Australian Open third-seed Grigor Dimitrov narrowly avoided a stunning upset against US qualifier Mackenzie McDonald, ranked 186th in the world, in the second round of the first tennis Grand Slam of the year, edging a late-night encounter on the Rod Laver Arena 4-6, 6-2, 6-4, 0-6, 8-6.
McDonald made the brighter start, converting his only break point and saving two on his own serve to take the first set. But Dimitrov quickly dug himself out of the hole, dominating the second set, recording 10 winners to McDonald’s two, and then took a 2-1 lead behind his strong serve in the third frame.
In the fourth, the Bulgarian appeared to suffer a collapse that plagued him so often earlier in his career, with double faults and unforced errors handing McDonald a clean sweep. It was the first time Dimitrov was on the receiving end of a “bagel” since Istanbul in May 2016 against Argentina’s Diego Schwartzman.
In the deciding set, Dimitrov appeared to have recovered, but McDonald held serve comfortably until 6-5, when the American went down 0-30 but managed to recover. Undaunted, Dimitrov pounced on the American’s next service game and wrapped up the encounter at his first match point, after close to three and a half hours of play shortly before midnight, local time.
Dimitrov will play Russia’s Andrey Rublev in the third round, a rematch of their 2017 US Open encounter, which the Russian won in straight sets. Rublev is unlikely to be more rested than Dimitrov, however, having played a five-setter against David Ferrer in his opener and then taking more than three hours to win against former finalist Marcos Baghdatis in four sets in the second round.
(Photo: Brad Touesnard/flickr.com)