See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: E2 elimination requires anti-periplanar geometry between the H being removed and the leaving group (Br). The reaction proceeds in a single concerted step with strong base (NaOH acting as base). Step 1 - Identify the starting material: The substrate is a trans-fused or trans-substituted bicyclic compound. From the drawing, it appears to be a system where a cyclohexane ring is connected to a cyclopentane ring bearing a CH3 group. The carbon bearing Br also has an H on a wedge, and the adjacent carbon (at the ring junction) also has an H on a wedge. The CH3 is on the cyclopentane ring. Step 2 - E2 requirement: For E2, the H and Br must be anti-periplanar (180° dihedral). In the depicted conformation, we must identify which H is anti to the Br. Step 3 - Analyze geometry: The starting material shows a trans relationship between the H on the carbon bearing Br and the H on the adjacent carbon. In the drawn conformation, the H at the ring junction (between cyclohexane and cyclopentane) is on a wedge bond, and the Br is also shown. For anti-periplanar elimination, the H that is anti to Br on the adjacent carbon must be used. Step 4 - Determine which H is anti to Br: Given the stereochemistry shown (both H atoms on wedge bonds at adjacent carbons, with Br also on a wedge-like position), the anti-periplanar H is the one at the ring junction between the two rings. Elimination of this H with Br gives a double bond at the ring junction. Step 5 - Product: Elimination gives a double bond between the two ring carbons (the junction carbon of cyclohexane and cyclopentane), producing a bicyclic alkene with CH3 on the cyclopentane portion. This corresponds to option (d): a fused bicyclic structure (cyclohexane fused to cyclopentane) with a double bond at the ring junction and a CH3 group. Step 6 - Why other options fail: - (a) and (b) are substitution (SN2) products, not elimination products; E2 with NaOH (a strong base, poor nucleophile conditions) favors elimination. - (c) shows an elimination product but with incorrect regiochemistry or stereochemistry - the double bond position or the ring geometry does not match the anti-periplanar requirement for the given stereochemistry. - (d) correctly shows the double bond at the ring junction resulting from anti-periplanar E2 elimination, consistent with the given stereochemistry of the starting material. Therefore, the correct answer is D.