Russian ironman Daniil Medevdev has ended Alex de Minaur’s US Open campaign with a signature fourth-round victory in New York.
De Minaur played some inspired attacking tennis early to raise hopes of an upset win before the third-seeded 2021 Open champion powered to a 2-6 6-4 6-1 6-2 triumph.
Australia’s world No.13 had been bidding to make the quarter-finals at Flushing Meadows for a second time but couldn’t stay with Medvedev after dropping a tight second set.
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De Minaur’s departure ends Australia’s participation in the singles, with the 24-year-old having to settle for a likely rankings rise to a career-high No.12 next Monday.
Runner-up in 2019 before denying Novak Djokovic a calendar-year grand slam sweep two years ago, Medvedev advances to an all-Russian quarter-final against eighth seed Andrey Rublev.
That didn’t look on the cards early against de Minaur, whose aggressive play had the former world No.1 rattled and made a repeat of his win over Medvedev in Toronto last month seem a distinct possibility.
Playing aggressive tennis, de Minaur won nine of nine points rushing the net and broke Medvedev in the fifth and seventh games to steam through the opening set in 32 minutes.
But a spectacular tweener from Medvedev during a 34-shot winning rally sparked the Russian into action midway through the second set.
De Minaur had to fight off five break points to hold for 2-2 but couldn’t recover from 15-40 down serving at 4-5 as Medvedev levelled the match at a set apiece.
The Russian shifted up a gear in the fourth, overpowering de Minaur from the back with two more breaks to take a two-sets-to-one lead.
It was all but over when de Minaur dropped serve twice more in the fourth set as Medvedev surged to 5-1 before closing out the contest after two hours and 40 minutes.
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