Well ordered Proper Classes. I am studying NBG set theory and am looking for a well-ordered proper class that is not $On$ (i.e the class of all ordinals). I am considering $p(On)$ due to the fact that for each $x\in p(On)$, $x \subseteq On$. So then each element $x$ of $p(On)$ has an (member relation) $E$ least element say $t_x$.
Now I'm considering a relation $S$ on $p(On)$ such that $xSy$ iff $t_xEt_y$ when $t_x \neq t_y$. If $t_x=t_y$ then $xSy$ iff $t_{(x\backslash t_x)}Et_{(y\backslash t_y)}$. So then each finite subclass of $On \in p(On)$can be well ordered by $S$.I do not know how to define $S$ so that it can well order infinite subclasses of $On$ though so any hints would be appreciated, if it is in fact possible to well-order $p(On)$.
I'm doing this to try find a well-ordered proper class that is not order isomorphic to $On$, so if the method I'm using is futile for that endeavour I'd appreciate being told and given a push in the right direction too.
 A: Consider the class $C = \{(\alpha, n) : \alpha\in\mathrm{On}\text{ and } n=0, 1\}$ and define an ordering $\prec$ over $C$ as lexicographical order. 
Our $C$ is well-ordered: let $B\subset C$ is a subclass. If the intersection of $B$ and the initial segment $S=\{x\in C : x\prec (0,1)\}$ is not empty, then $B$ has the minimal element since the initial segment is isomorphic to $\mathrm{On}$. Otherwise $B$ is a subclass of $\{x\in C : (0,1)\preceq x\}$, which is also isomorphic to $\mathrm{On}$, so $B$ has the minimal element.
You can see that the order-type of $C$ is $\mathrm{On+On}$ and it is not isomorphic to $\mathrm{On}$ with usual order. More direct proof of $C\not\cong \mathrm{On}$ comes from examining possible sizes of initial segments: $C$ has a proper-class sized initial segment whereas $\mathrm{On}$ is not.
However, unlike $\mathrm{On}$, no proper class with membership relation is isomorphic to $C$. This is because if $(B,\in)$ is well-ordered proper class then its initial segments must be a set.
A: A trivial example is $\operatorname{On} \cup \{*\}$ for some $* \not \in \operatorname{On}$ that is put "on top", i.e. we extend the order on $\operatorname{On}$ by letting $\alpha < *$ for all $\alpha \in \operatorname{On}$. Clearly $<$ remains to be a strict well order and its order type is "$\operatorname{On} + 1$", since $*$ has class many predecessors.
