Boston Red Sox catcher Danny Jansen will do the unthinkable Monday night and become the first player in MLB history to play for both teams in one game.
Red Sox catcher Danny Jansen looks on prior to the game between the Baltimore Orioles and the Boston Red Sox at Oriole Park at Camden Yards. / Reggie Hildred-USA TODAY Sports
Yes, you read that opening sentence correctly. How is this possible? Let's start at the beginning.
The Red Sox and Toronto Blue Jays were wrapping up a three-game series at Fenway Park on June 26 when play was suspended in the second inning due to rain. At the time, Jansen was playing for Toronto, the team that drafted him in 2013—starting behind the plate and batting seventh. He was actually in the batter's box when the game was officially suspended.
The game was scheduled to be made up Aug. 26, and will resume right where it left off June 26 as part of a split doubleheader between the Red Sox and Blue Jays.
However, Jansen won't be digging into the batter's box again for the Blue Jays. Toronto traded him to Boston in exchange for three prospects ahead of the trade deadline, giving him a chance to make some pretty wild history.
Red Sox manager Alex Cora confirmed Friday that Jansen will be in Boston's lineup when the game is rescheduled, meaning he'll be behind the plate as the catcher while the Blue Jays assign a pinch-hitter to take his place in the batter's box.
Woah. Let that sink in a moment or two.
Jansen has been playing well since arriving in Boston, batting .257/.366/.429 with two homers in 13 games.
