Are Kahler differentials the same on the affine closure on a quasi-affine scheme? Let $X$ be a quasi-affine scheme; that is, the natural map 
$$X\rightarrow \overline{X}:=Spec(\Gamma(X,\mathcal{O}_X))$$
 is an inclusion.  Each scheme has a quasi-coherent sheaf of Kahler differentials $\Omega$, and the above open inclusion induces a $\Gamma(X,\mathcal{O}_X)$-module map of global Kahler differentials
$$ \Gamma(\Omega_{\overline{X}})\rightarrow \Gamma(\Omega_{X})$$
Is this map always an isomorphism?
Edit: Changed $\mathcal{O}_X$ to $\Gamma(X,\mathcal{O}_X)$.
 A: The assumption implies that the natural embedding induces an isomorphism $\Gamma(X,\mathscr O_X)\simeq \Gamma(\overline X,\mathscr O_{\overline X})$. Then this means that  the complement of $X$ has at least codimension $2$. 
In addition assume that $X$ is noetherian and $S_2$ (for instance normal). 
In this case if $\Omega_{\overline X}$ is a reflexive sheaf, then the restriction 
$\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{\overline X})=\Gamma(X,\Omega_{X})$ is an isomorphism.
More generally, let $Z:=\overline X\setminus X$. If $\mathrm{depth}_Z\Omega_{\overline X}\geq 2$ then the restriction $\Gamma(\overline X,\Omega_{\overline X})\to \Gamma(X,\Omega_{X})$ is an isomorphism.
This certainly holds if $\Omega_{\overline X}$ is a reflexive sheaf, but obviously it could hold "by accident" even if one of the above conditions fail, so I am not claiming that these are necessary conditions, but at least they sure seem to provide a natural set of conditions under which the required map is an isomorphism.
Sketch that if $X$ is $S_2$, then a reflexive coherent sheaf $\mathscr F$ is also $S_2$:
First observe that by the argument in this answer to another MO question we may assume that $X$ is affine and it is enough to prove that $H^i_x(X,\mathscr F)=0$ for $i=0,1$ for all $x\in X$ and it also follows that $\mathrm{depth}_Z\mathscr F\geq 2$ even if $Z$ is not contained in an affine piece of $X$. To do that write $\mathscr F^\vee$ as the quotient of a (locally) free sheaf ($X$ is affine!). Then $\mathscr F$ is a submodule of the dual of this locally free sheaf, let's call it $\mathscr E$, and the quotient $\mathscr E/\mathscr F$ is torsion-free. Therefore none of them have torsion and so $H^0_x(X,\mathscr F)=0$ and $H^1_x(X,\mathscr F)$ embeds into $H^1_x(X,\mathscr E)$. But the latter is $0$ by the assumption that $X$ is $S_2$.
EDIT 1 removed intro paragraph about the starting assumption.
EDIT 2 added "more generally" paragraph.
EDIT 3 added noetherian assumption. this is probably not necessary but without this one should possibly be more careful about the other conditions.
EDIT 4 Added Sketch above.
