Here are four remarks about the homology and homotopy type of a compact, complex manifold $M$:

If $M$ is Kähler, then it is symplectic and thus $H^2(M,\mathbb{R}) \ne 0$. (Also, as explained in a blog posting by David Speyer, you still have $H^2(M,\mathbb{R}) \ne 0$ even if $M$ is non-projective but algebraic.)

An interesting first example of a non-Kähler manifold is a Hopf manifold, by definition $(\mathbb{C}^n\setminus 0)/\Gamma_r$, where $\Gamma_r$ is a rescaling by $r$ with $|r| \ne 0,1$. This example has $H^1(M,\mathbb{R}) \ne 0$.

On the other hand, even-dimensional, compact Lie groups have left-invariant complex structures. If $M$ is such a manifold and is simply connected, then it is also 2-connected. $H^1(M,\mathbb{Z}) = H^2(M,\mathbb{Z}) = 0$ and $M$ is manifestly not Kähler. On the other hand, no such example is 3-connected and you always have $H^3(M,\mathbb{R}) \ne 0$.

There is (or was) a long-standing conjecture that no even-dimensional sphere other than $S^2$ has a complex structure.

So, question: Is there for each $n$, a compact, complex manifold $M$ which is $n$-connected?