| Year: | 1991 |
| Dates: | 8 September – 24 September |
| Num Teams: | 3 |
| Associations: | 3 |
| Champion Other: |  Puebla |
| Count: | 1 |
| Second Other: |  Police |
| Matches: | 4 |
| Goals: | 7 |
| Prevseason: | 1990 |
| Nextseason: | 1992 |
The 1991 CONCACAF Champions' Cup was the 27th edition of the annual international club football competition held in the CONCACAF region (North America, Central America and the Caribbean), the CONCACAF Champions' Cup. It determined that year's club champion of association football in the CONCACAF region and was played from 7 April till 24 September 1991.
The teams were split in three zones (North, Central and Caribbean), each one qualifying the winner to the final tournament, where the winners of the Caribbean and Central zones played a semi-final to decide who was going to play against the Northern champion in the final. All the matches in the tournament were played under the home/away match system.
Mexican club Puebla beat Trinidarian Police 4–2 on aggregate.[1] Therefore, Puebla won their first CONCACAF championship, which was also their first international title.[2] [3]
Brooklyn Italians and Pembroke Hamilton Zebras withdrew. Puebla and Universidad de Guadalajara automatically advanced to third round.
Puebla advances to the CONCACAF Final Series.
Torneo Centroamericano de Concacaf 1991------------------------
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Alajuelense bye.
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Puebla advance to final (bye)----