Stan Wawrinka (ATP 88) ha battuto Aljaz Bedene (76) con il punteggio di 7-5 7-6 (7/5) nel primo turno del torneo di San Pietroburgo. Il losannese, al terzo successo in quattro confronti diretti con lo sloveno, raggiunge agli ottavi il numero 4 del tabellone Karen Khachanov (24).
Il vodese, che ha beneficiato di una wild card in Russia, è riuscito a far suo il primo set grazie a una rimonta in extremis. Dopo aver perso il proprio servizio al quinto game, il tennista di Saint-Barthélemy ha infatti infilato cinque giochi consecutivi per portarsi dal 3-5 al 7-5 1-0. Nella seconda frazione l'equilibrio è perdurato sino al tie-break, con il rossocrociato, finalista nel 2016 con Alexander Zverev, apparso non ancora sui suoi livelli migliori.
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