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Topic: Oxime formation question  (Read 3563 times)

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Offline ugradstudent

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Oxime formation question
« on: April 01, 2015, 10:48:43 AM »
Hi, when synthesizing oxime from a ketone using NH2OH·HCl and Pyridine dissolved in MeOH, is the base for neutralizing the HCl so hydrolysis of oxime doesn't occur to revert it back to the ketone? The base isn't there as a catalysis?


Offline orgopete

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Re: Oxime formation question
« Reply #1 on: April 02, 2015, 07:51:46 PM »
You should look up the mechanism and optimal pH for oxime formation. The reaction can seem a bit confusing (at least for me). How does the carbonyl oxygen leave?
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Offline ugradstudent

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Re: Oxime formation question
« Reply #2 on: April 03, 2015, 10:54:33 AM »
Doesn't the nitrogen of hydroxylamine attack the carbonyl carbon first, then two proton transfers occur as the carbon oxygen takes the hydrogens on the nitrogen to become water, a good leaving group?

Also, I looked up the optimal pH for oxime formation in http://onlinelibrary.wiley.com/doi/10.1002/ceat.200700448/pdf.
The reaction time is fastest when the pH is 10. How does basic conditions speed up the reaction in this situation?
« Last Edit: April 03, 2015, 11:24:33 AM by ugradstudent »

Offline orgopete

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Re: Oxime formation question
« Reply #3 on: April 03, 2015, 01:59:24 PM »
I was expecting a different answer. I thought the optimal pH was 4-5, http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/aldket1.htm. The rate limiting step is not attack by hydroxylamine, but the elimination step. The proton transfer steps need to be examined more closely.

If the oxygen were to leave as water, I would expect the ROH2(+) concentration to be very low at high concentrations. At low pH, the hydronium ion concentration may increase, but so would amine protonation. That can inhibit addition to the carbonyl group and/or elimination of water.
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Offline ugradstudent

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Re: Oxime formation question
« Reply #4 on: April 05, 2015, 11:47:05 AM »
What you're saying about the pH range being at 4-5 does make sense, so wouldn't that mean we have to use an amount of pyridine equivalent to or less than the hydroxylamine to create a slightly acidic environment?

Also, as for the mechanism during high pH, is it possible for carbonyl oxygen to leave as a hydroxide after one proton transfer and pick up another proton on the way out from MeOH or the hydroxylamine? This way an ROH2(+) intermediate wouldn't need to be form.

Offline orgopete

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Re: Oxime formation question
« Reply #5 on: April 05, 2015, 02:36:50 PM »
Now you are getting the conundrum. At low pH, the hydroxylamine and any intermediate amines would become protonated, thus retarding the reaction. It would make it more difficult to kick out hydroxide from an ammonium intermediate. I prefer to invoke an uphill protonation of the oxygen and in which the nitrogen is not. The option is to invoke a concerted mechanism. I still think a similar set of properties must be present to orient the reagents.
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