Regioselectivity of SN2 reaction on adenine

I found the following question in my examination:  

[![enter image description here](https://i.stack.imgur.com/fYLEj.jpg)](https://i.stack.imgur.com/fYLEj.jpg)




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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.




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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 ?




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](https://en.wikipedia.org/wiki/Indole) and [ethanol](https://en.wikipedia.org/wiki/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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