Hurewicz theorem states that if $X$ is a simply-connected CW-complex then $X$ is $(n-1)$-connected if and only if it is $(n-1)$-acyclic and that in this case $\pi_n(X)=H_n(X)$. Moreover, it is also known that if $X$ is path-connected then $H_n(X)=0\Rightarrow \pi_n(X)=0$. So there is a deep conection between homology and homotopy groups.
My question is whether there is a path-connected CW-complex $X$ such that $\pi_1(X)\neq0$ but $H_1(X)= 0$, i.e. it is $1$-acyclic but no simply connected. Moreover, is there any general condition or situation in which we know this will always happen? Or, which conditions make this situation imposible?
Thanks for your help.
The one general thing that can be said is simply to state the 1-dimensional Hurewicz theorem: $$H_1(X) \approx \pi_1(X) \, / \, [\pi_1(X),\pi_1(X)] $$ In words, $H_1(X)$ is isomorphic to the quotient of $\pi_1(X)$ by the commutator subgroup of $\pi_1(X)$; even more briefly, $H_1(X)$ is the abelianization of $\pi_1(X)$.
So to construct examples with trivial $H_1$ but nontrivial $\pi_1$, you merely pick your favorite nontrivial group $G$ having the property that $G$ is equal to its own commutator subgroup $[G,G]$, equivalently $G$ has trivial abelianization (one says that such a $G$ is a perfect group). You can then construct a CW complex whose fundamental group is isomorphic to $G$, using a presentation for $G$ and the standard method for constructing a CW complex from a presentation, as indicated in the comment of @user10354138.
But there are also other constructions and examples. Perhaps the famousest of all examples is the Poincare dodecahedral space whose fundamental group is the "binary icosahedral group of order $120$" as you'll see by following that link.