Botic van de Zandschulp beat top seed Andrey Rublev. (Zandschulp Instagram Photo)
Dutch qualifier Botic Van de Zandschulp upset defending champion Andrey Rublev 6-3, 6-4 to make St. Petersburg Open semi-finals.
- Associated Press St Petersburg (Russia)
- Last Updated:October 30, 2021, 10:04 IST
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Dutch qualifier Botic Van de Zandschulp upset defending champion Andrey Rublev 6-3, 6-4 in the quarterfinals on the St. Petersburg Open on Friday. Van de Zandschulp, ranked 69th, beat the top-seeded Rublev in 1 hour, 33 minutes for his 16th win from his past 19 matches at all levels. The 26-year-old Van de Zandschulp, who reached the U.S. Open quarterfinals last month, will play Marin Cilic in the semifinals. Cilic, who won the tournament in 2011, defeated third-seeded Roberto Bautista Agut 6-4, 3-6, 6-3.
This is what it takes to beat the World No. 6 ICYMI, Botic van de Zandschulp defeated top seed Andrey Rublev with stunning tennis in St. Petersburg! pic.twitter.com/yL9P4owqzb
— ATP Tour (@atptour) October 29, 2021
American Taylor Fritz will play Germany’s Jan-Lennard Struff in the other semifinal at the hard-court tournament.
Fritz, who celebrated his 24th birthday by beating compatriot Tommy Paul on Thursday, defeated John Millman 6-4, 6-2, and Struff upset second-seeded Denis Shapovalov 6-4, 6-3.
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