[Uberbane]

Hi everyone,

I've been trying to rationalize why the C-5 carbon is halogenated as opposed to the C-2 carbon when using N-halosuccinimides in DMF but I'm not sure if my rationale is correct.



I think the reaction proceeds by electrophilic aromatic substitution as opposed to a radical mechanism as is the case with some NXS reactions. I know that within the pyrimidine substituent, the C-4 amine is electron donating which activates the ring at the ortho and para positions, which are occupied either by nitrogens or connected to the second aromatic ring.

In terms of the pyrazole component, both the 4 and 5 positions are saturated which leaves only the third position available for halogenation (C5). Does this one occur while the other is unreactive due to the resonance stabilization offered by the lone pair on N7 (and N6)? I would assume that both reactions would hypothetically maintain the molecules overall aromaticity.

I'm just not sure which forces are responsible for the reaction at C5 occurring while C2 remains unreacted.

Thanks!

[clarkstill]

Think about the charged intermediates you would form in each reaction, and the factors that stabilize/destabilize them. Although this is a ground-state argument for a kinetically-controlled reaction, you can probably use the Hammond postulate to assert that the lower energy intermediate forms via the lower energy transition state.

[orgopete]

I agree the halogenation mechanism is ionic rather than radical. Since you know the reaction occurs at C-5 rather than C-2, what you need to do is use this result to tell you about the nature of the reactant.

This is how I might rationalize the result. If you were to react pyridine and pyrrole at the same time, which might react faster? I believe pyrrole will react faster and I might justify this as although pyrrole is aromatic via participation of the non-bonded electrons of the nitrogen, it might also be thought as a non-aromatic (and more reactive) dieneamine. I acknowledge that it is surprising that it should react at C-2 rather than C-3.

The sp2-nitrogen of pyridine (and pyramidine) cannot donate electrons as they are orthogonal to the aromatic electrons. Their donation would result in an antiaromatic form. The result is the pyridine nitrogen has an electron withdrawing and deactivating effect upon pyridine in electrophilic reactions. You may see this enables nucleophilic attacks on pyridine rings at C-2 and C-4.

I'm not saying this is the correct explanation and would satisfy a physical-organic chemist, but it is an explanation that might succeed in predicting what product you may observe.

[Uberbane]

Thanks for both of your help. So I attempted to draw out both scenarios with arrow pushing, and this is what I came up with (please let me know if anything is wrong):



So, if I understand correctly, the pyrimidine ring is pi-deficient due to the two nitrogens having an electron withdrawing and deactivating effect, which is further compounded by the presence of the amine which would direct electrophilic aromatic substitution at the ortho- and para- positions. As you mentioned, the lone pairs are orthogonal to the aromatic electrons and thus cannot be donated, which is in contrast to the pyrazole...

The pyrazole, on the other hand, is pi-excessive and the N1 lone pair is involved in aromaticity. The succinimide anion then picks up the C-5 proton, regenerating aromaticity and forming the expected product.

Would it be oversimplification to state that C-5 is the expected product due to the pi-excessive nature of the pyrazole in contrast to the pi-deficiency of the pyrimidine?

Thanks again for your help.

[orgopete]

You have drawn a mechanism consistent with how would have predicted the reaction to take place, well done.