Is a von Neumann algebra just a C*-algebra which is generated by its projections?
von Neumann algebras have the nice property that they are generated by their projections (the elements satisfying $e = e^{\ast} = e^2$) in the sense that they are the norm closure of the subspace generated by projections. This is a sensible property to require from the perspective of "noncommutative measure theory" where one thinks of von Neumann algebras as generalizations of algebras of the form $L^{\infty}(X)$ ($X$ a $\sigma$-finite measure space); here the projections are the indicator functions of measurable subsets of $X$ (modulo sets of measure zero) and the subspace generated by the projections are the simple functions.
Does this property characterize von Neumann algebras among $C^{\ast}$-algebras?
Solution 1:
No. Consider the commutative C* algebra $c$ of convergent sequences of complex numbers (with pointwise multiplication), corresponding to $C(X)$ where $X$ consists of a convergent sequence and its limit. It is also the norm closure of the span of its projections (which are the sequences of $0$'s and $1$'s that are eventually $0$ or $1$), but it is not a von Neumann algebra: the weak closure is $\ell^\infty$.
More generally, IIRC, $C(X)$ is the norm closure of the span of its projections iff $X$ is totally disconnected, but in order for it to be a von Neumann algebra $X$ must be extremally disconnected.
Solution 2:
Approximately finite dimensional C*-algebras are generated (as Banach spaces) by their projections, but they are not von Neumann algebras except in the finite dimensional case. $c$ is an example of this, and the algebra of compact operators on a separable Hilbert space is another.
Blackadar's survey article "Projections in C*-algebras" contains a section called "Existence of projection properties" (starting on page 138), where this property is listed as one among many axioms that are compared ("LP"). It is mentioned there that in the case of $C(X)$ with $X$ compact Hausdorff, most of the axioms are equivalent to total disconnectedness of $X$, which is also equivalent to $C(X)$ being approximately finite dimensional if $X$ is metrizable.
As Yemon mentioned, work of Kaplansky on AW*-algebras is relevant. Piecing together a bit from Blackadar's exposition, citing work of Kaplansky and Kadison (and possibly Sakai, see 6.3.1 and 6.3.3), the following three properties (all at once, not separately) characterize a C*-algebra $A$ as a W*-algebra:
Every maximal commutative C*-subalgebra of $A$ is generated by projections.
The projections of $A$ form a complete lattice.
$A$ has a separating family of normal states.
Adding to Robert Israel's last paragraph, in 6.3.4 it is stated that $C(X)$ is an AW* algebra if and only if $X$ is extremally disconnected (Stonean), and $C(X)$ is a W*-algebra if and only if $X$ is extremally disconnected with a separating family of normal measures (hyperstonean).
Solution 3:
Alfsen and Schultz in their book Geometry of state spaces of operator algebra's define a notion for a base norm space to be a spectral convex set, which turns out to be equivalent to its associated order unit space being the norm closure of its compression units (e.g. the projections in a C$^*$-algebra).
Not all spectral convex sets are state spaces of $C^*$-algebra's and as was noted in the previous answers, even if it is a $C^*$-algebra it does not have to be a von Neumann algebra.