lublypnay
Registered: 10/01/09
Posts: 12
|
|
|
Reply with quote | #1 |
When counting the number of electrons involved in the planar p orbital arrangement, why wouldn't you count the lone pair on N? In other words, why isn't it part of the p loop? |
| |
n0nst0p511
Registered: 10/03/09
Posts: 4
|
|
|
Reply with quote | #2 |
I am also a bit confused on this problem. Counting the # of pi electrons in the loop of pz orbitals, I counted 3 from Carbon, but I'm unsure how many are from Oxygen and Nitrogen. Any help I would definitely appreciate! |
| |
ayusam
Registered: 09/25/09
Posts: 20
|
|
|
Reply with quote | #3 |
I'm not entirely sure if I'm correct but here it goes...xP
Try building a model. It'll help out a lot at visualizing this. Oxazole is aromatic because it's in a ring, planar, and follows Huckel's Rule (6). 6 is from the 2 double bonds (4) and the lone pair on Nitrogen (2). The lone pairs on oxygen do not count as being part of the ring because they are not planar with the molecule. With a model, you notice that the 2 lone pairs are in the py and pz orbital, so they cannot be included in calculating Huckel's Rule.
Hopefully this helps! :]
|
| |
lublypnay
Registered: 10/01/09
Posts: 12
|
|
|
Reply with quote | #4 |
Actually, ONE of the lone pairs on Oxygen IS planar with the rest of the molecule since O is adopting an sp2 hybridization. So for clarification, you would count the 2 pi bonds within the ring and the lone pair on O, giving us a total of 6 electrons.
Furthermore, I finally found out why the lone pair on N is not counted as part of the p loop: the p orbital from the pi bond that N is attached to prevents its lone pair from being in the same plane (you can't have both regions occupy the same space).
An awesome tip for knowing when to count the lone pair on N as part of the loop is to see if there are only single bonds surrounding it. In that way, it's lone pair can act as a p orbital and therefore be a part of the p loop.
|
| |
