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Organic ChemistryMain MenuLecture 1: Introduction to Molecular OrbitalsLecture 10: Intermolecular Forces and Intro to ReactivityLecture 11: Alkenes as Nucleophiles ILectures 12 and 13: Alkenes II & IIILecture 14: Benzenes as Nucleophiles ILecture 15 objectives and skillsLecture 2: Molecular Orbitals in Organic StructuresLecture 3: Molecular Orbitals and ResonanceLecture 4: Molecular Orbitals and AromaticityLecture 5: Nomenclature and IR SpectroscopyLecture 6: Stereochemistry ILecture 7: Stereochemistry IILecture 8: Newman and Fisher ProjectionsLecture 9: Conformations and Cyclohexane RingsJulio Terra2fb22914f73f9e5a767bf7fb83987dca0e153e4a
12020-04-11T13:19:31-07:00SOLUTION 21.1b1plain2020-04-11T13:19:31-07:00The first reaction is faster. The substrate substitution and base are the same, but iodide is the better leaving group over chloride. I- is larger and the weaker base (more stable anion). In both E1 and E2 reactions, the type of leaving group affects the activation energy. Debatable that they are approximately the same rate. In both cases, the protic solvent is the base (both weak bases). The identity of the base is the only difference between the two reactions. Since the strength and concentration of base do not affect the rate of E1 reactions, they should proceed at the same rate. A slight difference in rate may come from the solvent effects. Water is the more polar of the two solvents, which could stabilize the carbocation enough MORE than NH3 to increase reaction rate in water.
