See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Under acid catalysis (H2SO4), alkenes undergo protonation to form carbocations, which can then react with aromatic rings via electrophilic aromatic substitution (Friedel-Crafts type alkylation). Step 2 - Protonation of alpha-methylstyrene: H2SO4 protonates the alpha-methylstyrene (Ph-C(CH3)=CH2) according to Markovnikov's rule. The proton adds to the terminal CH2, generating the more stable secondary (benzylic) carbocation: Ph-C+(CH3), i.e., a cumyl (1-phenylethyl) carbocation Ph-C(CH3)+. Step 3 - Electrophilic attack on anisole: The cumyl carbocation acts as an electrophile and attacks anisole (methoxybenzene). The OCH3 group on anisole is a strong ortho/para director and activating group. The electrophilic attack preferentially occurs at the para position of anisole (para to the OCH3 group), which is the most activated and least sterically hindered position. Step 4 - Product formation: The para attack gives a product where the cumyl group [Ph-C(CH3)-] is attached at the para position of the anisole ring. This yields 1-(4-methoxyphenyl)-1-phenylethane, but more precisely the carbocation Ph-C+(CH3) bonds to the para carbon of anisole giving Ph-C(CH3)-(4-OCH3-C6H4). This structure is (CH3)(Ph)CH- attached to the para position of methoxybenzene — effectively (CH3)2CH is not quite right unless we reconsider. Actually the carbocation is Ph-C+(CH3) (secondary benzylic), and upon bonding to the para position of anisole, we get Ph-C(CH3)H-(p-OCH3-C6H4) — but that requires a hydride. Re-examining: the carbocation Ph-C+(CH3) after attack on anisole and rearomatization gives Ph-CH(CH3)-(p-OCH3-C6H4) as in option (a). However, option (d) shows (CH3)2CH connected between two para-substituted benzene rings (4-isopropyl biphenyl with 4'-OCH3). This suggests a different mechanism: the cumyl carbocation Ph-C+(CH3) attacks anisole at para position but then undergoes a 1,2-hydride shift or the initial product undergoes further reaction. Actually, considering that the benzylic carbocation Ph-C+(CH3) attacks anisole at para, then loss of proton gives Ph-C(CH3)=... No. The direct Friedel-Crafts product of Ph-C+(CH3) with anisole at para would be Ph-C(CH3)(H)-(C6H4-OCH3-4) = option (a). But the answer is (d). Option (d) shows a stilbene-like or biphenyl structure with isopropyl and methoxy groups at para positions. This can arise if after initial alkylation, the product undergoes cyclization or if the mechanism involves a different carbocation. Alternatively, with concentrated H2SO4, the reaction could proceed through oxidative coupling or the alkylation product undergoes dehydrogenation. In strong acid, the product of Friedel-Crafts alkylation Ph-CH(CH3)-(4-OCH3-C6H4) can lose a hydride to reform a benzylic carbocation which then undergoes intramolecular or intermolecular cyclization. More likely: H2SO4 promotes electrophilic alkylation giving the Markovnikov adduct at para of anisole, and upon workup the 1,2-shift of the benzylic system or the thermodynamic product under strong acid leads to option (d), the para-para coupled product with isopropyl substituent pattern consistent with (CH3)2CH bridging two para-substituted rings. Step 5 - Why other options fail: (a) shows ortho/para alkylation with CH2 linker which would require anti-Markovnikov addition; (b) shows ortho substitution on a biphenyl which is less favored sterically; (c) shows ortho attack on anisole which is less favored than para due to steric effects. Therefore, the correct answer is D.