Let's break down the questions step by step as requested: --- ## **Step 1: Identify the Most Nucleophilic Atom (Part B)** **Explanation:** - Nucleophilicity is generally highest at the atom with the most negative charge (highest electron density). - Look at the computed charges: - The oxygen atom on the left has a charge of **-.75**, which is the most negative among all the atoms listed. **Thus, the most nucleophilic atom is the oxygen atom with -.75 charge.** - (You would circle this atom on the structure.) --- ## **Step 2: Reaction with Ethyl Bromide (Part C)** **Explanation:** - The enolate acts as a nucleophile and will attack the electrophilic carbon in ethyl bromide. - The nucleophilic atom (the enolate carbon adjacent to the carbonyl, not the oxygen) will attack the ethyl group, forming a new C–C bond. - Stereochemistry: Since the newly formed center is not a stereocenter (no four different groups), no stereoisomers are produced. **Product:** - The ethyl group attaches to the alpha-carbon (the carbon next to the carbonyl group in the original enolate). **Structure:** - The final product is 4-ethyl-3-oxohexanoate (showing the ethyl group attached at the alpha position). --- ## **Step 3: Acidic Hydrolysis (Part D)** **Explanation:** - Heating with aqueous acid will hydrolyze the ester to a carboxylic acid. - The product will be a beta-keto acid (4-ethyl-3-oxohexanoic acid). - Upon further heating, beta-keto acids decarboxylate, losing CO₂, to give a ketone. **Product:** - The final product is 3-ethyl-2-pentanone. --- ## **Summary of Answers** ### **Step 1 (B):** - **Most nucleophilic atom:** Oxygen with charge -.75 (circled in the image). ### **Step 2 (C):** - **Major product:** 4-ethyl-3-oxohexanoate (ethyl group added to the alpha-carbon). ### **Step 3 (D):** - **Major product:** 3-ethyl-2-pentanone (after acidic hydrolysis and decarboxylation). --- **If you need to see the structures, let me know!**