Let $\mathcal{C}$ be the $E_k$-monoidal $\infty$-category of left modules over a fixed connective $E_{k+1}$-ring spectrum $A$. Suppose that $M$ is an object of $\mathcal{C}$ which is an $E_0$-algebra, in other words $M$ admits a unit map $A\to M$. There is an adjunction $U\colon Alg_{E_k}(\mathcal{C})\leftrightarrows Alg_{E_0}(\mathcal{C})\colon F$ by which we can construct the free $E_k$-algebra on $M$. Note that the $E_k$ algebra generated by the $E_0$-algebra $M$, i.e. with fixed unit, will not generally be equivalent to the $E_k$-algebra generated by $M$ thought of as object of $\mathcal{C}$ (an algebra for the trivial operad).

The unit of the adjunction gives a map of $E_0$-algebras $u\colon M\to F(M)$. I do not have a good description of $F(M)$ in general, but I am interested in showing that if the cofiber of the unit map $A\to M$ is $d$-connected (so $\pi_\ast(M/A)=0$ for $\ast<d$ ) then $cof(u)$ is $2d$-connected.

**My rough (and I believe incorrect) attempt is as follows:** we can construct the $E_k$-algebra $F(M)$ as an operadic left Kan extension along the inclusion $E_0^\otimes\hookrightarrow E_k^\otimes$, which tells us that it is (I think!), just as an object of $\mathcal{C}$, i.e. ignoring multiplicative structure, equivalent to a colimit over the diagram of injections $\langle m\rangle\to \langle n\rangle$ in $Fin_\ast$ that live over $\langle 1\rangle$. This is a bit terse, but the idea is that you're supposed to take a colimit over the "active" morphisms in $E_0^\otimes$ that live over $\langle 1\rangle$. Recall that $E^\otimes_0=N(Fin^{inj}_\ast)$, i.e. it's the nerve of the category of finite pointed sets with morphisms $f$ such that $f^{-1}(i)$ has at most one element for each $i$ not the base point in the codomain. In other words, the base point goes to the base point, and everything else either goes to something which isn't hit by anything else, or goes to the base point. Also recall that the active morphisms therein are precisely those which *only* take the base point to the base point, so we've ruled out all the maps that take non-base-point elements to the base point. Thus an active morphism of $E_0^\otimes$ is exactly the data of an injection $\{1,\ldots,m\}\to\{1,\ldots,n\}$ (in particular necessitating that $m<n$).

This means then that $F(M)$ as an object of $\mathcal{C}$ is the colimit in $\mathcal{C}$ of a diagram whose objects are $M^{\otimes_A n}$ for $n\geq 0$, with $M^{\otimes_A 0}=A$, and with maps in between them coming from injections on the indexing sets using the unit of $M$. Let's call the colimit of this diagram $N$. It seems like $N$ is probably quite complicated, but it obviously admits a map from $M$ since $M$ is in the diagram constructing $N$. Then, using that colimits commute, we can take the cofiber of the inclusion of $M$ into $M^{\otimes_A n}$ for all $n>0$, and produce $N/M$ as $colim(M^{\otimes_An}/M)$ for $n>0$. And here I'm being really shady by just writing that as some kind of quotient, because there are a lot of maps from $M$ to $M^{\otimes_An}$ and we're equalizing all of them. Anyway, while $N/M$ is probably quite complicated, it seems like we can say that, at the very least, it is the colimit of a diagram of $2d$-connected spaces, since $M$ is $d$-connected and every object of that diagram is of the form $M^{\otimes_A n}$ for $n>1$, with possibly a bunch of cells attached when we mod out by all the maps from $M$. Thus at least up to degree $2d-1$, $M$ and $F(M)$ are equivalent.