See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: E2 elimination requires anti-periplanar geometry between the leaving group (Br) and the beta-hydrogen (or beta-deuterium) being removed. The reaction proceeds via a concerted syn-periplanar or anti-periplanar mechanism; for E2, anti-periplanar is strongly preferred. Step 1 - Identify the stereochemistry of the starting material: The starting material is a substituted cyclohexane with Br and D on the same carbon (C1) and adjacent carbons bearing H atoms. Reading the wedge-dash notation: at C1 (bearing Br and D), Br is on a bold wedge (axial or equatorial up) and D is on a wedge as well. The CH3 group is on a dash at the adjacent carbon. The H atoms on the neighboring carbons are shown with specific stereochemistry. Step 2 - Apply E2 anti-periplanar requirement: In E2 elimination, the base (KOH/alcohol) abstracts a beta-hydrogen that is anti-periplanar to the leaving group Br. The molecule must adopt a conformation where Br and the beta-H (or beta-D) are anti to each other (180 degrees dihedral). Step 3 - Determine which beta-hydrogen is anti-periplanar to Br: Given the stereochemical arrangement shown, the deuterium D is on the same carbon as Br (geminal), so it cannot be the beta-hydrogen. The neighboring H atoms must be evaluated. The H that is anti-periplanar to Br in the preferred ring conformation will be eliminated preferentially. However, because of the kinetic isotope effect, C-H bonds are broken faster than C-D bonds. But the question asks for the major product based on anti-periplanar geometry. Step 4 - Identify which elimination gives which product: - Elimination of Br with the H on one side gives a methylcyclohexene without D at the double bond position. - Elimination of Br with the H on the other side gives a product where D is retained at the allylic/vinylic adjacent position. - The anti-periplanar H relative to Br in the depicted stereochemistry leads to formation of a 3-deuterio-1-methylcyclohex-1-ene, where D is at C3 and the double bond is between C1 and C2 (with methyl at C1). This corresponds to option (c). Step 5 - Why other options fail: - Option (a): Shows D and H both present at C6 on the same carbon with defined stereochemistry, implying no D was lost but the geometry doesn't match the anti-periplanar elimination product. - Option (b): Shows no deuterium in the product, meaning D was eliminated along with Br; this would require C-D bond breaking which is slower due to kinetic isotope effect, so it is not the major product. - Option (d): Places methyl and D on different carbons in a pattern inconsistent with the regiochemistry expected from this elimination. The anti-periplanar requirement selects the H (not D) that is anti to Br, and due to the kinetic isotope effect D is retained. The double bond forms between the carbon bearing Br and the carbon that loses H, giving 1-methyl-3-deuteriocyclohex-1-ene (option c) as the major product. Therefore, the correct answer is C.