Baltimore Ravens are reportedly moving from corner Shaun Wade, their fifth-round selection in this year’s draft. According to NFL Network’s Ian Rapoport, the Ravens are closing on a deal to trade Wade to AFC rivals New England Patriots.
The Ravens will receive a draft compensation that includes a seventh-rounder in 2022 and a fifth-round pick the year after.
The decision to trade Wade may come as a surprise to some as it’s rare that a team gives up on a drafted player before even seeing him in action. However, it appears that the Ohio State product struggled in playing as a slot corner, which is where Baltimore intended to use him the most. Having more reliable options available on the roster, the organization jumped on the chance to recoup their original investment while picking up an additional pick.
The Patriots, on the other hand, could use help in the secondary, and Wade seems as good an option as any. They are possibly also betting that his previous high ceiling can lead to him becoming a solid NFL player for them.
Shaun Wade spent his first two seasons at Ohio State playing a nickel corner and excelling in the role. This made him a highly-rated prospect ahead of the 2020 NFL draft, where some projected him as a possible first-round pick. However, Wade decided to stay one more year at college to get more reps as an outside corner.
This proved to be a bad decision, as Wade didn’t have the best seasons in 2020 and had few games where he notably struggled in one-on-one coverage. This caused his draft stocks to significantly drop and allowed the Ravens to get him at No.160 overall.
