See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Dehydrohalogenation (E2 elimination) requires an anti-periplanar arrangement of the H and the leaving group (Br). The geometry of the resulting alkene (cis or trans) is determined by the spatial relationship of the substituents in the conformation that undergoes elimination. Step 1: Identify the substrate. 2-Bromobutane has the structure CH3-CHBr-CH2-CH3. For E2 elimination, we look at the Newman projection along the C2-C3 bond. C2 bears: CH3 (from C1), Br, H. C3 bears: CH3 (from C4), H, H. Step 2: E2 requires anti-periplanar geometry, meaning H (on C3) and Br (on C2) must be 180° apart (anti to each other). Step 3: Determine which product (cis or trans-2-butene) forms from which conformation. The two groups that end up on the double bond are CH3 (C1 side, on C2) and CH3 (C4 side, on C3). The relative orientation of these two CH3 groups in the eliminating conformation determines cis vs trans product. Step 4: For cis-2-butene, both CH3 groups must end up on the same side of the double bond. This means in the reacting conformation, the CH3 on C2 and the CH3 on C3 must be on the same side (syn/gauche to each other), while H (on C3) and Br (on C2) are anti-periplanar. Step 5: Examine option (a): Newman projection where front carbon (C2) has CH3 at top, H upper-left, H lower-left, and back carbon (C3) has CH3 upper-right, Br right, H lower-right. Here the H on front (C2) and Br on back... re-examining: in option (a), front has CH3, H, H and back has CH3, Br, H. The H on C2 (front, lower-left) and Br on C3 (back, right) — checking dihedral: lower-left front vs right back is approximately 120° apart, not anti. Step 6: Re-examining option (a) carefully: Front carbon C2 has CH3 (top), H (upper-left), H (lower-left). Back carbon C3 has CH3 (upper-right), Br (right), H (lower). For anti-periplanar elimination, we need H on C2 at bottom (180° from Br on C3 at top) — but Br is at right on C3. The H on C2 at lower-left is roughly anti to CH3 on C3 at upper-right. The H on C2 at upper-left is roughly anti to H on C3 at lower. Neither H on C2 is perfectly anti to Br. However, for option (a), the conformation shown has the two CH3 groups (C1-methyl on C2 at top, and C4-methyl on C3 at upper-right) — they are gauche to each other (60° apart). When H (on C2) and Br (on C3 back) are anti, the CH3 groups being gauche means they will be on the same face of the double bond → cis-2-butene. Step 7: In conformation (a), rotating to make H on C2 anti to Br on C3: the CH3 groups on C2 and C3 end up on the same side of the developing double bond, giving cis-2-butene. This matches option (a). Step 8: Option (b) would give trans-2-butene because in that conformation the two CH3 groups are anti to each other when H and Br are anti-periplanar. Options (c) and (d) are either eclipsed or have incorrect anti-periplanar arrangements. Therefore, the correct answer is A.