Let $A$ be a commutative algebra, not necessarily unital, over a field $k$ (of characteristic not equal to $2$, or even equal to $0$, if it helps). A **second-order formal deformation** of $A$ is a $k[h]/h^3$-bilinear associative product $\star$ on $A[h]/h^3$ such that quotienting by $h$, we obtain the original product on $A$. Writing such a product as

$$a \star b = ab + h m_1(a, b) + h^2 m_2(a, b), a, b \in A$$

it's not hard to verify that $\{ a, b \} = m_1(a, b) - m_1(b, a)$ is a **Poisson bracket** on $A$, that is, a Lie bracket satisfying the Leibniz rule $\{ a, bc \} = \{ a, b \} c + b \{ a, c \}$. Given a nonzero Poisson bracket on $A$, it is interesting to ask whether we can find a formal deformation (replace $k[h]/h^3$ with $k[[h]]$) which gives rise to it as above ("deformation quantization").

But of course we can't ask this question until we have a nonzero Poisson bracket in the first place. So:

Which commutative algebras admit a nonzero Poisson bracket?

If there is no reasonable description in general feel free to restrict to the finitely-generated case or smooth functions on manifolds etc.

What I know: any polynomial algebra in $2$ or more variables admits a nonzero Poisson bracket (take the symmetric algebra on a nonabelian Lie algebra). Any nonzero Poisson bracket gives a nonzero element of the alternating part of the second Hochschild cohomology $H^2(A, A)$, so if this group is trivial then no such brackets exist. I doubt this implication can be reversed in general, but I don't know a counterexample. If you do, I have a math.SE question you should answer!

itsmotivation, so I'll leave the comment here. You ask: does the antisymmetrization of a Hochschild 2-cocycle satisfy Jacobi? The answer is NO. Recall that a bivector field is certainly a 2-cocycle, and the Hamiltonian flows for a Poisson bivector field foliate the space into (even-dimensional) symplectic leaves. But there are non-integrable plane distributions in $\mathbb R^3$, the standard one being $\ker(dz+xdy)$. This distribution is generated by Hamiltonian vector fields for the bivector (continued) $\endgroup$ – Theo Johnson-Freyd Aug 4 '11 at 9:05