Ian Rapoport tweeted what many Bucs fans didn’t want to hear on Monday.
The Bucs defensive leader will have a new home in San Francisco.
Source: The #49ers are signing LB Kwon Alexander to a 4-year deal worth $54M.
— Ian Rapoport (@RapSheet) March 11, 2019
Alexander – who is coming off of ACL surgery – was forced to miss most of 2018, but accumulated 45 total tackles, with six going for loss, two forced fumbles, two passes defended and one sack in the six games he played last season.
The deal will reportedly pay him an average of $13.5 million over four years.
Since being drafted by the Bucs in the fourth round of the 2015 draft, Alexander has been a staple on the field and in the locker room for Tampa Bay, leading the league in solo tackles in 2016 and being named to the Pro Bowl in 2017.

Bucs LB Kwon Alexander – Photo by: Mark Cook/PR
In his four years in Tampa, Alexander has totaled 380 total tackles, seven sacks, six interceptions, six forced fumbles and 31 tackles for loss.
The Buccaneers had tried to negotiate with Alexander recently but when the terms his camp wanted were disclosed, Tampa Bay backed out to let Alexander test the free agent waters. It didn’t take long as former Bucs star safety and current 49ers GM John Lynch have reportedly negotiated a deal with Alexander and his agent Drew Rosenhaus to bring the former LSU star to the other Bay area.
Linebacker now moves up even higher on the Bucs needs list. With Alexander gone and Kendell Beckwith’s long-term future still unknown after missing all of 2018 after a care accident the team will be on the lookout in free agency and the draft to find Alexander’s replacement. Another linebacker, Jack Cichy was also injured last season, missing 10 games after tearing his ACL in the same game as Alexander against the Cleveland Browns.