1969 CONCACAF Championship qualification

A total of 12 CONCACAF teams entered the competition.  Costa Rica, as the hosts, and  Guatemala, as the defending champions, qualified automatically, leaving 4 spots open for competition. The 10 teams were divided into 5 groups of 2 in which one of them will advance to the final tournament.

According to different sources, only 9 teams entered the competition,[1] eventually reduced to 8 after Haiti renounced.[2] Later on, Haiti tried to join again but the late application was rejected.[3] Honduras, El Salvador and the USA[4] did not enter.
Costa Rica, Guatemala, Netherlands Antilles and Trinidad and Tobago qualified automatically. Play-offs were arranged for the remaining two spots.[5]

Preliminary round

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Series One

20 April 1969 Stade Sylvio Cator, Port-au-Prince, Haiti  Haiti 2 - 0 United States  

11 May 1969 San Diego Stadium, San Diego, United States  United States 0 - 1 Haiti  

Haiti qualifies with aggregate score of 3-0.

Series Two

  Mexico3–0Bermuda  
Mancilla   6', 30'
Langarica   52'
  Bermuda2–1Mexico  
Brangman   34'
Dowling   61'
Santana   56'

Mexico qualifies with aggregate score of 4-2.

Series Three

  Jamaica1–1Panama  
Ziadie   (pen.) Tapia   46'
Panama  1–2  Jamaica
Espinoza   20' Largie   27'
Hill   85'

Jamaica qualifies with aggregate score of 3-2.

Series Four

  Honduras was disqualified due to Football War with El Salvador, so   Netherlands Antilles advanced to the tournament automatically.

Series Five

  El Salvador was disqualified due to Football War with Honduras, so   Trinidad and Tobago advanced to the tournament automatically.

Teams qualified to the 1969 CONCACAF Championship

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References

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