Shaun Wade, a rookie cornerback, was traded by the Baltimore Ravens to the New England Patriots on Thursday, an unexpected move.
Wade was released by the franchise that had picked him before he could even put on an NFL uniform, which left many people perplexed.
Why would the Ravens lose on a man they chose 160th overall in the NFL Draft of 2021 less than four months ago? especially a player with Wade’s lofty potential.
Wade formerly held a reputation for being one of the top young cornerbacks in college football and a likely first-round draft pick. Sadly, a terrible junior season destroyed his draft prospects in 2020.
Wade was assigned to the outside cornerback position by Ohio State, a decision that backfired for both the organization and the player.
In 2020, Wade was the first FBS player to allow 30 or more completions, 500 or more receiving yards, and six or more touchdown catches.
Wade may return to his pre-2020 form, though, if he plays his native position in the NFL, according to the Ravens. Evidently, Baltimore didn’t experience the fulfillment of that idea.
Why did Shaun Wade get traded?
The Ravens decided to trade Wade because he was outplayed by his competition. And he wasn’t just outplayed by one player, he was outplayed by a number of other young defensive backs.
Even if he made the team, Wade probably wouldn’t have played much even if he did. In addition, the Ravens didn’t want to keep him on the team if they could avoid it because better players had been passed over.
Chris Westry, a recent free agency signing, made every effort to make the team this summer. Despite being shorter than Wade by an inch, the 6-foot-4 cornerback is another high-ceiling prospect.