Question

Write down the Koch reaction( hydro carboxylation of alkene)?

Answer 1

Koch reaction is a strongly acid catalyzed carbonylation reaction that occurs at high pressure and temperature. It is an additional reaction of carbon monoxide and water to the double bond of alkene to produce carboxylic acid. Generally, Carbonylation is the reaction that involves carbon monoxide. In this reaction, CO is introduced to the alkene substrate.

Complete step by step answer:
Now we will study about the reaction and the product formed in Koch reaction;
This Koch method is profoundly used in industries to produce propionic acid from ethylene which is used as preservative for food especially baked food and animal feed and grains in its salts form.

$C{H_2} = {\text{ }}C{H_2}\; + {\text{ }}CO{\text{ }} + {\text{ }}{H_2}O{\text{ }}\xrightarrow{{H + }}{\text{ }}C{H_3} - C{H_2}COOH$
$\text{ (Ethylene) (Propionic acid) }$

Chemical Reaction:
Propylene on hydro carboxylation is converted to carboxylic acid which is major product as follows:
C{H_3} - {\text{ }}CH{\text{ }} = {\text{ }}C{H_2}\; + {\text{ }}CO{\text{ }} + {\text{ }}{H_2}O\xrightarrow{{{H^ + }}}$$$$C{H_3} - CH\left( {COOH} \right) - C{H_3} + {\text{ }}C{H_3} - C{H_2} - C{H_2}\left( {COOH} \right)
(Propylene) (Tertiary carboxylic acid)
(Major product)
Mechanism:
Strong mineral acids like $H_2SO_4$, HF or $BF_3$ are used as catalysts in the reaction that protonate the alkene which is then attacked by the CO resulting into the formation of carbocation. Then the resultant carbocation produces tertiary carboxylic acid by hydrolysis. Alkyl and hydride shift in the carbocation occurs to produce the thermodynamically favored tertiary carbocation.
${H^ + }$
C{H_3} - {\text{ }}CH{\text{ }} = $$$$C{H_2} $\to$ $C{H_3} - {\text{ }}C{H^ + } - {\text{ }}C{H_3}\;$+${C^ - } \equiv {O^{ + \;\;}}$ $\to$
$\left[ {C{H_3} - {\text{ }}CH(C \equiv {O^ + }} \right)-C{H_3} \to \leftarrow \;\;$
$C{H_3} - {\text{ }}CH({C^ + } = O)-{\text{ }}C{H_3}]$
$H - O - H$
C{H_3} - CH\left( {COOH} \right) - {\text{ }}C{H_3}$$$$C{H_3} - {\text{ }}CH(C\left( {O{H_2}^ + } \right){\text{ }} = O)-{\text{ }}C{H_3}

The product in the reaction is tertiary carboxylic acid with an increase in one carbon atom from the substrate alkene. This increase in carbon number can be seen in the above given reaction where propene is giving butanoic acid as the product.

Note: an alkene is a hydrocarbon that has a double bond character between carbon–carbon. Alkenes have many applications in the industry. They are used in the manufacturing plastics, detergents, fuels alcohol and liquors.