Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is $C{H}_{3}C{H}_{2}CH=C{H}_2\underset{Rh\text{ catalyst}}{\overset{H_2/CO}{\to }}$

Imagine you are an industrial chemist tasked with creating a long-chain aldehyde from a simple alkene. How would you do it? The answer lies in a brilliant and widely used chemical process known as hydroformylation, or the oxo process.

In this reaction, an alkene is treated with a mixture of hydrogen gas ($H_2$) and carbon monoxide ($CO$), often referred to as synthesis gas or syngas.

To make this magic happen, a transition metal catalyst, typically Rhodium ($Rh$) or Cobalt ($Co$), is employed. Let's dive into the fascinating mechanism behind this transformation.

The reaction doesn't just happen spontaneously. The Rhodium catalyst plays a crucial role in activating the $H_2$ and $CO$ molecules.

Through a series of complex coordination steps, the catalyst helps generate the active electrophilic species: the formyl cation ($[H-\stackrel{+}{C}=O]$).

This highly reactive intermediate is now primed and ready to attack the electron-rich double bond of our starting material, 1-butene ($C{H}_{3}C{H}_{2}CH=C{H}_2$).

Now comes the critical step that determines our final product. The formyl cation approaches the alkene. But which carbon does it attach to?

According to Markovnikov's rule, the electrophile will add to the carbon atom that results in the formation of the most stable carbocation intermediate.

If the formyl group attaches to the terminal carbon (C1), a secondary carbocation is formed at C2 ($C{H}_{3}C{H}_2-\stackrel{+}{C}H-C{H}_2-CHO$). If it were to attach to C2, a less stable primary carbocation would form. Therefore, the reaction proceeds via the more stable secondary carbocation.

With our stable secondary carbocation formed, the final step is rapid and decisive.

A hydride ion ($H^{-}$), generated earlier in the catalytic cycle, swoops in and attacks the positively charged carbon.

This neutralizes the charge and completes the addition across the double bond, yielding our major product: pentanal ($C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}CHO$), a straight-chain aldehyde.

I know mechanisms can sometimes feel abstract, but visualizing the flow of electrons and the pursuit of stability makes it all click into place. Always trust the stability of your intermediates!