See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type and mechanism: The reaction is SN2 with NaI in acetone (a classic Finkelstein-type condition). SN2 proceeds with inversion of configuration (Walden inversion) at the carbon bearing the leaving group. Step 2 - Identify the stereocenter undergoing substitution: The leaving group Br is on the upper carbon. In the starting material, the upper carbon has: CH3 (top), Br (left, coming toward viewer on wedge bond), H (right, coming toward viewer on wedge bond), and a bond to the lower carbon going down. So Br is on a bold/wedge bond pointing to the left. Step 3 - Apply SN2 inversion: In SN2, the nucleophile (I-) attacks from the back side relative to the leaving group (Br). Since Br is on the left side (wedge, coming toward viewer), the nucleophile I- attacks from the right/back side, and inversion occurs. After inversion, I will occupy the position that was previously occupied by the back-lobe — meaning I ends up on a wedge bond on the RIGHT side of the upper carbon, and what was previously the H on the right wedge bond now moves to the left wedge bond position. Step 4 - Determine the configuration of the product upper carbon: Starting configuration at upper carbon: Br (left, wedge), H (right, wedge), CH3 (up), lower-C (down). After inversion: I (right, wedge), H (left, wedge), CH3 (up), lower-C (down). This means in the product, the upper carbon has H on the LEFT (bold/wedge) and I on the RIGHT (bold/wedge). Step 5 - The lower carbon is not involved in the SN2 reaction, so its configuration remains unchanged: H (left, bold/wedge), CH3 (right, bold/wedge), CH2-CH3 (down). Step 6 - Match with options: Option (b) shows exactly this: upper carbon has CH3 (up), H (left, bold/wedge), I (right, bold/wedge); lower carbon has H (left, bold/wedge), CH3 (right, bold/wedge), CH2-CH3 (down). This is the inverted product at the upper carbon with the lower carbon unchanged. Step 7 - Why other options fail: - Option (a): I is on the left wedge and H on the right wedge — this is retention of configuration, not inversion. Incorrect. - Option (c): CH3 replaces H on the upper carbon, which is structurally wrong (changes the connectivity). - Option (d): The lower carbon configuration is changed (CH3 and H swapped), but SN2 only inverts the carbon bearing the leaving group; the lower carbon is unaffected. Therefore, the correct answer is B.