Why do the projections of a von Neumann algebra not always constitute a Boolean algebra?

A Boolean algebra must also satisfy the distributive law $p\vee(q\wedge r)=(p\vee q)\wedge (p\vee r)$. This is not true for projections on a Hilbert space in general. For instance, imagine that $\mathscr{H}$ is $2$-dimensional and $p,q,$ and $r$ are three different rank $1$ projections. Then $p\vee q=p\vee r=1$ but $q\wedge r=0$ so $p\vee(q\wedge r)=p\neq 1=(p\vee q)\wedge(p\vee r)$.

(It is true for commuting projections, since if $p$ and $q$ commute then $p\wedge q=pq$ and $p\vee q=p+q-pq$, and then the distributive law above can be verified by a simple computation. As a result, the projections in a commutative von Neumann algebra form a complete Boolean algebra.)