A connected compact smooth manifold $M$ of dimension $n$ with nonempty boundary deformation to a simplicial complex where all cells have dimension at most $n-1$. See this answer by Lee Mosher for a proof. The basic idea, as far as I understand, is to remove the boundary to get a noncompact manifold, take a triangulation of the resulting manifold, and then shrink all the $n$-cells starting "from the outside" (like when you deformation retract $[0,\infty)$ onto $\{0\}$) – this last part is rather tricky and the first version of my answer was a bit naive, thanks to Lee Mosher for giving me a link to the full proof.
And now by cellular approximation, a map from a simplicial complex of dimension $\le n-1$ to $S^n$ is nullhomotopic, because you can consider a CW-complex structure on $S^n$ where the $(n-1)$-skeleton is just a point.
(I used a triangulation here, I'm not sure it's strictly necessary... Maybe there's an easier argument.)
For instance, you can start with the $E8\oplus E8$ manifold $M$: This is a 4-dimensional closed simply connected manifold with the intersection form isomorphic to $E8\oplus E8$. This manifold has vanishing Kirby-Siebenmann invariant $k(M)$ (the signature is divisible by 16), but is not smoothable (this was first proven by Donaldson). The existence of such manifolds is nontrivial: Freedman proved that you can realize any unimodular integer quadratic form as the intersection form of a closed simply-connected topological 4-manifold. Since $k(M)=0$, the manifold $M\times (0,1)$ is smoothable. (This is again nontrivial and is due to Kirby and Siebenmann: For a closed $4$-manifold $k(M)=0$ if and only if $M\times (0,1)$ is smoothable.)
In particular, taking $W=M\times [0,1]$ we obtain a compact 5-dimensional manifold whose interior is smoothable but the boundary $M\sqcup M$ is not.
You can read much more at the manifold atlas.
Addendum. One can say precisely what happens for (closed) simply-connected 4-manifolds $M$ which bound 5-manifolds $N$ with smoothable interior. By looking at the collar neighborhood of $M$ in $N$, we see that $M\times (0,1)$ embeds in $int(N)$ and, hence, is smoothable. Thus, $k(M)=0$. The homeomorphism type of $M$ is then determined only by its intersection form (Freedman), more precisely, by the lattice $(H_2(M), Q)$ where $Q$ is the intersection form.
The intersection form is unimodular (and even).
There are two cases:
(a) $Q$ is definite. Then, by Donaldson's theorem, $Q$ is diagonal, i.e. is given by the rank $r$ identity matrix $I_r$ or by $-I_r$. This form $Q$ is realized by the smooth manifold which is the connected sum of $r$ copies of $CP^2$ or the same manifold with opposite orientation.
(b) $Q$ is indefinite. Each indefinite unimodular form is the $r$-fold direct sum of rank 2 hyperbolic quadratic forms. This is realized by the connected sum of $r$ copies of $S^2\times S^2$.
To conclude:
$M$ is smoothable if and only if its intersection form is indefinite or is the form $\pm I_r$, of rank $r$.
Once you drop simple connectivity, there are some known obstructions to smoothability (again, the intersection form is diagonal if definite). But the overall picture is totally unclear.
Best Answer
Just to complete the details of Dan's hint. The top homology of a non-compact manifold vanishes, e.g. by Poincare duality $H_n(M) \cong H^0_c(M)$ where $H^*_c(M)$ is cohomology with compact support. But $H^0_c(M)$ is just the kernel of $d : C^0_c(M) \to C^1_c(M)$. Therefore $C^0_c(M)$ consists of constant functions and because $M$ is non-compact the only such function is the zero function.
Alternatively, one can see the result directly from Lefschetz duality, a generalization of Poincare duality. This gives $H_n(M) \cong H^0(M,N)$ where $N$ is the boundary of $M$. But $H^0(M,N) \cong \tilde H^0(M/N) = 0$.
I'm sure there are many other and probably simpler ways to see this. But the upshot is that non-compact manifolds and manifolds with boundary are simpler from the point of view of homotopy theory than their closed cousins (just thinking of ${\bf R}^n$ which has homototy type of a point).