See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When a halohydrin is treated with alcoholic KOH (a base), it can undergo either intramolecular SN2 (to form an epoxide) or E2 elimination depending on the stereochemistry of the substituents. Step 1: Identify the starting material. The starting material is trans-1-chloro-2-hydroxycyclohexane. The Cl and OH groups are on adjacent carbons (C1 and C2) in a trans (diaxial or diequatorial) relationship, as indicated by the bold wedge bonds showing both H atoms on the same face. Step 2: Consider the reaction with alcoholic KOH. Alcoholic KOH is a strong base. For epoxide formation via intramolecular SN2, the OH group (after deprotonation by KOH to form an alkoxide) must attack the carbon bearing Cl from the back side (anti attack). This requires the O and Cl to be anti-periplanar. Step 3: Assess stereochemistry for epoxide formation. In the trans isomer, for the alkoxide at C2 to attack C1 (which bears Cl) in an SN2 fashion, the O and Cl must be anti to each other. In the trans-1-chloro-2-hydroxycyclohexane, when both substituents are diaxial, the O and Cl are anti-periplanar, allowing backside attack. This would give the epoxide (option b). Step 4: However, the given correct answer is cyclohexanone (option c). This occurs because under the reaction conditions, the trans-halohydrin undergoes an E2-type elimination facilitated by the base, but more importantly, the sequence involves: first, the alkoxide formed attacks to give epoxide, and then under the basic conditions the epoxide undergoes rearrangement, OR the reaction proceeds via oxidation-like pathway. Actually, reconsidering: trans-2-chlorocyclohexanol with alcoholic KOH — the trans configuration means Cl and OH are anti, so intramolecular SN2 by alkoxide on C-Cl gives cyclohexene oxide (epoxide). But the answer is given as C (cyclohexanone). Step 5: Re-examining the structure. Looking more carefully at the image, the bold bonds show both H atoms on wedge (same side), meaning Cl and OH are also on the same side — this is actually the CIS isomer (cis-1-chloro-2-hydroxycyclohexane). In the cis isomer, Cl and OH are on the same face, so they are NOT anti-periplanar; intramolecular SN2 (epoxide formation) is not favored. Instead, alcoholic KOH acts as a base for E2 elimination: it deprotonates the OH (or acts on H), and elimination occurs. The cis-halohydrin with base undergoes elimination where KOH deprotonates the OH to form alkoxide, which then facilitates elimination of HCl — but since Cl and the adjacent H must be anti for E2, and simultaneously the alkoxide can perform E1cb or the base removes H anti to Cl. The elimination of HCl from this substrate gives cyclohex-2-en-1-ol (option a) or cyclohexanone. The cis-halohydrin can also tautomerize or the base abstracts the H on the carbon bearing OH, leading to an enol which tautomerizes to cyclohexanone. Specifically, KOH can cause dehydrohalogenation to give the enol (vinyl alcohol form), which tautomerizes to cyclohexanone. This is the Favorskii-type or halohydrin-to-ketone conversion: cis-2-chlorocyclohexanol + KOH → cyclohexanone + KCl + H2O via E2 on the C-H adjacent to OH, forming the enol, which tautomerizes to the ketone. Step 6: Why other options fail. (a) Cyclohex-2-en-1-ol would require specific elimination without tautomerization. (b) Epoxide requires anti-periplanar O and Cl — not possible in cis isomer under these conditions. (d) 3-Chlorocyclohexene is not a product of this reaction. Therefore, the correct answer is C.