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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
Primary Bromination, three possible primary C-H bonds for substitution Total hydrogens x “reactivity ratio” for primary = 3 x 1 = 3 Secondary Bromination (1,2), four possible secondary C-H bonds for substitution Total hydrogens x “reactivity ratio” for secondary = 4 x 80 = 320 Secondary Bromination (1,3), two possible secondary C-H bonds for substitution Total hydrogens x “reactivity ratio” for secondary = 2 x 80 = 160 Tertiary Bromination, one possible tertiary C-H bond for substitution Total hydrogens x “reactivity ratio” for tertiary = 1 x 1600 = 1600
% Primary = 3/2083 x 100% = 0.1% % Secondary (1,2) = 320/2083 x 100% = 15.4% % Secondary (1,3) = 160/2083 x 100% = 7.8% % Tertiary = 1600/2083 x 100% = 77%
When methylcyclobutane (below) is brominated in the presence of light, three different types of monobromination products can be formed (primary, secondary, and tertiary substitution). Draw the structures of all possible products. SOLUTION
Calculate the expected % abundance of each product, assuming the relative Br2 reactivities of 1600:80:1 towards tertiary, secondary, and primary C-H bonds, respectively. SOLUTION
