Cohen-Macaulay rings and Normal rings is there an example  that R is Cohen-Macaulay but not normal ring?
what about the converse example?
 A: Since CM rings which are regular in codimension one are normal, you should expect singularities in codimension one.  So, just take the $R = k[x, y]/y^3 - x^2$ [cuspidal singularity].
This is a complete intersection, so CM.  To see that the node is not normal, the fraction field is $k(t)$ where $x = t^3$ and $y = t^2$, and note that $t$ is in the integral closure but not in $R$.
A: Let $S_n$ denote Serre's condition, that is, a finitely generated $R$-module $M$ is $S_n$ if $$\operatorname{depth} M_p \geq \min(n,\dim M_p)$$ for every $p\in \operatorname{Spec}R$
and let $R_n$ mean that $R_p$ is regular for every $p\in \operatorname{Spec}R$ of height at least $n$. Then
$R$ is normal iff it is $S_2$ (as a module over itself) and $R_1$ and 
$R$ is CM iff it is $S_n$ for all $n$.
Now it is easy to give examples of one that's not the other:
An $R$ which is CM, but not $R_1$ will not be normal, so for instance anything with $1$-codimensional singular set (as the coordinate ring of the cuspidal cubic, or really any singular plane curve in user148177's answer). Other examples can be given (for instance) by hypersurfaces in arbitrary dimension.
To get something that's normal but not CM is perhaps even easier. Well, you may be thinking of the usual example for not CM of two planes meeting in a single point. That's not going to work. From the above description you can tell that in dimension at most $2$ normal implies CM. But as soon as you go to higher dimensions, probably the first non-CM example of dimension at least $3$ will be normal. 
For an explicit example, take the coordinate ring of a cone over an abelian variety of dimension at least $2$. That this is not CM follows for example from Lemma 4.3 of this paper of Patakfalvi. Using that lemma you can construct many other examples.
