I understand that every productive set is not recursively enumerable, but is the other way around also true?
If not, what is an example of a set which is not r.e. but not productive?
Update: The answer is no.
We know by Myhill's theorem: a set $A$ is productive $<=>$ $\overline A$ is r.e.-hard.
So to show that there exists a set which is not r.e. and its complementary is not r.e.-hard would be enough to know that it is also not productive.
By Post's theorem we know that there exists a simple set $S$. That is, a set which is r.e., not r.e.-hard and not recursive.
Consider $\overline S$.
It can't be r.e. or recursive, otherwise $S$ would be recursive.
It can't be (not r.e.)-hard: that would mean there exists a total recursive function $f$ such that $\overline K$ reduces to $\overline S$ by $f$. This can't be, because by the same $f$ we would show that $K$ reduces to $S$ and so $S$ would be r.e.-hard.
So $\overline S$ is not r.e. but not productive.