I found the following question in my examination:
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I was able to make out from the options that the usage of a base is to remove the supposedly acidic hydrogen from either the Nitrogen $A$ or $B$. Then the Nitrogen with a negative charge would attack $\ce{CH3I}$ through an $\ce{SN2}$ path and hence overall, the Hydrogen is replaced by methyl.

Now I got stuck on deciding which Nitrogen would be involved here? Both A and B have reduced electron density as their lone pairs are delocalised.
A point of difference maybe that $A$ isn't directly responsible for aromaticity in the 6 membered ring while $B$ is directly involved in aromaticity of the 5 membered ring.

But I don't know if that creates a difference. It would be great if someone could highlight where I'm going wrong and how was I supposed to judge in the examination as I did not have access to $\ce{pK_a}$ values and other experimental data.

Some papers online suggest existence of different paths than this, but considering the level of the exam (JEE-main) it came in, real experimental reaction isn't much considered and more of theoretical approach is considered. I'll reframe my question.

Say that attack had to take place by either Nitrogen 'A' or, 'B' (Which is what the question hints us by using a base and so are the options framed in such a way to confuse)
Which Nitrogen would be preferred?

The correct answer given is (4) that is, Nitrogen $B$ is considered. How can we justify and explain this ?

  • $\begingroup$ 1 is the major product according to this patent patents.google.com/patent/SU1100276A1/en $\endgroup$
    – Waylander
    Jul 6, 2022 at 7:07
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    $\begingroup$ @Waylander thanks for that source, but that doesn't match with the answer given in my exam... Now obviously, a practical experiment is clearly what really occurs. But then I believe there is a difference in the patent and the question. The patent just uses DMF solvent while the question also has a base in it. While in the first case the SN2 reaction would directly occur, and hence most basic Nitrogen will attack $\ce{CH3I}$ but in our case, the base would first extract an acidic H from 'A' or 'B' Nitrogens and then the attack would take place. I'm not sure if that's whats going on here. $\endgroup$ Jul 6, 2022 at 11:26
  • $\begingroup$ This is possibly solvent and base dependent i.e. whether you use a very strong base to deprotonate the starting adenine (and it would require a very strong base) or just use one to mop up the H+produced by the methylation - the question unfortunately does not specify. I have not be able to find further literature on this, but I do not have Scifinder access. I wonder if this question is based on a real experiment? $\endgroup$
    – Waylander
    Jul 6, 2022 at 11:57
  • $\begingroup$ I've made edits considering inputs given by you. Rather than looking at what would happen through experimental approach, is it possible to explain the given answer by considering theoretical approaches? $\endgroup$ Jul 6, 2022 at 16:23
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    $\begingroup$ Other posters may feel different about this, but I do not think I could say anything more than "you'll get a mixture" without knowing more details of the conditions. $\endgroup$
    – Waylander
    Jul 6, 2022 at 16:49

1 Answer 1


If a strong base (not just the methyl iodide and the solvent) is present, then it will tend to deprotonate the $\ce{N-H}$ nitrogen on the five-membered ring, as this proton is about as acidic as one on an alcohol hydroxyl group; compare the $\mathrm{pK_a}$ values given for indole and ethanol. This $\ce{N-H}$ proton is exceptionally acidic because the negative charge left behind by removing the proton is effectively delocalized around the aromatic five-membered ring.

So the base removes that proton, and the deprotonated nitrogen then replaces the proton with the electrophile forming product (4).

Incidentally, once the five-membered ring is deprotonated nothing really prevents the electrophile from (alternatively) attacking the other nitrogen atom on that ring. That would give a different product from any of the choices given.


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