Identify the product obtained when isopropyl | vedclass

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(C) Step $1$: Isopropyl magnesium chloride is $(CH_3)_2CHMgCl$.Step $2$: Reaction with dry ice $(CO_2)$ forms a complex: $(CH_3)_2CHMgCl + CO_2 \right

Vedclass · Accepted Answer

$2-	ext{Methylpropanoic acid}$

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(C) Step $1$: Isopropyl magnesium chloride is $(CH_3)_2CHMgCl$.Step $2$: Reaction with dry ice $(CO_2)$ forms a complex: $(CH_3)_2CHMgCl + CO_2 ightarrow (CH_3)_2CHCOOMgCl$.Step $3$: Acidic hydrolysis of the complex yields the carboxylic acid: $(CH_3)_2CHCOOMgCl + H_2O/H^+ ightarrow (CH_3)_2CHCOOH + Mg(OH)Cl$.Step $4$: The product $(CH_3)_2CHCOOH$ is $2-	ext{Methylpropanoic acid}$ (also known as isobutyric acid).

Identify the product obtained when isopropyl magnesium chloride in dry ether reacts with dry ice $(CO_2)$ forming a complex, which is further hydrolysed.

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