Let's solve the problem step-by-step: --- ## **Step 1: Analyze the Structure** Given: **(2R,3R)-2-chloro-3-methylpentane** **Structure:** - 5-carbon chain (pentane) - Cl at C-2 (with R stereochemistry) - Methyl at C-3 (with R stereochemistry) The provided structure confirms this. --- ## **Step 2: Identify the Reaction Type** - **NaOCH₃ (sodium methoxide)** is a strong base. - **Elimination reaction (E2 mechanism)** will occur. - For E2, elimination is anti-periplanar: the base removes a proton from a β-carbon **anti** to the leaving group (Cl). --- ## **Step 3: Identify β-hydrogens** - **Cl is at C-2.** - β-carbons are C-1 and C-3. ### **At C-1:** - Only has two H atoms. ### **At C-3:** - Methyl group at C-3 (one β-H). --- ## **Step 4: Stereochemistry & E2 Mechanism** For **E2** to occur, the eliminated hydrogen and the leaving group (**Cl**) must be **anti** to each other. - At **C-2**: Cl is on a wedge (up). - At **C-1**: The hydrogen anti to Cl must be on a dash (down). - At **C-3**: The methyl is on a wedge (up), so the hydrogen anti to Cl is on a dash (down). --- ## **Step 5: Draw the Major Product (Zaitsev Product)** The base abstracts the more substituted β-H (C-3), leading to the more substituted (Zaitsev) alkene: - Double bond formed between **C-2 and C-3**. ### **Draw the Product:** - Remove Cl from C-2 and H from C-3. - Double bond between C-2 and C-3. - Methyl at C-3. - The product is **3-methyl-2-pentene**. ### **Stereochemistry:** The double bond can be **E** or **Z**. Because the starting material is (2R,3R) and the elimination proceeds anti-periplanar, **the product will be the E-isomer** (the larger groups — ethyl and methyl — are on opposite sides). --- ## **Step 6: Final Answer** ### **Major Product Structure:** ``` CH3 | CH3-CH=CH-CH2-CH3 ``` Or explicitly: **(E)-3-methyl-2-pentene** --- ### **Summary Table** | Step | Details | |-------------------------|------------------------------------------------| | Starting material | (2R,3R)-2-chloro-3-methylpentane | | Reagent | NaOCH₃ (strong base, E2) | | β-H eliminated | From C-3 (more substituted, Zaitsev product) | | Double bond forms | Between C-2 and C-3 | | Stereochemistry | (E)-3-methyl-2-pentene (major product) | --- ## **Final Answer (Structure & Stereochemistry)** **The major product is (E)-3-methyl-2-pentene.** **Structure:** ``` CH3 | CH3-CH=CH-CH2-CH3 ``` - The double bond is E (ethyl and methyl groups are on opposite sides). --- **If you need a drawing of the product, let me know!**