上海交通大学：《大学化学》教学资源_加州理工课件学习_Series06ResonanceStructures

2 2市 Structures Reading:Gray:(2-14) OGC:(3.8) VI-1

Resonance Resonance Structures Reading: Gray: (2-14) OGC: (3.8) VI-1

Resonance Structures two structural possibilities for [C3H]: H H H H H H C一C= C= C: H H H H however,the actual molecule is neither of the two structures: H H H H H V-2

C C H H H H H C C C H H H H H C C C H H H H H C Resonance Structures [C3H5] : – two structural possibilities for - - however, the actual molecule is neither of the two structures: -δ -δ VI-2

Resonance Structures:Another Example H Two structural possibilities for benzene: H H H Each C-C bond can be thought of as a H H 1.5-electron bond in the actual structure V-3

H C C H H H H C H Resonance Structures: Another Example Two structural possibilities for benzene: C C C H C C H H H H C H C C C H C C H H H H C H C C C Each C - C bond can be thought of as a 1.5-electron bond in the actual structure VI-3

Formal Charge Formal Charge:the difference between the number of valence shell electrons for an element and the number of electrons surrounding it in a molecule (including half of the bond electrons and all the free electrons) Example: The leftmost carbon has 3 bonds H with 2 electrons each,plus 2 free electrons;so it has 5 electrons surrounding it.The formal charge H equals the valence electrons minus surrounding electrons: 4-5=-1 Formal Charge V-4

C C H H H H H C - Formal Charge Formal Charge: the difference between the number of valence shell electrons for an element and the number of electrons surrounding it in a molecule (including half of the bond electrons and all the free electrons) Example: The leftmost carbon has 3 bonds with 2 electrons each, plus 2 free electrons; so it has 5 electrons surrounding it. The formal charge equals the valence electrons minus surrounding electrons: 4 - 5 = –1 = Formal Charge VI-4

“Reasonable”Resonance Structures 1.Form octets if possible 2.Maximize bonding 3.Distribute formal charges in a reasonable way -minimize if possible -more negative on more electronegative atoms -avoid same sign on neighboring atoms V-5

“Reasonable” Resonance Structures 1. Form octets if possible 2. Maximize bonding 3. Distribute formal charges in a reasonable way -minimize if possible -more negative on more electronegative atoms -avoid same sign on neighboring atoms VI-5

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 4e￾O 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- = 16e￾this 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