# Reaction between Salicylaldehyde and 2-Benzoylpyridine

In a paper by Sheng et al (10.1016/j.dyepig.2018.07.036) the authors synthesize the compound MZC as shown in the scheme below.

I'm looking for information about what the reaction mechanism is that underpins this reaction. If anyone can share some insights it is much appreciated.

In the experimental section it is written

2-Benzoylpyridine (4 mmol, 0.732 g) and ammonium acetate (22 mmol, 1.70 g) was dissolved in acetic acid (30 mL) and followed by the addition of salicylaldehyde (6 mmol, 0.733 g). The reaction was refluxed for 6 h. The resulting solution mixture was cooled to room temperature and cold water was then added to the mixture. The solid was collected by filtration and washed with a small amount of cooled water, which was purified via recrystallization using acetonitrile to give MZC (0.687 g, 60 % yield).

• Treat it step by step. For example, step 1: what does a ketone form when mixed with ammonia? – Zhe Nov 12 '18 at 17:54
• I assume that ammonium acetate $\ce{H3COONH4}$ provides for $\ce{NH3}$ in the reaction mixture. Ammonia will react with the ketone group in 2-Benzoylpyridie and form an imine, and that gives a hint on how an additional nitrogen enters the ring system. So the question then is, does the imine reacts with the aldehyde group in Salicylaldehyde? – John Nov 12 '18 at 19:02
• The reaction is run in refluxing acetic acid so that is forcing conditions. Imines are not as nucleophilic as amines but they are stiil nucleophiles. – Waylander Nov 12 '18 at 20:04
• Technically, the reaction is one pot, so the order is not 100% certain. But you could also reasonably form the aldimine first on reactivity grounds (though the reaction is likely under Curtin-Hammett kinetics). Then keep in mind that the pyridine is relatively nucleophilic. I quickly tried drawing this out, and it seems feasible. – Zhe Nov 12 '18 at 20:25