For any finite type $\mathbb{Q}$-algebra $A$, i.e. $A = \mathbb{Q}[\mathbf{x}]/I$ is it true, that the total ring of fractions $Q(\mathbb{R} \otimes_\mathbb{Q} A) = Q(\mathbb{R}[\mathbf{x}]/I\mathbb{R}[\mathbf{x}])$ is isomorphic to $\mathbb{R} \otimes_\mathbb{Q} Q(A)$?
This is in relation to this question: Integral closure/normalization under base change.
It is not true. For instance, take $A=\mathbb{Q}[x]$ so that $Q(\mathbb{R}\otimes_{\mathbb{Q}} A)=\mathbb{R}(x)$ and $\mathbb{R}\otimes_{\mathbb{Q}}Q(A)=\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{Q}(x)$. They are different because $\frac{1}{x-\pi}$ belong to the former but not to the latter. In fact, they are even not isomorphic as rings because the former is a field while the latter isn't ($x-\pi$ is not invertible).