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Effect of KOH
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Topic: Effect of KOH (Read 3826 times)
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adinboy
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Effect of KOH
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on:
May 03, 2014, 12:03:41 AM »
So I'm confused as to what exactly KOH does. I first though it debrominates and leaves an alkene but then I saw a reaction where it adds an -OH group. Here are the two reactions in question:
So what exactly is the role KOH?
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discodermolide
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Re: Effect of KOH
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Reply #1 on:
May 03, 2014, 02:57:36 AM »
What is the difference in the type of reaction KOH undergoes with the molecules concerned?
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adinboy
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Re: Effect of KOH
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Reply #2 on:
May 03, 2014, 02:16:47 PM »
Are you asking like if it is SN1 or SN2?
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discodermolide
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Re: Effect of KOH
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Reply #3 on:
May 03, 2014, 02:35:28 PM »
Yes something like that.
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AlphaScent
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Re: Effect of KOH
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Reply #4 on:
May 03, 2014, 07:55:29 PM »
besides sn1and sn2, which are substitution rreactions. what are the other types of reactions that are either mono or bimolecular in determining the mechanism/rate??
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adinboy
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Re: Effect of KOH
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Reply #5 on:
May 03, 2014, 09:02:13 PM »
Well there's E1 and E2 and I believe this one would be an E2 meaning the C-H and C-X would break at the same time producing the alkene in a single step....but then the -OH addition I'm not sure
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discodermolide
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Re: Effect of KOH
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Reply #6 on:
May 03, 2014, 09:08:07 PM »
What is the OH doing to the alkyl bromide?
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adinboy
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Re: Effect of KOH
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Reply #7 on:
May 04, 2014, 12:26:59 AM »
Well the K is bonding with the Br while the OH bonds with 1 hydrogen of the CH3 group to make KBr and H2O thereby creating a double bond
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discodermolide
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Re: Effect of KOH
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Reply #8 on:
May 04, 2014, 01:51:03 AM »
Yoe are talking about this molecule
reacting with KOH?
You already said the answer above.
So what is the reaction type called?
http://www.chemicalforums.com/Themes/Borek/images/bbc/SMILES.gif
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adinboy
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Re: Effect of KOH
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Reply #9 on:
May 04, 2014, 03:13:28 AM »
E2?
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kriggy
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Re: Effect of KOH
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Reply #10 on:
May 04, 2014, 03:35:56 AM »
How can it be E2 if you get hydroxylated product?
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discodermolide
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Re: Effect of KOH
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Reply #11 on:
May 04, 2014, 03:49:21 AM »
Apart from the last comment, this molecule cannot undergo an E2 process.
You already had the answer above.
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