Do two II$_1$-factors with trivial intersection generate $B(H)$?

Question

Let $H$ be an infinite dim. separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $A$, $B \subset B(H)$ be II$_1$-factors such that $A \cap B = \mathbb{C}I$.

Examples:
(1) Take $B = A'$ then $A \cap B = \mathbb{C}I$ by definition of a factor.
(2) Take $(A' \subset B)$ an irreducible subfactor, then $A \cap B = \mathbb{C}I$ by definition of irreducibility.

Obviously $\langle A' , B' \rangle = (\mathbb{C}I)' = B(H)$.

Question: Is it also true that $\langle A , B \rangle = B(H)$ ?
Else, what are counterexamples?

Remark: It's true for the examples (1) and (2).

Notation: $\langle S \rangle= (S \cup S^* \cup \mathbb{C}I) ''$

Answer 1

You can phrase the question as whether $A\cap B=\mathbb C$ implies $A'\cap B'=\mathbb C$.

Let $M$ be some II$_1$-factor in $B(H) $, and let $A=M\otimes1$, $B=1\otimes M$ in $B(H\otimes H) $. Then $$ A'=M'\otimes B(H),\ \ \ B'=B(H)\otimes M', $$ so $A'\cap B' =M'\otimes M'$.