If $M$ is a <i>hyperfinite type $I\!I\!I_1$ factor</i>, 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:<br>
&bull; The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.<br>
&bull; The group of unitaries in $M$ is contractible.<br>
&bull; 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*}
<hr>

As a consequence, we also get that ${\operatorname B}{\operatorname{Out}(M)}\cong K(\mathbb Z,4)$.

I recommend my talk "[A <i>K</i>(&#8484;,4) in nature][1]"  (MSRI, April 2014), for an explanation of how to realize $\operatorname{Out}(M)$ as the automorphism group of a naturally occurring mathemtical object.

  [1]: http://www.msri.org/workshops/689/schedules/18246