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Mexico gets drawn in Group B of 2018 CONCACAF U-20 Championship

Mexico will share Group B with Jamaica, Aruba, Nicaragua, Saint Martin and Grenada.

Japan v Mexico: U23 Friendly Photo by Carlos Rodrigues/Getty Images

Mexico was seeded into Group B in the 2018 CONCACAF U-20 Championship, which will be held in Bradenton, Florida in the month of November. Unlike past editions, there was no qualification process this time, meaning all CONCACAF teams that entered were part of the tournament. This means Mexico will take part in a six-team group, different from past editions that only had 4. The other teams drawn in Mexico’s group were Jamaica, Aruba, Nicaragua, Grenada and Saint Martin. The winners of the group qualify to the second round in which they will be placed in Groups of 3. The top two teams of those groups will qualify to the U-20 World cup and the Group winners will play each other in the final.

Mexico’s group looks to be fairly easy, with the only true danger being Jamaica. Since only group winners get to qualify to the second round, it looks to be a battle with them for that sole ticket. Mexico will open the tournament facing Nicaragua, which looks to be the second toughest team after Jamaica, on November 2. Two days later, they will then face Saint Martin, who looks to be the weakest rival on paper. On November 6, they will face their toughest match against Jamaica. On November 8, they will play against Grenada before finishing the first round on November 10 against Aruba.

Should Mexico qualify, they will then be placed in Group B of the second round. There they will face Group D winners ( Panama and Canada should be the favored to win that group) and Group E winners (most likely El Salvador and Guatemala). The matches in those groups would be played on November 15 and 19. Only four teams get to qualify to the U-20 World Cup to be held in Poland in 2019.