Solutions_Manual_for_Organic_Chemistry_6th_Ed 41

Solutions_Manual_for_Organic_Chemistry_6th_Ed 41 - 2 -14...

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Unformatted text preview: 2 -14 continued (c) L �" c / n et dipole 0 �� F � = F (d) ( e) 0�, .�� 0 'x o " " o � ( g) ( i) ro or 0; 13 �1 1 C / H ' CH3 n et n et / N ,� ,/ Q1 o each end oxygen has one-half negative charge as it is the composite of two resonance forms; see n et solution to 1 -38(b) small dipole (0.52) net ( f) 1 l arge dipole ( 1 .70) T H-C I ( h) n et large dipole (2.95) l arge dipole �� N CD .. l arge dipole (2.72) H3C (k) 1 \) F \ � B I - net 1 \," ' CH 3 CH3 = small dipole (0.67) eC = large dipole (l .45) (m) 0 � N" " / n et 2- 1 5 With chlorines on the same side of the double bond, the bond dipole moments reinforce each other, resulting in a large net dipole. With chlorines on opposite sides of the double bond, the bond dipole moments exactly cancel each other, resulting in a zero net dipole. large net dipole n et dipole = 0 n et dipole 0 In (k) through (m), the symmetry of the molecule allows the individual bond dipoles to cancel. \\ � IFl-F (I) CI ----+ - n et dipole H "'" \\ " H H tt� C � �CI H / C == C '" H 1 31 n et dipole = 0 ...
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