The 2',3'-O-isopropylideneuridine (1) reacts with MeI in the presence of an excess of NaH in THF giving 2',3'-O-isopropylidene-5'-O-methyluridine (2). Prolonged reaction time gives rise to 2',3'-O-isopropylidene-3,5'-O-dimethyluridine (4). The use of an equimolar amount of base and alkylating agent results predominantly in methylation at N(3) (--> 3).