Drawing Resonance Structures Count the number of electrons available,and devise a reasonable structure having that many electrons example:[CO]2- C→4e 0→3(6e) 2 extra electrons 24 electrons total 8e in bonds +16e-as free electrons 24e total VI-6
Drawing Resonance Structures Count the number of electrons available, and devise a reasonable structure having that many electrons example: [CO3] 2 - 8e- in bonds +16e- as free electrons 24e- total - O C O O O C O O O C O O - C 4eO 3(6e- ) + 2 extra electrons 24 electrons total - - - - VI-6
Resonance Structures of N2O N20一2(5e)+6e=16e N-O-N this structure won't work:no octets on N's +2 ON No nor this one:high formal charges N N nor this one:such configurations are Q destabilized by“ring strain” ⊕ N :N- g一N= ⊕ N一O:⊙ best structures possible -7
N O N N O N N N O N N O N2O 2(5e- ) + 6e- = 16ethis structure won’t work: no octets on N’s nor this one: high formal charges +2 O N N nor this one: such configurations are destabilized by “ring strain” + best structures possible Resonance Structures of N2O - - - - + VI-7
Benzene: Much more stable and much less reactive than 3 isolated double bonds H H H H +Br— Br Br— Br H H kcal H but... △H=-29.2 mol H H No 1+Br—BrX产 Reaction AH=+2 kcal H mol
C C C C C C H H H H H H + Br Br Benzene: • Much more stable and much less reactive than 3 isolated double bonds but... X No Reaction ΔH = −29.2 kcal mol ΔH = +2 kcal mol C C H H H H + Br Br C C H H H H Br Br VI-8
Benzene's "Resonance Stabilization" kcal H=-28.4 mol △H=-26.5 kcal 州 mol kcal △H=+5.6 mol H kcal 3 △H=-49.3 H mol VI-9
+ H H X + H H + H H ΔH = −28.4 kcal mol Benzene’s “Resonance Stabilization” ΔH = −26.5 kcal mol ΔH = +5.6 kcal mol ΔH = −49.3 kcal mol + 3 H H VI-9
More Resonance Stability Propanol H3C-CH2-CH2-OH H3C-CH2-CH2-O +H+ K≈1015M1 Phenol 0-H +H+ K=10-10M-1 VL-10
More Resonance Stability O H O: - : : O - O - + H+ H3C-CH2-CH2-OH H3C-CH2-CH2-O- + H+ K ≈ 10-15 M-1 K = 10-10 M-1 Propanol Phenol : : : : : : : : VI-10