# Why is the nitrogen's lone pair part of sp2 orbital in pyridine but part of p orbital in pyrrole?

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I just can't understand this. Both molecules are aromatic, but the nitrogen's lone pair 'sticks out' in pyrridine and is delocalized as part of the pi-ring in pyrrole.

Why so? I've read half a dosen discriptions online, but am clearly missing something.

I just can't understand this. Both molecules are aromatic, but the nitrogen's lone pair 'sticks out' in pyrridine and is delocalized as part of the pi-ring in pyrrole.

Why so? I've read half a dosen discriptions online, but am clearly missing something.

DISCLAIMER: *VERY LONG ANSWER!
*

As a quick review, recall that there exist three

This is inherently a difficult phenomenon to answer, so we'll have to do a deep symmetry analysis as well as considering general aspects of VSEPR theory.

Ultimately we will find that the reasons why the lone pair in pyridine is outside the ring and in pyrrole it is inside the ring are:

- Pyrrole, a five-membered ring has the
*same*number of#pi# electrons as pyridine, a six-membered ring, which results in an extra, fourth electron group on pyrrole's#sp^2# nitrogen (yes, this is unusual). - Pyrrole, for purposes of energetic favorability (aromaticity), has
#sp^2# hybridization, despite having four electron groups, so the#2p_y# orbital stays*unhybridized*, allowing it to delocalize electron density throughout the ring. - If hydrogen is to be bonded to pyrrole, then it is the
#sp^2# orbital that must align towards hydrogen to overlap and form the bond. Therefore, the*unhybridized*#2p_y# orbital is the one that holds the lone pair.Below, I actually go into how I know this.

BACKGROUND INFORMATION: OVERVIEW OF ORBITAL SYMMETRY AND IRREDUCIBLE REPRESENTATIONS

This is a difficult topic, but I'll try to keep it not too complicated. Pyrrole and pyridine are classified as

#C_(2v)# molecules, and pertain to this character table:[

#A_1# ,#A_2# ,#B_1# , and#B_2# are called "irreducible representations". When I say "#A_1# symmetry" or "#B_2# symmetry", I am referring to these.]As for the rest of the table, focus on the first five columns (first column contains

#A_1, . . . , B_2# ). Basically:#hatE# is an identity operation that returns the same molecule back without doing anything (defined for the purpose of completeness).#hatC_2(z)# is a rotation operation on the#z# axis that returns the same molecule upon rotating#\mathbf(180^@)# about the#z# axis. For both molecules, this is the*only*rotation axis they have.#hatsigma_v(xz)# is the reflection plane along the#x# and#z# axes. This operation returns the molecule reflected through the#xz# plane (coplanar with the ring).#hatsigma_v'(yz)# is the reflection plane along the#y# and#z# axes. This operation returns the molecule reflected through the#yz# plane (perpendicular to the ring).The

#1# in the table tells you that the operation returned the same molecule back, with orbital lobes of one sign landing on orbital lobes of the*same*sign.The

#-1# in the table tells you that the operation returned the same molecule back, with orbital lobes of one sign landing on orbital lobes of the*opposite*sign.Example:

A

#2p_y# orbital perpendicular to the ring returns:#1# from#hatE# #-1# from#C_2(z)# (since it becomes vertically flipped)#-1# from#sigma_v(xz)# (since it becomes vertically flipped)#1# from#sigma_v'(yz)# (since it becomes the same orbital again)So, it has

#\mathbf(B_2)# symmetry.A similar analysis with the (spherical)

#1s# orbital of hydrogen would give you#\mathbf(A_1)# symmetry, the totally symmetric "irreducible representation".The rest of the answer assumes you know how to do this, so please familiarize yourself with the example above before moving on.

A rule you should keep in mind for the rest of this answer:

*Two orbitals transforming under different "irreducible representations" cannot overlap and make a bond. Example:*#A_1 ne B_2# ,*so an orbital of*#A_1# *symmetry is not compatible with an orbital of*#B_2# *symmetry*.PART 1: CONSIDERING THE NUMBER OF PI ELECTRONS: SKETCHING MO DIAGRAMS (SYMMETRY)

A quick-and-dirty way of

*sketching MO diagrams*for these molecules is using a Frost circle mnemonic (not to be confused with a Frost diagram).This is how it turns out for the

#pi# systems of each molecule (note that pyridine diagram has the two#sp^2# valence electrons that aren't in the ring,*omitted*):- The energy of each "group orbitals" increases with the number of
*nodes*(#0,1,2# ) in the "group orbitals". - The six
#pi# electrons are all capably delocalized throughout the ring for both molecules, because they are all in bonding orbitals. - None of the symmetries in the MO diagram are
#A_1# !! - Recall that the
#1s# orbital of hydrogen belongs to the#A_1# "irreducible representation"!!What we get from these Frost circles is that:

- Due to pyrrole having the
*same*number of electrons as would be needed on the six-membered pyridine ring for aromaticity, the pyrrole nitrogen has an extra electron group. - The
#sp^2# (#A_1# ) orbital that bonds with hydrogen in*pyrrole*cannot overlap with any#2p_y# orbital in the aromatic ring. These#sp^2# orbitals must be outside the ring. - The
#sp^2# (#A_1# ) orbital that contains the lone pair in*pyridine*cannot overlap with any#2p_y# orbital in the aromatic ring either. These#sp^2# orbitals must be outside the ring.PART 2: CONSIDERING THE NUMBER OF ELECTRON GROUPS (VSEPR THEORY)

Pyridine -

In pyridine, nitrogen only has three electron groups, with "true"

#sp^2# hybridization (*not*merely forced by the ring constraints and the hope for aromaticity), so the lone pair is in the third,*nonbonding*#sp^2# orbital (#A_1# symmetry).Since the

#sp^2# orbital is the wrong symmetry (#A_1# symmetry), it cannot delocalize electron density into the ring, which contains#2p_y# orbitals (individually#B_2# symmetry, but#A_2# or#B_2# as a group; either way,#A_1 ne A_2 ne B_2# ).The result of this incompatibility is shown by its sticking out of the ring.

Pyrrole -

In pyrrole, nitrogen actually has four electron groups (three

#sigma# bonding, one#pi# -bonding lone pair), but...- the constraints of the ring...
- the hope for aromaticity...
...these conditions make it

*more energetically favorable*to ideally have#sp^2# -hybridized orbitals (#A_1# symmetry) instead of#sp^3# -hybridized orbitals (also#A_1# symmetry). The third#sp^2# orbital is used to bond with hydrogen.If pyrrole's nitrogen did ideally have

#sp^3# hybridization (#A_1# symmetry), it could still bond with hydrogen, but the#sp^3# orbital would*not*be able to delocalize electron density into the ring because it would be the wrong symmetry to overlap with the#2p_y# orbitals (#B_2# symmetry, and#A_1 ne B_2# ).Sign up to view the whole answerBy signing up, you agree to our Terms of Service and Privacy Policy

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- Due to pyrrole having the

- The energy of each "group orbitals" increases with the number of

*Answer 2Sign up to view the whole answerSign up with email*

In pyridine, the lone pair of nitrogen is part of an sp2 hybrid orbital because the nitrogen atom is bonded to three other atoms, resulting in trigonal planar geometry. In pyrrole, the lone pair of nitrogen is part of a p orbital because the nitrogen atom is involved in a five-membered ring with four carbon atoms, resulting in sp2 hybridization and a delocalized pi system.

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*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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