2015 Challenger Ciudad de Guayaquil – Doubles

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Doubles
2015 Challenger Ciudad de Guayaquil
ChampionArgentina Guillermo Durán
Argentina Andrés Molteni
Runner-upPortugal Gastão Elias
Brazil Fabrício Neis
Final score6–3, 6–4
Events
Singles Doubles
← 2014 · Challenger Ciudad de Guayaquil · 2016 →

Guillermo Durán and Andrés Molteni won the title, beating Gastão Elias and Fabrício Neis 6–3, 6–4

Seeds[]

Draw[]

Key[]

  • 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

Draw[]

First round Quarterfinals Semifinals Final
1 Argentina G Durán
Argentina A Molteni
6 3 [10]
Argentina JI Londero
Brazil P Sakamoto
3 6 [4] 1 Argentina G Durán
Argentina A Molteni
6 6
Austria M Linzer
Chile JC Sáez
Italy R Bellotti
Austria D Novak
1 1
Italy R Bellotti
Austria D Novak
w/o 1 Argentina G Durán
Argentina A Molteni
6 4 [10]
3 Spain R Carballés Baena
Netherlands M Vervoort
6 6 3 Spain R Carballés Baena
Netherlands M Vervoort
0 6 [7]
WC Argentina F Argüello
Ecuador AC March
4 2 3 Spain R Carballés Baena
Netherlands M Vervoort
w/o
France M Bourgue
France A Michon
2 2 Brazil A Ghem
Brazil T Monteiro
Brazil A Ghem
Brazil T Monteiro
6 6 1 Argentina G Durán
Argentina A Molteni
6 6
Argentina N Kicker
Sweden C Lindell
4 4 2 Portugal G Elias
Brazil F Neis
3 4
Bosnia and Herzegovina D Džumhur
Poland K Majchrzak
6 6 Bosnia and Herzegovina D Džumhur
Poland K Majchrzak
2 1
WC Argentina G Andreozzi
Argentina Valentin Florez
2 6 [10] WC Argentina G Andreozzi
Argentina V Florez
6 6
4 Argentina P Cachín
Argentina R Olivo
6 3 [8] WC Argentina G Andreozzi
Argentina V Florez
2 4
Uruguay M Cuevas
Bolivia H Dellien
2 Portugal G Elias
Brazil F Neis
6 6
Ecuador I Endara
Ecuador G Escobar
w/o Ecuador I Endara
Ecuador G Escobar
1 1
WC Ecuador E Gómez
Ecuador R Quiroz
612 3 2 Portugal G Elias
Brazil F Neis
6 6
2 Portugal G Elias
Brazil Fabrício Neis
714 6

References[]

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