# Alkyl insertion with cobalt and CO

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)?

• I would think that at least one of the products would be acetone due to $\alpha$ migratory insertion and reductive elimination. – Zhe May 20 at 23:27
• 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. – orthocresol May 20 at 23:28