The nitrogen marked in the purine ring is the most electron-rich one and still carries a hydrogen. Base (OHX−) removes that N-H proton, and the resulting nucleophilic nitrogen then attacks the methyl carbon of CHX3I in a substitution step. The methyl group ends up on that nitrogen, giving the N-methylated purine as the major product.
JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option