See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the two olefinic (alkene) bonds in the starting material: (1) The endocyclic double bond in the cyclohexene ring, which is conjugated with the aldehyde (CHO) group, making it an alpha,beta-unsaturated aldehyde (enal). (2) The exocyclic isopropenyl double bond (CH2=C(CH3)-) at C4, which is a simple, isolated terminal alkene. Step 2 - Determine which double bond reacts preferentially with HBr (a mineral acid, electrophilic addition): The problem states 'addition of a mineral acid to an olefin bond.' The endocyclic double bond is conjugated with the electron-withdrawing CHO group (alpha,beta-unsaturated system), making it less electron-rich and less reactive toward electrophilic addition. The isolated exocyclic isopropenyl double bond (CH2=C(CH3)-) is more electron-rich and more reactive toward electrophilic addition with HBr. Step 3 - Apply Markovnikov's rule to the isopropenyl group (CH2=C(CH3)-): H+ adds to the terminal CH2= (less substituted carbon) and Br- adds to the internal carbon bearing the methyl group (more substituted, giving tertiary carbocation intermediate). This gives -C(CH3)2Br (a tertiary bromide) at C4. Step 4 - The ring double bond (conjugated with CHO) remains intact because it is deactivated by the electron-withdrawing aldehyde and the isolated alkene reacts preferentially. Step 5 - The product therefore has: the cyclohexene ring double bond still present (with CHO), and the isopropenyl group converted to -C(CH3)2Br. This matches option (c): a cyclohexene ring bearing CHO, with -C(CH3)2Br at C4. Step 6 - Why other options fail: (a) places Br on the ring adjacent to CHO - wrong regiochemistry and wrong double bond selectivity. (b) reacts the ring double bond and gives secondary bromide from isopropenyl - violates both selectivity and Markovnikov. (d) reacts the ring double bond and leaves isopropenyl intact - wrong double bond selectivity. Therefore, the correct answer is C.