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Topic: saponification of ethyl-4-nitobenzoate lab  (Read 5986 times)

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Offline Elphabalives

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saponification of ethyl-4-nitobenzoate lab
« on: August 09, 2010, 02:01:50 PM »
I need some help figuring out the mechanism for this reaction.  We took ethyl-4-nitrobenzoate, added NaOH and HCL to get 4-nitrobenzoic acid. I would assume that the OH attacks the carbonyl C, pushing the pi electrons to the oxygen. The maybe the H from HCL protonates that CH3CH2O on the left, the C=O is reformed and CH3CH2O leaves? Am i on the right track?

Offline Jorriss

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Re: saponification of ethyl-4-nitobenzoate lab
« Reply #1 on: August 09, 2010, 04:27:40 PM »
That's pretty much it, but the way you've drawn it, as far as I can tell - in some places you're acting like it's acidic conditions and in some places, basic. I'd go over it one more time and make sure you're drawing each step as if it's in basic conditions.

I might be off but that's the impression I got.

Offline discodermolide

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Re: saponification of ethyl-4-nitobenzoate lab
« Reply #2 on: August 09, 2010, 07:31:08 PM »
I need some help figuring out the mechanism for this reaction.  We took ethyl-4-nitrobenzoate, added NaOH and HCL to get 4-nitrobenzoic acid. I would assume that the OH attacks the carbonyl C, pushing the pi electrons to the oxygen. The maybe the H from HCL protonates that CH3CH2O on the left, the C=O is reformed and CH3CH2O leaves? Am i on the right track?

Simple ester hydrolysis. Gives the sodium salt of the acid. The HCl is used to re form the carboxylic acid, it's that simple
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