Monoidal Equivalence for Drinfeld--Jimbo Quantum Groups For $U_q(\frak{g})$ the Drinfeld--Jimbo quantum group, its category of representations is equivalent to the category of representations of $U(\frak{g})$, or equivalently the category of Lie algebra representations of $\frak{g}$. Both categories have an obvious monoidal structure, what is not obvious is if this is an equivalence of monoidal categories.
Edit: As Phil's comment below, this is not a monodial equivalence. As the linked answer says, the problem is that the associators in both cases are different. What is the easiest way to see that this is so? The discussion about 6j symbols and coordinates on stacks is unfortunately lost on me. In fact, for modules over a Hopf algebra the assoicators look easy to the naive untrained eye, why is it about the assoiators on either category that is non-trivial?
 A: For simplicity let’s just do the $\mathfrak{sl}(2)$ case.
Let X be the 2-dimensional natural representation.  From the fusion rules, $\mathrm{Hom}(X \otimes X,1)$ and $\mathrm{Hom}(1, X \otimes X)$ are one-dimensional.  Choose a map in each normalized such that the zig-zag is the identity:
$$X = X \otimes 1 \rightarrow X \otimes (X \otimes X) \rightarrow (X \otimes X) \otimes X \rightarrow 1 \otimes X = X.$$
Note that this fixes the maps up to rescaling them in opposite ways.  Now compute the quantum dimension (aka the circle):
$$1 \rightarrow X \otimes X \rightarrow 1.$$
This is a numerical invariant of such a monoidal category and a calculation shows that it is $q+q^{-1}$.  Thus the monoidal categories are not equivalent except for replacing q by its inverse.
In the classical case these maps are the determinant map and the copairing associated to the determinant map.  I.e. since $\det(e_1, e_1) = 0$, $\det(e_1, e_2) = 1$, $\det(e_2, e_1) = -1$, $\det(e_2, e_2) = 0$, the copairing sends 1 to $e_1 \otimes e_2 - e_2 \otimes e_1$.  Thus the value of the circle is $$\det(e_1 \otimes e_2 - e_2 \otimes e_1) = 1+1 = 2.$$
(Some side notes for experts.  


*

*If $X$ were not self-dual you'd have to replace some of the $X$'s above with $X^{*}$ which results in some trickiness, the upshot of which is you either need a pivotal structure, or (following ENO) you compute $\mathrm{qdim}(X)\mathrm{qdim}(X^{*})$ to get a monoidal invariant.  This wouldn't quite let you distinguish everything because of switching q and -q.  

*When $X$ is self-dual something very tricksy is going on with the signs here...  I will update this bullet when I understand it better.)

A: First of all, the fact that those are equivalent as mere categories is true only for (locally-) finite dimensional representations, and when $q$ is generic enough. This part is somewhat "easy".
Then, this is a deep theorem by Drinfeld and Kazhdan--Lusztig that they become equivalent as (braided) monoidal categories, if on the $U(\mathfrak g)$ side one keeps the same tensor product, but introduces a highly non trivial associator coming from the so-called KZ equation. Hence you're right that the associator for modules over an Hopf algebra is trivial, but then on the quantum side the coproduct, hence the formula defining the tensor product of modules, is complicated. On the non-quantum side, one keeps the same simple coproduct/tensor product, but at the cost of having to put a highly non-trivial associator, which of course is not equivalent to the identity. 
A nice exposition can be found there: https://arxiv.org/abs/0711.4302
Edit: to clarify a point raised in OP's edit: your first use of the word "associator" here means what people also call fusion rules which are non-trivial even if the associator in the sense of associativity constraints of monoidal categories (which is what your second use of the word "associator" means) are identities. Hence the bottom line is that the fusion rules of $U_q$ and $U$ are different, because their coproducts are different, and that the former is obtained from the latter by plugging this KZ associator.
