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I'm trying to solve this problem:

reaction

Specifically, how does one know which nitrogen is more basic? I'm assuming that for the first molecule, that it's the nitrogen without the proton, since it seems like the other nitrogen has already been protonated (and already has acted as a base.)

Is my reasoning correct? How does one solve the other molecules?

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Use a combination of

  • the exclusion principle: sort out the unlikely centres with low electron density and

  • paper work to figure out those nitrogen atoms for which the additional positive charge can be distributed over a larger part of the molecule without moving the proton.

enter image description here

In the case of caffeine (1) the amide and the imide are rather unlikely centres for protonation! You have probably already ruled out the amide nitrogen in the benzamide case (your second molecule) in favour of the aniline site.

From an initial number of four possible centres in 1, only two remain.

Now, let's see whether the positive charge resulting from protonation can be stabilized.

protonation of tertiary amine

This looks fair, but creates a quarternary ammonium ion for which no obvious charge distribution seems to be available.

Let's examine the last possibility:

protonation with charge delocalisation

Would you agree that this looks much more promising?

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  • $\begingroup$ My pleasure! I reckon that you figured out the other ones easily! $\endgroup$ – Klaus-Dieter Warzecha Apr 8 '15 at 21:47
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For amine basicity:

Basicity decreases with increasing s character

Basicity decreases with increasing stability, due to resonance/ delocalization

https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/amine1.htm (taken from this webpage)

For the first molecule for example, the double bonded nitrogen is less basic due to increasing s character (closer to nucleus, less likely to bind to H).

I think most of your examples can be reasonably thought out with these two key points.

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  • $\begingroup$ double bonded nitrogen is more basic, because the other one has delocalized lone pair - your second point... $\endgroup$ – Mithoron Apr 8 '15 at 13:45

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