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LB Kwon Alexander expected to sign $54M pact with 49ers

By Alex Butler
Former Tampa Bay Buccaneers linebacker Kwon Alexander (R) will become the highest-paid linebacker in the NFL, on an annual basis, when his deal with the San Francisco 49ers becomes official Wednesday. File Photo by David Tulis/UPI | <a href="/News_Photos/lp/365d9819fc6ae45941cabd3c9677aae5/" target="_blank">License Photo</a>
Former Tampa Bay Buccaneers linebacker Kwon Alexander (R) will become the highest-paid linebacker in the NFL, on an annual basis, when his deal with the San Francisco 49ers becomes official Wednesday. File Photo by David Tulis/UPI | License Photo

March 11 (UPI) -- Free agent Kwon Alexander is expected to sign a four-year contract with the San Francisco 49ers.

Sources told NFL Network that Alexander will sign a $54 million deal with the 49ers, including $27 million in guaranteed cash, at the start of the new league year Wednesday.

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Alexander, 24, had 45 tackles, two passes defensed, two forced fumbles and a sack in six starts in 2018 for the Tampa Bay Buccaneers. The fourth round pick in the 2015 NFL Draft made his first Pro Bowl in 2017, after piling up 97 tackles, four passes defensed, three interceptions and a forced fumble in 12 starts that season. Alexander led the NFL with 108 solo tackles during his 2016 campaign with Tampa Bay.

The linebacker sustained a torn ACL in 2018.

Alexander is expected to slide in next to Fred Warner and serve as a replacement for Reuben Foster. The $13.5 million average annual value of the contract tops that of Carolina Panthers linebacker Luke Kuechly, who cashes in for $12.36 million annually. The deal makes Alexander the highest-paid inside linebacker in the league.

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