The given reaction involves a Grignard reagent (\(\text{C}_2\text{H}_5\text{MgBr}\)) reacting with acetone (\(\text{(CH}_3\text{)}_2\text{C=O}\)) to form an alcohol (\(X\)). The alcohol undergoes a substitution reaction with hydrogen bromide (\(\text{HBr}\)) to form a bromoalkane. The final step involves heating with copper at 573 K, which leads to the formation of an alkene. The major product is 2-Bromo-3-methylbutane.
Hence, the correct answer is (3) 2-Bromo-3-methylbutane