I believe that there isn't a commutative model for the DGA of cochains on a space, because of cohomology operations. This question has some nice explanations of why this is so. For example, there is a way of constructing Steenrod operations on the cohomology of any $E_{\infty}$ DGA over $\mathbb{F}_p$, and if this DGA is strictly graded commutative then most of the Steenrod operations, including $Sq^0$, vanish.

However, I recently became aware of a result of Guggenheim and May (Differential Torsion Products, Theorem 4.1) which states that for any commutative ring $R$, there is a quasi-isomorphism of DGAs $g:C^* (BT,R) \to H^*(BT,R)$, where $T$ is a torus.

**Question 1** Why does this result not contradict that there isn't a commutative model for cochains on a space?

I can guess what the answer will be: by "model" I should mean an equivalent $E_{\infty}$ DGA, and Guggenheim and May state that this equivalence $g$ kills all $\smile_1$ products, showing that this $g$ is not a map of $E_{\infty}$ DGAs, but I would appreciate more detail, especially in regards to why this fact about $\smile_1$ products shows that the equivalence is not $E_{\infty}$.

**Question 2** This is a softer question: what, if anything, do I lose if I replace $C^*(BT,\mathbb{F}_p)$ with $H^*(BT,\mathbb{F}_p)$?

Of course this depends on what I am trying to do; I am asking this question in the context of trying to understand the singular Cartan model for torus-equivariant cohomology, but I would appreciate any wisdom.