Rings with isomorphic center not necessarily Morita equivalent a friend asked me this question; after a bit of searching, I'm still unable to answer it. 

Question
  Are there two rings with isomorphic centers which are not Morita equivalent?

Our first method of attack was to use that Morita equivalent rings have isomorphic $K_0,K_1,$ and $K_2$. However, Bass proved in '64 that f.g. projective modules over the monoid ring (over a PID) on a free monoid are free, so $K_0$ doesn't distinguish $F\langle x,y\rangle$ from $F$. I know that $K_1(F)$ is the units of $F$, but I'm having a hard time finding any computations of the $K$-theory of noncommutative rings like $F\langle x,y\rangle$.
It would be awesome if there was a counter-example proved using low degree $K$-theory, but any counter-example would work (assuming one exists).
 A: The $\mathbb R-\mathbb H$ example is very accessible, but I thought I would add a few more that come to mind where the differences in the rings are more striking.
For example, the center of the upper triangular matrices $U_n(\mathbb F)$ for a field $\mathbb F$ and $n >1$ consists of the constant diagonal matrices (just like in the full matrix ring), so its center is isomorphic to $\mathbb F$. But the ring of upper triangular matrices has a nontrivial Jacobson radical, unlike that of $\mathbb F$. Since having Jacobson radical zero is a Morita invariant property, these two rings can't be Morita equivalent.
For another example, consider the endomorphism ring $E$ of a countably infinite dimensional $\mathbb F$ vector space $V$. This can be thought of as the row finite matrices in the set of matrices with sides indexed by $\mathbb N$, and just like for finite square matrix rings, the center is the set of constant diagonal matrices, so the center is isomorphic to $\mathbb F$. This ring is Jacobson semisimple, but now it is not Noetherian on either side, nor is it even simple. For both these reasons, it can't be Morita equivalent to $\mathbb F$. In fact, you can even show that $E$ is the only ring Morita equivalent to $E$!
A: Let $k$ be a field with nontrivial Brauer group, and let $A$ be a central simple algebra over $k$ representing a nonzero class in its Brauer group. Then $k$ and $A$ are not Morita equivalent, but both have center $k$. For example you can take $k = \mathbb{R}, A = \mathbb{H}$. 
