#2 Seth Rollins
![Rollins has never faced Undertaker one-on-one.](https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=190 190w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=720 720w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=640 640w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=1045 1045w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=1200 1200w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=1460 1460w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg?w=1600 1600w, https://statico.sportskeeda.com/editor/2020/06/10dbe-15916985514871-800.jpg 1920w)
Now, we come to Superstars who are more realistic candidates to be among The Undertaker's final few opponents in WWE. Seth Rollins is currently a super-villain on RAW, leading his group of followers as they rise up the card. And while the group might need expanding, there might be some huge plans for the 'Monday Night Messiah' in the near future.
Once Rollins concludes his feud with Rey Mysterio, he could go in a number of different directions. One of them could possibly be to take The Undertaker out, along with his Disciples. It would be intriguing, to say the least, with Rollins looking to sacrifice Undertaker on his way to gaining even more power as the 'Messiah'.
This could even be a Boneyard Match, with a format and story similar to the one between 'Taker and AJ Styles. The 'Unholy Trinity' of the Deadman, the American Badass, and Mark Calaway himself would take out Rollins, Murphy, Austin Theory, and any other Superstar who joins the faction.
Hopefully, Seth Rollins does have a match with The Undertaker. His former Shield partners, Dean Ambrose and Roman Reigns have both faced the Deadman in singles action, while The Shield took on 'Taker and Team Hell No in a 6-man tag team match back in 2013. But we are yet to see 'Taker and Rollins go at it one-on-one.