As vector spaces, the Lie algebra $\mathfrak{so}(n)$ is isomorphic $\bigwedge\nolimits^2 \mathbb{R}^n$ with the isomorphism given on simple bivectors $$\Phi: \bigwedge\nolimits^2 \mathbb{R}^n \to \mathfrak{so}(n): u \wedge v \mapsto \langle u, \cdot\rangle v - \langle v, \cdot \rangle u$$ and extended by bilinearity, where $\langle \cdot, \cdot \rangle$ denotes the standard inner product on $\mathbb{R}^n$.
Does $\bigwedge\nolimits^2 \mathbb{R}^n$ a priori have any natural Lie algebra structure so that the above map is a Lie algebra isomorphism?
Trivially, we can use $\Phi$ to pullback the Lie bracket from $\mathfrak{so}(n)$ to $\bigwedge\nolimits^2 \mathbb{R}^n$, but this is a structure that is put by hand a posteriori.