See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type and site: The reaction is SN2 with NaOH (hydroxide as nucleophile). SN2 reactions occur with inversion of configuration at the carbon bearing the leaving group. The leaving group is Cl at the left stereocenter. Step 2 - Identify which stereocenter reacts: Only the C-Cl bond is susceptible to SN2 attack. The C-OEt bond is not a leaving group under these mild basic conditions (ethoxide is not a good leaving group with NaOH). Therefore, only the carbon bearing Cl undergoes substitution. Step 3 - Determine stereochemical outcome at the reacting center: In the starting material, Cl is on a bold wedge at the left stereocenter (coming toward the viewer). SN2 proceeds with inversion, so the incoming OH will end up on a dash (going away from the viewer) at that carbon. This means the configuration at the left carbon is inverted. Step 4 - Determine stereochemical outcome at the non-reacting center: The right stereocenter (bearing OEt) is not involved in the reaction. Its configuration is retained. In the starting material, OEt is on a bold wedge at the right carbon, so it remains on a bold wedge in the product. Step 5 - Match to answer choices: The product has OH on a dash at the left carbon (inverted from Cl on wedge) and OEt on a bold wedge at the right carbon (retained). This matches option (b): left stereocenter has OH on dash, right stereocenter has OEt on bold wedge. Step 6 - Eliminate other options: - Option (a): Both groups on dashes, meaning the OEt center also changed configuration - incorrect, OEt is not displaced. - Option (c): Cl is retained (no reaction occurred) and OH appears at right carbon - incorrect, Cl is the leaving group not OEt. - Option (d): Both groups are OH on dashes, implying OEt was also displaced - NaOH cannot displace OEt under these conditions. Therefore, the correct answer is B.