2.6.
     
  5. Demps

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    my subscription expired, but I took good notes on them. let me look at it hold on
     
  6. nondescriptive

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    bulky base will yield the hoffman product; just go with it. they won't try to trick you or do anything that can be contested.
     
  7. Demps

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    yo bro I don't quite see any example with terbutoxide with 2ndary halide... -_-;
     
  8. LetsGo2DSchool

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    Alright, I went back to re-watch that part of the video to make sure I wasn't mixing simple things up like a dildo head lol.

    In OChem video #2.4, around the 22:45 mark, Chad says verbatim:

    "If you use a bulky base there's some teachers who teach that 'Every time you use a bulky base you're gonna get Hoffman' and that's not true. Technically, if you have a bulky halide I like to say it, tertiary halide and a bulky base, every time you will get the Hoffman product. Now if your halide is secondary, well if it's secondary you don't always get the Hoffman product, you often get the Zaitsev product in that case. So the real rule when you have a bulky base, is you'll only typically for sure get the Hoffman product if it's a bulky base and a tertiary halide.... Cool...cool"

    Nondescriptive makes a good point that it may be safer to go with Hoffman on the DAT since they won't present something that's debatable.
     
  9. invictusx

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    There are certain leaving groups that indicate a Hoffman elimination; just be familiar with those.
     
  10. invictusx

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    This is also true because with a secondary halide, you can produce a disubstituted product, which is far more stable than a mono-substituted product.
     
  11. wired202808

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    remember this chart and you can never go wrong
    Strong Weak
    1 degree Carbon SN2 or E2 (only with bulky base) Nothing happens
    2 degree Carbon SN2 or E2 (only with bulky base) SN1 or E1
    3 degree Carbon E2 only !!! SN1 or E1
     
  12. SLYgUy2098

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    Agreeing with everyone here, I assume a bulky base yields a Hoffman product. As mentioned before, t-butoxide is a nice example.
     
  13. Demps

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    LetsGo2DSchool,

    I found the example that we were looking for on Destroyer.

    2011 edition #78

    we can clearly see that bulky base + 2ndary halide combo yields Zeitsev product - more stable alkene

    So what Chad said was true I guess. never doubt the man lol.

    I'm glad we got this one of the way
     
  14. Demps

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    LOL I must be tired. I clearly misread it lol..

    I apologize for my mistake. its not terbutoxide, they used ethoxide argh -_-;
     
  15. LetsGo2DSchool

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    :laugh: Get some rest bro. :laugh:
     
  16. Demps

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    bro I found it
    #133
     

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