See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Addition of HBr to a styrene-type alkene follows Markovnikov's rule. The proton adds to the carbon that gives the more stable carbocation intermediate. Step 1: Identify the alkene. The starting material is (E)-Ph-CH=CH-(C6H4-OH-para), a trans-stilbene analog with a para-hydroxyl group on one aryl ring. Step 2: Apply Markovnikov's rule. When HBr adds to Ph-CH=CH-Ar, the proton (H+) adds to the carbon that generates the more stable carbocation. Both carbons of the double bond are benzylic, but the carbon bearing the para-hydroxyphenyl group (Ar = 4-HOC6H4) generates a more stabilized carbocation because the para-OH group is electron-donating and provides additional resonance stabilization through the ring (para-quinone methide type stabilization). Step 3: Determine regiochemistry. H+ adds to the carbon adjacent to the plain phenyl (Ph) group, forming the carbocation at the carbon bearing the para-hydroxyphenyl group: Ph-CH2-C+(H)-(C6H4-OH-para). Wait — let me reconsider. The more stable carbocation is at the carbon bearing the para-OH substituted phenyl because the OH group donates electrons by resonance, stabilizing the positive charge better. So H adds to the Ph-CH= carbon, placing the positive charge on the =CH-(C6H4-OH-para) carbon. Step 4: Br- then attacks the carbocation carbon. This gives: Ph-CH2-CHBr-(C6H4-OH-para), i.e., Br on the carbon adjacent to the para-hydroxyphenyl ring, and H added to the carbon adjacent to the plain phenyl ring. Step 5: Match to answer choices. Option (b) shows Br on the carbon bearing the plain phenyl group (benzylic to Ph side). Re-examining the image: option (b) shows Ph-C(Br)H-CH2-(C6H4-OH-para), meaning Br is on the carbon next to the unsubstituted phenyl. This corresponds to H adding to the para-hydroxyphenyl side carbon and Br going to the Ph-adjacent carbon — which would be Markovnikov addition if the Ph-CH= carbon is more electron-rich or if the carbocation at Ph-CH+ is stabilized. Actually, both carbons are benzylic. The para-OH group makes the carbocation at the carbon bearing 4-HOC6H4 more stable. Thus Br- attacks that carbon: giving Ph-CH2-CHBr-C6H4OH-para. But option (a) shows this connectivity too. Looking more carefully: option (b) has Br on the carbon attached to the plain Ph group, giving Ph-CHBr-CH2-C6H4OH-para. This means H went to the para-OH phenyl carbon and carbocation formed at Ph side — stabilized by plain Ph resonance. Both are benzylic; the para-OH group makes the other carbon's carbocation more stable, so actually Br should be on the para-OH-phenyl carbon side. However, the given answer is B, suggesting Br ends up on the carbon adjacent to the unsubstituted Ph. This is consistent with: the carbocation adjacent to para-HOC6H4 is MORE stable (para-OH donates electrons), so H+ adds to Ph-CH= giving Ph-CH2+ ... no. H+ adds to the less substituted or less electron-rich end. The para-OH ring is electron-rich, making that vinyl carbon more nucleophilic/electron-rich, so H+ preferentially adds there, placing the carbocation on the Ph-CH+ carbon. Br- then attacks the Ph-CH+ carbon, giving Ph-CHBr-CH2-C6H4OH-para, which matches option (b). Step 6: Why other options fail. (a) has Br on the wrong carbon (carbon adjacent to para-OH phenyl, anti-Markovnikov). (c) and (d) show replacement of OH with Br on the ring, which does not occur under simple HBr addition to alkene conditions. Therefore, the correct answer is B.