The major product of the following reaction is :
CH$_3$CH$_2$CH=CH$_2$ + H$_2$/CO $\xrightarrow{\text{Rh catalyst}}$
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Hydroformylation (Oxo process) converts alkenes into aldehydes. Remember that with modern catalysts like those based on rhodium (Rh), the reaction favors the formation of the straight-chain (n-alkanal) product over the branched (iso-alkanal) product, especially with terminal alkenes.
The given reaction is the hydroformylation of an alkene (but-1-ene) using synthesis gas (a mixture of CO and H$_2$) in the presence of a rhodium catalyst. This reaction is also known as the Oxo process.
In this reaction, a hydrogen atom (H) and a formyl group (-CHO) are added across the double bond.
For an unsymmetrical alkene like but-1-ene, the addition can occur in two ways:
1. Addition to the terminal carbon: This results in a linear aldehyde.
$\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$ (pentanal)
2. Addition to the second carbon: This results in a branched aldehyde.
$\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}(\text{CHO})\text{CH}_3$ (2-methylbutanal)
The choice of catalyst and reaction conditions influences the ratio of the linear to branched product (n/iso ratio).
Rhodium-based catalysts, especially those with phosphine ligands, are known to be highly selective for the formation of the linear aldehyde. The linear product is generally the desired major product in industrial applications.
Therefore, the major product of this reaction is the linear aldehyde, pentanal.
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