See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify starting material (I)** $CH_3CHOHCH_3$ (propan-2-ol) reacts with $PBr_3$ to give the alkyl bromide: $$CH_3CHOHCH_3 \xrightarrow{PBr_3} CH_3CHBrCH_3 \text{ (I)}$$ **Step 2: Grignard formation (II)** The alkyl bromide reacts with $Mg$ in ether to form the Grignard reagent: $$CH_3CHBrCH_3 + Mg \rightarrow CH_3CHMgCH_3 \text{ (II)}$$ **Step 3: Epoxide ring opening (III)** The Grignard reagent attacks the epoxide (oxirane). Nucleophilic attack occurs at the less hindered carbon of the epoxide ring: $$CH_3CHMgCH_3 + \text{(epoxide)} \rightarrow CH_3CH(OCH_2CH_3)CH_3 \text{ (III)}$$ The Grignard adds, and the ring opens to form a secondary alkoxide intermediate. **Step 4: Hydrolysis (IV)** Water workup protonates the alkoxide: $$CH_3CH(OCH_2CH_3)CH_3 + H_2O \rightarrow CH_3CH(OCH_2CH_3)CH_3 \text{ (IV)}$$ This gives **isopropyl ethyl ether**: $CH_3-CH(OCH_2CH_3)-CH_3$ **Answer: (D)** represents the correct final product with the ether functional group formed.