Reaction of $\ce{(Cp)Co(Me)2(CO)}$ in the presence of excess $\ce{CO}$ yields two different organic products, $\ce{A}$ and $\ce{B}.$ The only metal-containing product is $\ce{(Cp)Co(CO)2}.$ The IR spectrum of $\ce{A}$ contains a peak at $\pu{1731 cm-1}$ whereas that of $\ce{B}$ has no peak in the carbonyl region. Write down mechanisms that explain the formation of $\ce{A}$ and $\ce{B}$.

From the IR, I think it is likely that an aldehyde is being produced, and looking at the reaction I assume it is the alkyl insertion of the methyl into the $\ce{CO}$ forming ethanal $(\ce{A}).$

$\ce{B}$ is confusing me however as the only other reactants are $\ce{Me}$ and $\ce{CO}$ (as the $\ce{Cp}$ remains in the metal product). Is some other reaction occurring (is $\ce{B}$ simply the enolised form of $\ce{A}$ formed from a β-hydride shift)?

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    $\begingroup$ I would think that at least one of the products would be acetone due to $\alpha$ migratory insertion and reductive elimination. $\endgroup$ – Zhe May 20 at 23:27
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    $\begingroup$ I concur with @Zhe. The other one, I would guess, is ethane. To form acetaldehyde you need a proton from somewhere to make the aldehyde C–H bond, and this system doesn't have it. $\endgroup$ – orthocresol May 20 at 23:28

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