H_2 + CO + alkene xrightarrow{text{catalyst}} | vedclass

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Question

(B) The reaction described is the hydroformylation (or oxo process) of an alkene. Taking the alkene as $R-CH=CH_2$,the catalyst used is $Co_2(CO)_8$.

Vedclass · Accepted Answer

aldehyde,oxidation

Vedclass · Answer

(B) The reaction described is the hydroformylation (or oxo process) of an alkene. Taking the alkene as $R-CH=CH_2$,the catalyst used is $Co_2(CO)_8$. In the first step,the alkene reacts with $CO$ and $H_2$ to form an aldehyde as a stable intermediate. The conversion of an alkene to an aldehyde involves the addition of a formyl group $(-CHO)$,which is an oxidation process relative to the alkene. Subsequently,the aldehyde undergoes hydrogenation to form a primary alcohol. The reaction sequence is: $R-CH=CH_2 + CO + H_2$ $\xrightarrow{Co_2(CO)_8} R-CH_2-CH_2-CHO$ $\xrightarrow{H_2} R-CH_2-CH_2-CH_2OH$. Thus,the stable intermediate is an aldehyde and the nature of the initial step is oxidation.

$H_2 + CO +$ alkene $\xrightarrow{\text{catalyst}} 1^\circ$ alcohol. What is the stable intermediate and the nature of the reaction?

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