Let's break down each multi-step synthesis problem and fill in the missing reactants, conditions, or products. --- ## **Problem 1** ### **Step-by-step Solution** #### **Step 1: Benzene to Iodobenzene** - **Starting Material:** Benzene - **Reagent/Reaction:** To introduce an iodine atom onto benzene, we typically use **I₂** and an **oxidizing agent** (such as HNO₃ or H₂O₂) to generate the electrophile. - **Product:** Iodobenzene **Equation:** C₆H₆ + I₂ / HNO₃ → C₆H₅I --- #### **Step 2: Iodobenzene to Phenylhydrazine** - **Reagent/Reaction:** Iodobenzene reacts with **hydrazine (NH₂NH₂)** to substitute the iodine with a hydrazine group. - **Product:** Phenylhydrazine (C₆H₅NHNH₂) + HI **Equation:** C₆H₅I + NH₂NH₂ → C₆H₅NHNH₂ + HI --- #### **Step 3: Phenylhydrazine to p-Nitrophenylhydrazine** - **Reagent/Reaction:** Nitration (HNO₃, H₂SO₄) introduces a nitro group, mainly para to the NHNH₂ group due to its activating nature. - **Product:** p-Nitrophenylhydrazine **Equation:** C₆H₅NHNH₂ + HNO₃/H₂SO₄ → p-Nitrophenylhydrazine --- #### **Step 4: Friedel-Crafts Alkylation** - **Reagent/Reaction:** The compound reacts with **CH₃CCl₃ (tert-butyl chloride)** and **AlCl₃** to introduce a tert-butyl group via Friedel-Crafts alkylation. - **Product:** para- or ortho-tert-butyl-p-nitrophenylhydrazine (depending on substituent positions). --- ### **Summary Table** | Step | Starting Material | Reagent/Condition | Product | |------|----------------------|------------------------|------------------------------| | 1 | Benzene | I₂/HNO₃ | Iodobenzene | | 2 | Iodobenzene | NH₂NH₂ | Phenylhydrazine | | 3 | Phenylhydrazine | HNO₃, H₂SO₄ | p-Nitrophenylhydrazine | | 4 | p-Nitrophenylhydrazine | (CH₃)₃CCl, AlCl₃ | tert-butyl-p-nitrophenylhydrazine | --- ## **Problem 2** ### **Step-by-step Solution** #### **Step 1: Toluene to Benzoic Acid** - **Starting Material:** Toluene (C₆H₅CH₃) - **Reagent/Reaction:** Oxidation using **H₂CrO₄** converts methyl to carboxylic acid. - **Product:** Benzoic acid (C₆H₅COOH) --- #### **Step 2: Benzoic Acid to Benzenesulfonic Acid** - **Reagent/Reaction:** Sulfonation using **conc. H₂SO₄ (2 eq)** introduces SO₃H. - **Product:** Benzenesulfonic acid (C₆H₅SO₃H) --- #### **Step 3: Toluene to Tribromotoluene** - **Reagent/Reaction:** **Br₂/FeBr₃** (excess) brominates methylbenzene on all ortho and para positions. - **Product:** 2,4,6-Tribromotoluene --- ### **Summary Table** | Step | Starting Material | Reagent/Condition | Product | |------|------------------|---------------------|------------------------| | 1 | Toluene | H₂CrO₄ | Benzoic acid | | 2 | Benzoic acid | 2 eq H₂SO₄ | Benzenesulfonic acid | | 3 | Toluene | 3 eq Br₂/FeBr₃ | 2,4,6-Tribromotoluene | --- ## **Final Answers (Filled-in):** ### **Problem 1:** 1. **I₂, HNO₃** (or I₂, H₂O₂) 2. **NH₂NH₂** (hydrazine) 3. **p-Nitrophenylhydrazine** 4. **(CH₃)₃CCl, AlCl₃** ### **Problem 2:** 1. **Benzoic acid (C₆H₅COOH)** 2. **Benzenesulfonic acid (C₆H₅SO₃H)** 3. **2,4,6-Tribromotoluene** --- If you need the structures or more detailed explanations, let me know!