The Ravens have agreed to trade cornerback Shaun Wade to the Patriots for a seventh-round pick in 2022 and a fifth-round pick in 2023 (Twitter links via NFL.com’s Ian Rapoport and Jeff Zrebiec of The Athletic). The seventh-rounder is originally a Texans pick, per ESPN.com’s Jamison Hensley (on Twitter). The deal will free up space in the Ravens’ secondary while adding a once highly-touted prospect to the Pats’ unit.
Wade was in the late first-round/early second-round conversation back in 2020. However, he chose to return to Ohio State for one more year. In hindsight, he probably regrets that decision. Wade had a rough season with the Buckeyes, including a disastrous performance in the national championship game. After DeVonta Smith went off for 12/215/3 at his expense, he tumbled all the way to the fifth-round of the 2021 draft.
It’s a low-risk move for the Patriots, who aren’t giving up much to acquire Wade and his modest contract. As the No. 160 overall pick, Wade is set to make less than $3.5MM over the next four years. This season, he’ll make just $660K in base salary with a $739K cap hit.
Wade will provide depth for the Patriots as they await word on Stephon Gilmore‘s status. He remains on the PUP list while also pushing for a better contract.
Photo courtesy of USA Today Sports Images.