We wish to show that the center of $\mathbb H$ is just $\mathbb R$. Let $z=a+bI+cJ+dK$. If $z$ were in the center, we could subtract off the real part to get something else in the center, so we might as well assume that $a=0$ (just to simplify computations).
We can compute the commutator $[z,I]=zI-Iz$, and similarly $[z,J],[z,K]$ by first computing the commutators of $I,J,K$ pairwise with each other, and then extending by linearity.
Since these basis elements anti-commute, we have $[I,J]=2k,[J,K]=2I,[K,I]=2J$. Therefore $[z,I]=-2cK+2dJ$. Thus, for $z$ to be in the center we must have $c=d=0$. Taking the commutator with either $J$ or $K$ shows that $b=0$. Thus, the center is just $\mathbb R$, and so we cannot have a copy of $\mathbb C$ in the center.