See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the reaction type. KCN in EtOH is a classic SN2 reaction. CN⁻ is the nucleophile and Br is the leaving group. SN2 proceeds with inversion of configuration (Walden inversion) at the stereocenter. Step 2 – Assign configuration of the reactant. The reactant has a central carbon with four groups: CO2CH3, D, Br, and H. From the drawing: D is on a wedge (coming toward viewer, left), Br is on a wedge (coming toward viewer, right), H is going down (into page, dashed), and CO2CH3 is up. To assign R/S, we need the priority order of the four groups: 1. Br (highest, atomic number 35) 2. CO2CH3 (oxygen atoms attached directly) 3. D (deuterium, mass number 2 > H) 4. H (lowest) With H pointing away from the viewer (on a dash going down/back), we look at the arrangement of Br → CO2CH3 → D. In the structure, with the given spatial arrangement (D on left wedge, Br on right wedge, H on dash downward, CO2CH3 up), placing the lowest priority (H) going away from us: the sequence Br(1) → CO2CH3(2) → D(3) traces a clockwise (R) direction. Reactant configuration = S. Wait – let me re-examine carefully. H is shown going downward (into page = dash). With H pointing away, we look at Br, CO2CH3, D in decreasing priority order: Br (1) is on the right (bold bond, toward viewer) CO2CH3 (2) is at top D (3) is on the left (bold bond, toward viewer) Since H is pointing away (into the page), this is a valid direct reading. Looking from the front with H back: going from Br (right) → CO2CH3 (top) → D (left) is counterclockwise = S. Reactant configuration = S. Step 3 – Assign configuration of the product after SN2 (inversion). SN2 inverts the configuration. If the reactant is S, the product (before re-prioritization) would be R. However, we must re-assign priorities because the leaving group Br is replaced by CN. New groups on the product carbon: CO2CH3, CN, D, H. Priority order: 1. CO2CH3 (carbon bonded to two oxygens) 2. CN (carbon triple-bonded to nitrogen; expanded: C(N,N,N)) Actually CN: the nitrile carbon is attached to N (triple bond, so counts as C bonded to N,N,N by phantom atoms). CO2CH3: the carbonyl carbon is attached to O,O,O (phantom atoms). So CO2CH3 > CN. 1. CO2CH3 2. CN 3. D 4. H Inversion of the center changes S → but we must re-evaluate with new priorities. In the product, CN replaces Br with inversion: the spatial arrangement is inverted relative to the reactant. With the new priority order CO2CH3 > CN > D > H, the inverted center gives S configuration. Product configuration = S. Step 4 – Conclusion. Reactant = S, Product = S → answer is (d) S, S. Step 5 – Why other options fail. (a) R, R – incorrect because reactant is S not R. (b) R, S – incorrect reactant assignment. (c) S, R – incorrect product assignment; even though SN2 inverts geometry, re-prioritization with CN instead of Br gives S not R. Therefore, the correct answer is D.