The San Francisco 49ers have come to an agreement with Tampa Bay Buccaneers linebacker Kwon Alexander. The four-year deal will pay him up to $54 million, according to Ian Rapoport of NFL Network.


News of the signing was first reported by Pro Football Talk.


Adam Schefter of ESPN reports that the deal includes $27 million in guarantees.


Rick Stroud of the Tampa Bay Times was the first to report the 49ers' interest in Alexander. While deals can be agreed upon, nothing can become official until Wednesday at 1 p.m. PT.

The Buccaneers made Alexander a fourth-round selection out of LSU in 2015. He registered 45 combined tackles, a sack, two passes defensed, and two forced fumbles through six starts with the Buccaneers in 2018, which ended early due to a torn ACL. The Buccaneers placed Alexander on injured reserve on October 22, 2018.

Alexander, 24, has recorded 380 combined tackles, seven sacks, 22 passes defensed, six interceptions, a touchdown, and six forced fumbles through his four NFL seasons. His best statistical season came in 2016 when he registered 145 combined tackles, three sacks, seven passes defensed, an interception, a touchdown, and a forced fumble.

Alexander was selected to the Pro Bowl in 2017.

Career Statistics
Season Team G GS Comb Total Ast Sck SFTY PDef Int
2018 Buccaneers 6 6 45 34 11 1.0 0 2 0
2017 Buccaneers 12 12 97 70 27 0.0 0 4 3
2016 Buccaneers 16 16 145 108 37 3.0 0 7 1
2015 Buccaneers 12 12 93 59 34 3.0 0 9 2
Total 380 271 109 7.0 0 22 6



Related News