[Uberbane]

Hi everyone,

I've had a question come up about my research come up a few times, and I'm not sure how to justify the result, as to me it seems as though there should be a mixture of products in the case of this reaction due to tautomerization.

However, only the first product is observed in this case as determined by NMR is the top product. I thought that this might be due to aromaticity, but tautomerization would not break the overall aromaticity of the heterocycle. My other thought is that the nitrogen-rich nature of the pyrazolopyrimidine ring somehow prevents the tautomerization from occurring.

Any insight or hints in the right direction would greatly be appreciated.

Cheers!

[pgk]

Although this is an issue of quantum chemistry rather than pure organic chemistry alone, let’s try to make it simple.
1). Roughly, the various tautomers may not be equal, regarding their energy content, as well as their contribution in the tautomerism.
2). The same happens to the various aromatic forms that might not be equal, regarding their energy content, as well as their contribution in the overall aromatic hybridism (especially, in fused aromatic rings).
3). In this case, both products are aromatic but the top product has three endocyclic double bonds in the pyrimidine ring and two endocyclic double bonds in the pyrazole ring and therefore, it has a slightly lower energy content than the bottom form, which has two endocyclic double bonds and two exocyclic double bonds in the pyrimidine ring, together with two endocyclic double bonds in the pyrazole ring.
4). As a consequence, the top product can easily be obtained by kinetic control (e.g. lower temperatures, shorter reaction times), in contrast to the bottom product that can only be obtained by thermodynamic control (e.g. reflux, longer reaction times), even without Lewis catalysis.

[rolnor]

Is it not the other way around, to get the thermodynamic most stable product (the top one) you can need longer reaction time?
I think you need a catalyst to get any reaction at all in any case.
I usually silylate the nucleoside base first but that is maby not nescesary in this case.

[pgk]

1). Roughly, thermodynamic control means offering larger amounts of energy and kinetic control means a quick reaction under milder conditions. However, thermodynamic vs kinetic control mainly has to do with the relative stability of the transition state rather than the relative stability of the final product. Thus:
In this case and regarding the hemiaminal ether formation from the corresponding hemiacetal ester, both transition states are energetically equal but the formation of the bottom product needs additional energy, due to the higher energy content of the aromatic form and therefore the top product is more easily and quickly obtained.
In other cases, it may be different. E.g., in 1,4- conjugative Grignard you need more energy (thermodynamic control), in order to achieve to the more extended but more stable 6-member transition state (or Cu(I) catalysis) than the less extended and less stable 4-member transition state of the 1,2-Grignard addition that can quickly be obtained under milder conditions (kinetic contol). But both 1,4- and 1,2- Grignard final products are almost equally stable.
2). “Even without catalyst” is a figure of speech that may lead to confusions and I fully apologize for this. Anyway, catalysis just decreases the activation energy of the reaction and theoretically, it is not necessary when working at high temperatures but this does not always happen, in practice.

[rolnor]

I have done some chemistry with 2-deoxyribose. To make the 1-O-methylfuranoside I use a short reaction time (HCl in MeOH). To get the 1-O-methylpyranoside I have a longer reactiontime. The five membered furanoside is the less stable but is formed faster (kinetical product). The six membered pyranoside is the thermodynamic more stable product and but is formed more slowly. If the transition states can be more or less thermodynamical stable it can be different, I understand.

[pgk]

1-O-methylation of sugars occurs via ring opening to the linear form, followed by ring closure via dehydration of the formed hemiaminal group (it is not a SN1 or SN2 substitution, as frivolously figures in some books). As a consequence, pyranoside formation needs a more extended transition state and therefore, it is thermodynamically controlled; in contrast to the furanoside formation that needs a less extended transition state and therefore, it is kinetically controlled.

[rolnor]

I think the stability of the products are important, its possible to convert the 1-O-methylfuranoside with acid to the
1-O-methylpyranoside, but not the other way around.

[pgk]

Indeed, it is very important because the products selective formation under thermodynamic control, depends on their relative thermodynamic stability. But:
“It is (also) important to emphasize that the relative proportions of alternative products formed will be defined by their relative thermodynamic stabilities under the conditions of the reaction, which may possibly differ from those of the isolated products.”
Peter Sykes: A Guidebook to Mechanism in Organic Chemistry, 6th Edition, Longman Ltd, London, (1985), Chapter 6.8: Kinetic versus Thermodynamic Control, pp.163-164

[rolnor]

I think I understand, but in my example with furanoside vs pyranoside it is clear that the six-membered ring is more stable both under the conditions of the reaction as well as under neutral conditions.

[pgk]

Exactly, it may happen that the most stable final product to be the most stable under the reaction conditions, too; but it may also be not.

[pgk]

My post above: “1-O-methylation of sugars occurs via ring opening to the linear form, followed by ring closure via dehydration of the formed HEMIAMINAL group” is wrong.
The right is: “1-O-methylation of sugars occurs via ring opening to the linear form, followed by ring closure via dehydration of the formed HEMIACETAL group”
I fully apologize for that misspelling that leads to serious confusions.

[Uberbane]

Thanks for all of your enlightening comments. I had a conversation with one of my colleagues about this reaction and the unexpected result earlier today, and she suggested that it may be related to steric hindrance.

In most cases this reaction was done with X = I, in which case it might be true, but in some cases X = Br and in either case the result was strictly the top product, so I'm not sure if this is a major contributing factor because the Br would not be as sterically hindering, if at all.

[pgk]

1). If you draw the substituted pyrazole ring with a pencil on a piece of paper, by respecting the bond length, the bond angles, as well as the halogens atomic radii and the help of simple geometry, you will see that that the steric hindrance of the halogen substitution with both iodine or bromine, is insignificant.
Pyrazole bond lengths:
Aromatic C-C ≈ 1.40 (average)
Aromatic C-N = 1.37
C-Br = 1.76
C-I = 2.14
N-H = 1.01
N-C (substituent) = 1.47
Pyrazole bond angles:
Ring C-C-C = 108
Ring C-N-N = 109
Substituent C-X = 126
Substituent N-C = 125.5
Halogen’s atomic radius:
Br = 1.14
I = 1.33
Bond lengths and atomic radii are in Å, bond angles are in degrees and all are taken from various sources.
2). If you draw the electrons flow in the overall molecule on a piece of paper, you will see that the halogen’s electron donating effect, thermodynamically favors the formation of the bottom product, in both aromatic forms.
3) As a conclusion, neither the halogen's steric hindrance nor the halogen’s electron donating effect can explain the experimentally obtained result. Consequently, the experimentally obtained result can be explained by the kinetic control of the reaction, in addition to the higher contribution of the top form in the aromatic hybridism, as initially assumed. 

[kriggy]

I would need to dig some literature but AFAIK in case of indazoles (benzopyrazole) the tautomers are  in favor of 1H over 2H. And while you might get some alkylation of N2, its not syntheticaly usefull (or at least, was not for me) so if you want to make the N2-glycosilated product then you probably need another direction like glycosilating a pyrazole derivative that can possibly be reacted on N2 and then run some ring closing reactions to finish the 6-membered ring.
Another method that could work is to protect the N1 of your compound, reduce the double bond, glycosilate, deprotect and oxidize. Works for purines might work here as well