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2024 Kiskút Open – Doubles

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Doubles
2024 Kiskút Open
Final
ChampionsFrance Titouan Droguet
France Matteo Martineau
Runners-upSweden André Göransson
Ukraine Denys Molchanov
Score4–6, 7–5, [10–8]
Events
Singles Doubles
← 2023 · Kiskút Open · 2025 →

Bogdan Bobrov and Sergey Fomin were the defending champions[1] but chose not to defend their title.

Titouan Droguet and Matteo Martineau won the title after defeating André Göransson and Denys Molchanov 4–6, 7–5, [10–8] in the final.

Seeds

Draw

Key

First round Quarterfinals Semifinals Final
1 Sweden A Göransson
Ukraine D Molchanov
4 77 [10]
France M Guinard
France G Jacq
6 62 [8] 1 Sweden A Göransson
Ukraine D Molchanov
7 6
Alt Romania C Crețu
Romania A Jecan
6 77 Alt Romania C Crețu
Romania A Jecan
5 2
Croatia N Serdarušić
Czech Republic M Vrbenský
1 63 1 Sweden A Göransson
Ukraine D Molchanov
7 6
3 Netherlands B Stevens
Greece P Tsitsipas
3 65 Alt Italy F Agamenone
Argentina S Rodríguez Taverna
5 2
Alt Italy F Agamenone
Argentina S Rodríguez Taverna
6 77 Alt Italy F Agamenone
Argentina S Rodríguez Taverna
77 2 [10]
WC Hungary Attila Boros
Hungary Botond Kistantal
4 4 India RC Bollipalli
India N Kaliyanda Poonacha
65 6 [8]
India RC Bollipalli
India N Kaliyanda Poonacha
6 6 1 Sweden A Göransson
Ukraine D Molchanov
6 5 [8]
Slovakia J Kovalík
Slovakia I Zelenay
79 5 [9] France T Droguet
France M Martineau
4 7 [10]
France T Droguet
France M Martineau
67 7 [11] France T Droguet
France M Martineau
4 6 [12]
WC Hungary Mátyás Füle
Hungary G Madarász
6 3 [7] 4 France J Eysseric
Italy A Pellegrino
6 0 [10]
4 France J Eysseric
Italy A Pellegrino
3 6 [10] France T Droguet
France M Martineau
3 6 [10]
Czech Republic P Nouza
Czech Republic P Rikl
77 7 2 France T Arribagé
Romania VV Cornea
6 3 [8]
Czech Republic Z Kolář
Ukraine V Manafov
63 5 Czech Republic P Nouza
Czech Republic P Rikl
77 5 [8]
Serbia I Sabanov
Serbia M Sabanov
2 63 2 France T Arribagé
Romania VV Cornea
63 7 [10]
2 France T Arribagé
Romania VV Cornea
6 77

References