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abb0t

  • one year ago

Challenge: What is the product of the reaction given?

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  1. abb0t
    • one year ago
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    |dw:1439871598305:dw|

  2. anonymous
    • one year ago
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    What an interesting coincidence, I was just thinking about this the other day: http://chemistry.stackexchange.com/questions/31937/grubbs-wittig-hydroboration-reactions-and-woodward-hoffmann-rules

  3. abb0t
    • one year ago
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    it wouldn't be a challenge without a little trickery ;)

  4. anonymous
    • one year ago
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    Can I ask if it polymerizes in any way?

  5. abb0t
    • one year ago
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    Nope.

  6. anonymous
    • one year ago
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    Including forming catenanes?

  7. abb0t
    • one year ago
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    yep. No polymerization, co-polymerization, no coordinate complex formation.

  8. anonymous
    • one year ago
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    I didn't excel at organic chemistry... do you mind walking me through the fundamentals?

  9. anonymous
    • one year ago
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    I took the course and passed it, however by no means did I consider myself "proficient" at all.

  10. abb0t
    • one year ago
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    You can challenge yourself by reviewing the fundamentals yourself, @Robert136

  11. aaronq
    • one year ago
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    Alright, i'm bored so here's a shot. I remember doing the mechanism for Inorganic II, but this was like 2+ years ago, so it's possibly incorrect. If it's wrong, give me a hint

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  12. abb0t
    • one year ago
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    what program did you use to draw that? and you're on the right track. very close tho. what about the triple bond? and how odd, i didn't think they taught this is inorganic. this is organic :P

  13. abb0t
    • one year ago
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    |dw:1440030476184:dw|

  14. aaronq
    • one year ago
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    I used Chemdraw 15.0 Oh true so it's twice against the triple bond.. I didn't think the triple bond would react with the catalyst. Nice Yeah, my prof for inorganic was.. eccentric.. he talked about such random things, from like how to build a ruby laser to giving a full repertoire on inorganic pigments lol smart dude though, he was the chair of the dept. anyway, thanks for posting this q

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