What is the hybridization of the nitrogen in pyrrole?:

http://upload.wikimedia.org/wikipedia/commons/thumb/1/15/Pyrrole.png/300px-Pyrrole.png

Is it sp3 or sp2 because it is a conjugated system?

sp3

No, it's sp2. The nitrogen is bound to two carbons, one hydrogen, and the lone pair is in the empty p-orbital.

sp2 is what I figured too. Kaplan says it is sp3 for some reason, and that the sp3 hybridized orbitals overlap with the p-orbitals of the other carbons to form the conjugated ring. That just sounds wrong.

Its sp3, NOT sp2. There are 4 regions of electron density due to the lone pair of electrons on the N.

Its sp3, NOT sp2. There are 4 regions of electron density due to the lone pair of electrons on the N.

First, you don't count lone pairs when figuring out the hybridization. Secondly, the nitrogen is attached to two carbons which are both sp2 hybridized and therefore must also be sp2 hybridized if it is to overlap for maximum stability.

sp3

Since pyrrole is conjugated, it also must be planar. My reasoning is that the nitrogen is sp2 hybridized (2 sp2 orbitals are bonded to an adjacent carbon and 1 sp2 orbital is bonded to the hydrogen) forming a planar geometry. The lone pair of electrons is in a nonhybridized p-orbital perpendicular to this plane and conjugated with the p-orbitals of the other carbons (which are also sp2).

to calculate hybridization you do: # of lone pairs + # of sigma bonds. Therefore, 1+3=4 --> sp3

Okay, guys. We know pyrrole is aromatic so it needs the lone pair on the nitrogen to get the necessary six electrons. To form the conjugated system, the lone pair has to be in the plane above the atoms. This cannot happen if the nitrogen is sp3 hybridized. It MUST be sp2 hybridized to be aromatic.

Okay, guys. We know pyrrole is aromatic so it needs the lone pair on the nitrogen to get the necessary six electrons. To form the conjugated system, the lone pair has to be in the plane above the atoms. This cannot happen if the nitrogen is sp3 hybridized. It MUST be sp2 hybridized to be aromatic.


It is SP3 no doubt, but can undergo sp2 hybridization, which is what it does. Who the hell said that you don't count lone pairs when doing hybridization? What would that make ammonia then?

Roycer, you are correct, it is sp2. If you still have your organic chem text book you can check it out, this is a fairly common example in any aromatic compound chapter.

It is SP3 no doubt, but can undergo sp2 hybridization, which is what it does. Who the hell said that you don't count lone pairs when doing hybridization? What would that make ammonia then?

Yes, the four things attached to nitrogen would give an sp3 hybridization but pyrrole is aromatic and so the nitrogen has to become sp2 hybridized. So when talking about the aromatic molecule wouldn't we say it's sp2?

When I said you don't count the lone pair, I only meant in this case because it has to be up in the plane above the atoms. I was just trying to conceptually explain how the three OTHER things contributed to the trigonal planar character. I admit, I didn't word it right.

The nitrogen in the ring is clearly sp2 hybridized. Normally, the nitrogen would be sp3 hybridized, but it must rehybridize to sp2 so that the lone pair of electrons can be delocalized into the pi-system of the ring and that the molecule can be aromatic. What we see here is that aromaticity is highly favored.

ok, i just googled it and checked a few different websites. The ones i checked seem to agree that BENZENE is sp2. I assume because of its aromaticity, that pyrrole would also be sp2. Good thing i learned this now and not after taking the mcat:laugh:

Thank you. sp2 it is. :)

First, you don't count lone pairs when figuring out the hybridization. Secondly, the nitrogen is attached to two carbons which are both sp2 hybridized and therefore must also be sp2 hybridized if it is to overlap for maximum stability.

1. yes you do

2. that doesn't make sense. if you had propene, one of the carbons is still sp3 hybridized



lastly, someone said something about pyrrole being planar so it had to be sp2... the nitrogen could have the two carbon bonds in one plane and the lone pair and the hydrogen in another making it Sp3.

however, aromaticity always prevails and for this reason alone, the nitrogen must donate its e- into the ring making it Sp2.

1. yes you do

2. that doesn't make sense. if you had propene, one of the carbons is still sp3 hybridized

1. Yep. I know you do. Check up a few posts.
2. Propene is not an aromatic ring. We're talking ring stability.

sp3

lastly, someone said something about pyrrole being planar so it had to be sp2... the nitrogen could have the two carbon bonds in one plane and the lone pair and the hydrogen in another making it Sp3.

Even if you had it that way, the lone pair wouldn't be sticking straight up at 90 degrees so I don't think it'd work. It needs to be sp2 for it to be aromatic.

Even if you had it that way, the lone pair wouldn't be sticking straight up at 90 degrees so I don't think it'd work. It needs to be sp2 for it to be aromatic.


I completely agree with you saqrfaraj. Aromaticity gives molecules special stability due to the parallel overlap of p-orbitals. It just wouldn't work if the nitrogen was sp3 hybridized.

The nitrogen in the ring is clearly sp2 hybridized. Normally, the nitrogen would be sp3 hybridized, but it must rehybridize to sp2 so that the lone pair of electrons can be delocalized into the pi-system of the ring and that the molecule can be aromatic. What we see here is that aromaticity is highly favored.

Correct answer above: sp2

since the lone pair of electrons on the nitrogen delocalize into the aromatic ring, if it were not aromatic, then it would be sp3.

What is the hybridization of the nitrogen in pyrrole?:

http://upload.wikimedia.org/wikipedia/commons/thumb/1/15/Pyrrole.png/300px-Pyrrole.png

Is it sp3 or sp2 because it is a conjugated system?

sp2 without a doubt.