See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the product configuration. The product shown is 2-methoxybutane. The wedge-dash drawing shows the central carbon (C2) with: - OCH3 on a wedge (coming toward viewer) - Et on a dashed wedge (going away from viewer, to the left) - Me on a plain bond (to the right) - H at the bottom (in the plane) To assign configuration: OCH3 (wedge, front), Et (dash, back), Me (right, plane), H (bottom, plane). Priority order: OCH3 > Et > Me > H. With H pointing roughly in the plane (not away), we need to carefully assign R/S. In the product drawing, Et is on a dash (going back), OCH3 is on a wedge (coming forward), Me is to the right, H is at the bottom. Arranging priorities: OCH3 (1) > Et (2) > Me (3) > H (4). H is at the bottom in the plane. Going 1->2->3 (OCH3->Et->Me): OCH3 is front-top, Et is back-left, Me is right. This traces a clockwise arc when viewed with H away... we need to check H's position. H is in the plane at bottom, so it's partially toward viewer. The rotation OCH3(wedge/front) -> Et(dash/back) -> Me(right) appears counterclockwise, giving S configuration. So the product is (S)-2-methoxybutane. Step 2 - Determine the configuration of the reactant. SN2 reaction proceeds with inversion of configuration (Walden inversion). Therefore, the reactant 2-bromobutane must have the opposite (R) configuration at C2. Step 3 - Determine which Fischer projection corresponds to (R)-2-bromobutane. In a Fischer projection of 2-bromobutane, the vertical chain runs top (CH3, C1) to bottom (CH2CH3, C4), with C2 in the middle. Horizontal bonds point toward the viewer. For option (d): Br on top... wait, the vertical axis in a Fischer projection for 2-bromobutane should have the main carbon chain vertical. In option (d): Br is on top, Me is on the left horizontal, H is on the right horizontal, Et is on the bottom. But the top and bottom of a Fischer projection are the chain carbons (Me = C1, Et = C4), so having Br on top is unusual. Re-reading option (d): the central carbon has Br at top, Me on left, H on right, Et at bottom. Here the vertical bonds (Br top, Et bottom) go away from viewer and horizontal bonds (Me left, H right) come toward viewer. Actually in option (d), the Fischer projection has the chain arranged as: top = Br, left = Me, right = H, bottom = Et. This means the chain is Me (C1, left horizontal, toward viewer) and Et (C4, bottom, away from viewer), with Br and H on vertical... This seems non-standard. Re-interpreting: In option (d), the central C2 has: top bond to Br, left horizontal bond to Me, right horizontal bond to H, bottom bond to Et. Horizontal bonds come toward viewer, vertical go away. So: Br is going away (up), Et is going away (down), Me comes toward viewer (left), H comes toward viewer (right). Assigning R/S: priorities at C2: Br(1) > Et(2) > Me(3) > H(4). Br is going away (back-up), H is coming toward viewer (right). With H coming toward viewer, we invert at the end. Going Br->Et->Me: Br is at top-back, Et is at bottom-back, Me is at left-front. The rotation Br(top-back)->Et(bottom-back)->Me(left-front) - projecting onto the plane, from top to bottom to left = counterclockwise, so S; but since H is toward us (not away), we invert: actual configuration = R. Step 4 - Confirm: (R)-2-bromobutane undergoes SN2 with inversion to give (S)-2-methoxybutane, consistent with the product shown. Options (a), (b), (c) would give the wrong enantiomer or wrong connectivity. Therefore, the correct answer is D.