See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: In an S_N2 reaction, the nucleophile attacks the electrophilic carbon from the back side (180° to the leaving group), and the transition state is a pentacoordinate species where both the incoming nucleophile and the leaving group are partially bonded to the central carbon. Step 1 - Identify the charges in the transition state. The nucleophile here is R-O^- (an alkoxide, carrying a full negative charge). As it begins to form a bond with the carbon, it partially loses its negative charge — so in the transition state, OR carries a partial negative charge, δ(-). The leaving group is Br^- (bromide). As the C-Br bond begins to break heterolytically (with both electrons going to Br), Br begins to acquire negative character — so in the transition state, Br also carries a partial negative charge, δ(-). Step 2 - Overall charge balance. The reactant side has one full negative charge (on R-O^-). In the transition state, that negative charge is distributed between the incoming OR and the departing Br, both bearing δ(-). This is consistent with charge conservation. Step 3 - Evaluate the options. (a) OR is δ(-) and Br is δ(+): Incorrect. Br cannot become δ(+) while breaking a C-Br bond heterolytically toward Br. (b) OR is δ(+) and Br is δ(+): Incorrect. The nucleophile R-O^- cannot become δ(+) as it donates electrons. (c) OR is δ(+) and Br is δ(-): Incorrect. The nucleophile cannot become δ(+). (d) OR is δ(-) and Br is δ(-): Correct. The incoming nucleophile retains partial negative character, and the leaving group acquires partial negative character as the bond breaks. Step 4 - Geometry check. Option (d) also shows the correct trigonal bipyramidal-like transition state with OR and Br on opposite sides (backside attack), consistent with S_N2 stereochemistry. Therefore, the correct answer is D.