Let's address each question step-by-step, referencing **Scheme 1** and **Scheme 2** as shown in your images. --- # 1. Identify Each Step (NA, LGL, PT, Re) in Scheme 2 Let's break down the steps in **Scheme 2** (Aldol Addition Mechanism): ## Step 1 - **Action:** The base (OH⁻) abstracts an α-hydrogen from acetophenone, forming an enolate ion. - **Type:** **Proton Transfer (PT)** ## Step 2 - **Action:** The enolate (nucleophile) attacks the carbonyl carbon of benzaldehyde (electrophile). - **Type:** **Nucleophilic Attack (NA)** ## Step 3 - **Action:** The resulting alkoxide ion is protonated by water, forming the β-hydroxy ketone (aldol). - **Type:** **Proton Transfer (PT)** ## Step 4 - **Action:** Under heat, the β-hydroxy ketone eliminates water to form the α,β-unsaturated ketone (chalcone). - **Type:** **Leaving Group Loss (LGL)** (since water leaves; this step is part of the E1cb mechanism) --- # 2. Why Is Aldehyde More Susceptible to Nucleophilic Attack Than Ketone? **Two main reasons:** ### a) **Steric Factors:** - **Aldehyde:** Has one hydrogen and one alkyl/aryl group attached to the carbonyl carbon, so less steric hindrance. - **Ketone:** Has two alkyl/aryl groups, causing more crowding and hindrance for the nucleophile. ### b) **Electronic Factors:** - **Aldehyde:** Only one electron-donating group (R group), so carbonyl carbon is more **electron-deficient** and more electrophilic. - **Ketone:** Two electron-donating groups reduce the partial positive charge on carbonyl carbon, making it less electrophilic. --- # 3. E1cb vs. E1: Similarities and Differences ### **Commonality:** - Both involve the **loss of a leaving group** (in this case, water) and formation of a double bond. ### **Differences:** - **E1cb:** The **base removes a proton first** to generate a carbanion intermediate, then the leaving group departs. - **E1:** The leaving group leaves first, forming a carbocation intermediate, which then loses a proton. **Summary Table:** | | E1cb | E1 | |------------------|-------------------------------|------------------------------| | Intermediate | Carbanion | Carbocation | | First Step | Proton is removed | Leaving group departs | | Second Step | Leaving group departs | Proton is removed | | Example Here | Aldol condensation dehydration | Simple alcohol dehydration | --- # 4. (a) Physical Properties Table for 1-Indanone, Piperonal, NaOH | Compound | Molecular Formula | Molar Mass (g/mol) | Melting Point (°C) | Boiling Point (°C) | Appearance | |---------------|------------------|--------------------|--------------------|--------------------|-------------------| | 1-Indanone | C9H8O | 132.16 | 60-62 | 260-262 | White solid | | Piperonal | C8H6O3 | 150.13 | 36-38 | 263 | White crystalline | | Sodium Hydroxide | NaOH | 40.00 | 318 | 139 (decomposes) | White solid | --- # 4. (b) Experimental Procedure Outline 1. **Mix Reactants:** Combine 1-indanone and piperonal in a flask. 2. **Add Base:** Add aqueous sodium hydroxide dropwise while stirring. 3. **Reaction:** Stir the mixture at room temperature (or gently heat if needed) for the required time. 4. **Workup:** Quench the reaction with water and acidify if needed. 5. **Isolation:** Collect the precipitate by filtration. 6. **Purification:** Recrystallize the product from ethanol or another suitable solvent. 7. **Dry Product:** Dry the purified enone and record yield and melting point. --- ## **Summary Table of Steps for Scheme 2** | Step | Mechanism Step | Type | |------|-----------------------|--------------------| | 1 | Enolate formation | PT | | 2 | Attack on aldehyde | NA | | 3 | Protonation | PT | | 4 | Dehydration | LGL (E1cb) | --- **Let me know if you need a full mechanism drawing or more details for any step!**