1996 CONCACAF Champions' Cup

The 1996 CONCACAF Champions' Cup was the 32nd edition of the annual international club football competition held in the CONCACAF region, the CONCACAF Champions' Cup. It determined that year's club champion of association football in the CONCACAF region and was played from March 9, 1996, through July 20, 1997.
The teams were split into 2 zones. The North/Central zone was split into 3 groups, qualifying each winner to the final tournament. The winner of the Caribbean zone, earned a place in a playoff against the U.S. representative for a spot in the final tournament. All qualifying matches in the tournament were played under the home/away match system, while the final tournament was played in Guatemala City.
That final stage composed of four teams which played each other in a single round-robin tournament. Mexican team Cruz Azul won their fourth CONCACAF trophy, after finishing 1st. in the final table with 7 points over 3 matches played.

North/Central American Zone

Group 1

First Round
Cruz Azul on bye to the Third round.Alajuelense won 6–3 on aggregated.
----Árabe Unido won 2–1 on aggregated.
----Victoria won 7–3 on aggregated.
Second Round
Alajuelense on bye to the Fourth round.Victoria won 3–2 on aggregated.
Third RoundCruz Azul won 2–1 on aggregated.
Fourth Round

''Cruz Azul won 5–2 on aggregated.''

Group 2

First RoundNecaxa on bye to second round.
Second Round

''Necaxa advanced to the CONCACAF Final Group stage.''

Group 3

Final group stage

Final stage was played in Guatemala City, Guatemala. July 15–20, 1997.

Matches

All the games played in Guatemala City:
----
----
----
----
----
----