Van de Zandschulp is the only remaining Dutch player in the Australian Open singles draws after Jesper de Jong, Tallon Griekspoor and Suzan Lamens were all eliminated in the first round in Melbourne.
After upsetting 30th-ranked Brandon Nakashima in the first round, Botic van de Zandschulp was drawn against Chinese 318th-ranked Juncheng Shang, who bested Roberto Bautista Agut in four sets in the opening round.
The two fought out an interesting battle with strong service games in the first set, which Shang lit up by winning 19 of 20 first service points. A lack of returning quality saw the Chinese 20-year-old reach break point just once, however, as he and Van de Zandschulp needed a tiebreak to determine a winner.
The tiebreak started with a quartet of breaks, with Shang initially coming out on top to reach set point after 11 points. Van de Zandschulp held onto his two next service points, however, and broke Shang to win a crucial third-straight point that won the Dutchman the first set.
His late turnaround in the tiebreak fired Van de Zandschulp up for the second set, which he started emphatically with four consecutive won games while conceding just three points to Shang. The Chinese youngster survived four break points in the fifth game, but could do little to prevent Van de Zandschulp from clinching a 6-2 set win.
Errors started to mount for Shang, who committed 14 errors on 51 points in the second set, and continued this trend in the third set. Despite three double faults and 11 unforced errors, however, Shang survived four games before Van de Zandschulp broke his Chinese opponent again.
Sheng failed to get to break point more than once throughout the match and fell to Van de Zandschulp on the Dutchman's fourth match point, which won him the set and thus the match in straight sets.
For the first time since 2022, Botic van de Zandschulp reached the third round of the Australian Open. It will not be an easy task to reach the fourth round, with Novak Djokovic his next opponent in Melbourne.