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Topic: Delocalization of nitrogen lone pair  (Read 9122 times)

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Offline kekie

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Delocalization of nitrogen lone pair
« on: May 10, 2014, 03:47:59 PM »
So, I heard that the lone pairs on nitrogens in thymine can delocalize/partisipate in resonance, while the lone pairs on the nitrogen of, for instance, pyridine, can't.

For reference,
Pyridine;

Thymine;

 

I would guess it's because the electron config of the nitrogen on the pyridine is like so;



And that the electron config of the nitrogens on the thymine are like so;


Because each of the nitrogens are bonded to three other things, so it has to have three unpaired electrons at the top.

(Lone pair in red, electrons participating in sigma bonds in blue)

In the second config (Thymine) the lone pair is in a p-orbital, letting it participate in resonance.

Am I correct?
« Last Edit: May 10, 2014, 04:20:02 PM by kekie »

Offline kriggy

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Re: Delocalization of nitrogen lone pair
« Reply #1 on: May 10, 2014, 04:07:34 PM »
I think the difference is that in pyridine, when you would add the electron pair into the system, you would disturb the aromaticity, however in thymine by addit the 2 pairs the system becomes aromatic.
edit: and btw why do you think the orbitals are like this? I mean, there is 1 s orbital and 3 p orbitals in nitrogen valence shell. So your schemes are not correct. The hybrid SP2 orbitals are made of combination on 1 s and 2p orbitals so if you create 3 SP2 orbitals then you have only 1 p orbital left. But you have 3 p orbitals adn 3 sp2 orbitals and 1 s orbital in valence shell.
« Last Edit: May 10, 2014, 04:59:01 PM by kriggy »

Offline Nescafe

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Re: Delocalization of nitrogen lone pair
« Reply #2 on: May 10, 2014, 07:27:09 PM »
So, I heard that the lone pairs on nitrogens in thymine can delocalize/partisipate in resonance, while the lone pairs on the nitrogen of, for instance, pyridine, can't.

For reference,
Pyridine;

Thymine;

 

I would guess it's because the electron config of the nitrogen on the pyridine is like so;



And that the electron config of the nitrogens on the thymine are like so;


Because each of the nitrogens are bonded to three other things, so it has to have three unpaired electrons at the top.

(Lone pair in red, electrons participating in sigma bonds in blue)

In the second config (Thymine) the lone pair is in a p-orbital, letting it participate in resonance.

Am I correct?

I thought this had to do with the fact that the electron pair on the nitrogen of pyridine is in a pi orbital which is perpendicular to the orbitals of the ring and that is why, due to the geometry it does not delocalize within the ring.

Offline kekie

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Re: Delocalization of nitrogen lone pair
« Reply #3 on: May 10, 2014, 10:51:28 PM »
That might be the case- I don't know.

You tell me - the above idea was just my guess.

Also, you don't need to quote the entire post like that.  :P

Offline Dan

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Re: Delocalization of nitrogen lone pair
« Reply #4 on: May 11, 2014, 09:23:26 AM »
I thought this had to do with the fact that the electron pair on the nitrogen of pyridine is in a pi orbital which is perpendicular to the orbitals of the ring and that is why, due to the geometry it does not delocalize within the ring.

This is correct. The N lone pair in pyridine cannot overlap with the π system - it's geometrically impossible.
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Offline goops

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Re: Delocalization of nitrogen lone pair
« Reply #5 on: May 11, 2014, 12:07:27 PM »
In pyridene, the  N is sp2 hybridised. If it tries to delocalize its e-s to form a pi bond, it gets sp hybridised. =N= tends to get linear due to sp hybridisation, this will distort the geometry and creates stress in the compound which further leads to unstability. So it prefers not delocalising its e-s.
Hope it would help..  :)

Offline Sam_

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Re: Delocalization of nitrogen lone pair
« Reply #6 on: May 11, 2014, 08:02:25 PM »
Here's your problem:

(Lone pair in red, electrons participating in sigma bonds in blue)

Single atom orbitals like sp2 and sp3 are no longer very useful for analysis once you start considering molecular orbitals.

Rather than trying to assign the nitrogen's electrons to nitrogen atomic orbitals, consider the molecular orbitals of the pyridine system that nitrogen can participate in.  The pyridine nitrogen is sp2 hybridized, so it contributes 2 sp2 orbitals to form a σ bond with each of the neighboring carbon atoms, one sp2 orbital is left for the lone pair, and it contributes its unhybridized p orbital to the aromatic π system.  The nitrogen contributes one valence electron to each of the σ bonding orbitals, one to the π system, and two to the nonbonding lone pair orbital.  The core electrons remain in the 1s orbital.

Even in the case of the nonbonding lone pair orbital, though, if you calculated the energies of the various orbitals, you would find some contributions from other atomic orbitals.

Offline kekie

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Re: Delocalization of nitrogen lone pair
« Reply #7 on: May 11, 2014, 11:13:48 PM »
Thanks!

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