First-order EM Feynman diagram?

A diagram which is first order in $alpha_text{EM}$ would have to have one vertex, because $alpha_text{EM}propto g^2$ where $g$ is the factor associated with each vertex (and the amplitude corresponding to the diagram gets squared). There's only one possible vertex in QED, namely the photon-electron-positron vertex, and it's impossible to arrange this in any way that conserves energy and momentum. There are only two really distinct possibilities:

• initial state photon, final state electron and positron: the final state has a rest frame and the initial state doesn't. Or vice-versa (initial state $e^-e^+$, final state $gamma$)
• initial state photon and electron, final state electron, or vice versa: in the rest frame of the final state, the total energy is $m_e c^2$, whereas the initial state has energy at least $m_e c^2 + E_gamma$

So no, there is no first-order diagram. The same argument goes to show that the corresponding process ($gammato e^+e^-$, $e^-gammato e^-$, or any variant) is kinematically forbidden. If you replace the photon with a sufficiently massive particle, like a Z boson, then it's totally fine. Of course, Z bosons aren't stable, so whatever diagram you draw that includes the $Zto e^+e^-$ vertex should probably also include whatever interaction produced the Z boson in the first place, but in theory if you had a free Z boson propagating through space, it could undergo this decay.

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Note the distinction I've drawn between a diagram and a process, which is defined by its initial and final states but incorporates many different diagrams.