Let's break down both questions step by step, focusing on the chemistry involved and using diagrams for clarity. --- # **Hydroboration-Oxidation Reaction** The given reagents for both questions are: 1. **BH₃** (Borane) 2. **H₂O₂, aq. NaOH** (Hydrogen peroxide in basic medium) This combination is known as **hydroboration-oxidation**, which converts **alkenes or alkynes to alcohols** with **anti-Markovnikov** regioselectivity (the OH group attaches to the less substituted carbon). --- ## **Question 4 Explanation** ### **Starting Material** - A cyclopentyl ring with an alkene and a methoxy (OMe) group as shown. ### **Step 1: Hydroboration** - Borane adds **syn** (on the same side) across the double bond. - **Boron** attaches to the **less substituted carbon** (anti-Markovnikov). - **Hydrogen** attaches to the more substituted carbon. **Diagram:** ``` OMe | Cyclopentane-CH=CH2 | (double bond) ``` → After hydroboration, boron at the less substituted end. ### **Step 2: Oxidation** - The boron is replaced by an **OH group**. ### **Key Points** - **Anti-Markovnikov:** OH ends up on the less substituted carbon. - **Syn addition:** Both H and OH add to the same face of the double bond (can lead to stereochemistry). ### **Analyzing the Options** - **First structure:** OH and OMe on the same carbon (not possible here). - **Second structure:** OH and OMe on adjacent carbons, OH on the less substituted carbon (this matches anti-Markovnikov). - **Third and fourth:** Show stereoisomers (enantiomers). Both are possible, but only one will be the "major" product unless the starting alkene is not chiral. **Correct Answer:** The product with the OH on the less substituted carbon, adjacent to the carbon with the OMe group, is the correct answer. **So, the answer is the second option (from the top).** --- ## **Question 5 Explanation** ### **Starting Material** - 1-Butyne (an alkyne). ### **Hydroboration-Oxidation with Alkynes** - Converts terminal alkynes to **aldehydes** (anti-Markovnikov). - The boron adds to the terminal carbon, and after oxidation, you get an aldehyde at the terminal position. **Diagram:** ``` CH≡C-CH2-CH3 ``` → After hydroboration-oxidation: ``` O || CH3-CH2-CH2-CH \ H ``` (Butanal) ### **Analyzing the Options** - **First structure:** Alcohol on the terminal position, but no alkene (not correct). - **Second structure:** Ketone at the second carbon (not correct for terminal alkyne). - **Third structure:** Two OH groups (incorrect). - **Fourth structure:** Dihydroxy compound (incorrect). - **Fifth structure:** Aldehyde at terminal position (correct). **Correct Answer:** The last structure, butanal (aldehyde at the end), is correct. --- ## **Summary Table** | Reaction Type | Reagents | Product Type | Regioselectivity | |---------------------------|--------------------------|----------------------|----------------------| | Hydroboration-Oxidation | BH₃, H₂O₂/NaOH | Alcohol (from alkene)| Anti-Markovnikov | | Hydroboration-Oxidation | BH₃, H₂O₂/NaOH (alkyne) | Aldehyde (terminal) | Anti-Markovnikov | --- ## **Visual Summary** ### **Question 4 Key Step**  *OH ends up on the less substituted carbon (anti-Markovnikov).* ### **Question 5 Key Step**  *Terminal alkyne yields an aldehyde.* --- If you need more details or have a specific option you want to discuss, let me know!