Quinnen Williams Wins AFC Defensive Player of the Week
Former Alabama defensive tackle and current New York Jets player, Quinnen Williams, was named the AFC Defensive Player of the Week for Week 6.
In his matchup against two-time reigning NFL MVP Aaron Rodgers and the Green Bay Packers, Williams recorded five tackles, three QB hits, two sacks, a forced fumble, and a blocked field goal. The two sacks on Sunday were a career high in a game.
The Jets beat the Packers at Lambeau Field 27-10 on Sunday, winning their third straight game and improving to a surprising 4-2 record on the season. The win puts New York among the top four teams in the AFC record wise, but still behind their AFC East rivals, the Buffalo Bills, who leads the AFC with a 5-1 record.
The former No. 3 overall pick currently leads the Jets this season in sacks (five) and QB hits (11). His five sacks tie him in eleventh place for the most in the league, and only puts him a sack and a half out of first place. The five sacks this season also only puts him two sacks behind tying his career high in an NFL season, seven, which he recorded in 2020.
Williams and the Jets play on the road against the Denver Broncos on Sunday, October 23, at 3:05 p.m. CT. The game will be aired on CBS in certain locations.
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