1. Chapter 2 Alkanes and Cycloalkanes: Introduction to Hydrocarbons Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2.1 Classes of Hydrocarbons

3. HydrocarbonsHydrocarbons AromaticAromaticAliphaticAliphatic AlkanesAlkanes AlkynesAlkynesAlkenesAlkenes

4. HydrocarbonsHydrocarbons AliphaticAliphatic AlkanesAlkanes Alkanes are hydrocarbons in which all of the bonds are single bonds. C C H H H H H H

5. HydrocarbonsHydrocarbons AliphaticAliphatic AlkenesAlkenes Alkenes are hydrocarbons that contain a carbon- carbon double bond. C C H H H H

6. HydrocarbonsHydrocarbons AliphaticAliphatic AlkynesAlkynes Alkynes are hydrocarbons that contain a carbon- carbon triple bond. HC CH

7. HydrocarbonsHydrocarbons AromaticAromatic The most common aromatic hydrocarbons are those that contain a benzene ring. H H H H H H

8. 2.2 Electron Waves and Chemical Bonds

9. Valence Bond Theory Molecular Orbital Theory The Lewis model of chemical bonding predates the idea that electrons have wave properties. There are two other widely used theories of bonding that are based on the wave nature of an electron. Models for Chemical Bonding

10. Examine how the electrostatic forces change as two hydrogen atoms are brought together. These electrostatic forces are: attractions between the electrons and the nuclei repulsions between the two nuclei repulsions between the two electrons + e–e– + e–e– Formation of H 2 from Two Hydrogen Atoms

11. Potential energy H + H Internuclear distance H H weak net attraction at long distances Figure 2.1

12. Potential energy H + H Internuclear distance H H H H H H attractive forces increase faster than repulsive forces as atoms approach each other Figure 2.1

13. Potential energy H + H H2H2 Internuclear distance 74 pm H H H H H H -436 kJ/mol maximum net attraction (minimum potential energy) at 74 pm internuclear distance Figure 2.1

14. 1s1s 1s1s H H 2 H atoms: each electron "feels" attractive force of one proton H 2 molecule: each electron "feels" attractive force of both protons H H

15. Potential energy H + H H2H2 Internuclear distance 74 pm H H H H H H -436 kJ/mol repulsive forces increase faster than attractive forces at distances closer than 74 pm Figure 2.1

16. Valence Bond Theory constructive interference between electron waves of two half-filled atomic orbitals is basis of shared-electron bond Molecular Orbital Theory derive wave functions of molecules by combining wave functions of atoms Models for Chemical Bonding

17. 2.3 Bonding in H 2 : The Valence Bond Model

18. 1s1s 1s1s H H in-phase overlap of two half-filled hydrogen 1s orbitals  bond of H 2 H H Valence Bond Model An electron pair can be shared when a half- filled orbital of one atom overlaps in-phase with half-filled orbital of another.

19.  Bond: orbitals overlap along the internuclear axis Cross section of orbital perpendicular to internuclear axis is circular. H H Valence Bond Model

20. Figure 2.4(a) The 1s orbitals of two separated hydrogen atoms are far apart. There is essentially no interaction. Each electron is associated with a single proton. Valence Bond Model of H 2

21. Figure 2.4(b) As the hydrogen atoms approach each other, their 1s orbitals begin to overlap and each electron begins to feel the attractive force of both protons. Valence Bond Model of H 2

22. Figure 2.4(c) The hydrogen atoms are close enough so that appreciable overlap of the two 1s orbitals occurs. The concentration of electron density in the region between the two protons is more readily apparent. Valence Bond Model of H 2

23. Figure 2.4(d) A molecule of H 2. The two hydrogen 1s orbitals have been replaced by a new orbital that encompasses both hydrogens and contains both electrons. Valence Bond Model of H 2

24. 2.4 Bonding in H 2 : The Molecular Orbital Model

25. Electrons in a molecule occupy molecular orbitals (MOs) just as electrons in an atom occupy atomic orbitals (AOs). Only two electrons are allowed in any one orbital, so, there can be two per MO, just as two could be in an AO. MOs are expressed as combinations of AOs. Main Ideas

26. Two AOs combine to yield two MOs Bonding combinationAntibonding combination  MO =  (H) 1s +  (H') 1s  ' MO =  (H) 1s -  (H') 1s The linear combination of atomic orbitals (LCAO) method expresses wave functions of molecular orbitals as sums and differences of wave functions of atomic orbitals. MO Picture of Bonding in H 2

27. bonding antibonding  ** MO H 2 MO H 2 Fig. 2.6: Energy-Level Diagram for H 2 MOs 1s1s 1s1s AO H AO H

28. 2.5 Introduction to Alkanes: Methane, Ethane and Propane

29. Methane(CH 4 ) CH 4 Ethane(C 2 H 6 )CH 3 CH 3 Propane(C 3 H 8 )CH 3 CH 2 CH 3 bp -160°Cbp -89°C bp -42°C The Simplest Alkanes

30. 2.6 sp 3 Hybridization and Bonding in Methane

31. Tetrahedral shape bond angles = 109.5° bond distances = 110 pm but structure seems inconsistent with electron configuration of atomic carbon Structure of Methane

32. Electron configuration of carbon 2s 2p only two unpaired electrons should form  bonds to only two hydrogen atoms bonds should be at right angles to one another

33. 2s2s 2p2p Promoting an electron from the 2s to the 2p orbital gives: sp 3 Orbital Hybridization 2p2p 2s2s Now mix (hybridize) the 2s orbital and the three 2p orbitals. Promotion of e:Orbital hybridization:

34. 2p2p 2s2s 2 sp 3 4 equivalent half-filled orbitals are consistent with four bonds and tetrahedral geometry. sp 3 Orbital Hybridization 4 new hybrid orbitals result. Each has ¼ s character and ¾ p character.  4 new hybrids

35. sp 3 Orbital Hybridization Another representation:

36. s p + – + take the s orbital and place it on top of the p orbital Shape of sp 3 hybrid orbitals

37. s + p + – + reinforcement of electron wave in regions where sign is the same destructive interference in regions of opposite sign Shape of sp 3 hybrid orbitals

38. sp hybrid The orbital shown is sp hybrid analogous procedure using three s orbitals and one p orbital gives sp 3 hybrid shape of sp 3 hybrid is similar. + – Shape of sp 3 hybrid orbitals

39. sp 3 hybrid The hybrid orbital is not symmetrical. The higher probability of finding an electron on one side of the nucleus than the other leads to stronger bonds. + –

40. – +– The C—H  Bond in Methane sp 3 s C H H—C  C H gives a  bond. In-phase overlap of a half-filled 1s orbital of hydrogen with a half-filled sp 3 hybrid orbital of carbon: + +

41. Justification for Orbital Hybridization It is consistent with structure of methane. It allows for formation of 4 bonds rather than 2. Bonds involving sp 3 hybrid orbitals are stronger than those involving s-s overlap or p-p overlap.

42. 2.7 Bonding in Ethane

43. Structure of Ethane CH 3 C2H6C2H6 tetrahedral geometry at each carbon C—H bond distance = 110 pm C—C bond distance = 153 pm

44. In-phase overlap of half-filled sp 3 hybrid orbital of one carbon with half-filled sp 3 hybrid orbital of another. Overlap is along internuclear axis to give a  bond. The C—C  Bond in Ethane

45. Formation of the sp 3 - sp 3  bond in ethane. The C—C  Bond in Ethane

46. 2.8 Isomeric Alkanes: The Butanes C 4 H 10

47. n-ButaneCH 3 CH 2 CH 2 CH 3 Isobutane(CH 3 ) 3 CH bp -0.4°Cbp -10.2°C

48. 2.9 Higher n-Alkanes

49. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 n-Pentane n-Hexane CH 3 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 n-Heptane

50. 2.10 The C 5 H 12 Isomers

51. n-Pentane CH 3 CH 2 CH 2 CH 2 CH 3 Isopentane (CH 3 ) 2 CHCH 2 CH 3 Neopentane (CH 3 ) 4 C C 5 H 12

52. How many isomers? The number of isomeric alkanes increases as the number of carbons increase. There is no simple way to predict how many isomers there are for a particular molecular formula. Table 2.1 Constitutionally Isomeric Alkanes CH 4 1 C 8 H 18 18 C 2 H 6 1 C 9 H 20 35 C 3 H 8 1 C 10 H 22 75 C 4 H 10 2 C 15 H 32 4,347 C 5 H 12 3 C 20 H 42 366,319 C 6 H 14 5 C 40 H 82 62,491,178,805,831 C 7 H 16 9

53. 2.11 IUPAC Nomenclature of Unbranched Alkanes

54. Table 2.2 IUPAC Names of Unbranched Alkanes Retained: methaneCH 4 ethaneCH 3 CH 3 propaneCH 3 CH 2 CH 3 butaneCH 3 CH 2 CH 2 CH 3

55. Note: n-prefix is not part of IUPAC name of any alkane. For example: n-butane is "common name" for CH 3 CH 2 CH 2 CH 3 ; butane is "IUPAC name." Others: Latin or Greek prefix for number of carbons + ane suffix Table 2.2 IUPAC Names of Unbranched Alkanes

56. Number of carbons NameStructure 5pentaneCH 3 (CH 2 ) 3 CH 3 6hexaneCH 3 (CH 2 ) 4 CH 3 7heptaneCH 3 (CH 2 ) 5 CH 3 8octaneCH 3 (CH 2 ) 6 CH 3 9nonaneCH 3 (CH 2 ) 7 CH 3 10decaneCH 3 (CH 2 ) 8 CH 3 Table 2.2 IUPAC Names of Unbranched Alkanes

57. Number of carbons NameStructure 11undecaneCH 3 (CH 2 ) 9 CH 3 12dodecaneCH 3 (CH 2 ) 10 CH 3 13tridecaneCH 3 (CH 2 ) 11 CH 3 14tetradecaneCH 3 (CH 2 ) 12 CH 3 15pentadecaneCH 3 (CH 2 ) 13 CH 3 16hexadecaneCH 3 (CH 2 ) 14 CH 3 Table 2.2 IUPAC Names of Unbranched Alkanes

58. Number of carbons NameStructure 17heptadecaneCH 3 (CH 2 ) 15 CH 3 18octadecaneCH 3 (CH 2 ) 16 CH 3 19nonadecaneCH 3 (CH 2 ) 17 CH 3 20icosaneCH 3 (CH 2 ) 18 CH 3 25pentacosaneCH 3 (CH 2 ) 23 CH 3 30triacontaneCH 3 (CH 2 ) 28 CH 3 Table 2.2 IUPAC Names of Unbranched Alkanes

59. 2.12 Applying the IUPAC Rules: Names of the C 6 H 14 Isomers

60. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 (CH 3 ) 2 CHCH 2 CH 2 CH 3 CH 3 CH 2 CH(CH 3 )CH 2 CH 3 (CH 3 ) 2 CHCH(CH 3 ) 2 (CH 3 ) 3 CCH 2 CH 3 The C 6 H 14 Isomers

61. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 The IUPAC name of the unbranched alkane with a chain of 6 carbons is hexane. Hexane The C 6 H 14 Isomers

62. IUPAC Nomenclature of Branched Alkanes (Table 2.5) Step 1) Find the longest continuous carbon chain and use the IUPAC name of the unbranched alkane as the basis (parent). If there are more than one longest chain of equal length, select the one having the most substituents as the parent. Step 2) Add the name of the substituent as a prefix. Substituents are named by dropping –ane and adding –yl.

63. IUPAC Nomenclature of Branched Alkanes (Table 2.5) Step 3) Number the chain from the end nearest to a substituent, and identify the carbons to which substituents are attached by number. If dissimilar substituents are equidistant from either end, number from the end having the lower alphabetical substituent. Step 4) Write the name listing the substituents alphabetically, separating numbers from numbers with a comma and numbers from words with a dash & parent name at the end.

64. (CH 3 ) 2 CHCH 2 CH 2 CH 3 CH 3 CH 2 CH(CH 3 )CH 2 CH 3 2-Methylpentane 3-Methylpentane The C 6 H 14 Isomers

65. (CH 3 ) 2 CHCH(CH 3 ) 2 (CH 3 ) 3 CCH 2 CH 3 2,3-Dimethylbutane 2,2-Dimethylbutane Use replicating prefixes (di-, tri-, tetra-, etc.) according to the number of identical substituents attached to the main chain. The C 6 H 14 Isomers

66. 2.13 Alkyl Groups

67. Methyl and Ethyl Groups Methyl Ethyl C C H HH H H CH 3 CH 2 C H H H CH 3 or

68. Unbranched Alkyl Groups If potential point of attachment is at the end of the chain, take the IUPAC name of the corresponding unbranched alkane and replace the -ane ending with -yl. R H R

69. Butyl If potential point of attachment is at the end of the chain, take the IUPAC name of the corresponding unbranched alkane and replace the -ane ending with -yl. R H R CH 3 CH 2 CH 2 CH 2 C C H HH H H C C HH H H or Unbranched Alkyl Groups

70. HexylCH 3 (CH 2 ) 4 CH 2 CH 3 (CH 2 ) 5 CH 2 CH 3 (CH 2 ) 16 CH 2 Heptyl Octadecyl Unbranched Alkyl Groups

71. The C 3 H 7 Alkyl Groups CH 3 CH 2 CH 2 C C H HH H H C H H or and CH 3 CHCH 3 C C H HH H H C H H or

72. CH 3 CH 2 CH 2 C C H HH H H C H H or The C 3 H 7 Alkyl Groups IUPAC name: Propyl Common name: n-Propyl

73. Naming Alkyl Groups (Table 2.6) Step 1:Identify longest continuous chain starting at point of attachment. Step 2: Drop -ane ending from name of unbranched alkane having same number of carbons as longest continuous chain and replace with -yl. Step 3:Identify substituents on longest continuous chain. Step 4:Chain is always numbered starting at point of attachment.

74. IUPAC name: 1-Methylethyl Common name: Isopropyl CH 3 CHCH 3 C C H HH H C H H or H The C 3 H 7 Alkyl Groups

75. CH 3 CH 2 CH 2 C C H HH H H C H H or Classification: Primary alkyl group Alkyl groups are classified according to the degree of substitution at the carbon that bears the point of attachment. A carbon that is directly attached to one other carbon is a primary carbon. The C 3 H 7 Alkyl Groups

76. Classification: Secondary alkyl group Alkyl groups are classified according to the degree of substitution at the carbon that bears the point of attachment. A carbon that is directly attached to two other carbons is a secondary carbon. CH 3 CHCH 3 C C H HH H C H H or H The C 3 H 7 Alkyl Groups

77. The C 4 H 9 Alkyl Groups IUPAC name: Butyl Common name: n-Butyl Classification: Primary alkyl group CH 3 CH 2 CH 2 CH 2 C C H HH H H C C HH H H or

78. IUPAC name: 1-Methylpropyl Common name: sec-Butyl Classification: Secondary alkyl group CH 3 CHCH 2 CH 3 C C H HH H H C C HH H H or 1 2 3 The C 4 H 9 Alkyl Groups

79. IUPAC name: 2-Methylpropyl Common name: Isobutyl Classification: Primary alkyl group 1 2 3 C H CH 2 CH 3 The C 4 H 9 Alkyl Groups

80. IUPAC name: 1,1-Dimethylethyl Common name: tert-Butyl Classification: Tertiary alkyl group 1 2 CCH 3 The C 4 H 9 Alkyl Groups

81. 2.14 IUPAC Names of Highly Branched Alkanes

82. Branched Alkanes Octane

83. Branched Alkanes 4-Ethyloctane

84. Branched Alkanes 4-Ethyl-3-methyloctane List substituents in alphabetical order.

85. Branched Alkanes 4-Ethyl-3,5-dimethyloctane List substituents in alphabetical order. But don't alphabetize di-, tri-, tetra-, etc.

86. First Point of Difference Rule The chain is numbered in the direction that gives the lower locant to the substituent at the first point of difference in the names. Don't add locants! 2,2,6,6,7-Pentamethyloctane ? 2,3,3,7,7-Pentamethyloctane ? What is the correct name ? 1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1

87. First Point of Difference Rule The chain is numbered in the direction that gives the lower locant to the substituent at the first point of difference in the names. Don't add locants! 2,2,6,6,7-Pentamethyloctane What is the correct name ? 8 7 6 5 4 3 2 1

88. 2.15 Cycloalkane Nomencalture CnH2nCnH2nCnH2nCnH2n

89. Cycloalkanes Cycloalkanes are alkanes that contain a ring of three or more carbons. Count the number of carbons in the ring, and add the prefix cyclo to the IUPAC name of the unbranched alkane that has that number of carbons. CyclopentaneCyclohexane

90. Ethylcyclopentane Cycloalkanes Name any alkyl groups on the ring in the usual way. CH 2 CH 3

91. Cycloalkanes Name any alkyl groups on the ring in the usual way. List substituents in alphabetical order and count in the direction that gives the lowest numerical locant at the first point of difference. 3-Ethyl-1,1-dimethylcyclohexane CH 2 CH 3 H3CH3CCH 3

92. 2.16 Sources of Alkanes and Cycloalkanes

93. Crude oil

94. Refinery gas C 1 -C 4 Light gasoline (bp: 25-95 °C) Light gasoline (bp: 25-95 °C) C 5 -C 12 Naphtha (bp 95-150 °C) Naphtha Kerosene (bp: 150-230 °C) Kerosene C 12 -C 15 Gas oil (bp: 230-340 °C) Gas oil (bp: 230-340 °C) C 15 -C 25 ResidueResidue

95. Cracking converts high molecular weight hydrocarbons to more useful, low molecular weight ones Reforming increases branching of hydrocarbon chains branched hydrocarbons have better burning characteristics for automobile engines Petroleum Refining

96. 2.17 Physical Properties of Alkanes and Cycloalkanes

97. Boiling Points of Alkanes Boiling points are governed by strength of intermolecular attractive forces. Alkanes are nonpolar, so dipole-dipole and dipole-induced dipole forces are absent. Only forces of intermolecular attraction are induced dipole-induced dipole forces.

98. Induced Dipole-Induced Dipole Attractive Forces + – + – two nonpolar molecules center of positive charge and center of negative charge coincide in each

99. + – + – movement of electrons creates an instantaneous dipole in one molecule (left) Induced dipole-Induced dipole Attractive Forces

100. + – + – temporary dipole in one molecule (left) induces a complementary dipole in other molecule (right) Induced dipole-Induced dipole Attractive Forces

101. + – + – temporary dipole in one molecule (left) induces a complementary dipole in other molecule (right) Induced dipole-Induced dipole Attractive Forces

102. + – + – the result is a small attractive force between the two molecules Induced dipole-Induced dipole Attractive Forces

103. + – + – the result is a small attractive force between the two molecules Induced dipole-Induced dipole Attractive Forces

104. Increase with increasing number of carbons more atoms, more electrons, more opportunities for induced dipole-induced dipole forces Decrease with chain branching branched molecules are more compact with smaller surface area—fewer points of contact with other molecules Boiling Points

105. Increase with increasing number of carbons more atoms, more electrons, more opportunities for induced dipole-induced dipole forces Heptane bp 98°C Octane bp 125°C Nonane bp 150°C Boiling Points

106. Decrease with chain branching branched molecules are more compact with smaller surface area—fewer points of contact with other molecules Octane: bp 125°C 2-Methylheptane: bp 118°C 2,2,3,3-Tetramethylbutane: bp 107°C Boiling Points

107. 2.18 Chemical Properties: Combustion of Alkanes All alkanes burn in air to give carbon dioxide and water.

108. Increase with increasing number of carbons more moles of O 2 consumed, more moles of CO 2 and H 2 O formed Heats of Combustion

109. 4817 kJ/mol 5471 kJ/mol 6125 kJ/mol 654 kJ/mol Heptane Octane Nonane Heats of Combustion

110. Increase with increasing number of carbons more moles of O 2 consumed, more moles of CO 2 and H 2 O formed Decrease with chain branching branched molecules are more stable (have less potential energy) than their unbranched isomers Heats of Combustion

111. 5471 kJ/mol 5466 kJ/mol 5458 kJ/mol 5452 kJ/mol 5 kJ/mol8 kJ/mol6 kJ/mol Heats of Combustion

112. Isomers can differ in respect to their stability. Equivalent statement: Isomers differ in respect to their potential energy. Differences in potential energy can be measured by comparing heats of combustion. Important Point

113. 8CO 2 + 9H 2 O 5452 kJ/mol 5458 kJ/mol 5471 kJ/mol 5466 kJ/mol O2O2O2O2 + 25 2 O2O2O2O2 + 25 2 O2O2O2O2 + 25 2 O2O2O2O2 + 25 2 Figure 2.17

114. 2.19 Oxidation-Reduction in Organic Chemistry Oxidation of carbon corresponds to an increase in the number of bonds between carbon and oxygen and/or a decrease in the number of carbon-hydrogen bonds.

115. increasing oxidation state of carbon -4-20+2+4 H H H C H H H H C OH O C H H O C H O C HO

116. increasing oxidation state of carbon -3-2 HC CH C C H H H H C C H H H H H H

117. But most compounds contain several (or many) carbons, and these can be in different oxidation states. Working from the molecular formula gives the average oxidation state. CH 3 CH 2 OHC2H6OC2H6O Average oxidation state of C = -2 How can we calculate the oxidation state of each carbon in a molecule that contains carbons in different oxidation states?

118. How to Calculate Oxidation Numbers 1. 1.Write the Lewis structure and include unshared electron pairs. 2. Assign the electrons in a covalent bond between two atoms to the more electronegative partner. H C H H H O H C H H O H C H H H C H

119. How to Calculate Oxidation Numbers 3. For a bond between two atoms of the same element, assign the electrons in the bond equally. H O H C H H H C H H O H C H H H C H

120. How to Calculate Oxidation Numbers 4.Count the number of electrons assigned to each atom and subtract that number from the number of valence electrons in the neutral atom; the result is the oxidation number. H O H C H H H C H Each H =+1 C of CH 3 =-3 C of CH 2 O =-1 O =-2

121. Fortunately, we rarely need to calculate the oxidation state of individual carbons in a molecule. We often have to decide whether a process is an oxidation or a reduction. CH 3 CH 2 OH + Cr 2 O 7 =  CH 3 CH 2 CO 2 H + Cr 3+ H 2 SO 4 Reducing Oxidizingoxidized reduced agent is agent is oxidized reduced This shows oxidation of a carbon compound.

122. Oxidation of carbon occurs when a bond between carbon and an atom which is less electronegative than carbon is replaced by a bond to an atom that is more electronegative than carbon. The reverse process is reduction. X Y X less electronegative than carbon Y more electronegative than carbon oxidation reduction CC Generalization

123. CH 3 Cl HCl CH 4 Cl 2 + + Oxidation + 2LiLiCl CH 3 Cl CH 3 Li + Reduction Examples

124. 2.20 sp 2 Hybridization and Bonding in Ethylene

125. C 2 H 4 H 2 C=CH 2 planar bond angles: close to 120° bond distances: C—H = 110 pm C=C = 134 pm Structure of Ethylene

126. 2s2s 2p2p Promote an electron from the 2s to the 2p orbital sp 2 Orbital Hybridization

127. 2s2s 2p2p 2p2p 2s2s

128. 2p2p 2s2s Mix together (hybridize) the 2s orbital and two of the three 2p orbitals

129. 2p2p 2s2s sp 2 Orbital Hybridization 2 sp 2 3 equivalent half-filled sp 2 hybrid orbitals plus 1 p orbital left unhybridized

130. sp 2 Orbital Hybridization

131. 2 sp 2 p     

132.  Bonding in Ethylene 2 sp 2 the unhybridized p orbital of carbon is involved in  bonding to the other carbon p

133.  Bonding in Ethylene

135. 2.21 sp Hybridization and Bonding in Acetylene

136. C2H2C2H2 linear bond angles: 180° bond distances: C—H = 106 pm CC = 120 pm Structure of Acetylene HC CH

137. 2s2s 2p2p Promote an electron from the 2s to the 2p orbital sp Orbital Hybridization

138. 2s2s 2p2p 2p2p 2s2s

139. 2p2p 2s2s Mix together (hybridize) the 2s orbital and one of the three 2p orbitals

140. 2p2p 2s2s sp Orbital Hybridization 2 sp 2 equivalent half-filled sp hybrid orbitals plus 2 p orbitals left unhybridized 2 p

141. sp Orbital Hybridization

142.    2 sp 2 p

143.  Bonding in Acetylene the unhybridized p orbitals of carbon are involved in separate  bonds to the other carbon 2 sp 2 p

144.  Bonding in Acetylene   Bonding in Acetylene one   bond involves one of the p orbitals on each carbon there is a second   bond perpendicular to this one 2 2 sp 2 2 p p

145.  Bonding in Acetylene   Bonding in Acetylene 2 2 sp 2 2 p p

146.  Bonding in Acetylene   Bonding in Acetylene 2 2 sp 2 2 p p

147. 2.22 Bonding in Water and Ammonia: Hybridization of Oxygen and Nitrogen

148. trigonal pyramidal geometry H—N—H angle = 107° but notice the tetrahedral arrangement of electron pairs N H H H : Previously: Table 1.7 Ammonia sp 3

149. bent geometry H—O—H angle = 105° but notice the tetrahedral arrangement of electron pairs O H.. H : Previously: Table 1.7 Water sp 3

150. 2p2p 2s2s sp 3 Hybridization of Nitrogen Mix together (hybridize) the 2s orbital and the three 2p orbitals

151. 2p2p 2s2s sp 3 Hybridization of Nitrogen 2 sp 3 3 equivalent half-filled orbitals and a lone pair are consistent with three bonds and tetrahedral geometry

152. sp 3 Hybridization of Nitrogen nitrogen atom nitrogen Get Fig. 2.24

153. End of Chapter 2 Alkanes and Cycloalkanes: Introduction to Hydrocarbons