Is there an example of a variety over the complex numbers with no embedding into a smooth variety?

Yes, there are even toric varieties. In fact, there are complete toric varieties with trivial picard group (see e.g., Eickelberg "Picard groups of compact toric varieties..." 1993). We have the following simple observations: 1) Any variety which can be embedded into a smooth variety has an ample family of line bundles. 2) A proper variety with trivial picard group cannot have an ample family of line bundles. 


A really neat well known example is as follows: Choose a conic $C_1$ and a tangential line $C_2$ in $\mathbb{P}^2$ and asssociate to a point $P$ on $C_1$ the point of intersection $Q$ of $C_2$ and the tangent line to $C_1$ at $P$. This gives a birational isomorphism from $C_1$ to $C_2$. Identify the curves by this map to get the quotient variety $\phi:\mathbb{P}^2\rightarrow{X}$ with $C:=\phi(C_1)$. Now if there was an embedding of $X$ in a smooth scheme then, there would surely exist an effective line bundle on $X$, say $L$ whose pull back to $\mathbb{P}^2$ will obviously be effective. Let us see why this is a contradiction. Let $L'$ be the pullback of $L$ to $\mathbb{P}^2$. Note that the degrees of $L'C_1$ and $L'C_2$ both coincide with the degree of $LC$ and are therefore equal. But $L'\cong\mathcal{O}(k)$ and therefore the degrees in question are $2k$ and $k$ respectively for $C_1$ and $C_2$. Therefore $k=0$ and $L'\cong\mathcal{O}$, which is noneffective! A contradiction! In view of VA's comment, I give a complete proof here for constructing $X$ as a scheme. In our special case it is a trivial pushout construction: Here I am thinking of $Y$ as $C_1\amalg{C_2}$, $Y'=C$ (the quotient by the birational isomorphism above), and $Z=\mathbb{P}^2$, but the argument is more general provided any finite set of closed points in $Z$ is contained in an affine open set. $X$ will denote the quotient. Claim: Suppose $j:Y\rightarrow{Z}$ is a closed subscheme of a scheme $Z$, and $g:Y\rightarrow{Y'}$ is a finite surjective morphism which induces monomorphism on coordinate rings. Then there is a unique commutative diagram (which I don't know how to draw here, but one visualize it easily): $Y\xrightarrow{j}{Z}$, $Y\xrightarrow{g}Y'$, $Y'\rightarrow{X}$, $Z\xrightarrow{h}X$ where $X$ is a scheme, $h$ is finite and induces monomorphisms on coordinate rings and $Y'\rightarrow{X}$ is a closed immersion. Proof: First assume that $Z$ is affine, in which case $Y,Y'$ are both affine too. Let $A,A/I,B$ be their respective coordinate rings. Then $B\subset{A/I}$ in a natural way. Let's use $j$ again to denote the natural map $A\rightarrow{A/I}$. Put $A'=j^{1}(B)$ and $Spec(A')=X$. The claim is clear for $X$. Also if $Z$ is replaced by an open subset $U$ such that $g^{1}g(U\cap{Y})=U\cap{Y}$, $X$ would be replaced by $U'=h(U)$ which is an open subset. Now this guarantees the existence of $X$ once it has been shown that $Z$ can be covered by affine open subsets $U$ such that $g^{1}g(U\cap{Y})=U\cap{Y}$. But this is obvious in our example. For our example it is also clear from the construction of $X$ that it is actually reduced and irreducible. QED. I hope this is satisfactory. 


See example 4.9 in Ravi Vakil's paper http://math.stanford.edu/~vakil/files/asnjun18.pdf , and the references therein. 

