Let $A, B$ be the C*-algebras and $\|\cdot\|_\alpha$ be a C*-norm on $A\odot B$, $\xi$ be a state on $A\otimes_\alpha B$. (Here, $\odot$ denotes the algebraic tensor product and $A\otimes_\alpha B$ denotes the completion of $A\odot B$ with $\|\cdot\|_{\alpha}$).
If $A$ is abelian and $\xi$ is a pure state on $A\otimes_{\alpha} B$, and $(\pi_{\xi}, H_{\xi}, v_{\xi})$ is the GNS triplet, can we verify the weak closure of $\pi_{\xi}(A\otimes_{\alpha} B)$ is all of $B(H_{\xi})$?
Certainly not. The condition you want is that the commutant of $A\odot B$ is trivial inside $B(H_\xi)$. The fact that $A$ is a abelian makes this usually impossible (elements of the form $a\otimes 1$ commute with $A$).
For instace you could take $A=\mathbb C\oplus \mathbb C$, $B=\mathbb C$. Then $A\odot B=A$, and $A$, being abelian, can only be irreducible if $H_\xi$ is one-dimensional. If we take $\xi(a,b)=(a+b)/2$, which is faithful, $H_\xi$ will have dimension $2$ and $A$ will not be irreducible.