[stephanopolous]

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.

[discodermolide]

What role does the ethoxide/ethanol play here?
I don't think that C₉H₇N is correct.

[stephanopolous]

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.

[discodermolide]

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.

[stephanopolous]

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.

[discodermolide]

What is your suggestion for the final product?
What happens to nitriles when you treat them with acid?

[stephanopolous]

The product is a carboxylic acid C6H5CH=CHCO2H.

[stephanopolous]

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? 

[stephanopolous]

Ohhhh wait I see- so it does turn a nitrile into a carboxylic acid, so is that intermediate still considered a nitrile?

[stephanopolous]

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 ...?

[stephanopolous]

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?

[discodermolide]

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.

[critzz]

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.

[stephanopolous]

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!!!

[stephanopolous]

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.