Mexico placed in Group E as CONCACAF announce 2019 U-17 CONCACAF Championship

On Tuesday, CONCACAF announced a new format for the 2019 U-17 CONCACAF Championship, as well as conducted the draw that seeded Mexico in Group E of the competition.

Like the U-20 Championship, the tournament will once again involve all 35 teams that entered, but the tournament itself will be divided in two rounds.

The lowest ranked 19 teams will be divided in four groups (A to D), with Groups A and D playing March 16 to 24 and Group B and C playing April 1 to 9. The group winners will qualify to the knockout stages. The other 16 higher ranked teams will play in the Group Stages in Groups E to H. The top three teams of each group (12 in total) will qualify to the knockout stages, along with the group winners of the first round. They will then play the Round of 16, Quarterfinals and Semifinals, where the four semifinalists will qualify to the U-17 World Cup.

The Group Stage and Knockout rounds will be played from May 1 to the 16 and all matches will take place at IMG Academy in Bradenton, Florida.

As the current U-17 CONCACAF champion, Mexico was seeded for the second round in Group E. Mexico will be in the group along with Jamaica, Trinidad and Tobago and Bermuda. While it’s a tough group for CONCACAF, the fact that the top three teams all qualify should make it that, barring a massive surprise, Mexico should qualify to the knockout stages. Although Mexico has had trouble with Caribbean teams, it has been mostly in higher age categories and not in the U-17, where Mexico had until recently been a worldwide powerhouse.

Mexico will debut against Jamaica, who should be the toughest team in the group. They will then face Bermuda, the weakest team who should be favored to not make it out of the group, and then close with their match against Trinidad and Tobago.

The U-17 World Cup will be played in Peru—the site of Mexico’s first U-17 World Cup title in 2005—which will be played from October 5-27.