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Topic: Which amine is more basic  (Read 4821 times)

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Offline Altered State

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Which amine is more basic
« on: April 03, 2013, 10:45:10 AM »
The question is, which amine of those 2 is a stronger base?



I know the more basic will be the one which protonated form is more stable, so the one in which the possitive charge on the N is more stable.

The final question is obvious. How does that double bond carbon bonded to the N affect stabilizing or destabilizing the possitive charge?
I know that Csp2 retire charge from near Csp3, but I do not know how it would act bonded to a N.
I suppose that sp2 carbon, being more electronegative than sp3 carbon, will retire more charge, and it will destabilize the protoned form, so azocyclohexene will be less basic than azocyclohexane. Is this right?

Offline Dan

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Re: Which amine is more basic
« Reply #1 on: April 03, 2013, 11:31:24 AM »
Think about aniline vs cyclohexylamine. This is is easier to argue on the basis of lone pair availability/hybridisation.
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Offline Altered State

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Re: Which amine is more basic
« Reply #2 on: April 03, 2013, 11:43:57 AM »
Think about aniline vs cyclohexylamine. This is is easier to argue on the basis of lone pair availability/hybridisation.

I have that couple clear. Aniline will be less basic because the lone pair is delocalizated all along the π system of the aromatic ring (I can write 3 resonant forms with possitive charge on N).
Is this case the same? I don't get it very well...

Offline Dan

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Re: Which amine is more basic
« Reply #3 on: April 03, 2013, 01:02:55 PM »
Hint:
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Offline Altered State

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Re: Which amine is more basic
« Reply #4 on: April 03, 2013, 01:42:33 PM »
Hint:

Seeing that I still think the same. Any pi system will be conjugated with the N lone pair and retire charge, making the compound less basic.

If I'm wrong seems to be something I'm losing.

Offline Dan

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Re: Which amine is more basic
« Reply #5 on: April 03, 2013, 02:32:40 PM »
Any pi system will be conjugated with the N lone pair and retire charge, making the compound less basic.

Yes, any π-system conjugated to the lone pair will reduce the basicity of the lone pair.
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Offline Altered State

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Re: Which amine is more basic
« Reply #6 on: April 03, 2013, 03:22:27 PM »
Any pi system will be conjugated with the N lone pair and retire charge, making the compound less basic.

Yes, any π-system conjugated to the lone pair will reduce the basicity of the lone pair.

Nice, thank you.

Offline opsomath

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Re: Which amine is more basic
« Reply #7 on: April 04, 2013, 11:13:56 AM »
The other answer on this thread is excellent, but I just wanted to commend you on your correct reasoning in your initial post although it didn't directly lead you to the answer you wanted. You said that the more basic one would be the one where the conjugate acid (the protonated compound) is more stable. This is, of course, correct. Unfortunately, it is a harder route to see the difference when you think about it that way versus just comparing the bases.

To follow your initial line of reasoning, if you protonate the one with the double bond, it no longer has conjugation between the double bond and the nitrogen lone pair. So, the protonated double-bonded compound takes an energy penalty for protonation, while the all-single-bonds compound does not. That's why the protonated form of the double bonded thing is less stable relative to the unprotonated form. Make sense?

Offline Altered State

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Re: Which amine is more basic
« Reply #8 on: April 04, 2013, 04:11:59 PM »
The other answer on this thread is excellent, but I just wanted to commend you on your correct reasoning in your initial post although it didn't directly lead you to the answer you wanted. You said that the more basic one would be the one where the conjugate acid (the protonated compound) is more stable. This is, of course, correct. Unfortunately, it is a harder route to see the difference when you think about it that way versus just comparing the bases.

To follow your initial line of reasoning, if you protonate the one with the double bond, it no longer has conjugation between the double bond and the nitrogen lone pair. So, the protonated double-bonded compound takes an energy penalty for protonation, while the all-single-bonds compound does not. That's why the protonated form of the double bonded thing is less stable relative to the unprotonated form. Make sense?

Excelent explanation!! I really loved it!
Can this be applyed to cases like cyclohexilamine/aniline or butylamine/butylamide aswell?
Protonating aniline makes that the lone pair is no more there, so pi-system delocalization (the 3 resonant forms on the aromatic ring) is not possible anymore, and the same way, if butylamide N is protonated, there is no longer conjugation with the carbonyl group.

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