Let's solve each reaction in **2 steps** and explain each step. --- ## **1. Reaction 1: Cyclopentane-1,1-dicarboxylic acid in acidic conditions** ### **Step 1: Identify the Functional Groups and Reagent** - The molecule has two carboxylic acid groups (one is a hydroxy acid). - The reagent is acid (**H₃O⁺**), which suggests a **decarboxylation** reaction, especially if a β-keto acid or β-hydroxy acid is present. **Explanation:** Under acidic conditions, β-hydroxy acids readily decarboxylate to give a ketone or aldehyde and CO₂. --- ### **Step 2: Predict the Product** - The β-hydroxy group (relative to the carboxylic acid) facilitates decarboxylation. - Removal of CO₂ from the carboxyl group adjacent to the hydroxy yields **cyclopentanone**. **Explanation:** The carboxyl group is lost as CO₂, and the remaining molecule is cyclopentanone. **Final Answer:** **Cyclopentanone + CO₂** --- ## **2. Reaction 2: Cyclopentane-1,1-dicarbonyl dichloride + Ethylenediamine** ### **Step 1: Identify the Functional Groups and Reagent** - Dicarbonyl dichloride (diacid chloride) reacts with a diamine (ethylenediamine). - This is a classic **polyamide (nylon-type) formation** reaction. **Explanation:** Acid chlorides react with amines to form amides, releasing HCl. --- ### **Step 2: Predict the Product** - Each acid chloride reacts with an amine group, forming a cyclic **imide** if intramolecular, or a diamide if only one equivalent is used. **Explanation:** Since there are two acid chlorides and a diamine, you will get a **bicyclic diamide** (a cyclopentane ring with two amide linkages to the ethylene bridge). **Final Answer:** **Bicyclic diamide (with cyclopentane core and ethylene bridges via amide bonds) + 2 HCl** --- ## **3. Reaction 3: Phthaloyl chloride + Hydroquinone and 1,2-ethanediol** ### **Step 1: First Reaction—Phthaloyl Chloride + Hydroquinone** - Both are di-functional: phthaloyl chloride (diacid chloride), hydroquinone (diol). - They form a **polyester** (polyarylate) via condensation, releasing HCl. **Explanation:** Each -Cl reacts with an -OH, forming ester bonds. --- ### **Step 2: Second Reaction—Polyester + 1,2-Ethanediol** - The polyester can further react with more diol (glycolysis), **breaking ester bonds** and forming bis(hydroxyethyl) phthalate. **Explanation:** Transesterification with ethylene glycol yields **bis(hydroxyethyl) phthalate**. **Final Answer:** **Bis(hydroxyethyl) phthalate + Hydroquinone (if complete transesterification)** --- ## **4. Reaction 4: Polyethylene Glycol + Terephthaloyl Diisocyanate** ### **Step 1: Identify the Functional Groups and Reagent** - Polyethylene glycol (polyol) reacts with diisocyanate. - This forms a **polyurethane**. **Explanation:** Isocyanate reacts with alcohol to form a **urethane (carbamate) linkage**. --- ### **Step 2: Predict the Product** - The repeating unit will join via urethane bonds between the glycol and terephthaloyl units. **Explanation:** You get a **polyurethane** chain with alternating polyethylene glycol and terephthaloyl units. **Final Answer:** **Polyurethane (from polyethylene glycol and terephthaloyl diisocyanate)** --- ## **Summary Table** | Reaction | Step 1 (Explanation) | Step 2 (Product) | Final Answer | |----------|---------------------|------------------|--------------| | 1 | Acidic decarboxylation of β-hydroxy acid | Cyclopentanone + CO₂ | Cyclopentanone + CO₂ | | 2 | Diamine + diacid chloride → diamide | Bicyclic diamide + 2 HCl | Bicyclic diamide + 2 HCl | | 3 | Diacid chloride + diol → polyester | Transesterification with glycol | Bis(hydroxyethyl) phthalate + hydroquinone | | 4 | Polyol + diisocyanate → polyurethane | Polyurethane formation | Polyurethane |