Alexander spent the first four seasons of his career with the Tampa Bay Buccaneers.

By Kaelen Jones
March 11, 2019

The 49ers and linebacker Kwon Alexander have reportedly agreed to a four-year, $54 million deal, according to NFL Network's Ian Rapoport.

Alexander, 24, is entering his fifth season in the league. He spent each of the first four campaigns with the Buccaneers. In 2017, he was named to his first career Pro Bowl.

Last season, Alexander appeared in only six games, racking up 45 total tackles, six tackles for loss and one sack. He suffered a season-ending torn ACL against the Browns in Week 7.

The former fourth-round pick led the league in solo tackles with 108 in 2016.

The MMQB's Conor Orr gives 49ers' signing an B- grade on our 2019 NFL Free Agency Grades 2019 that examines the best and worst moves by teams.

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