# Reaction Mechanism Identification (E1 or E2) and Products Let's solve each reaction stepwise, determining the mechanism and giving the product. --- ## **1.** ### **Step 1: Mechanism Identification** - **Reactant:** 2-bromo-2-methylpropane (tert-butyl bromide) - **Reagent:** NaOH (strong base, strong nucleophile) - **Substrate:** Tertiary alkyl halide **Explanation:** Tertiary halides with strong bases can go E2, but E1 is favored in polar protic solvents and with weak bases. However, NaOH is a strong base, but because the substrate is tertiary and can form a stable carbocation, both could occur, but E2 is sterically hindered. So, E1 is favored. ### **Step 2: Product** - **Mechanism:** **E1** - **Product:** 2-methylpropene (isobutylene) **Explanation:** The Br leaves, forming a tertiary carbocation, followed by elimination of a proton to give the alkene. --- ## **2.** ### **Step 1: Mechanism Identification** - **Reactant:** 2-butanol - **Reagent:** H₂O (acidic, likely under acidic conditions) - **Substrate:** Secondary alcohol **Explanation:** Secondary alcohols typically undergo E1 elimination under acidic conditions (dehydration), forming a carbocation intermediate. ### **Step 2: Product** - **Mechanism:** **E1** - **Product:** 2-butene (major, more substituted alkene) **Explanation:** Alcohol is protonated, water leaves, carbocation forms, then elimination gives the alkene. --- ## **3.** ### **Step 1: Mechanism Identification** - **Reactant:** 2-chloro-2-methylbutane - **Reagent:** MeOH (methanol, weak base, weak nucleophile) - **Substrate:** Tertiary alkyl halide **Explanation:** Weak base, polar protic solvent, tertiary halide — favors **E1**. ### **Step 2: Product** - **Mechanism:** **E1** - **Product:** 2-methyl-2-butene (major, more substituted alkene) --- ## **4.** ### **Step 1: Mechanism Identification** - **Reactant:** Bromocyclohexane - **Reagent:** Sodium ethoxide (NaOEt, strong base) - **Substrate:** Secondary alkyl halide **Explanation:** Strong base, secondary halide — **E2** is favored. ### **Step 2: Product** - **Mechanism:** **E2** - **Product:** Cyclohexene --- ## **5.** ### **Step 1: Mechanism Identification** - **Reactant:** Bromocyclohexane - **Reagent:** Sodium ethoxide (NaOEt, strong base) - **Substrate:** Secondary alkyl halide **Explanation:** Same as above: strong base, secondary halide — **E2**. ### **Step 2: Product** - **Mechanism:** **E2** - **Product:** Cyclohexene --- ## **6.** ### **Step 1: Mechanism Identification** - **Reactant:** 1-bromo-1,2,3,4-tetrahydronaphthalene - **Reagent:** MeOH (methanol, weak base) - **Substrate:** Benzylic bromide **Explanation:** Benzylic carbocations are stabilized, methanol is a weak base, so **E1** is favored. ### **Step 2: Product** - **Mechanism:** **E1** - **Product:** 1,2-dihydronaphthalene (alkene at the benzylic position) --- ## **Summary Table** | # | Substrate & Reagent | Mechanism | Product | |---|--------------------------------------|-----------|-------------------------------| | 1 | tert-butyl bromide + NaOH | E1 | 2-methylpropene | | 2 | 2-butanol + H₂O | E1 | 2-butene | | 3 | 2-chloro-2-methylbutane + MeOH | E1 | 2-methyl-2-butene | | 4 | bromocyclohexane + NaOEt | E2 | cyclohexene | | 5 | bromocyclohexane + NaOEt | E2 | cyclohexene | | 6 | 1-bromo-1,2,3,4-tetrahydronaphthalene + MeOH | E1 | 1,2-dihydronaphthalene | --- ### **Final Answer** - Reactions 1, 2, 3, and 6 proceed via **E1**. - Reactions 4 and 5 proceed via **E2**. - The products are the corresponding alkenes as indicated above.