Step 1: Write the Grignard reaction between cyclopentanone and $CH_3MgBr$. Cyclopentanone has a carbonyl group ($C{=}O$) in the ring. The Grignard reagent $CH_3MgBr$ acts as a source of the methyl carbanion ($CH_3^-$), which attacks the electrophilic carbonyl carbon. Step 2: Draw the product after acidic hydrolysis. After nucleophilic addition and hydrolysis with dilute acid ($H_3O^+$), a tertiary alcohol is formed: 1-methylcyclopentanol. The ring carbon that had the $C{=}O$ now bears $-OH$ and $-CH_3$, making it tertiary. Step 3: Dehydrate 1-methylcyclopentanol. On heating with concentrated $H_2SO_4$, the tertiary alcohol undergoes E1 dehydration. The most stable alkene (endocyclic) is formed: \[ \text{1-methylcyclopentanol} \xrightarrow{\text{Conc. }H_2SO_4,\ \Delta} \text{1-methylcyclopentene} + H_2O \] Step 4: Hydrogenate 1-methylcyclopentene. Treatment with $H_2$ over a platinum catalyst ($Pt$) adds hydrogen across the double bond: \[ \text{1-methylcyclopentene} + H_2 \xrightarrow{Pt} \text{methylcyclopentane} \] Step 5: Identify the final product. Methylcyclopentane is a cyclopentane ring with a methyl group. No functional groups or double bonds remain. This is option (1). Step 6: Summarize the sequence. Cyclopentanone $\xrightarrow{CH_3MgBr,\ H_3O^+}$ 1-methylcyclopentanol $\xrightarrow{H_2SO_4}$ 1-methylcyclopentene $\xrightarrow{H_2/Pt}$ methylcyclopentane. Step 7: State the final answer. \[ \boxed{\text{Methylcyclopentane}} \]
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