Let $\Omega$ be a bounded domain. I am not sure what a function in
$H($div$) = \{ u\in L^2(\Omega):\nabla\cdot u \in L^2(\Omega) \}$
but not in
$H^1 = \{ u\in L^2(\Omega): \frac{\partial u}{\partial x_i}\in L^2(\Omega) \}$
would look like. I know that at least two space dimensions are needed, but I haven't figured out how to construct an example.
Edit: I forgot to specify for a bounded domain.
Example: For concreteness we suppose our bounded domain contains the origin in the interior, but in the general case we can do a translation.
Take $$ u(x,y) = \begin{pmatrix} H(y) \\ 0 \end{pmatrix} $$ where $H$ is the Heaviside function. $u$ is a bounded function defined on a bounded domain, so $u$ is in $L^2$. The divergence of $u$ can be computed explicitly to be 0. But $\partial_y u$ is equal to the Dirac $\delta$ along the $x$-axis, and is not in $L^2$.
Example: One may say that the previous example is too singular. We can give a slightly less singular example. assume the origin is on the boundary of your domain. Let $$ u(x,y) = \begin{pmatrix} \frac{-y}{r} \\ \frac{x}{r} \end{pmatrix} $$ Again $u$ is a bounded (smooth) function on a bounded domain so is in $L^2$. You can directly compute that $\nabla \cdot u = 0$.
But $$ \partial_y u = \begin{pmatrix} - x^2 / r^3 \\ - xy/r^3 \end{pmatrix} $$ within the sector $\{x^2 \geq y^2\}$ you have that the first factor is bounded below by $1/r$, which is not square integrable.