I consider Morava K-theory at the prime $p=2$ and height $n$. $K(n)^*$ is multiplicative and complex-oriented, but the multiplication is not commutative. Suppose I have a complex bundle $E$ of rank m over a CW complex $N$. Following
https://www.math.ias.edu/~lurie/252xnotes/Lecture5.pdf
I can define an Euler class $e(E) \in K(n)^{2m}(N)$.
Question: Does $e(E)$ lie in the center of the ring $K(n)^{*}(N)$?
Yes, there are various ways to see this. One approach is to say that in $K(n)$-theory we always have $ab-ba=v_nQ(a)Q(b)$ for a certain Bockstein-type operation $Q$ of odd degree $2^n-1$, so any element $a$ with $Q(a)=0$ is central. The element $e(N)$ is the image of an element in $K(n)^{2m}(BU(m))$, which is concentrated in even degrees, so $Q(e(N))=0$.
Alternatively, the best way to construct $K(n)$ is as a ring object in the category of strict $MU$-modules (as in https://arxiv.org/abs/math/0011122). If we construct $K(n)$ that way, it is clear that everything in the image of the natural map $MU^*(X)\to K(n)^*(X)$ is central. In particular, the Morava $K$-theory Euler class is the image of the $MU$-theory Euler class and so is central.
Also, the Euler class is the same as the top Chern class (or perhaps the top Chern class of the dual bundle, depending on conventions) and in fact all the Chern classes are central (by either of the two arguments given above).
