Three To See, Rome: May 15

Tuesday, May 14, 2013 /by

Victoria Azarenka vs. Julia Goerges
Head-to-head: Azarenka leads 4-1

One of tennis’ top returners reached a mental break point in Madrid: Azarenka imploded, breaking her racquet and snapping at chair umpire Mariana Alves in a 1-6, 6-2, 6-3 loss to 24th-ranked Ekaterina Makarova. It was Vika’s first loss of the year and her first defeat to a woman ranked outside the Top 10 since she retired from her opening match in Montreal to 43rd-ranked Tamira Paszek last August.

The 25th-ranked Goerges can pose problems on clay. She won her first meeting with Azarenka at the 2011 Stuttgart event and went on to beat Samantha Stosur and then-No. 1 Caroline Wozniacki to win the title. She’s won both of her tour titles on clay and can create angles with her heavy topspin shots.

Clay is not Azarenka’s best surface, but she will be eager to move on after the Madrid misstep and get more match play on clay in preparation for the French Open. Azarenka tends to play close to the baseline and take the ball earlier, is a bit more accurate—particularly when stretched—is 18-1 on the season, and should advance here.

The Pick: Azarenka in two sets


Alexandr Dolgopolov vs. Stanislas Wawrinka
Head-to-head: Tied, 1-1

It’s been a sluggish start to the clay-court season for Dolgopolov, who is 4-3 on the slow stuff. In contrast, Wawrinka is on a roll. Stan boasts an 18-5 clay-court record this season; he's won nine of his last 10 matches to take the Oeiras title, reach the Madrid final, and return to the Top 10.

Dolgopolov is so quick off the mark, he accelerates with the abandon of kamikaze Vespa drivers who treat red lights in Rome more as suggestions rather than road rules. Look for Dolgo, who can hammer his forehand with authority and has a deceptively fast serve thanks to a quick-action motion, to try using his favored drop shot to try to exploit Wawrinka’s deeper court positioning. Given the fact that Dolgopolov won their lone clay-court meeting and cruised through the first round, while Wawrinka went the distance against qualifier Carlos Berlocq, you might assert the Swiss is vulnerable here.

I think Dolgopolov is dangerous, but I don’t see Wawrinka faltering. If Wawrinka serves well and uses his versatile one-handed backhand to break down the Ukranian’s backhand, I see him winning to set up a round-of-16 match with Novak Djokovic, which would be a rematch of the 2008 Rome final.

The Pick: Wawrinka in two sets


David Ferrer vs. Fernando Verdasco
Head-to-head: Ferrer leads 8-7

If results determine self-belief, then these two are operating at opposite ends of the confidence scale. Ferrer has contested two finals in his last four tournaments; Verdasco has not reached a quarterfinal since last September in Bangkok.

The left-handed Verdasco has shown signs of life, though. He edged No. 12 Milos Raonic, 6-4, 2-6, 7-6 (7) last week in Madrid, marking just the second time this year he’s won back-to-back matches. But the 5’9” Ferrer looms as a large obstacle. Ferrer was runner-up to Rafael Nadal in the 2010 Rome final and was two points from upsetting Rafa in the Madrid quarterfinals last week, only to suffer a gut-wrenching loss.

These Spaniards know each other well and know what patterns are coming. Verdasco will try to lash his favored left-handed forehand cross-court to Ferrer’s weaker wing, while Ferrer, who is so accurate opening the court with the inside-out forehand, will try to go down the line at times to force Verdasco into hitting his weaker backhand on the run.

Whenever I see those two together, I think of that 2010 U.S. Open night thriller when Verdasco fought back from a two-set hole, then rallied from 1-4 down in the fifth-set tiebreaker to beat Ferrer, 5-7, 6-7 (8), 6-3, 6-3, 7-6 (4), squeezing a running forehand up the line to reach the quarterfinals and later revealing that he apologized to his countryman for the heart-break. If I can’t get that match out of my head, I’m pretty sure Ferrer won’t forget it either, and I can’t see him looking past Verdasco.

The Pick: Ferrer in two sets

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