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Preparation of Carboxylic Acids and Their Derivatives Questions in English
Class 12 Chemistry · 8-2.Carboxylic acids and Their derivative · Preparation of Carboxylic Acids and Their Derivatives
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Identify the product obtained when isopropyl magnesium chloride in dry ether reacts with dry ice $(CO_2)$ forming a complex, which is further hydrolysed.
A
Propanoic acid
B
Propanal
C
$2-\text{Methylpropanoic acid}$
D
$2-\text{Methylpropanol}$
Solution
(C) Step $1$: Isopropyl magnesium chloride is $(CH_3)_2CHMgCl$.
Step $2$: Reaction with dry ice $(CO_2)$ forms a complex: $(CH_3)_2CHMgCl + CO_2 \rightarrow (CH_3)_2CHCOOMgCl$.
Step $3$: Acidic hydrolysis of the complex yields the carboxylic acid: $(CH_3)_2CHCOOMgCl + H_2O/H^+ \rightarrow (CH_3)_2CHCOOH + Mg(OH)Cl$.
Step $4$: The product $(CH_3)_2CHCOOH$ is $2-\text{Methylpropanoic acid}$ (also known as isobutyric acid).
Step $2$: Reaction with dry ice $(CO_2)$ forms a complex: $(CH_3)_2CHMgCl + CO_2 \rightarrow (CH_3)_2CHCOOMgCl$.
Step $3$: Acidic hydrolysis of the complex yields the carboxylic acid: $(CH_3)_2CHCOOMgCl + H_2O/H^+ \rightarrow (CH_3)_2CHCOOH + Mg(OH)Cl$.
Step $4$: The product $(CH_3)_2CHCOOH$ is $2-\text{Methylpropanoic acid}$ (also known as isobutyric acid).
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