1991 Arizona Classic – Singles
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| Singles | |
|---|---|
| 1991 Arizona Classic | |
| Final | |
| Champion | Belgium Sabine Appelmans |
| Runner-up | United States Chanda Rubin |
| Score | 7–5, 6–1 |
| Details | |
| Draw | 32 (4Q) |
| Seeds | 8 |
| Events | |
1991 tennis event results
Main article: 1991 Arizona Classic
Conchita Martínez was the defending champion, but did not compete this year.
Sabine Appelmans won the title by defeating Chanda Rubin 7–5, 6–1 in the final.[1]
Seeds
[edit ]- United States Gigi Fernández (second round)
- France Julie Halard (semifinals)
- Belgium Sabine Appelmans (champion)
- United States Amy Frazier (second round)
- United States Marianne Werdel (first round)
- Indonesia Yayuk Basuki (first round)
- Germany Barbara Rittner (first round)
- United States Debbie Graham (first round)
Draw
[edit ]Key
[edit ]- Q = Qualifier
- WC = Wild card
- LL = Lucky loser
- Alt = Alternate
- SE = Special exempt
- PR = Protected ranking
- ITF = ITF entry
- JE = Junior exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
- SR = Special ranking
Finals
[edit ] Semifinals
Final
Q
United States Chanda Rubin
6
6
Q
United States Chanda Rubin
5
1
3
Belgium Sabine Appelmans
7
6
3
Belgium Sabine Appelmans
3
6
6
2
France Julie Halard
6
3
3
Top half
[edit ] First round
Second round
Quarterfinals
Semifinals
1
United States G Fernández
6
6
8
United States D Graham
3
6
5
Bottom half
[edit ] First round
Second round
Quarterfinals
Semifinals
References
[edit ]- ^ "Results Plus, Tennis: Rubin Loses to Appelmans". The New York Times . Associated Press. 4 November 1991. Retrieved 27 November 2023.
Third-seeded Sabine Appelmans of Belgium won 11 of the last 13 games to capture her first professional title with a 7-5, 6-1 victory over 15-year-old American Chanda Rubin in the 150,000ドル Arizona Tennis Classic in Scottsdale, Ariz.