2006 BCR Open Romania – Singles
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Singles | |
---|---|
2006 BCR Open Romania | |
Final | |
Champion | Austria Jürgen Melzer |
Runner-up | Italy Filippo Volandri |
Score | 6–1, 7–5 |
Details | |
Draw | 32 (4Q / 3WC) |
Seeds | 8 |
Events | |
2006 tennis event results
The Men’s Singles tournament of the 2006 BCR Open Romania tennis championship took place in Bucharest, Romania, between 11 September and 17 September 2006. 32 players from 13 countries competed in the 5-round tournament. The final winner was Jürgen Melzer of Austria, who defeated Filippo Volandri of Italy.[1] The defending champion from 2005, Florent Serra, lost in the semifinals to Volandri.[2]
Seeds
[edit ]- Russia Dmitry Tursunov (second round)
- France Florent Serra (semifinals)
- Spain Carlos Moyá (quarterfinals)
- France Gilles Simon (second round)
- Italy Filippo Volandri (final)
- Spain Rubén Ramírez Hidalgo (quarterfinals)
- Germany Florian Mayer (quarterfinals)
- France Paul-Henri Mathieu (semifinals, retired)
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
8
France Paul-Henri Mathieu
4
0r
Austria Jürgen Melzer
6
0
Austria Jürgen Melzer
6
7
5
Italy Filippo Volandri
1
5
5
Italy Filippo Volandri
78
6
2
France Florent Serra
66
4
Top half
[edit ] First round
Second round
Quarterfinals
Semifinals
1
Russia D Tursunov
1
77
6
6
Spain R Ramírez Hidalgo
6
6
Bottom half
[edit ] First round
Second round
Quarterfinals
Semifinals
5
Italy F Volandri
3
6
6
References
[edit ]- ^ "Melzer claims Romanian Open title". BBC Sport. 17 September 2006. Retrieved 2 March 2018.
- ^ "Serra eases to maiden ATP title". BBC Sport. 18 September 2005. Retrieved 2 March 2018.