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2019 NFL Draft Order: Seahawks don’t get any compensatory picks, as expected

2018 NFL Draft Photo by Ronald Martinez/Getty Images

The NFL officially announced the compensatory picks for the 2019 NFL Draft, and as John P. Gilbert noted months ago, the Seattle Seahawks ain’t getting any. Many of the Seahawks’ 2018 free agents such as Luke Joeckel, Blair Walsh and Eddie Lacy went unsigned, and they made more free agent signings than had free agents go elsewhere, so that meant bad news for the 2019 draft.

Here’s the full list:


Round 3

Washington (No. 96 overall)

Patriots (No. 97)

Rams (No. 98)

Rams (No. 99)

Panthers (No. 100)

Patriots (No. 101)

Ravens (No. 102)

Round 4

Colts (No. 135)

Cowboys (No. 136)

Falcons (No. 137)

Eagles (No. 138)

Round 5

Giants (No. 171)

Falcons (No. 172)

Washington (No. 173)

Round 6

Patriots (No. 205)

Washington (No. 206)

Cardinals (No. 207)

Eagles (No. 208)

Vikings (No. 209)

Bengals (No. 210)

Bengals (No. 211)

49ers (No. 212)

Bengals (No. 213)

Chiefs (No. 214)

Round 7

Vikings (No. 247)

Cardinals (No. 248)

Cardinals (No. 249)

Vikings (No. 250)

Rams (No. 251)

Patriots (No. 252)

Washington (No. 253)


Needless to say, not John Schneider’s finest work that they put themselves in a position where they somehow didn’t get a single comp pick. Add in the picks they’ve lost due to trades made over the past couple of seasons — Duane Brown, Shalom Luani, and Brett Hundley — and the Seattle Seahawks draft order is as follows:

Round 1: 21st overall

Round 3: 84th overall

Round 4: 124th overall

Round 5: 159th overall

Four draft picks is the fewest in franchise history as things stand. If it wasn’t obvious enough that the Seahawks would be trading down in round one, this should seal the deal. And it’s also not out of the question that the Seahawks try to gain draft picks by trading players. Which ones? I don’t know, but there’s no way this sits well with the front office that their draft capital is barren right now.