See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In 1,4-addition (conjugate addition) of HCl to a conjugated diene, the proton adds to C1 and the chloride adds to C4, generating an allylic intermediate where the double bond migrates to the 2,3-position. Step 1: Identify the diene. From the skeletal structure shown, the diene appears to be 2-methyl-1,3-pentadiene (or a similar substituted conjugated diene). Numbering the conjugated system as C1=C2-C3=C4. Step 2: In 1,4-addition, H+ adds to C1 (the less substituted end) and Cl- adds to C4 (the more substituted end), or H+ adds to C4 and Cl- adds to C1, depending on which gives the more stable carbocation intermediate. The proton preferentially adds to give the more stable allylic carbocation. Step 3: H+ adds to C1, generating an allylic carbocation delocalized over C2-C3-C4. The resulting double bond after 1,4-addition sits between C2 and C3 (internal position), and Cl attaches at C4. Step 4: Examining option (a): It shows an internal double bond with the Cl placed at the terminal carbon bearing geminal methyl groups. This corresponds to the 1,4-addition product where H adds to one end of the conjugated system and Cl adds to the other end (C4), with the remaining double bond now internal between C2-C3. This is consistent with 1,4-addition regiochemistry. Step 5: Why other options fail: - Option (b): Shows Cl on a secondary (not the terminal gem-dimethyl) carbon, inconsistent with 1,4-addition regiochemistry for this diene. - Option (c): Shows a terminal (exo) double bond remaining, which would be a 1,2-addition product (direct addition across one of the double bonds). - Option (d): Shows a fully shifted product with Cl at C1 and a terminal isopropylidene group, not consistent with standard 1,4-HCl addition. Therefore, option (a) correctly represents the 1,4-addition product with the internal double bond and chlorine at C4. Therefore, the correct answer is A.