How smooth can non-nice associative operations on the reals be? Suppose ${*}:\mathbb R\times\mathbb R\to\mathbb R$ is $\mathcal C^k$ and associative. Does it necessarily satisfy the identity $a * b * c * d = a * c * b * d$?
For $k=0$ the answer is "no" -- a counterexample would be to let $x_1*x_2*\cdots*x_n$ to be the product of the $x_i$s up to and including the leftmost negative one (and the product of all of them if they are all nonnegative).
However, I can't find any counterexample for $k=1$. Will high enough $k$ guarantee that $a*b*c*d=a*c*b*d$? If not, will $\mathcal C^\infty$? Will "analytic"?
Bonus question: If there is a $k$ that does force the identity, does the same $k$ work with $\mathbb R^n$ instead of $\mathbb R$?
(Edit: obviously not; multiplication of $[\,{}^{\strut x}_0\;{}^{\strut y}_1\,]$ matrices is very smooth.)
The condition that $a*b*c*d\ne a*c*b*d$ for some $a,b,c,d$ is meant to force $*$ to be noncommutative while at the same time avoiding trivial solutions where $a*b*c$ doesn't depend on $b$.
(This is a variant of a question posed by Nikhil Mahajan which unfortunately failed to get interesting answers due to a technicality.)
 A: This is a partial answer assuming $\ast$ has a formal power series, that is, 
$$X\ast Y=\sum_{p,q\geq 0}\alpha_{p,q}X^p Y^q$$
In order to be associative, we must have $\alpha_{0,0}=0$ (why?), so assume $\alpha_{0,0}=0$. 
It is first shown in Lemma 3.2 of this article that $\alpha_{0,1},\alpha_{1,0}\in\left\{0,1\right\}$. 
It is then shown in Proposition 3.3 that if $\alpha_{1,0}=0$ and $\alpha_{0,1}=1$, then $X\ast Y=Y$. 
Theorem 3.6 then shows that if $\alpha_{1,1}=0$, then $\ast=0$. 
The case $\alpha_{1,0}=\alpha_{0,1}=0$ is shown to make $\ast$ commutative in Proposition 3.10. (side remark: Theorem 4.3 gives the following beautiful representation theorem: $\ast$ is associative iff it is of the form $a\ast b=f^{-1}(f(a)f(b))$ for some invertible formal power series $f$. )
The case $\alpha_{1,0}=\alpha_{0,1}=1$ is described on page 38 of this book, and is also shown to make $\ast$ commutative. 
Side remark on rational associative functions: Theorem C2 of this article gives the following representation theorem for associative rational functions: $X\ast Y=f(G(f^{-1}(X),f^{-1}(Y)))$, for some homografic function $f$ and $G$ one of $+$, $\cdot$, and $(X,Y)\mapsto\frac{X+Y}{1-XY}$. 
