The worst performance of his career! Shota Tominaga allowed 7 runs in 3 innings and gave up three home runs, leaving the game early.

The Chicago Cubs faced off against the White Sox today in a Chicago rivalry match, with starting pitcher Shota Tominaga only going 3+ innings. In three of those innings, the leadoff batter hit a home run. Ultimately, he pitched 3 innings, gave up 12 hits, and allowed 7 runs, marking his worst outing of the season. The Cubs lost the game 5-12.

In the first inning, Shota Tominaga was hit with a home run by the White Sox's leadoff hitter Chase Meidroth, putting him in a tough spot. He then allowed 4 more hits, resulting in 3 runs. In the second inning, Tominaga's struggles continued as he gave up another home run to leadoff batter Colson Montgomery, followed by consecutive hits that cost him 2 more runs. His pitch count exceeded 50, prompting the bullpen to start warming up.

In the third inning, although Tominaga allowed a double, he managed to get through unscathed. However, in the fourth inning, he gave up a home run to leadoff batter Austin Slater, tying his personal record for most home runs allowed in a game. Last season, during his rookie year, he pitched 3 innings against the Mets on June 21, giving up 11 hits, including 3 home runs, and allowing 10 runs before being taken out. After allowing another double, Tominaga was pulled early by the head coach.

This marks the first time this season that Tominaga has been pulled after just 3 innings, with 12 hits allowed and 7 earned runs, both setting season highs. He tied the season record for most home runs allowed in a game with three. The Cubs ended up losing 5-12, with Tominaga suffering his fourth loss of the season.