See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction: 2-butene (CH3-CH=CH-CH3) reacts with HBr under peroxide conditions (R2O2, heat), which is a free radical (Anti-Markovnikov) addition. In Anti-Markovnikov addition of HBr to 2-butene, the Br adds to the less substituted carbon. However, since both carbons of the double bond in 2-butene are equivalent (both are internal carbons bearing one CH3 and one H), Anti-Markovnikov and Markovnikov addition give the same product: 2-bromobutane (CH3-CHBr-CH2-CH3). Wait - let me reconsider. 2-butene is CH3-CH=CH-CH3. Both carbons of the double bond are secondary. Anti-Markovnikov addition places Br on C1 or C4... Actually for a symmetrical internal alkene like 2-butene, both carbons are equivalent, so HBr addition (either Markovnikov or anti-Markovnikov) gives 2-bromobutane regardless. Step 2 - Identify the product: The product is 2-bromobutane (CH3CHBrCH2CH3). C2 is a chiral center (bonded to CH3, Br, H, and CH2CH3 - all four different groups). Step 3 - Stereochemical analysis of free radical addition: Free radical addition proceeds via a planar (sp2) carbon radical intermediate. The bromine radical can attack from either face of the intermediate carbon radical with equal probability. This means both R and S configurations at C2 are formed in equal amounts. Step 4 - Conclusion on optical activity: Since both enantiomers (R-2-bromobutane and S-2-bromobutane) are formed in equal amounts, the product is a racemic mixture. A racemic mixture has no net optical activity because the two enantiomers cancel each other out. Option (b) Diastereomer is incorrect because there is only one chiral center, so diastereomers cannot exist. Option (c) Meso is incorrect because a meso compound requires at least two chiral centers with internal symmetry. Option (d) Optically pure enantiomer is incorrect because the radical mechanism gives equal attack from both faces. Therefore, the correct answer is A.