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Topic: Simple Aromatic Substitution  (Read 5869 times)

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Offline GregRC

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Simple Aromatic Substitution
« on: August 15, 2013, 01:14:27 PM »
Hey guys, I'm new here, and I'm going to be doing some basic research this semester in synthesis. Thing is, it's been about 4 years since i've been in an O-chem class.



I'm still looking for my old textbook, so right now I'm not much help to myself.

Offline Dan

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Re: Simple Aromatic Substitution
« Reply #1 on: August 15, 2013, 01:49:03 PM »
I'm still looking for my old textbook, so right now I'm not much help to myself.

This is not a good excuse. Please read the Forum Rules - you must demonstrate some kind of attempt at a question before you will get much help, and forum policy is not to dump answers.

There is a wealth of information online about the reactions of aromatic amines that you can look at to get started. E.g.

http://www.chem.uky.edu/courses/che232/JEA/ln/3.%20Aromatic%20Chemistry.pdf
http://www.cliffsnotes.com/sciences/chemistry/organic-chemistry-ii/amines/reactions-of-amines
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Offline kriggy

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Re: Simple Aromatic Substitution
« Reply #2 on: August 15, 2013, 01:52:00 PM »
Do you have any ideas how to proceed?
Try to look on each function group alone and think about how to convert it to the other

Offline GregRC

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Re: Simple Aromatic Substitution
« Reply #3 on: August 15, 2013, 02:24:31 PM »
What about the sandmeyer reaction?  NaNO2 and HCl.  Then CuCl?  Would that do anything to the primary amine? What about adding H2O after to get rid of the methyl?

Offline kriggy

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Re: Simple Aromatic Substitution
« Reply #4 on: August 15, 2013, 02:47:29 PM »
Yes, sadmayer should work. You cant actualy get rid of the -CH3, you need that carbon. What about KMnO4?

Offline GregRC

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Re: Simple Aromatic Substitution
« Reply #5 on: August 15, 2013, 02:53:10 PM »
Wouldn't that add 2 oxygens though?

Offline kriggy

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Re: Simple Aromatic Substitution
« Reply #6 on: August 15, 2013, 03:09:57 PM »
indeed. look at the function group it creates and then you can continue further. And btw, it could also react with another function group in your molecule

Offline GregRC

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Re: Simple Aromatic Substitution
« Reply #7 on: August 15, 2013, 03:19:28 PM »
So H2/Pd then?  Would i have to use a protecting group?

Offline kriggy

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Re: Simple Aromatic Substitution
« Reply #8 on: August 15, 2013, 03:29:06 PM »
Im not sure on H2/Pd you would probably need something else.
I think you need to use protecting group im not sure which one. Its getting more interesting than I thought before lol :D

Offline GregRC

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Re: Simple Aromatic Substitution
« Reply #9 on: August 16, 2013, 09:31:09 AM »
Anyone have any thoughts on how to proceed with the methyl group?  Apparently this is my first project on Monday.

Offline discodermolide

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Re: Simple Aromatic Substitution
« Reply #10 on: August 16, 2013, 09:45:41 AM »
What does permanganate do to aromatic methyl groups? Which product is formed?
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Offline AlphaScent

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Re: Simple Aromatic Substitution
« Reply #11 on: August 16, 2013, 09:52:41 AM »
Sandmeyer>Oxidize>??>??>product

This is where you are at.  Think about what discodermolide said.  Then go from there.

Sometimes you have to take two steps forward and then one back in o-chem...
If you're not part of the solution, then you're part of the precipitate

Offline kriggy

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Re: Simple Aromatic Substitution
« Reply #12 on: August 16, 2013, 02:24:31 PM »
Sandmeyer>Oxidize>??>??>product

This is where you are at.  Think about what discodermolide said.  Then go from there.

I think you need to oxidize before sandmayer otherwise it could convert amino group into nitro.

Sometimes you have to take two steps forward and then one back in o-chem...
Exactly this. I cant say how much truth is in this sentence :D

Offline discodermolide

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Re: Simple Aromatic Substitution
« Reply #13 on: August 16, 2013, 02:44:30 PM »
The nitro group will not convert to an amine under oxidising conditions. Reduction of nitro groups provides amines.
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Offline AlphaScent

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Re: Simple Aromatic Substitution
« Reply #14 on: August 16, 2013, 03:13:56 PM »
Sandmeyer is done to produce the halide.  In this case a chloride.  You have to do this first and then oxidize.  With a halide and nitro group present, the methyl would react to form an oxidized product.  Presumably an acid.
If you're not part of the solution, then you're part of the precipitate

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