MOSCOW, February 16. /TASS/. Russia’s Andrey Rublev lost to Australia’s Alex de Minaur in a first-round match at the Rotterdam Open, an Association of Tennis Professionals tournament, on Wednesday.
The match ended with the score of 6:4, 6:4 in favor of de Minaur. Rublev was seeded as No. 2 at the tournament. The Australian is now set to play with America’s Maxime Cressy.
Rublev, 25, is ranked as world’s fifth-best player by ATP.
De Minaur, 23, is world’s No. 25.