This was a tricky one. The answer is actually aromatic. You might be scratching your brains how that happened. It's not that difficult. First of all, we have to figure out what does this molecule even look like. What kind of lone pairs does it have? Well, nitrogen has one lone pair. Does boron have a lone pair? No. Boron has an empty p orbital. Boron and aluminum are special for always having that empty p orbital. That's kind of like a cation, right? It doesn't have a charge, but it's an empty orbital. Can an empty orbital participate in resonance? Sure. I could totally put my electrons into it, and that wouldn't be a problem. Meaning that this molecule is fully conjugated if the nitrogen will donate its lone pair. Let's see if it will.
- 1, we're looking at the nitrogen. What kind of hybridization does the nitrogen have? sp3.
- 2, will it help to create a 4+2 number if my lone pair donates? Well, let's count it up. I've got 2. I've got 4. This orbital here that is just sitting here, it does participate in conjugation but it doesn't add any electrons. With that orbital, I still just have 4 from the 2 double bonds. Now if I flip this lone pair into the ring, that becomes my 6th electron, my 5th and 6th pi electrons.
The answer is that yes, I would get 4n+2, so one lone pair will donate. Interesting. That's how I tricked you a little bit with boron because you probably weren't thinking that boron could be part of conjugation. But it has an empty orbital. So that's the same as saying a carbocation basically. Empty orbital that you can stuff electrons into.
Alright. So let's move on to problem g.
