As Qiaochu points out, $\mathbb{C}\mathbb{P}^2$ is a simply-connected elliptic space (its only non-zero rational homotopy groups being $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$ and $\pi_5(\mathbb{C}\mathbb{P}^2) \otimes \mathbb{Q} = \mathbb{Q}$) that is not homotopy equivalent to a suspension, since it has non-trivial cup product. It is also not rationally homotopy equivalent to a loop space. Indeed, a loop space has the rational homotopy type of a product of Eilenberg-MacLane spaces $K(\mathbb{Q}, k)$. Since $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$, there is would be a factor of $K(\mathbb{Q}, 2)$ in the rational homotopy type of $\mathbb{C}\mathbb{P}^2$, but $K(\mathbb{Q}, 2)$ has infinite cohomology.

As Qiaochu points out, $\mathbb{C}\mathbb{P}^2$ is a simply-connected elliptic space (its only non-zero rational homotopy groups being $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$ and $\pi_5(\mathbb{C}\mathbb{P}^2) \otimes \mathbb{Q} = \mathbb{Q}$) that is not homotopy equivalent to a suspension, since it has non-trivial cup product. It is also not rationally homotopy equivalent to a loop space. Indeed, a loop space has the rational homotopy type of a product of Eilenberg-MacLane spaces $K(\mathbb{Q}, k)$. Since $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$, there would be a factor of $K(\mathbb{Q}, 2)$ in the rational homotopy type of $\mathbb{C}\mathbb{P}^2$, but $K(\mathbb{Q}, 2)$ has infinite cohomology.