If $M$ is a hyperfinite type $I\!I\!I_1$ factor, then (at least conjecturally), its group of outer automorphisms is a $K(\mathbb Z,3)$.
This is based on the following three properties of that von Neumann algebra:
• The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.
• The group of unitaries in $M$ is contractible.
• The automorphism group of $M$ is contractible (conjectural).
To see that $\operatorname{Out}(M)\cong K(\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: \begin{gather*} U(Z(M)) \to U(M) \to \operatorname{Inn}(M) \\ \operatorname{Inn}(M) \to \operatorname{Aut}(M) \to \operatorname{Out}(M). \end{gather*}
As a consequence, we also get that ${\operatorname B}{\operatorname{Out}(M)}\cong K(\mathbb Z,4)$.
I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $\operatorname{Out}(M)$ as the automorphism group of a naturally occurring mathemtical object.
