Let's solve each step-by-step: --- ## **2.1** **Given:** Benzene ring–CH₂–CO–CH₂Br + 1 eq. NaH → **Product** ### **Step-by-Step Solution:** - **Step 1**: NaH is a strong base, it will deprotonate the most acidic hydrogen. - The hydrogen alpha to both the carbonyl and the bromine is the most acidic (on the CH₂ between CO and Br). - Deprotonation produces a stabilized enolate or carbanion. ### **Product Structure:** - The product is the deprotonated form (enolate or carbanion) at the alpha position: -  **Final Product:** Benzene ring–CH₂–CO–CH(–)–CH₂Br (with a negative charge on the carbon between CO and Br). --- ## **2.2** **Given:** Unknown starting material → (1) NaOMe/MeOH (2) Dilute H₃O⁺ → Acetone + Methyl acetate ### **Step-by-Step Solution:** - **Step 1:** The products are acetone and methyl acetate, which suggests a retro-aldol or retro-Claisen reaction. - **Step 2:** NaOMe/MeOH and acidic workup suggest a Claisen condensation in reverse. - **Step 3:** Combine acetone and methyl acetate to see what starting material could yield these by retro-Claisen. **Retrosynthetic analysis:** - Combine acetone (CH₃COCH₃) and methyl acetate (CH₃COOCH₃) at their alpha positions: - Product: CH₃COCH₂COOCH₃ (methyl acetoacetate) **Final Starting Material:** **Methyl acetoacetate** (CH₃COCH₂COOCH₃) --- ## **2.3** **Given:** Diethyl malonate + Ethyl acrylate → [Missing reagents] → Diethyl 2-(3-ethoxy-3-oxopropyl)malonate ### **Step-by-Step Solution:** - **Step 1:** Diethyl malonate is alkylated at the alpha position. - **Step 2:** Ethyl acrylate is an α,β-unsaturated ester (Michael acceptor). - **Step 3:** The reaction is a Michael addition (or conjugate addition). **Reagents:** 1. NaOEt/EtOH (to generate the enolate from diethyl malonate) 2. Ethyl acrylate (Michael addition) **Final Reagents:** **1) NaOEt/EtOH** **2) Ethyl acrylate** --- ## **2.4** **Given:** Cyclopentanone → 1 eq. NaH, then 1 eq. 1-chloropropane → **Product** ### **Step-by-Step Solution:** - **Step 1:** NaH deprotonates the alpha hydrogen of cyclopentanone, generating an enolate. - **Step 2:** The enolate attacks 1-chloropropane (an SN2 reaction), alkylating the alpha carbon. **Product Structure:** Cyclopentanone with a propyl group at the alpha position. **Final Product:** **2-propylcyclopentanone** --- ## **Summary Table** | # | Missing Item | Structure/Name | |-------|------------------------|------------------------------------------------| | 2.1 | Product | Benzene ring–CH₂–CO–CH(–)–CH₂Br (enolate) | | 2.2 | Starting Material | Methyl acetoacetate | | 2.3 | Reagents | 1) NaOEt/EtOH 2) Ethyl acrylate | | 2.4 | Product | 2-propylcyclopentanone | --- ### **Detailed Structures** **2.1:** Product:  (Negative charge on carbon alpha to both CO and Br) **2.2:** Starting Material:  **2.3:** Reagents: 1. NaOEt/EtOH 2. Ethyl acrylate **2.4:** Product:  --- Let me know if you need detailed mechanisms for any step!