Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

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

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the notation $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

Question: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?
Else, what are counterexamples?

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

Edit (August 3, 2014):
After the answer of Nik, giving the counterexample $(\mathcal{A} \otimes I_2)' \cap (\mathcal{B} \otimes I_2)' \simeq M_2(\mathbb{C})$, for $\mathcal{A}$, $\mathcal{A'}$, $\mathcal{B}$, $\mathcal{B'}$ ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$, it appears that there are also counterexamples with hyperfinite infinite dimensional intersection, by putting a state on $(\mathcal{A} \otimes I_2)^{\otimes \infty}$ and on $(\mathcal{B} \otimes I_2)^{\otimes \infty}$ generating ${\rm II}_1$ factors $\mathcal{M}=\overline{(\mathcal{A} \otimes I_2)^{\otimes \infty}}$ and $\mathcal{N}=\overline{(\mathcal{B} \otimes I_2)^{\otimes \infty}}$ such that $\mathcal{M'}=\overline{(\mathcal{A'} \otimes M_2(\mathbb{C}))^{\otimes \infty}}$ and $\mathcal{N'}=\overline{(\mathcal{B'} \otimes M_2(\mathbb{C}))^{\otimes \infty}}$ are ${\rm II}_1$ factors, $\mathcal{M} \cap \mathcal{N} = \mathbb{C} $ and $\mathcal{M'} \cap \mathcal{N'} \simeq \overline{M_2(\mathbb{C})^{\otimes \infty}} \simeq R$ the hyperfinite ${\rm II}_1$-factor.
For more details about infinite tensor product of von Neumann algebras, see this answer of Nik.

Generalization (posted here):
Let $\mathcal{A}_1 \dots \mathcal{A}_n \subset B(H)$ be ${\rm II}_1$-factors such that $\forall i \, \, \mathcal{A}_i'$ is also a ${\rm II}_1$-factor and $\bigcap_i \mathcal{A}_i = \mathbb{C}$.
Is it true that $\bigcap_i \mathcal{A}'_i$ is hyperfinite? Else, what are counterexamples?

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

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

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the notation $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

Question: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?
Else, what are counterexamples?

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

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be II$_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also II$_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

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

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the notation $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

Question: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?
Else, what are counterexamples?

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

Generalization
Let $\mathcal{A}_1 \dots \mathcal{A}_n \subset B(H)$ be II$_1$-factors such that $\forall i \, \, \mathcal{A}_i'$ is also a II$_1$-factor and $\bigcap_i \mathcal{A}_i = \mathbb{C}$.
Is it true that $\bigcap_i \mathcal{A}'_i$ is hyperfinite? Else, what are counterexamples?

As advised by Nik Weaver, I will spend a couple of days working on it myself before opening a new post.

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

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

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the notation $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

Question: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?
Else, what are counterexamples?

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

Edit (August 3, 2014):
After the answer of Nik, giving the counterexample $(\mathcal{A} \otimes I_2)' \cap (\mathcal{B} \otimes I_2)' \simeq M_2(\mathbb{C})$, for $\mathcal{A}$, $\mathcal{A'}$, $\mathcal{B}$, $\mathcal{B'}$ ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$, it appears that there are also counterexamples with hyperfinite infinite dimensional intersection, by putting a state on $(\mathcal{A} \otimes I_2)^{\otimes \infty}$ and on $(\mathcal{B} \otimes I_2)^{\otimes \infty}$ generating ${\rm II}_1$ factors $\mathcal{M}=\overline{(\mathcal{A} \otimes I_2)^{\otimes \infty}}$ and $\mathcal{N}=\overline{(\mathcal{B} \otimes I_2)^{\otimes \infty}}$ such that $\mathcal{M'}=\overline{(\mathcal{A'} \otimes M_2(\mathbb{C}))^{\otimes \infty}}$ and $\mathcal{N'}=\overline{(\mathcal{B'} \otimes M_2(\mathbb{C}))^{\otimes \infty}}$ are ${\rm II}_1$ factors, $\mathcal{M} \cap \mathcal{N} = \mathbb{C} $ and $\mathcal{M'} \cap \mathcal{N'} \simeq \overline{M_2(\mathbb{C})^{\otimes \infty}} \simeq R$ the hyperfinite ${\rm II}_1$-factor.
For more details about infinite tensor product of von Neumann algebras, see this answer of Nik.

Generalization (posted here):
Let $\mathcal{A}_1 \dots \mathcal{A}_n \subset B(H)$ be ${\rm II}_1$-factors such that $\forall i \, \, \mathcal{A}_i'$ is also a ${\rm II}_1$-factor and $\bigcap_i \mathcal{A}_i = \mathbb{C}$.
Is it true that $\bigcap_i \mathcal{A}'_i$ is hyperfinite? Else, what are counterexamples?

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be II$_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also II$_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

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

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the notation $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

Question: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?
Else, what are counterexamples?

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

Generalization
Let $\mathcal{A}_1 \dots \mathcal{A}_n \subset B(H)$ be II$_1$-factors such that $\forall i \, \, \mathcal{A}_i'$ is also a II$_1$-factor and $\bigcap_i \mathcal{A}_i = \mathbb{C}$.
Is it true that $\bigcap_i \mathcal{A}'_i$ is finite dimensional? Else, what are counterexamples?

As advised by Nik Weaver, I will spend a couple of days working on it myself before opening a new post.

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be II$_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also II$_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

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

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the notation $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

Question: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?
Else, what are counterexamples?

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

Generalization
Let $\mathcal{A}_1 \dots \mathcal{A}_n \subset B(H)$ be II$_1$-factors such that $\forall i \, \, \mathcal{A}_i'$ is also a II$_1$-factor and $\bigcap_i \mathcal{A}_i = \mathbb{C}$.
Is it true that $\bigcap_i \mathcal{A}'_i$ is hyperfinite? Else, what are counterexamples?

As advised by Nik Weaver, I will spend a couple of days working on it myself before opening a new post.