The Chern classes give a map f:BU \to \prod_n K(Z,2n), which is a rational equivalence. However, it is not an equivalence over Z because the cohomology of BU is just a polynomial algebra and has no Steenrod operations. In particular, the generators of the homotopy groups \pi_{2n}(BU)=Z will not map to generators of the homotopy groups \pi_{2n}(K(Z,2n))=Z. Another way to say this is that the duals of the Chern classes in homology are not in the image of Hurewicz; only certain multiples of them are. What multiples you have to take is determined by the order of the k-invariants of BU, which are certain Steenrod operations of the fundamental classes of K(Z,2n).

Steenrod operations can be understood as obstructions to the cup product on ordinary cohomology being strictly commutative. On the other hand, the fact that f is not an equivalence can also be understood as an obstruction to addition in K-theory being strictly commutative. Indeed, any space with a strictly commutative group structure is a product of K(\pi_n,n)s under the map given by a right inverse of the Hurewicz map.

So in some sense you could say that products in cohomology are only homotopy-commutative and sums in K-theory are only homotopy-commutative "for the same reason". Is there some deeper story behind this? I don't know exactly what I'm asking for, but I'd like to get a better understanding of what's going on in this picture.