Let $\mathfrak{h}$ be a Lie algebra (not necessarily finite dimensional). Does there necessarily exist a Lie group $G$ such that for the Lie algebra corresponding to $G$, denoted $\mathfrak{g}$, we have $\mathfrak{g} \cong \mathfrak{h}$?
I thought of this earlier today and I think the answer is no. This is not based in any intuition, the reason I think this is due to the following statement on Wikipedia (paraphrased): "If a Lie algebra corresponds to a Lie group $G$, it is typically denoted $\mathfrak{g}$." This seems to imply that given a Lie algebra, we don't necessarily have an associated Lie group.
There is the example of Van Est and Korthagen from $1964$, giving an infinite-dimensional (Banach) Lie algebra $\mathfrak{g}$, so that there is no Banach Lie group $G$ with Lie algebra $\mathfrak{g}$. Hence Lie's third theorem fails in the infinite-dimensional case, i.e., more precisely, it depends on how we define the concept of infinite-dimensional Lie groups. For a discussion on this, see the article A remark on non-enlargable Lie algebras by H. Omari.