See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the alkene structure. The alkene shown is (CH3)(CH2CH3)C=C(CH2CH3)(CH3). Both carbons of the double bond bear identical substituents (one methyl and one ethyl group each), making this a symmetrically substituted internal alkene. Step 2 - Reaction conditions. The reaction uses HBr with peroxide (R2O2), which initiates a free-radical (anti-Markovnikov) addition mechanism. Under radical conditions, Br• adds to the less substituted end first (anti-Markovnikov), but since both carbons are identically substituted, Br can add to either carbon with equal probability giving the same carbon skeleton. Step 3 - Identify possible products and stereochemistry. Even though the double bond is symmetrical (both carbons have same groups), upon addition of HBr the two carbons of the double bond become stereocenters. The product is: CH3-CH(Br)-CH(CH2CH3)-... wait, let me re-examine. The alkene is: C2H5(CH3)C=C(CH3)(C2H5). Adding HBr across this double bond gives: CH3(C2H5)CH-CH(CH3)(C2H5) with Br on one carbon. The product is 3-bromo-2-methyl... Actually the product is: (C2H5)(CH3)CH-CBr... No. Adding HBr: one carbon gets H, the other gets Br. Product: (CH3)(C2H5)CH-CBr(CH3)(C2H5) or (CH3)(C2H5)CBr-CH(CH3)(C2H5). Due to symmetry of the alkene, both give the same compound. The product has two adjacent stereocenters: C* bearing (H, CH3, C2H5, and the adjacent C) and C* bearing (Br, CH3, C2H5, and the adjacent C). Each stereocenter is a chiral center. Two stereocenters give up to 4 stereoisomers: (R,R), (S,S), (R,S), (S,R). The (R,R) and (S,S) are enantiomers (one pair), and (R,S) and (S,R) — we must check if a meso form exists. Since the two stereocenters have different substituents (one has Br, one has H), there is no meso form, so all 4 stereoisomers are distinct: 2 enantiomers from the syn addition and 2 enantiomers from the anti addition. Radical addition proceeds non-stereospecifically, giving a mixture of all stereoisomeric products. Thus 4 stereoisomeric products are formed. Step 4 - Why other options fail. Option (a) 2 ignores the two stereocenters. Option (c) 3 would require a meso compound which doesn't exist here since the two carbons bear different groups (H vs Br). Option (d) 6 overcounts. Therefore, the correct answer is B.