Does every Lie algebra come from commutator of some associative product operation? Suppose $\mathfrak{g}$ is an Lie algebra. Is it possible to define an associative product operation $\star$ on $\mathfrak{g}$ such  that $[A,B]=A\star B - B \star A$ ?  If it is not possible to do so in general, how do we know which Lie algebras come as commutator of some product operation ? (Are there any useful criteria ?)
I am aware (without proof) that every Lie algebra has an embedding $f$ into an associative algebra such that the Lie bracket $[A,B]$ corresponds to $f(A)f(B)-f(B)f(A)$,but the question that I ask above demands something stronger.
 A: The answer is no in general, and I suspect it's no for all simple Lie algebras.  Here's an argument that works for many simple Lie algebras including $\mathfrak{sl}_2$.  (Dollars to doughnuts there's a more direct elementary proof for $\mathfrak{sl}_2$.)
Suppose that $\mathfrak{g}$ is a non-commutative Lie algebra which arises as the commutators for some algebra.  Then there's a map of $\mathfrak{g}$-modules $\mathfrak{g} \otimes \mathfrak{g} \rightarrow \mathfrak{g}$ given by multiplication $x \otimes y \mapsto xy$ and another map given by opposite multiplication $x \otimes y \mapsto yx$.  Since $\mathfrak{g}$ is non-commutative these two maps are linearly independent, so $\dim \mathrm{Hom}_{\mathfrak{g}\text{-mod}}(\mathfrak{g} \otimes \mathfrak{g}, \mathfrak{g}) \geq 2$.  So any non-commutative Lie algebra with the property that $\dim \mathrm{Hom}_{\mathfrak{g}\text{-mod}} (\mathfrak{g} \otimes \mathfrak{g}, \mathfrak{g}) = 1$ cannot come from an associative algebra.  Examples include $\mathfrak{sl}_2$ and all simple Lie algebras outside the A series.
A: The answer is indeed no for all semisimple Lie algebras. Such an associative product would determine a left-symmetric and a right-symmetric structure on the Lie algebra. It is known that this cannot exist for semisimple Lie algebras over a field of characteristic zero because of Whitehead's lemma. In fact, there is no left-symmetric structure. If $L$ denotes the Lie algebra, the left-multiplication defines an $L$-module $M_L$ which must satisfy $H^1(L,M_L)=0$. Since the identity $id$ is a $1$-cocylce of $L$ with values in $M$, it must be a coboundary, i.e., there exists an element $e$ such that the right-multiplication $R(e)$ by $e$ is the identity. Because of $[L,L]=L$ however, all right multiplications have zero trace. Hence the identity has zero trace - a contradiction.
