Ochem , base question

[Marmar tala]

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http://en.wikipedia.org/wiki/File:Imidazole_chemical_structure.png

which Nitrogen is more basic?
1 or 3?

 

[UndergradGuy7]

I say the one with no hydrogen on it, so 3. Because it's lone pair of electrons does not do resonance with the ring because the nitrogen already has a double bond.

 

[jirotrom]

I would say 3... but I'm not sure.

 

[Marmar tala]

why do you think this reasoning is wrong:
the N#3 is sp2 hybridized and the lone pair is in the p orbital so if it bonds to a Hydrogen then it will change the aromaticity of the molecule therefore it does not do that.
I am kind of confused😕

 

[Baylor2011]

If you donate the electrons on the #3 N to a proton, it's still sp2 hybridized (cation in an empty p orbital), and the ring is still aromatic - therefore most basic (most likely to donate protons). On the other hand, if you try to add a proton to the #1 N, the compound is no longer aromatic - which is bad.

I attached a hypothetical reaction (orgohacks will probably prove me wrong) - Don't chastise me...just throwing in my thought

 

[UndergradGuy7]

My guess...
So say if the double bond on 3 were to attack a hydrogen. Yes it will lose aromaticity, if it loses the double bond, but then the lone pair on N can be used to regain aromaticity... (hypotetical)

But the double bond electrons would not be used for the base reaction... the lone pair would.

that your question?


in for orgohacks explanation...

 

[DQLEUCD]

I'm confused, both look to be the same molecule. Both are aromatic since they are in conjugation and obey huckle's rule.

 

[Marmar tala]

Thank you everybody, I get it now 🙂
I mistakenly assigned sp3 to the new molecule at N #3, it is still sp2 and the molecule still is aromatic.

 

[orgohacks]

If you donate the electrons on the #3 N to a proton, it's still sp2 hybridized (cation in an empty p orbital), and the ring is still aromatic - therefore most basic (most likely to donate protons). On the other hand, if you try to add a proton to the #1 N, the compound is no longer aromatic - which is bad.

I attached a hypothetical reaction (orgohacks will probably prove me wrong) - Don't chastise me...just throwing in my thought

That's exactly right actually! 🙂

The imidazole sp2 nitrogen has a free lone pair at right angles to the pi system, therefore it can protonate. The other imidazole nitrogen's lone pair is tied up in the aromatic system and is therefore not basic, in the drawing shown here.