Is there a condition for the nilradical to equal the derived subalgebra?

Question

Suppose $\mathfrak{g}$ is a finite dimensional Lie algebra over $\mathbb{C}$. Is there a condition for when $\mathfrak{r}=NR(\mathfrak{g})=[\mathfrak{g},\mathfrak{g}]$?

I realize that $[\mathfrak{g},\mathfrak{g}] \subset \mathfrak{r}$, so I suppose when I'm asking is for when $\mathfrak{r} \subset [\mathfrak{g},\mathfrak{g}]$.

Answer 1

As mentioned above, $[\mathfrak{g},\mathfrak{g}] \subset NR(\mathfrak{g})$ says that $\mathfrak{g}$ is solvable (over characteristic zero), and $[\mathfrak{g},\mathfrak{g}] = NR(\mathfrak{g})$ is a genuine condition for a solvable Lie algebra. The oscillator Lie algebra does satisfy this, see here, but the Heisenberg Lie algebra does not, of course (no nilpotent Lie algebra does). This is a start to find some sufficient conditions for having equality.