Does there exist a manifold with finitely generated homology groups that is not homotopy equivalent to a compact manifold with boundary?

If the finitely generated of the homology groups implies the finitely generated of the fundamental group (noticed in the comments that this is not true), then for 3-manifolds, by Scott's compact kernel theorem, there are no such manifolds: a three-dimensional manifold admits a smooth structure, then it is triangulable, hence homeomorphic to a CW-complex, and CW-complexes with isomorphic homotopy groups are homotopy equivalent by Whithead's theorem.

I am also interested in several variations of this question. Does it exist..

1. ..a smooth manifold without boundary
2. .. a smooth manifold (possibly with boundary)
3. ..a manifold without boundary
4. .. a manifold (possibly with boundary)

is not homotopically equivalent

A)..a compact manifold without boundary

B)..a compact manifold (possibly with boundary)

A minimal example for all type A questions that I know of is four-dimensional: the configuration space of two-element subsets of the plane. Its first homology group is isomorphic to Z (abelinization of the braid group), and the third cohomology group is trivial (proved by Vasiliev see "Topology of complements to discriminants" § Cohomology of braid groups with constant coefficients, Proposition 1), which contradicts Poincaré duality.