d A compound with molecular formula C 24 H 48 must have either one double bond

D a compound with molecular formula c 24 h 48 must

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d) A compound with molecular formula C 24 H 48 must have either one double bond or one ring. It cannot have a triple bond, but it may have a double bond. e) 2.47. a) an sp 2 hybridized atomic orbital b) a p orbital c) a p orbital 2.48. a) N OH H N OH H N OH H b) O O O c) 2.49. a) (CH 3 ) 3 CCH 2 CH 2 CH(CH 3 ) 2 b) (CH 3 ) 2 CHCH 2 CH 2 CH 2 OH c) CH 3 CH 2 CH=C(CH 2 CH 3 ) 2
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CHAPTER 2 33 2.50. a) C 9 H 20 b) C 6 H 14 O c) C 8 H 16 2.51. (d) is not a valid resonance structure, because it violates the octet rule. The nitrogen atom has five bonds in this drawing, which is not possible, because the nitrogen atom only has four orbitals with which it can form bonds. 2.52. 15 carbon atoms and 18 hydrogen atoms: C C C C C C C C C C C C C C C H H H H H H H H H H H H H H H H H H 2.53. O N N N a) b) c) d) O 2.54. Cl Cl Cl Cl 2.55. a) O O b) O O c) d) N N
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34 CHAPTER 2 e) N H N H N H N H N H H H H H H f) g) O O O O O O O h) O O O O O O
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CHAPTER 2 35 i) OH OH OH OH j) O O O O O 2.56. These structures do not differ in their connectivity of atoms. They differ only in the placement of electrons, and are therefore resonance structures. 2.57. a) constitutional isomers b) same compound c) different compounds that are not isomeric d) constitutional isomers 2.58. a) b) OH c) O d) O O e) Br f)
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36 CHAPTER 2 2.59. The nitronium ion does not have any significant resonance structures because any attempts to draw a resonance structure will either 1) exceed an octet for the nitrogen atom or 2) generate a nitrogen atom with less than an octet of electrons, or 3) generate a structure with three charges. The first of these would not be a valid resonance structure, and the latter two would not give significant resonance structures. 2.60. O O O O O O 2.61. Both nitrogen atoms are sp 2 hybridized and trigonal planar, because in each case, the lone pair participates in resonance. 2.62. HO OH H H H HO OH H H H HO OH H H H HO OH H H H HO OH H H H O OH H H H O OH H H H O OH H H H
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CHAPTER 2 37 2.63. a) The molecular formula is C 3 H 6 N 2 O 2 b) There are two sp 3 hybridized carbon atoms c) There is one sp 2 hybridized carbon atom d) There are no sp hybridized carbon atoms e) There are six lone pairs (each nitrogen atom has one lone pair and each oxygen atom has two lone pairs) f) O N O H NH 2 localized localized localized delocalized g) O N O H NH 2 not relevant (only connected to one other atom) trigonal planar trigonal pyramidal tetrahedral tetrahedral bent trigonal planar h) O N O H NH 2 O N O H NH 2 O N O H NH 2 2.64. a) The molecular formula is C 16 H 21 NO 2 b) There are nine sp 3 hybridized carbon atoms c) There is seven sp 2 hybridized carbon atoms d) There are no sp hybridized carbon atoms e) There are five lone pairs (the nitrogen atom has one lone pair and each oxygen atom has two lone pairs) f) The lone pairs on the oxygen of the C=O bond are localized. One of the lone pairs on the other oxygen atom is delocalized. The lone pair on the nitrogen atom is delocalized.
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  • Fall '11
  • RAHMAN
  • Atom, Mole, Lewis Structure, Chemical bond, C C H H H C C C H H H H H C H H C C O H H H H H H C H H C C C H H H C H O H C H H

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