Can you just explain this more in-depth? I don’t understand why the lone pair is in a p orbital at the top but not at the bottom? Maybe I need to brush up on MO theory but I don’t get what the criteria is to determine that.

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you just explain this more in-depth? I don’t understand why the lone pair is in a p orbital at the top but not at the bottom? Maybe I need to brush up on MO theory but I don’t get what the criteria is to determine that.
1.4 LONE PAIRS
Compare the following two structures:
We saw in the previous section that the first structure is aromatic. The second structure, called pyrrole, is also aromatic, for the same reason.
The nitrogen atom adopts an sp² hybridized state, which places the lone pair in a p orbital, thereby establishing a continuous system of
overlapping p orbitals,
H
Pyrrole
and there are six x electrons (two from the lone pair + another two from each of the bonds = 6). Both criteria for aromaticity have been
satisfied, so the compound is aromatic. Notice that the lone pair is part of the aromatic system, As such, the lone pair is less available to
function as a base:
Aromatic
sp² hybridized
(the lone pair occupies
a p orbital)
Aromatic
HH
A
Nonaromatic
H-CI
1.4 LONE PAIRS 9
This doesn't occur, because the ring would lose aromaticity. If the nitrogen atom were to be protonated, the resulting nitrogen atom (with
a positive charge) would be sp³ hybridized. It would no longer have a p orbital, so the first criterion for aromaticity would not be satisfied
(thus, nonaromatic). Protonation of the nitrogen atom would be extremely uphill in energy and is not observed.
In contrast, consider the nitrogen atom of pyridine:
Pyridine
In this case, the lone pair is NOT part of the aromatic system. This localized lone pair is not participating in resonance, and it does not
occupy a p orbital. The nitrogen atom is sp2 hybridized, and it does have a p orbital, but the p orbital is occupied by an electron (as
illustrated by the double bond that is drawn on the nitrogen atom). The lone pair actually occupies an sp2 hybridized orbital, and it is
therefore not contributing to the aromatic system. As such, it is available to function as a base, because protonation does not destroy
aromaticity:
ol
20
Still aromatic
Transcribed Image Text:1.4 LONE PAIRS Compare the following two structures: We saw in the previous section that the first structure is aromatic. The second structure, called pyrrole, is also aromatic, for the same reason. The nitrogen atom adopts an sp² hybridized state, which places the lone pair in a p orbital, thereby establishing a continuous system of overlapping p orbitals, H Pyrrole and there are six x electrons (two from the lone pair + another two from each of the bonds = 6). Both criteria for aromaticity have been satisfied, so the compound is aromatic. Notice that the lone pair is part of the aromatic system, As such, the lone pair is less available to function as a base: Aromatic sp² hybridized (the lone pair occupies a p orbital) Aromatic HH A Nonaromatic H-CI 1.4 LONE PAIRS 9 This doesn't occur, because the ring would lose aromaticity. If the nitrogen atom were to be protonated, the resulting nitrogen atom (with a positive charge) would be sp³ hybridized. It would no longer have a p orbital, so the first criterion for aromaticity would not be satisfied (thus, nonaromatic). Protonation of the nitrogen atom would be extremely uphill in energy and is not observed. In contrast, consider the nitrogen atom of pyridine: Pyridine In this case, the lone pair is NOT part of the aromatic system. This localized lone pair is not participating in resonance, and it does not occupy a p orbital. The nitrogen atom is sp2 hybridized, and it does have a p orbital, but the p orbital is occupied by an electron (as illustrated by the double bond that is drawn on the nitrogen atom). The lone pair actually occupies an sp2 hybridized orbital, and it is therefore not contributing to the aromatic system. As such, it is available to function as a base, because protonation does not destroy aromaticity: ol 20 Still aromatic
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