Let's solve the first reaction as an example: --- ### **Reaction 1: Lactone (Cyclic Ester) + CH₃NH₂ (Methylamine) with Acid or Base Catalyst** #### **Step 1: Nucleophilic Attack (Aminolysis of Lactone)** - The nucleophile, methylamine (CH₃NH₂), attacks the carbonyl carbon of the lactone. - This opens the lactone ring, breaking the ester bond and forming a tetrahedral intermediate. **Explanation:** The lone pair on the nitrogen of methylamine attacks the electrophilic carbonyl carbon, pushing electrons up to the oxygen, which opens the ring. --- #### **Step 2: Proton Transfers and Product Formation** - Proton transfers (catalyzed by acid or base) lead to the collapse of the tetrahedral intermediate. - The leaving group (the alkoxide from the ring) grabs a proton, forming a hydroxy group, and the nitrogen becomes bonded to the carbonyl carbon, yielding a **cyclic amide (lactam)**. **Explanation:** The intermediate collapses to form an amide bond, with the former ester oxygen becoming an alcohol (ring-opened). --- ### **Final Answer:** The **major product** is a ring-opened amide (lactam) with a terminal hydroxy group and a methylamino group where the ester was. **Product Structure:** ``` HO–(CH₂)₄–C(=O)–NHCH₃ ``` This is 5-hydroxy-N-methylpentanamide. --- Would you like to see the other reactions solved in the same format?

### **Reaction: Aminolysis of Lactone with Methylamine** --- #### **Step 1: Nucleophilic Attack** - The nucleophile, **methylamine (CH₃NH₂)**, attacks the **carbonyl carbon** of the lactone. - This reaction opens the lactone ring, resulting in a **tetrahedral intermediate**. **Explanation:** Methylamine's nitrogen, rich in electrons, attacks the electrophilic carbon in the carbonyl group, causing the carbon-oxygen double bond to break and forming an intermediate where the oxygen bears a negative charge. --- #### **Step 2: Proton Transfers and Product Formation** - The tetrahedral intermediate undergoes proton transfers, facilitated by acid or base, leading to the collapse of the intermediate. - The departing alkoxide group (from the lactone) captures a proton to form a hydroxy group, while the nitrogen becomes bonded to the carbon, resulting in a **cyclic amide (lactam)**. **Explanation:** The collapse of the tetrahedral structure forms an amide bond, with the lactone's original oxygen becoming a hydroxy group. This yields a product that contains both an amide and an alcohol functional group. --- ### **Final Answer:** The major product formed is a **ring-opened amide (lactam)** with a terminal hydroxy group and a methylamino group at the site of the original ester bond. **Product Structure:** ``` HO–(CH₂)₄–C(=O)–NHCH₃ ``` This compound is **5-hydroxy-N-methylpentanamide**.