See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When an alkene reacts with a halogen (Br2), the pi electrons of the double bond act as a nucleophile and attack the electrophilic bromine molecule. The bromine molecule is polarized (Br-Br with induced or permanent partial charges), making it an electrophile. Step 1: Br2 approaches the electron-rich double bond of cyclohexene. The pi electrons attack one bromine atom, forming a bromonium ion intermediate (a cyclic, three-membered ring with a positive bromine) and releasing bromide ion (Br-). Step 2: The bromide ion (nucleophile) attacks one of the two carbons of the bromonium ion from the back side (anti addition), opening the ring and producing the trans-1,2-dibromocyclohexane product. Step 3: Since two atoms (both bromines) are added across the double bond and the reaction is initiated by an electrophile (Br2 acting as an electrophile toward the pi bond), this is classified as electrophilic addition. Why other options fail: (a) Nucleophilic addition would require the attacking species to be a nucleophile attacking an electrophilic carbon (e.g., carbonyl), not applicable here. (b) Nucleophilic substitution involves replacement of a leaving group by a nucleophile, not addition across a double bond. (d) Electrophilic substitution occurs in aromatic systems where a hydrogen is replaced, not in simple alkenes. (e) Free radical substitution involves homolytic cleavage and radical intermediates, not the ionic bromonium mechanism seen here. Therefore, the correct answer is C.