If X is a quasiprojective variety with condition S_2, and if its normalization Y is Cohen-Macaulay, is X necessarily Cohen-Macaulay? Suppose that $X$ is a quasiprojective variety satisfying $S_2$, and that the normalization ${X}^{\mathrm{N}}$ of $X$ is Cohen-Macaulay.  
Question: Is $X$ Cohen-Macaulay?
This already fails for surfaces without the $S_2$ condition.  I'm wondering if there is a proof using dualizing complexes in case $X$ is $S_2$.
 A: No. We can find a simplicial complex $\Delta$ such that the Stanley-Reisner ring $SR(\Delta)$ is $S_2$ but not CM. The normalization of $SR(\Delta)$ will be a direct product of polynomial rings, so CM. 
To see that such a $\Delta$ exists, recall that $SR(\Delta)$ is $S_2$ if $\Delta$ is pure of dimension $d$ (for some $d$) and $\mathrm{link}_F(\Delta)$ is connected for any $F$ of dimension $\leq d-2$, while $SR(\Delta)$ is CM if the link of every face (including the empty face) has the homology of a wedge of spheres of appropriate dimension. So, if we take $\Delta$ to be a triangulation of a $2$-torus or a cylinder, we should get an example.

If you want a domain, here is a more complicated answer. Suppose we can find the following: A surface $Y$, which is $S_2$ but not $R_1$, with normalization $\pi: \tilde{Y} \to Y$, such that $H^1(\tilde{Y}, \mathcal{O}) = H^2(\tilde{Y}, \mathcal{O}) = 0$ but $H^1(Y, \mathcal{O}) \neq 0$. Take a sufficiently ample line bundle $L$ on $Y$ and let $X$ be the cone on the projective embedding of $Y$ by $L$. The pullback $\pi^{\ast} L$ will be ample on $Y$ (see discussion here) and it seems to me that the normalization of $X$ will be the cone on the projective embedding of $\tilde{Y}$ by powers of $\pi^{\ast} L$. The local cohomology groups of a projective cone are related to the sheaf cohomology of $\mathcal{O}(k)$. In our setting, $H^1(Y, \mathcal{O}) \neq 0$, which will make $X$ not $S_3$, but $H^1(\tilde{Y}, \mathcal{O}) = H^2(\tilde{Y}, \mathcal{O})=0$ and I believe all higher powers of $\pi^{\ast} L$ will vanish by Serre vanishing if $L$ is chosen sufficiently ample.
I haven't checked through every claim above in detail, but I have checked the following example. Take $\tilde{Y} = \mathbb{P}^1 \times \mathbb{P}^1$ and take $Y = N \times \mathbb{P}^1$ where $N$ is the nodal cubic $x^3+y^3=xyz$ in $\mathbb{P}^2$. We'll take $L$ to correspond to the Segre embedding $N \times \mathbb{P}^1 \subset \mathbb{P}^2 \times \mathbb{P}^1 \subset \mathbb{P}^5$.
Let's restate that in terms of commutative algebra. Set $A = k[p,q]$, where $k$ is a field of characteristic not $3$. Let $B$ be the subring of $A$ in degrees divisible by $3$, so generated by $p^3$, $p^2 q$, $p q^2$, $q^3$. Let $C$ be the subring of $B$ generated by $x = p^2 q$, $y = p q^2$ and $z = p^3+q^3$. Note that $xyz = x^3+y^3$. Let $R$ be the Segre product of $B$ and $k[u,v]$ (i.e. the subring of $B \otimes k[u,v]$ generated by $p^3 u$, $(p^2 q) u$, $(p q^2) u$, $q^3 u$, $p^3 v$, $(p^2 q) v$, $(p q^2) v$, $q^3 v$.) Let $S$ be the Segre product of $C$ and $k[u,v]$. 
We claim that $R$ is the normalization of $S$. They clearly have the same fraction field. To see that $R$ is integral over $S$, observe that $p^3$ and $q^3$ obey the monic polynomial $\theta^2 - z \theta + xy=0$. Also $R$ is a Segre product of normal rings, hence normal.
We must now check that $S$ is $S_2$ but not CM. It is easy to see that it is not CM: The Hilbert series is $1+\sum_{n \geq 1} 3(n^2+n) t^n = 1+\frac{6 t}{(1-t)^3} = \frac{1+3t+3t^2 - t^3}{(1-t)^3}$, while the numerator of the Hilbert series for a CM ring always has positive coefficients. 
Since $S$ is graded, it is enough to check that it is $S_2$ at the origin. This is the same as checking at the local cohomology $H^1_{\mathfrak{m}}(S)$ vanishes, which is the same as to check that the map from $S$ to the section ring $\bigoplus_{k \geq 0} H^0(N \times \mathbb{P}^1, L^{\otimes k})$ is surjective. This is true.
See also discussion here about what rings that are $S_2$ but not $S_3$ look like.
