Let's go step by step for each part of the given problem: --- ## **1. Draw product(s) and mechanisms** ### **a.** **Reagents:** HCl **Reaction:** Electrophilic addition of HCl to an alkene (Markovnikov addition). **Starting compound:** 3-methyl-1-butene #### **Mechanism:** 1. The double bond attacks the H of HCl, forming a carbocation at the more substituted carbon (Markovnikov). 2. Cl⁻ attacks the carbocation. #### **Product:** 2-chloro-3-methylbutane **Structure:** ``` CH3 | CH2—C—CH2—CH3 | Cl ``` --- ### **b.** **Reagents:** H₂O, H₂SO₄ **Reaction:** Acid-catalyzed hydration of alkene (Markovnikov addition of water). #### **Mechanism:** 1. Protonate the alkene to form the more stable carbocation. 2. Water attacks the carbocation. 3. Deprotonate to form the alcohol. #### **Product:** 2-methyl-2-butanol **Structure:** ``` CH3 | CH3—C—CH2—CH3 | OH ``` --- ### **c.** **Reagents:** 1) BH₃ 2) H₂O₂, HO⁻ **Reaction:** Hydroboration-oxidation (anti-Markovnikov addition of water). #### **Mechanism:** 1. BH₃ adds to the less substituted carbon of the double bond. 2. Oxidation replaces B with OH. #### **Product:** 1-butanol **Structure:** ``` CH2—CH2—CH2—CH2OH ``` --- ### **d.** **Reagents:** Br₂, H₂O **Reaction:** Halohydrin formation. #### **Mechanism:** 1. Alkene attacks Br₂, forming a bromonium ion. 2. H₂O attacks the more substituted carbon. 3. Deprotonation leads to halohydrin. #### **Product:** Trans-2-bromo-1-phenylethanol **Structure:** ``` OH | Ph—CH—CH2—Br ``` (OH on more substituted carbon, Br on less substituted) --- ### **e.** **Reagents:** 1) 9-BBN 2) H₂O₂, HO⁻ **Reaction:** Hydroboration-oxidation (for a cyclohexene ring). #### **Product:** Cyclohexanol (OH on less substituted carbon of the double bond) --- ### **f.** **Reagents:** Br₂ **Reaction:** Anti addition of Br₂ to alkene. #### **Product:** Trans-1,2-dibromo-2-methylbutane **Structure:** ``` Br Br \ / CH3—C—C—CH3 | CH3 ``` --- ## **2. Draw a stepwise mechanism for three alcohols from the bicyclic alkene** **Reagents:** H₂O, H₂SO₄ **Mechanism:** 1. Protonation of the double bond forms a carbocation. 2. Water attacks the carbocation. 3. Deprotonation yields the alcohol. 4. Rearrangements can occur, leading to different alcohols. **You need to draw the mechanism for each alcohol formed, showing protonation, possible rearrangement, water attack, and deprotonation.** --- ### **Summary Table** | Part | Product | Key Mechanism Step(s) | |------|------------------------------------|-------------------------------| | a | 2-chloro-3-methylbutane | Markovnikov, carbocation | | b | 2-methyl-2-butanol | Markovnikov, hydration | | c | 1-butanol | anti-Markovnikov, hydroboration| | d | trans-2-bromo-1-phenylethanol | bromonium, water attack | | e | cyclohexanol | hydroboration-oxidation | | f | trans-1,2-dibromo-2-methylbutane | anti addition of Br₂ | --- If you want detailed arrow-pushing mechanisms for parts b, c, and d, **please specify which one you'd like to see first**, and I'll provide a step-by-step drawing and explanation!