Compound A and C in the following reaction ar | vedclass

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(B) $CH_3CHO$ reacts with $CH_3MgBr$ followed by $H_2O$ to form $CH_3-CH(OH)-CH_3$ ($A$,Propan$-2-$ol).$A$ undergoes acid-catalyzed dehydration with $

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Positional Isomer

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(B) $CH_3CHO$ reacts with $CH_3MgBr$ followed by $H_2O$ to form $CH_3-CH(OH)-CH_3$ ($A$,Propan$-2-$ol).$A$ undergoes acid-catalyzed dehydration with $H_2SO_4$ at $\Delta$ to form $CH_3-CH=CH_2$ ($B$,Propene).$B$ undergoes Hydroboration-Oxidation to form $CH_3-CH_2-CH_2OH$ ($C$,Propan$-1-$ol).Since $A$ is Propan$-2-$ol and $C$ is Propan$-1-$ol,they differ in the position of the hydroxyl group.Therefore,$A$ and $C$ are positional isomers.

Compound $A$ and $C$ in the following reaction are
$CH_3CHO$ $\xrightarrow[(i) \, CH_3MgBr]{(ii) \, H_2O} A$ $\xrightarrow{H_2SO_4, \Delta} B$ $\xrightarrow{\text{Hydroboration-Oxidation}} C$

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Compound $A$ and $C$ in the following reaction are$CH_3CHO$ $\xrightarrow[(i) \, CH_3MgBr]{(ii) \, H_2O} A$ $\xrightarrow{H_2SO_4, \Delta} B$ $\xrightarrow{\text{Hydroboration-Oxidation}} C$