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), <s> 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.</s>

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 that I know for all type A questions 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 Poincare duality.