Bridge obstruction to regular edge colouring

An $r$-regular graph with $r\ge 2$ and a bridge in a graph separating an odd number of vertices from the rest cannot have an $r$-colour proper edge colouring. Such a colouring would partition the edges into $r$ perfect matchings. Each perfect matching must cross the odd-sided cut of a graph, so each would have to contain its unique crossing edge, the bridge in a graph. An edge cannot belong to two colour classes.

For odd $r$, every bridge in a graph in an $r$-regular graph automatically has odd-sized sides: if $S$ is either resulting connected component of a graph, double counting gives $r|S|=2|E(S)|+1$, so parity forces $|S|$ odd. Thus an odd-$r$ regular graph with a bridge in a graph is a class two regular graph.