$\mathbb{R}(\mathbb{C})$ would require $\mathbb{R}$ to be a vector space over $\mathbb{C}$, meaning scalar multiplication by complex numbers on $\mathbb{R}$. But $i\cdot 1 = i \notin\mathbb{R}$, so $\mathbb{R}$ is not closed under scalar multiplication by $\mathbb{C}$. Therefore $\mathbb{R}(\mathbb{C})$ is NOT a vector space. $\mathbb{R}(\mathbb{R})$, $\mathbb{R}(\mathbb{Q})$, and $\mathbb{C}(\mathbb{Q})$ are valid vector spaces.