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anonymous

  • one year ago

Can someone PLEASE help me out with orgo?!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @ganeshie8

  3. Astrophysics
    • one year ago
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    @Empty

  4. anonymous
    • one year ago
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    also. |dw:1436510848405:dw|

  5. Astrophysics
    • one year ago
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    Alright, well show your attempt

  6. anonymous
    • one year ago
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    is it anti markonikov or markovnikov?

  7. anonymous
    • one year ago
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    I believe it is Markovnikov

  8. Astrophysics
    • one year ago
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    I'm not russian

  9. anonymous
    • one year ago
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    me neither,. but that's the scientist of which created the law of substituents movements

  10. Astrophysics
    • one year ago
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    Ah interesting, well empty is good at chemistry he can help you with this if he chooses to haha.

  11. anonymous
    • one year ago
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    |dw:1436511004422:dw|

  12. Astrophysics
    • one year ago
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    esp o chem

  13. anonymous
    • one year ago
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    lol thanks though

  14. Empty
    • one year ago
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    I don't think this is Markovnikov or antiMarkovnikov, that doesn't really apply here because we're replacing the alcohol with this amine group. Markovnikov sorta stuff is more like how things place on a double bond.

  15. anonymous
    • one year ago
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    true

  16. anonymous
    • one year ago
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    but I have a lewis acid here, are there any rules I can apply here?

  17. anonymous
    • one year ago
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    and I would really appreciate if you could help me out with the first question as well

  18. Empty
    • one year ago
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    The second question I can answer really well I'm not too familiar with amine reactions anymore kinda rusty but I can kinda guess what you'll do. Like I can draw arrows for you.

  19. anonymous
    • one year ago
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    whatever fits you I would really appreciate any kind of help here

  20. anonymous
    • one year ago
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    if you prefer the second one then great

  21. Empty
    • one year ago
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    So for the first one, we oxidize it into a keytone to get: |dw:1436511257479:dw| That's just sorta me brainstorming on steps, you still have to remove the alcohol group which won't be very easy so I don't know, I need to go review my Ochem. The second one though I'll explain in a sec.

  22. anonymous
    • one year ago
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    great thank you so much! and you said you're familiar more with the second question, can you help me with that as well? please?

  23. Empty
    • one year ago
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    For the second one, you have toluene, which really just means methylbenzene. As you may recall, carbon is an electron donating group which is why tertiary carbocations are more stable than secondary carbocations. Since carbon doesn't have any free electrons, I can't exactly draw the resonance properly, but I can do it with aniline which is very similar in that it is electron donating but has NH2 instead of CH3. So we can push electrons around like this: |dw:1436511629053:dw| So aniline and toluene both have higher electron densities at these locations which are called "ortho" and "para" which are relative to the group attached, have you heard of these before?

  24. anonymous
    • one year ago
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    yes of course

  25. Empty
    • one year ago
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    So how in depth would you like me to go, I can draw out the whole reaction step by step or just say that your major product will be 2-bromotoluene because the reaction is much more common there since it's basically a 2/3 ratio of the possible spots you can hit randomly and the methyl group isn't large enough for stearic bulk to really matter here. You will also get the para product as well as a minor product though, so if you want one or the other you'll have to do other steps like add a protecting group. or purify it afterwards.

  26. anonymous
    • one year ago
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    can you show the major product?

  27. anonymous
    • one year ago
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    I just need to see it visually so I could understand the reaction better

  28. Empty
    • one year ago
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    Draw what you think it is, I have already named it and explained it, and if you mess up I can help you out.

  29. anonymous
    • one year ago
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    cool one sec

  30. anonymous
    • one year ago
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    |dw:1436512240299:dw|

  31. anonymous
    • one year ago
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    is that correct?

  32. anonymous
    • one year ago
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    @Empty ?

  33. Empty
    • one year ago
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    Yeah perfect! Also I just realized how to do your first question easily.

  34. anonymous
    • one year ago
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    you're brilliant! thank you!

  35. Empty
    • one year ago
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    We add thionyl chloride to turn the alcohol into a good leaving group: |dw:1436512375879:dw|

  36. anonymous
    • one year ago
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    thank you so much!!!!!

  37. anonymous
    • one year ago
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    wow you really helped me out here. Thank you su much!

  38. Empty
    • one year ago
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    I can write out the first step better: |dw:1436512464217:dw| Well even that is sorta questionable, the problem is there's such a wide range of people's knowledge in ochem that it's hard to say if my shorthan dis understood or not, but the best part abot Ochem is being able to figure them out by drawing arrows just like you can solve a math problem with algebra. Glad I could help!

  39. anonymous
    • one year ago
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    help an awesome day!

  40. anonymous
    • one year ago
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    have*

  41. Empty
    • one year ago
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    Yeah you too, ochem is pretty awesome haha

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