Stephen Drew To Make 2014 Debut, Bat Eighth For Red Sox vs. Indians

Stephen DrewNearly two weeks after signing a one-year, $10 million contract with the Boston Red Sox, shortstop Stephen Drew will make his 2014 debut Monday night against the Cleveland Indians.

After officially signing with the Sox on May 21, Drew played six games in the minor leagues — three with the Single-A Greenville Drive and three with the Triple-A Pawtucket Red Sox — before meeting up with the team in Cleveland on Monday.

Drew will bat eighth and play shortstop, meaning that Xander Bogaerts will move to third base to make his first start of the season at that position.

The white-hot Brock Holt will remain in the lineup, playing first base and batting in the leadoff spot. Holt went 4-for-4 with four doubles in the Sox’s 4-0 win over the Tampa Bay Rays on Sunday, their seventh in a row.

A.J. Pierzynski will return to the lineup, batting fifth and resuming catching duties for starting pitcher John Lackey, who leads the team with six wins.

Drew and the Boston lineup will face former Red Sox pitcher Justin Masterson, who is 2-4 on the season with a 5.21 ERA.

Check out both full lineups below.

Boston Red Sox (27-29)
Brock Holt, 1B
Xander Bogaerts, 3B
Dustin Pedroia, 2B
David Ortiz, DH
A.J. Pierzynski, C
Jonny Gomes, LF
Grady Sizemore, RF
Stephen Drew, SS
Jackie Bradley Jr., CF

John Lackey, RHP (6-3, 3.27)

Cleveland Indians, (27-30)
Michael Bourn, CF
Asdrubal Cabrera, SS
Michael Brantley, LF
Jason Kipnis, 2B
Lonnie Chisenhall, 1B
Jason Giambi, DH
Yan Gomes, C
David Murphy, RF
Mike Aviles, 3B

Justin Masterson, RHP (2-4, 5.21 ERA)

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