Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: stephanopolous on March 31, 2014, 07:00:04 PM
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I'm ready to pull my hair out trying to figure out this problem. I believe it's a combination of steps we've learned but I can't figure out the complete mechanism.
Here is the problem:
C6H5COH + CH3CN------>EtO-, EtOH-----> (C9H7N)---->H3O+/heat---->
I believe the C9H7N is telling you that's what you have after the first steps, not telling you to add that.
I have the answer (guessed it correctly) but I want to understand why it's that.
Any help would be greatly appreciated. So far all I did was a nucleophilic attack on the aldehyde by the nitrile (electrons on the N) to form an intermediate, but I'm not sure where to go from there.
I looked through all of our power point slides, all of the book chapters we covered, and googled 17,000 different ways that it may come up - nothing.
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What role does the ethoxide/ethanol play here?
I don't think that C9H7N is correct.
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I'm not sure what the ethanol/ethoxide do. I copied this problem straight from the exam and double checked it- that's what the full problem is. I thought maybe it served as a base to turn the nitrile into something else like an amide but when I wrote that out it didn't work.
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How about this as a first step?
The acetonitrile is deprotonated by the base to give a resonance stabilised anion. This attacks the aldehyde to give the product I drew.
You take it from there.
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Ok thank you, I'll give it a try. The final product has no nitrogen in it- so I'm wondering what part of the work-up is taking that off. The acid wouldn't do that alone right?
Thanks again for drawing that out, that was really nice of you.
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What is your suggestion for the final product?
What happens to nitriles when you treat them with acid?
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The product is a carboxylic acid C6H5CH=CHCO2H.
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Well a nitrile treated with acid forms a carboxylic acid but when the nitrile adds to the aldehyde it's no longer a nitrile correct?
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Ohhhh wait I see- so it does turn a nitrile into a carboxylic acid, so is that intermediate still considered a nitrile?
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I'm also wondering how the double bond in the product is formed. Does the acid protonate the oh, making it a good leaving group, and then ...?
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Does the acid do that too? Ok so there's a secondary alcohol- it will protonate, the acid will come in and take off the hydrogen on that same carbon, the electrons will come up, kick off the water, and form the double bond?
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So in summary of that lot:
The first product is still a nitrile.
The acid hydrolyses it to the carboxylic acid and causes the elimination of water to give the product you drew.
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Hi,
In the attachment I drew the full mechanism of the first addition + elemination.
I didn't draw the hydrolisis of the nitrile group, but you can look that up in your book (well known reaction). :)
Hope it helps.
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Wow I just realized that I was drawing the CH3CN wrong- I was drawing it as CH3N and getting really confused. OK I'm going to try and draw it all out again and see it if makes sense now. Thank you!!!
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It's ok if there's three arrows on the nitrile right? What I mean is, base takes a hydrogen off of methyl group, electrons move and make a double bond with the next carbon and then there's a third arrow moving the electrons of one of the triple bonds up onto the nitrogen to make it a nucleophile.
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Hi,
In the attachment I drew the full mechanism of the first addition + elemination.
I didn't draw the hydrolisis of the nitrile group, but you can look that up in your book (well known reaction). :)
Hope it helps.
Thank you very much!!!
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So in summary of that lot:
The first product is still a nitrile.
The acid hydrolyses it to the carboxylic acid and causes the elimination of water to give the product you drew.
Thanks so much for all your help
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I don't think that C9H7N is correct.
Seems right to me. It is the reactants minus H2O.