Certainly! Here’s a **stepwise answer** for each question, with explanations at each step and final answers at the end. --- ## 1. 9-BBN and 1-Octene ### **Step 1:** Understand the reaction stoichiometry - 9-BBN hydroborates alkenes in a **1:1 molar ratio** (one molecule of 9-BBN adds across one double bond of the alkene). - 1-Octene has one double bond per molecule. **Explanation:** Each molecule of 9-BBN can react with one molecule of 1-octene. --- ### **Step 2:** Relate to the amount given - To hydroborate **one mole** of 1-octene, you need **one mole** of 9-BBN. **Explanation:** Since the reaction is 1:1, one mole of 9-BBN will hydroborate one mole of 1-octene. --- **Final Answer:** **You would need one mole of 9-BBN to hydroborate one mole of 1-octene.** --- ## 2. BH₄⁻ Reaction with Water (Mechanism) ### **Step 1:** First reaction with water - **BH₄⁻ + H₂O → BH₃ + H₂ + OH⁻** - BH₄⁻ donates a hydride ion (H⁻) to the water, producing hydrogen gas and BH₃. **Explanation:** Borohydride acts as a nucleophile, attacking the proton of water, releasing H₂ gas. --- ### **Step 2:** Further reaction with two more water molecules - **BH₃ + 3H₂O → B(OH)₃ + 3H₂** - Each water molecule reacts with boron, fully hydrolyzing it. **Explanation:** BH₃ is hydrolyzed stepwise by water, eventually forming boric acid (B(OH)₃) and more hydrogen gas. --- **Final Answer:** **Products after full hydrolysis: B(OH)₃ (boric acid) and H₂ gas.** --- ## 3. Lewis Acid-Base Adduct: BH₃ and Diethyl Ether ### **Step 1:** Draw the Lewis structures - **BH₃**: Boron has 3 valence electrons, forming 3 single bonds with H, has an empty p orbital (Lewis acid). - **Diethyl ether (Et₂O):** Oxygen has two lone pairs (Lewis base). **Explanation:** Boron is electron-deficient and accepts a lone pair from oxygen. --- ### **Step 2:** Show formal charges - Oxygen donates a lone pair to boron, forming a coordinate bond. - Both atoms remain neutral, as no electrons are lost or gained, just shared. **Explanation:** The adduct is neutral, as the donation is a coordinate covalent bond, not ionization. --- **Final Answer:** **The adduct is Et₂O→BH₃, with a coordinate bond from O (Et₂O) to B (BH₃); all atoms are neutral.** --- ## 4. Hydroboration-Oxidation of the Given Alkene (Stereochemistry) ### **Step 1:** Identify the alkene and hydroboration product - The alkene is 3-methyl-1-butene. - Hydroboration-oxidation (BH₃, then H₂O₂/NaOH) adds **OH to the less substituted carbon** (anti-Markovnikov) and **H to the more substituted**. **Explanation:** Hydroboration-oxidation proceeds via syn addition, both H and OH add on the same face. --- ### **Step 2:** Predict products and stereochemistry - The **alcohol** will be at the terminal carbon (position 1), giving **3-methyl-1-butanol**. - **Stereochemistry:** Since the product is a primary alcohol (no chiral center at C-1), there’s **no stereoisomerism** in this case. **Explanation:** With no new chiral center, only one product is formed. --- **Final Answer:** **The only product is 3-methyl-1-butanol; no stereoisomers are formed because there is no chiral center.** --- Let me know if you want the full mechanisms or structures drawn out!

Certainly! Below is the structured Markdown response addressing all parts of the question based on the provided information. --- # Pre-Lab Questions Solutions ## 1. Hydroboration of 1-Octene with 9-BBN ### **Step 1:** Reaction Stoichiometry - **Reaction Type:** Hydroboration - **Molar Ratio:** 9-BBN reacts with 1-octene in a **1:1** ratio. **Explanation:** Each molecule of 9-BBN can react with one molecule of 1-octene due to the presence of one double bond in 1-octene. ### **Step 2:** Amount Required - To hydroborate **one mole** of 1-octene, you need **one mole** of 9-BBN. **Explanation:** Since the reaction is 1:1, the total amount of 9-BBN required is equal to the amount of 1-octene. **Final Answer:** **You would need one mole of 9-BBN to hydroborate one mole of 1-octene.** --- ## 2. Reaction of BH₄⁻ with Water ### **Step 1:** Initial Reaction - **Reaction:** BH₄⁻ + H₂O → BH₃ + H₂ + OH⁻ **Explanation:** Borohydride (BH₄⁻) donates a hydride ion (H⁻) to water, resulting in the formation of hydrogen gas (H₂) and borane (BH₃). ### **Step 2:** Hydrolysis of Borane - **Further Reaction:** BH₃ + 3H₂O → B(OH)₃ + 3H₂ **Explanation:** BH₃ reacts with water in a stepwise manner to produce boric acid (B(OH)₃) and additional hydrogen gas. **Final Answer:** **After full hydrolysis, the products are B(OH)₃ (boric acid) and H₂ gas.** --- ## 3. Lewis Acid-Base Adduct of BH₃ and Diethyl Ether ### **Step 1:** Lewis Structures - **BH₃:** Boron has 3 valence electrons, forming three bonds with hydrogen atoms, and has an empty p orbital. - **Diethyl Ether (Et₂O):** Oxygen has two lone pairs of electrons. **Explanation:** Boron, being electron-deficient, acts as a Lewis acid and can accept a lone pair from the oxygen in diethyl ether. ### **Step 2:** Formation of the Adduct - The lone pair from the oxygen forms a coordinate bond with the boron atom. **Explanation:** The adduct (Et₂O→BH₃) comprises a coordinate bond without changing the formal charges of the involved atoms. **Final Answer:** **The adduct formed is Et₂O→BH₃, with a coordinate bond from O (Et₂O) to B (BH₃); all atoms remain neutral.** --- ## 4. Hydroboration-Oxidation of 3-Methyl-1-Butene ### **Step 1:** Hydroboration Reaction - The alkene is **3-methyl-1-butene**. - Hydroboration results in the addition of **OH to the less substituted carbon** (anti-Markovnikov). **Explanation:** The hydroboration-oxidation process involves syn addition, where both H and OH add to the same side of the double bond. ### **Step 2:** Predicting Stereochemistry - The product after hydroboration-oxidation is **3-methyl-1-butanol**. - There is **no stereoisomerism** as no new chiral centers are generated. **Explanation:** The product formed is a primary alcohol without chiral centers leading to only one possible structure. **Final Answer:** **The only product is 3-methyl-1-butanol; no stereoisomers are formed because there is no chiral center.** --- This concludes the solutions to the pre-lab questions. Let me know if you need further details or explanations!