Let's break down this reaction step by step using curved-arrow (electron pushing) notation. --- ## **Step 1: Analyze the Reactants** - **Reactant:** Cyclohexane ring with a Br and an OH₂ (hydronium) group on adjacent carbons. - **Reagents:** Water (H₂O) and heat. This setup suggests an **E1 elimination reaction** (since Br is a good leaving group, and water can act as a base). --- ## **Step 2: Mechanism Steps** ### **Step 1: Loss of Leaving Group (Br⁻)** - The electron pair from the C–Br bond moves onto Br, causing Br to leave as Br⁻. - This generates a **carbocation** at the carbon that was attached to Br. **Curved Arrow:** From the bond between the carbon and Br → onto Br. ### **Step 2: Deprotonation to Form Alkene** - Water (H₂O) acts as a base and removes a proton (H⁺) from the carbon adjacent to the carbocation (the carbon that had the OH₂ group). - The electron pair from the C–H bond forms a double bond (π bond) between the two carbons. **Curved Arrow:** From the lone pair on H₂O → to the hydrogen on the carbon adjacent to the carbocation. From the C–H bond → to form a double bond between the two carbons. --- ## **Step 3: Draw the Product** - The product will be a cyclohexene (with a double bond between the two previously substituted carbons), and water (H₃O⁺) and Br⁻ as byproducts. --- ## **Step-by-Step Mechanism with Curved Arrows** ### **Step 1: Loss of Br⁻** ``` Br | —C—C—OH2+ | (rest of ring) Curved Arrow: C—Br bond → Br ``` ### **Step 2: Deprotonation and Double Bond Formation** ``` (+) | —C═C— | (rest of ring) Curved Arrow: Lone pair on H₂O → H on neighboring C; C—H bond → C=C double bond ``` --- ## **Final Answer: Summary** ### **Mechanism Steps:** 1. **Br leaves, forming a carbocation.** 2. **Water deprotonates adjacent carbon, forming a double bond (alkene).** ### **Product:** - **Cyclohexene** (double bond where Br and OH₂⁺ were initially attached) --- ### **Overall Curved Arrow Mechanism** 1. **C–Br bond → Br** (Br leaves) 2. **H₂O lone pair → H** (on carbon next to carbocation) 3. **C–H bond → C=C** (double bond forms) --- ### **Visual Summary** ``` Step 1: Step 2: Br H | | —C—C—OH2+ → —C═C— | | (rest of ring) (rest of ring) | Curved arrows as described above. ``` --- **This is a classic E1 elimination mechanism. Let me know if you'd like a drawn arrow diagram!**