A von neumann algebra $R\subseteq B(H)$ is called algebraically cyclic if there exists $\zeta\in H$ with $H=R\zeta$.
Example. If $R$ acts irreducibly on $H$ then $R$ is clearly algebraically cyclic.
Q. Make an example of a cyclic von Neuman algebra which is not algebraic.
Let $H=\ell^2$ and $R=\ell^\infty$ (acting as multiplication operators on $H$). Any function $\phi\in \ell^2$ with full support is cyclic, i.e. $\overline{R\phi}=\ell^2$, because $\frac1{\phi(k)}\delta_k\cdot\phi=\delta_k$ and the linear combinations of $\delta_k$, $k\in \mathbb{N}$, are dense in $\ell^2$.
Assume that $\phi\in H$ is an algebraically cyclic vector, i.e. $R\phi=H$. It is clear that $\phi$ has full support. We will construct an unbounded function $\psi$ such that $\psi\phi\in\ell^2$. Since $\phi$ has no zeroes, it follows that $\psi\phi\notin R\phi$, yielding a contradiction.
Construction of $\psi$: Let $A_n=\{k\in\mathbb{N}\mid 2^{-(n+1)}< |\phi(k)|\leq 2^{-n}\}$. Since $\phi\in \ell^2$, we have $\phi(k)\to 0$. Thus, the set $N=\{n\in\mathbb{N}\mid A_n\neq \emptyset\}$ is infinite. Pick a sequence $(k_j)$ such that for every $j\in\mathbb{N}$ there is an $n_j\in N$ such that $k_j\in A_{n_j}$ and $n_i\neq n_j$ for $i\neq j$. Let $$ \psi=\sum_{j=1}^\infty n_j \delta_{k_j}. $$ Since $j\mapsto n_j$ is injective, $\psi$ is unbounded. Morevover, $$ \sum_{k=1}^\infty|\phi(k)\psi(k)|^2=\sum_{j=1}^\infty |\phi(k_j)|^2 n_j^2\leq \sum_{j=1}^\infty n_j^2 2^{-2n_j}\leq \sum_{n=1}^\infty n^2 4^{-n}<\infty. $$ Hence $\psi\phi\in\ell^2$.