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Stan Wawrinka (ATP 3) giocherà domenica contro Alexander Zverev (27) per il titolo nel torneo di San Pietroburgo. Il vodese si è qualificato per la finale sconfiggendo nel penultimo atto Roberto Bautista Agut (16) con il punteggio di 7-6 (10/8) 6-3 in 1h52', guadagnando così la possibilità di cogliere il suo quinto titolo stagionale dopo Chennai, Dubai, Ginevra e US Open (25a finale in carriera).
L'incontro è vissuto su un primo set molto combattuto dove nessuno dei due è stato capace di rubare il servizio all'avversario. Nell'appendice del tie-break il rossocrociato ha rischiato, salvando due palle set, prima di conquistarlo alla seconda possibilità a sua disposizione. La seconda frazione è invece iniziata in modo molto diverso. Lo spagnolo, accusando forse il colpo, ha subito il break in entrata, mostrandosi molto più debole al servizio. Wawrinka ne ha approfittato anche nel settimo game, per rubargli un'altra volta la battuta e volare a servire per la partita, chiusa al secondo match point.
ATP St. Pietroburgo, il servizio su Wawringa - Batista Agut (Sportsera 24.09.2016)
RSI Sport 24.09.2016, 23:15
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