## Step-by-Step Solutions --- ### a. **Draw the Mechanism** **Step 1: Protonation of Alcohol (OH)** - The hydroxyl group on 4-methyl-2-pentanol is protonated by sulfuric acid (H₂SO₄), converting it to a good leaving group (water). **Step 2: Loss of Water, Formation of Carbocation** - The protonated alcohol loses water, generating a secondary carbocation at C-2. **Step 3: Elimination (E1 Mechanism)** - A base (HSO₄⁻ or H₂O) abstracts a proton from an adjacent carbon (C-1 or C-3), resulting in the formation of a double bond. - In this case, the product shown is 4-methyl-1-pentene, where the double bond is formed between C-1 and C-2 (less substituted alkene). **Explanation:** The mechanism is an **E1 elimination**: protonation → loss of water (carbocation) → deprotonation (alkene formation). --- ### b. **Why Is This the Minor Product?** **Step 1: Analyze Alkene Substitution** - The product shown (4-methyl-1-pentene) is a **less substituted alkene** (double bond at the end, terminal alkene). **Step 2: Zaitsev's Rule** - Zaitsev's rule states that **the more substituted alkene (internal alkene)** is typically the major product in E1 eliminations because it's more stable. - The major product would be 4-methyl-2-pentene (double bond between C-2 and C-3), which is more substituted and thus more stable. **Explanation:** The given product is the **minor product** because it is the less substituted (and therefore less stable) alkene. The reaction prefers to form the more substituted, more stable alkene as the major product. --- ### c. **Propose a 2-Step Synthesis for the Major Product as Shown** **Step 1: Make OH a Better Leaving Group** - Convert the alcohol to a good leaving group (e.g., use tosyl chloride (TsCl) to make a tosylate, or PBr₃ to make an alkyl bromide). **Step 2: Use a Strong, Bulky Base for E2 Elimination (Hofmann Product)** - Treat the resulting tosylate or bromide with a **strong, bulky base** such as potassium tert-butoxide (t-BuOK). - Bulky bases favor elimination at the **least substituted position** (Hofmann product), giving the terminal alkene (4-methyl-1-pentene) as the major product. **Explanation:** By making the alcohol a better leaving group and then using a bulky base, you favor **Hofmann elimination** (least substituted alkene), making 4-methyl-1-pentene the major product. --- ## **Final Answers** **a.** Draw the E1 mechanism: 1. Protonate OH → H₂O leaves → carbocation at C-2 → deprotonate C-1 → form 4-methyl-1-pentene. **b.** The product is a minor product because it is the less substituted alkene. Zaitsev's rule favors formation of the more substituted and stable alkene (4-methyl-2-pentene). **c.** 1. Convert OH to a good leaving group (e.g., tosylate or bromide). 2. React with a strong, bulky base (e.g., t-BuOK) to favor E2 Hofmann elimination, making the terminal alkene (4-methyl-1-pentene) the major product.