Former Tampa Bay Buccaneers linebacker Kwon Alexander has agreed to a deal with the San Francisco 49ers according to ESPN’s Adam Schefter. The deal will be for four years worth $54 million, with $27 million guaranteed.
49ers intend to sign former Bucs LB Kwon Alexander to a 4-year, $54 million deal that includes $27 million guaranteed, per source.
— Adam Schefter (@AdamSchefter) March 11, 2019
San Francisco was in the market for a linebacker after releasing Reuben Foster after a domestic violence allegation midseason last year.
Alexander spent the first four seasons of his career with the Tampa Bay Buccaneers, reaching one Pro Bowl over that span. He has 271 total tackles and 31 tackles for loss in his career.