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 $L|C$ 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 non-effective! 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.