In the book *Hopf algebras* by Eiichi Abe (1980), there is a theorem on page 57-58 that reads
that 

>Given a $k$-linear space $H$, suppose that there are $k$-linear maps $\mu:H\otimes H\rightarrow H$, $\eta:k\rightarrow H$, $\Delta:H\rightarrow H\otimes H$, $\varepsilon:H\rightarrow k$ such that $(H,\mu,\eta)$ is a $k$-algebra and $(H,\Delta,\varepsilon)$ is a $k$-coalgebra. Then the following conditions are equivalent. (i) $\mu,\eta$ are $k$-coalgebra morphisms. (ii) $\Delta,\varepsilon$ are $k$-algebra morphisms ...

and that 

>When a $k$-linear space $H$ together with $k$-linear maps $\mu,\eta,\Delta,\varepsilon$ satisfy one of the equivalent conditions of Theorem 2.1.1. (quote above) then $(H,\mu,\eta,\Delta,\varepsilon)$ or simply $H$ is called a $k$-bialgebra.

Since a Hopf algebra is a bialgebra with an antipodal map, if you have an algebra such that $\mu$ and $\eta$ are not coalgebra morphisms, or $\Delta,\varepsilon$ are not algebra morphisms then it would fail to be a bialgebra and thus a Hopf algebra.

It won't tell you for every algebra since there are bialgebras that are not Hopf algebras, but it is a start.



After looking into it further I found in the book *Hopf Algebras: An Introduction* by Sorin Dascalescu, et al. (2001), there is a remark on page 151 that reads

>In a Hopf algebra, the antipode is unique, being the inverse of the element I in the algebra $Hom(H^c,H^a)$. The fact that $S:H\rightarrow H$ is the antipode is written as $S*I=I*S=\eta\varepsilon$, and using the sigma notation $\sum S(h_1)h_2=\sum h_1S(h_2)=\varepsilon(h)1$ 

[Note: $H^c$ is the bialgebra looking at the coalgebra structure on it, or $(H,\Delta,\varepsilon)$ from above. $H^a$ is the bialgebra looking at the algebra structure on it, or $(H,\mu,\eta)$ from above. Also I changed the names of the multiplication, comultiplication, unit, and counit maps to correspond with what I had above.]

So if you cannot find an antipodal map for your bialgebra, it fails to be a Hopf algebra. Again, this doesn't completely answer the question because I do not know how you would go about figuring this out, but it should help you take a step in the right direction.