Sure! Let's solve the reaction sequence **step by step**, explaining each transformation and drawing the intermediates/products. --- ### **Step 1: Hydroboration-Oxidation of Alkene** **Reagents:** - **BH₃·THF, °C–60°C** (Hydroboration) - **30% H₂O₂, 3M NaOH, THF** (Oxidation) **Transformation:** - The hydroboration-oxidation sequence converts the terminal alkene into a primary alcohol via **anti-Markovnikov addition** of water (OH to less substituted carbon). **Intermediate after Step 1:** - The product is a secondary alcohol at the position where the double bond was. **Structure:** - The phenyl ring, original alcohol, and the cyclic amine remain unchanged. - The double bond is replaced by a –CH₂CH₂OH at the end. **Explanation:** Hydroboration-oxidation adds an OH group to the less substituted carbon of the alkene, converting it to a primary alcohol. --- ### **Step 2: Hydrogenation (Reduction of Imine/Double Bonds)** **Reagents:** - **H₂, Pd(OH)₂, MeOH/HCl** **Transformation:** - This step **reduces the imine (C=N)** to an amine (C–NH). **Intermediate after Step 2:** - The imine group in the ring is reduced to an amine (–NH–). **Explanation:** Hydrogenation with Pd(OH)₂ under acidic conditions reduces C=N to C–NH. --- ## **Summary of Steps and Intermediates** ### **Step 1 Product:** - The alkene is converted to a primary alcohol at the terminal position (anti-Markovnikov). - Imine and existing alcohol remain unchanged. ### **Step 2 Product:** - The imine is reduced to an amine. - The primary alcohol (from step 1) and existing alcohol remain. --- Let's draw the structures for both steps: --- #### **Step 1 Product:** ``` Ph \ C(OH) | (cyclohexane) | (piperidine ring) | CH2-CH2-OH ``` #### **Step 2 Product:** ``` Ph \ C(OH) | (cyclohexane) | (piperidine ring with NH) | CH2-CH2-OH ``` --- **Proceed to the next steps only after confirming each transformation.** Let me know if you want to continue with the next steps!