SAINT PETERSBURG, Russia (AP) – Botic Van de Zandschulp of the Netherlands created an upset on Friday by defeating defending champion Andrey Rublev 6-3, 6-4 in the quarterfinals of the St. Petersburg Open.
Van de Zandschulp, 69th in the world rankings, eliminated first-seeded Rublev in one hour and 33 minutes to reach 16 wins in his last 19 matches at all levels.RELATED
Van de Zandschulp, 26, who reached the US Open quarterfinals last month, will face Marin Cilic in the semifinals. Cilic, who conquered the St. Petersburg tournament in 2011, beat third-seeded Roberto Bautista Agut 6-4, 3-6, 6-3.
American Taylor Fritz will have a duel with German Jan-Lennard Struff in the other semi-final of the tournament on hard courts.
Fritz, who celebrated his birthday by defeating compatriot Tommy Paul on Thursday, knocked out John Millman 6-4, 6-2, while Struff knocked out second-seeded Denis Shapovalov 6-4, 6-3.