Mets Trade Outfielder Jay Bruce To Cleveland Indians
NEW YORK (CBSNewYork/AP) -- The Mets finally traded Jay Bruce on Wednesday night, sending the outfielder to the Cleveland Indians in exchange for a pitching prospect.
The team confirmed the trade on Twitter. The Mets are receiving right-hander Ryder Ryan, who will be assigned to the Class A Columbia (S.C.) Fireflies. Cleveland also has agreed to pay the remaining $4.2 million of Bruce's $13 million salary this season, WFAN baseball insider Jon Heyman reported.
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According to Heyman, the Yankees were also involved in discussions for Bruce, but they did not want to pay his full salary.
Bruce is hitting .256 with 29 homers and 75 RBIs for the Mets, who at 50-61 are 11 games under .500 for the first time in three years. New York is 16½ games behind Washington in the NL East.
Bruce was acquired from Cincinnati at last year's trade deadline. He will become a free agent after the World Series.
The Mets tried to trade Bruce after picking up his option last offseason but were unable to find a partner. They again unsuccessfully attempted to deal him before last week's non-waiver trade deadline.
While Major League Baseball's trade deadline was July 31, a player can be dealt if he goes through waivers unclaimed.
Ryan was a 30th-round draft pick by the Indians in 2016 out of the University of North Carolina. In 33 relief appearances this season with the Class A Lake County (Ohio) Captains, he had a 3-4 record and 4.79 ERA.
Cleveland, seeking its first World Series title since 1948, lost Game 7 of last year's World Series to the Chicago Cubs and leads the AL Central.
(© Copyright 2017 CBS Broadcasting Inc. All Rights Reserved. The Associated Press contributed to this report.)
