See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The substrate is (1-phenylbut-3-en-1-yl) or more precisely a benzene ring bearing a -CH(CH3)-CH=CH2 group (i.e., the double bond is terminal, between C3 and C4 of the side chain). The product is a benzene ring bearing -CH(CH3)-CH(OH)-CH3, meaning the OH has been added to the internal carbon (C3, Markovnikov position) and a methyl group appears at the terminal carbon - wait, let me re-examine. Looking carefully: Starting material has CH3CHCH=CH2 on benzene (a butenyl group: C1 of chain is CH3, C2 is CH bearing benzene, C3=C4 with CH=CH2). The product is CH3CH-CHCH3 with OH on the second carbon of the product side chain. So the double bond CH=CH2 has been hydrated to give -CH(OH)-CH3, which is Markovnikov addition of water. Option (a): H2O/H+ - acid-catalyzed hydration follows Markovnikov's rule, adding OH to the more substituted carbon of the double bond. The double bond is terminal (CH=CH2), so OH would go to the internal carbon (CH) giving -CH(OH)-CH3. This WOULD give the Markovnikov product. However, under acidic conditions with a nearby benzylic position, carbocation rearrangements or other complications can occur. The benzylic carbocation at C2 is more stable than the secondary carbocation at C3, so hydride shift could occur giving a different regiochemistry. This makes option (a) unreliable or giving a mixture/wrong product. Option (b): BH3.THF followed by H2O2/OH- is hydroboration-oxidation, which gives anti-Markovnikov addition of water. BH3 would add boron to the terminal carbon (CH2) giving OH at the terminal carbon after oxidation, yielding -CH2-CH2OH or -CH=CH2 -> -CH2CH2OH. This would give the anti-Markovnikov alcohol (OH at terminal carbon), NOT the observed product with OH at the internal carbon. So option (b) does not give the correct product. Option (c): Hg(OCOCH3)2/H2O followed by NaBH4/NaOH is oxymercuration-demercuration. This reaction gives Markovnikov addition of water without carbocation rearrangements (because the mercurinium ion mechanism avoids free carbocations). The OH adds to the more substituted (internal) carbon of the double bond: CH=CH2 -> -CH(OH)-CH3. This gives exactly the observed product reliably and without rearrangement. Why (a) fails: Acid-catalyzed hydration via carbocation can lead to rearrangements due to the adjacent benzylic carbon being more stable; the desired product may not be the major product. Why (b) fails: Hydroboration-oxidation gives anti-Markovnikov product (OH at terminal carbon), which is not the observed product. Why (d) fails: Since (a) and (b) do not reliably give the correct product, 'all are possible' is incorrect. Only option (c) gives the correct Markovnikov product without rearrangement. Therefore, the correct answer is C.