See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is (CH3)2CHCH2OH, which is 2-methyl-1-propanol (isobutanol). Step 2: Reaction with PBr3. PBr3 converts the primary alcohol to the corresponding alkyl bromide: (CH3)2CHCH2Br (1-bromo-2-methylpropane). Step 3: Reaction with Mg (in dry ether). Mg converts the alkyl bromide into a Grignard reagent: (CH3)2CHCH2MgBr. Step 4: Reaction with ethylene oxide (H2C-CH2 with an oxygen bridge, i.e., oxirane). The Grignard reagent acts as a nucleophile and attacks ethylene oxide. The carbon nucleophile (CH3)2CHCH2- opens the epoxide ring, adding a -CH2CH2O- unit. After ring opening, the oxygen carries a negative charge (as the magnesium alkoxide). The product of ring opening: (CH3)2CHCH2-CH2CH2-OMgBr. Step 5: Aqueous acid workup (H3O+). The magnesium alkoxide is protonated to give the primary alcohol: (CH3)2CHCH2CH2CH2OH. This is 4-methyl-1-pentanol, which corresponds to option (d): (CH3)2CHCH2CH2CH2OH. Why other options fail: - Option (a) adds only one carbon and introduces a secondary OH — inconsistent with ethylene oxide ring opening adding two carbons. - Option (b) (CH3)2CHCH2CH2OH adds only two carbons total from start but the Grignard carbon plus ethylene oxide should give a chain extended by 2 carbons beyond the original alkyl group, giving 6 carbons total — option (b) has only 5 carbons and doesn't account for the full chain. - Option (c) appears to be a branched product inconsistent with the straightforward Grignard/ethylene oxide reaction. - Option (d) correctly shows the Grignard carbon (4 carbons from isobutyl) plus two carbons from ethylene oxide, giving a 6-carbon primary alcohol: (CH3)2CHCH2CH2CH2OH. Therefore, the correct answer is D.