[arsenal]

Propose mechanism 

[orgopete]

Without writing anything down, I think you should open the aminal, and if you have a cyano imine (like a cyano ketone), the cyanide can be a leaving group. A cyano added to an aldehyde (or imine) can form an anion, as in a cyanide catalyzed benzoin condensation. Let's see what you come up with.

[arsenal]

What do you think about it ? 
I can't continue, help me ! thanks 

[Dan]

As Orgopete said, you need to break the hemiaminal up. The first thing you did was deprotonate the hemiaminal OH, which is good, but I think that anion will then collapse - use the O^- lone pair to kick out R-^-NAc (a resonance stabilised anionic leaving group). Formation of your four-membered cyclic intermediate looks like a very high energy option in comparison.

From there it's plain sailing - Orgopete's Benzoin clue is key. See what you can come up with.

[orgopete]

Okay, another clue, if you open the aminal as Dan suggested, you will have an aldehyde (and there is an aldehyde in your product).

[opsomath]

I just want to say that when I saw this originally, I thought there had to be a mistake. Nice mechanism problem. I have never seen a step quite analogous to the re-cyclization before, and I think that re-ring opening to an imine must be a significant side reaction?

[arsenal]

i can't think anymore, i hope you all help me, please 

[opsomath]

You've already done all the hard parts! The ring you've made is identical to the ring in the product. You just have to show elimination of some leaving groups to form double bonds, making the ring into the aromatic pyridine in the product.

[orgopete]

I don't know the mechanism, but you have identified the steps for the bonds in the product. Hydroxide or water loss from the new aminal should be easy. The one step you have changed from what I was thinking was to eliminate the CN early. I was suggesting the CN would be important to eliminate the chloride as it increased the CH acidity.

NB, cyanide cannot deprotonate ethanol.

[arsenal]

as you think, what'll happen next ? 

[orgopete]

Well, I guess I can see what you were trying to do. Phenylacetaldehyde has a pKa of 13. It should be virtually completely ionized. The cyano group is still the problem in step 2. Don't eliminate it at that stage as Dan suggested. You can use the C=O to absorb electrons and make the nitrogen a sufficiently good leaving group. If you make an anion of the aldehyde, you can add to the C=O and lose water with the electrons of the nitrogen.

That should leave you with two problems, get rid of the chlorine and replace the CN with OEt. The CH of the phenylacetonitrile are acidic. Use them to eliminate the chloride. you will have a 2-cyanopyridine. Substitute the cyano group with ethoxide.

I don't know the order that all of those exchanges must take place. I am simply suggesting how the carbon skeleton can plausibly form from the starting material.

[orgopete]

…Don't eliminate it at that stage as Dan suggested. You can use the C=O to absorb electrons and make the nitrogen a sufficiently good leaving group.…

The anion of NAc should be a sufficiently good leaving group. After protonation, you can do the intramolecular cyclization with the enolate of the aldehyde.

[zaphraud]

could you flip the molecule about the x axis 180 degrees and consider different removal and re-addition products? (not sure myself, just smells like a trick question to me.)

[Oxy]

My solution: