Let's break down the reaction sequence step by step: ## **Step 1: 1 eq. HBr (Hydrobromic Acid)** - The starting compound is a cyclopentyl group attached to a terminal alkyne. - Addition of 1 equivalent of HBr to a terminal alkyne gives a **Markovnikov addition** of Br, so Br adds to the more substituted carbon of the alkyne, forming a bromo-alkene. **Product after Step 1:** Cyclopentyl–CH=CH–Br (bromo-alkene, Br at the internal carbon, H at the terminal carbon) --- ## **Step 2: BH₃ (Hydroboration)** - Hydroboration-oxidation of an alkene is **anti-Markovnikov** and adds an OH to the less substituted carbon (the terminal carbon in this case). --- ## **Step 3: H₂O₂, aq. NaOH (Oxidation)** - This step replaces the boron with an OH group, maintaining the anti-Markovnikov addition. **Final Product:** - The product will have a Br at the internal carbon and an OH at the terminal carbon, both on the former alkyne chain. - The structure: Cyclopentyl–CH(OH)–CH₂Br. --- ## **Matching With the Choices** From the image, the **correct structure is:** - Cyclopentyl group attached to a carbon with **OH** (secondary alcohol), which is attached to a carbon with **Br** (primary bromo group). So, the selected answer at the bottom of the image is **correct**. ## **Final Answer** The major product is: ``` Br | Cyclopentyl—CH—CH2—OH ``` Or more clearly, as shown in the selected answer: - The **OH** is on the carbon directly attached to cyclopentane. - The **Br** is on the next carbon. **Explanation Recap:** 1. HBr adds Br to the internal carbon (Markovnikov). 2. Hydroboration-oxidation adds OH to the terminal carbon (anti-Markovnikov). 3. The product is cyclopentyl-CH(OH)-CH₂Br. **Correct Option:** The one with **OH on the cyclopentyl carbon and Br on the next carbon** (selected option).