Let $N \subset M'$ be an irreducible inclusion of subfactors, meaning that $N \neq M'$ but the relative commutant is trivial, i.e., $N' \cap M' = \mathbb{C}\cdot I$. There are lots of examples of such things. Then the commutant of the von Neumann algebra generated by $M \cup N$ is contained in both $M'$ (since it contains $M$) and $N'$ (since it contains $N$), so it must be trivial. Therefore this von Neumann algebra must be $B(H)$.

Let $N \subset M'$ be an irreducible inclusion of subfactors, meaning that $N \neq M'$ but the relative commutant is trivial, i.e., $N' \cap M' = \mathbb{C}\cdot I$. There are lots of examples of such things (Google "example of irreducible subfactor"). Then the commutant of the von Neumann algebra generated by $M \cup N$ is contained in both $M'$ (since it contains $M$) and $N'$ (since it contains $N$), so it must be trivial. Therefore this von Neumann algebra must be $B(H)$.