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

If the finiteness of the homology groups implies the finiteness of the fundamental group, 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