If a Lie algebra L decomposes as a direct sum of its derived subalgebra and its center, is L reductive?

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A Lie algebra $L$ is said to be reductive if for any ideal $\mathfrak{a}$ of $L$, there is an ideal $\mathfrak{b}$ of $L$ such that $L=\mathfrak{a}\oplus\mathfrak{b}$.

It is known that a reductive Lie algebra decomposes as $L = L'\oplus Z(L)$, where $L'$ is the derived subalgebra of $L$ and $Z(L)$ is the center of $L$. Is the converse true?

The answer is "yes" if $L'$ is semisimple; can we deduce that $L'$ is semisimple just from the decomposition $L = L'\oplus Z(L)$?