Shaun Wade’s stint as a Baltimore Raven is officially over just a few months after it began. The former Ohio State standout and fifth-round pick of this year’s NFL draft was traded to the New England Patriots today.
Wade who was once considered a first-round lock, fell after a subpar 2020 season. He signed his rookie contract with the Ravens on May 13, a four-year deal worth nearly 3.8 million.
There seemed to be a logjam of defensive backs in Baltimore and New England needed some depth in the secondary. The two teams struck a deal sending Wade to Bean Town in exchange for a seventh-round pick in 2022 and a fifth-round pick in 2023.
Even though Wade has played well during the pre-season, there was some speculation that he would not make the 53 man roster in Baltimore. So the Ravens made a move to get something instead of potentially nothing, knowing Wade would most likely not make it through waivers unclaimed. With so many defensive backs, they simply could not keep everyone.
Compensation for the Shaun Wade trade. https://t.co/vl2OvAnDFU
— Ian Rapoport (@RapSheet) August 26, 2021
This is a low-risk, high-reward move by the Pats. When Wade was selected, many experts thought the Ravens got a steal that late in the draft. Although it has to be difficult to be moving on just a few weeks before the season kicks off, New England may now be the one getting a bargain. There is no doubt, the former Buckeye has the potential to be a star in the NFL.
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