Let's answer each question step-by-step, with explanations after each step and a final answer at the end. --- ### (e) **Reagent:** Br₂, CH₂Cl₂ **Product:** 2,3-dibromobutane **Type:** Halogenation of alkene **Step 1:** Identify the reaction type based on the reagent. Br₂ in CH₂Cl₂ adds Br atoms across a double bond (anti addition). **Step 2:** The product shown is 2,3-dibromobutane, so the starting material must be 2-butene. **Final Answer:** - **Reactant:** 2-butene (CH₃CH=CHCH₃) - **Type:** Alkene halogenation --- ### (f) **Reagents:** 1. Hg(OAc)₂, H₂O, THF 2. NaBH₄, HO⁻ **Step 1:** This is an oxymercuration-demercuration reaction, which adds H and OH across a double bond (Markovnikov, no rearrangement). **Step 2:** The double bond is between C2 and C3. OH goes to the more substituted carbon (C3), H to C2. **Final Answer:** - **Product:** CH₃CH(OH)CH₂CH₃ (2-butanol) - **Type:** Oxymercuration-demercuration (hydration of alkene) --- ### (g) **Reagents:** 1. mCPBA 2. H₂SO₄, H₂O **Step 1:** mCPBA forms an epoxide from cyclohexene. **Step 2:** Acidic hydrolysis opens the epoxide to give a trans-1,2-diol. **Final Answer:** - **Product:** trans-1,2-cyclohexanediol - **Type:** Anti dihydroxylation --- ### (h) **Reagents:** H₂, Pd⁰ **Step 1:** Catalytic hydrogenation reduces double bonds to single bonds. **Step 2:** The product is 1,2,3,4-tetramethylcyclopentane. **Final Answer:** - **Product:** 1,2,3,4-tetramethylcyclopentane - **Type:** Catalytic hydrogenation (alkene to alkane) --- ### (i) **Reagents:** 1. BH₃, THF 2. H₂O₂, H₂O, NaOH **Step 1:** Hydroboration-oxidation adds H and OH across the triple bond in an anti-Markovnikov fashion. **Step 2:** The product is CH₃CH₂CH₂COH=CH₂ (an aldehyde: pentanal). **Final Answer:** - **Product:** Pentanal (CH₃CH₂CH₂CH₂CHO) - **Type:** Hydroboration-oxidation of terminal alkyne (anti-Markovnikov addition) --- ### (j) **Product Provided:** trans-1,2-cyclohexanediol (both enantiomers shown) **Step 1:** Identify the reaction that forms a vicinal diol from cyclohexene. **Step 2:** Osmium tetroxide (OsO₄) or cold KMnO₄ gives syn dihydroxylation. The question likely used OsO₄ or cold KMnO₄. **Final Answer:** - **Reagent:** OsO₄, H₂O₂ or KMnO₄ (cold, dilute) - **Type:** Syn dihydroxylation --- ### (k) **Reagents:** 1. KMnO₄ (hot) 2. H₃O⁺, H₂O **Step 1:** Hot KMnO₄ cleaves double bonds, fully oxidizing them. **Step 2:** For a methyl-substituted cyclohexene, this will result in ring opening and formation of a carboxylic acid. **Final Answer:** - **Product:** Adipic acid (hexanedioic acid) - **Type:** Oxidative cleavage --- ### (l) **Reagents:** 1. NaNH₂, NH₃ 2. CH₃Br, THF **Step 1:** NaNH₂ deprotonates the terminal alkyne to form an acetylide ion. **Step 2:** Alkylation with CH₃Br adds a methyl group to the terminal carbon. **Final Answer:** - **Product:** Phenyl-1-butyne (C₆H₅C≡CCH₃) - **Type:** Alkylation of terminal alkyne --- Let me know if you want these in a table or need a specific chemical structure drawn!