To solve this problem, we need to understand the reaction of acetone with ethyl magnesium bromide ($C_2H_5MgBr$) followed by hydrolysis.$\newline$This is a Grignard reaction, which involves the following steps:$\newline$1. Reaction of acetone with $C_2H_5MgBr: \newline$• Acetone is a ketone with the structure $(CH_3)_2C=O. \newline$• The Grignard reagent $C_2H_5MgBr$ acts as a nucleophile, attacking the carbonyl carbon of acetone.$\newline$• This results in the formation of an alkoxide intermediate: $(CH_3)_2C(OMgBr)C_2H_5. \newline$2. Hydrolysis of the alkoxide intermediate:$\newline$• The alkoxide is then treated with water and an acid ($H^+$), which protonates the alkoxide to form an alcohol.$\newline$• The final product is $(CH_3)_2C(OH)C_2H_5,$ which is 2-methylbutan-2-ol.$\newline$Analyzing the options:$\newline$• Option 1: 2-methylpropan-2-ol is incorrect because it has one less carbon atom.$\newline$• Option 2: Pentan-2-ol is incorrect because the methyl group is not at the correct position.$\newline$• Option 3: Pentan-3-ol is incorrect because the hydroxyl group is not at the correct position.$\newline$• Option 4: 2-methylbutan-2-ol is correct as it matches the structure of the product.$\newline$Therefore, the correct IUPAC name of the product is 2-methylbutan-2-ol, corresponding to Option 4.