= Solution
The anionic <nitrogen> donor replaces iodide to give
$$
\boxed{A=\mathrm{CpFe(CO)_2(\kappa N\!\text{-}C_4H_4N)},\qquad \text{byproduct}=\mathrm{KI}.}
$$
Here Cp denotes $\eta^5$ cyclopentadienyl and $\kappa N$ specifies attachment through the ring <nitrogen> of the pyrrolyl <anion>, $\mathrm{C_4H_4N^-}$. Fe(II), $d^6$, receives six <electrons> from Cp, four from the two CO ligands and two from an N-bound <pyrrolyl ligand>: $6+6+4+2=18$.
Two coupled CO stretching modes, approximately symmetric and antisymmetric combinations, account for the two strong carbonyl-region absorptions. The bent two-CO arrangement of this <piano-stool complex> permits both combinations to be infrared active. They are not evidence for two different chemical products.
Heating expels \b[two CO molecules] and changes the binding of the <pyrrolyl ligand> from $\eta^1$ to $\eta^5$, giving the sandwich \b[<azaferrocene>],
$$
\boxed{B=\mathrm{Fe(\eta^5-C_5H_5)(\eta^5-C_4H_4N)}.}
$$
The four <electrons> lost with CO are replaced by the four additional <electrons> donated on changing the <hapticity> of the <pyrrolyl ligand>. Thus $B$ is also Fe(II), $d^6$, and 18-electron: $6+6+6=18$. Each complex has one time-averaged Cp proton environment, two pairs of protons from the <pyrrolyl ligand>, one adjacent to N and one farther from N. Hence \b[three proton environments, with integrals $5:2:2$], fit both structures. For <proton nuclear magnetic resonance>, three signals means three <chemical shift> environments; the individual resonances can still show coupling.
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