One of this offseason's top defensive free agents has been taken off the market. According to several reports, the San Francisco 49ers are signing former Buccaneers linebacker Kwon Alexander to a four-year deal worth $54 million.
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
The average annual value of Alexander's contract ($13.5 million) is the highest of any inside linebacker in the league, according to Spotrac, surpassing Luke Kuechly's by over $1 million. The $27 million guarantee for Alexander is similar what Kuechly was guaranteed on his deal.
Alexander does not turn 25 years old until later this offseason, and during his first three years in the league he played very well for the Bucs, even making the Pro Bowl in 2017 despite playing in only 12 games. Alexander averaged 111 tackles (8 for loss), two sacks, and seven passes defensed per year during those three seasons, displaying high-level talent. He suffered a torn ACL early in the 2018 season, though, and sat out for the rest of the year.
The contract for Alexander continues a trend of the 49ers getting out there with big, surprising deals early in free agency, which they did with fullback Kyle Juszcyk two years ago and running back Jerick McKinnon last year. None of the three players plays at what is considered to be a premium position in the modern NFL, so it's certainly an interesting trend that is developing here.
The 49ers are expected to be major players on the open market and could still add someone like star Seahawks safety Earl Thomas, who is reportedly seeking a deal that makes him the highest-paid safety in the league. It remains to be seen whether the Niners or Thomas' supposed preferred team -- the Cowboys -- will match his price.
