Why there is no $5$-dimensional analogous of Quaternions?

Why there is no $5$-dimensional analogous of Quaternions? Why the following definition is not well-defined? $$i^2=j^2=k^2=\ell^2=ijk=jk\ell=k\ell i=-1,\quad ijkl=1.$$

If you are looking for a real vector space $V$ with basis $\{1,i,j,k,\ell\}$ and an associative product such that $V$ becomes an $\Bbb{R}$-algebra, then we run into the following difficulty.
The rule that $i^2=-1$ means that $V$ also has a structure as a vector space over the field $\Bbb{C}=\Bbb{R}(i)$.
But a vector space over $\Bbb{C}$ necessarily has an even dimension as a vector space over $\Bbb{R}$.
• In the same vein, if $\Bbb{H}$ is the usual division algebra of quaternions then any associative algebra $A$ containing $\Bbb{H}$ as a subalgebra is also a free left (or right) $\Bbb{H}$-module. Therefore $\dim_{\Bbb{R}}A$ is necessarily divisible by four. – Jyrki Lahtonen Nov 5 '17 at 8:10
• Slightly more relaxed: if $A$ is a real alternative algebra with $n$ anticommuting square roots of negative one, then $A$ is a representation of the Clifford algebra ${\rm Cliff}(n)$. Since $\mathrm{Cliff}(n)\cong M_k(\Bbb K)$ for some $k$, the dimensions of its reps are all multiples of $km$ where $m=1,2,2,4,4,8$ and $\Bbb K=\Bbb R,\Bbb R^2,\Bbb C,\Bbb C^2,\Bbb H,\Bbb H^2$. Note $k$ and $\Bbb K$ can be determined from $n$ using the "Clifford clock." – anon Nov 5 '17 at 21:52