See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Bromine addition to alkenes proceeds via anti addition through a bromonium ion intermediate, meaning the two bromine atoms add to opposite faces of the double bond. Step 1: Identify the substrate. Trans-2-butene has the two methyl groups on opposite sides of the double bond (E configuration). The two carbons of the double bond each bear one methyl and one hydrogen. Step 2: Mechanism. Br2 attacks the pi bond to form a cyclic bromonium ion intermediate. The nucleophilic bromide then attacks from the back face (anti addition). Step 3: Stereochemical outcome. Trans-2-butene is achiral and symmetric. Anti addition of Br2 generates two new stereocenters at C2 and C3. Due to the trans geometry of the starting alkene, anti addition produces the (2R,3R) and (2S,3S) dibromide... wait, let me reconsider. Actually, for trans-2-butene undergoing anti addition: the two faces of the bromonium ion are equivalent due to the symmetry of trans-2-butene. Attack on either face of the bromonium ion gives the same product. The product formed is (2R,3S)-2,3-dibromobutane, which is the meso compound. Both faces of attack give the same meso compound because the molecule has an internal plane of symmetry. Step 4: The meso compound is achiral (it has an internal mirror plane making the two stereocenters cancel each other out), so it is not optically active and is not a racemic mixture. Step 5: Why other options fail: - (b) racemic: A racemic mixture would require formation of equal amounts of (R,R) and (S,S) enantiomers, which does not occur here; only the meso compound forms. - (d) optically active: The meso compound is optically inactive due to internal symmetry. - (a) achiral and (c) meso: The meso compound IS achiral (by definition, a meso compound is achiral despite having stereocenters), so both (a) and (c) correctly describe the product. Therefore, the correct answer is A,C.