12. Estructura del Metano Tetrahédrica ángulos de enlace = 109.5° longitud de enlace = 110 pm sin embargo la estructurapareceinconsistentecon la configuraciónelectrónica del carbono
13. ConfiguraciónElectrónica del carbono solo dos electronesdesapareados debeformar enlaces con solo dos átomos de hidrógeno los enlaces debenestar en ángulo recto uno con respectoal otro 2p 2s
14. Hibridación Orbital sp3 30´s Linus Pauling 2p Se promueve un electrón del orbital 2s al 2p 2s
20. Forma de los orbitaleshíbridossp3 p + – Toma el orbital s y colócalo en la parte superior del orbital p + s
21. + Forma de los orbitaleshíbridos sp3 s + p + – Complemento de ondaelectrónica en regionesdonde el signoes el mismo Interferenciadestructiva en regiones de signoopuesto
22. – híbridosp + el orbital mostradoeshíbridosp procesoanalogousandotresorbitalesp y uno sdahíbridossp3 la forma de los híbridossp3es similar Forma de los orbitaleshíbridossp3
23. + – híbridosp - el orbital híbrido no essimétrico - mayor probabilidad de encontrar un electrón en un lado del núcleoque en otro - produce enlaces másfuertes Forma de los orbitaleshíbridossp3
24. + – – El enlace C—H en el Metano Traslape en fase de un orbital semilleno 1s de hidrógeno con un orbital híbridosemillenosp3 de carbono: + sp3 s H C produce un enlace . + H—C C H
25. Justificaciónpara la Hibridación Orbital consistente con la estructura del metano permite la formación de 4 enlaces en lugar de 2 los enlaces involucrados en los orbitaleshíbridossp3 son másfuertesque los involucrados en el traslapes-s o p-p
27. Estructura del Etano C2H6 CH3CH3 geometríatetrahédrica en cadacarbono distancia de enlace C—H = 110 pm distancia de enlace C—C = 153 pm
28. El enlace C—C en el Etano Traslape en fase de un orbital híbridosemillenosp3 de un carbono con un orbital híbridosemillenosp3de otro. El traslapees a lo largo del ejeinternuclearparadar un enlace .
29. El enlace C—C en el Etano Traslape en fase de un orbital híbridosemillenosp3 de un carbono con un orbital híbridosemillenosp3de otro. El traslapees a lo largo del ejeinternuclearparadar un enlace .
36. ¿Cuántosisómeros? El número de isómeros se incrementa al incrementar el número de carbonos. No hay unamanerasencilla de predecircuántosisómeros hay paraunafórmula molecular en particular.
37. Tabla 1 Número de IsómerosConstitucionales de Alcanos CH4 1 C2H6 1 C3H8 1 C4H10 2 C5H12 3 C6H14 5 C7H16 9
38. Tabla 1 Número de IsómerosConstitucionales de Alcanos CH4 1 C8H18 18 C2H6 1 C9H20 35 C3H8 1 C10H22 75 C4H10 2 C15H32 4,347 C5H12 3 C20H42 366,319 C6H14 5 C40H82 62,491,178,805,831 C7H16 9
40. Boiling Points of Alkanes 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
41. Induced dipole-Induced dipole attractive forces + – + – two nonpolar molecules center of positive charge and center of negative charge coincide in each
42. Induced dipole-Induced dipole attractive forces + – + – movement of electrons creates an instantaneous dipole in one molecule (left)
43. Induced dipole-Induced dipole attractive forces – + – + temporary dipole in one molecule (left) induces a complementary dipole in other molecule (right)
44. Induced dipole-Induced dipole attractive forces – – + + temporary dipole in one molecule (left) induces a complementary dipole in other molecule (right)
45. Induced dipole-Induced dipole attractive forces – – + + the result is a small attractive force between the two molecules
46. Induced dipole-Induced dipole attractive forces – – + + the result is a small attractive force between the two molecules
47. Boiling Points 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
48. Boiling Points increase with increasing number of carbons more atoms, more electrons, more opportunities for induced dipole-induced dipole forces Heptanebp 98°C Octanebp 125°C Nonanebp 150°C
49. Boiling Points 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
53. Heats of Combustion increase with increasing number of carbons more moles of O2 consumed, more moles of CO2 and H2O formed decrease with chain branching branched molecules are more stable (have less potential energy) than their unbranched isomers
56. Alkenes Alkenes are hydrocarbons that contain a carbon-carbon double bond also called "olefins" characterized by molecular formula CnH2n said to be "unsaturated"
80. C C Stereochemical Notation CH2(CH2)6CO2H CH3(CH2)6CH2 Oleic acid H H cis and trans are useful when substituents are identical or analogous (oleic acid has a cis double bond) cis and trans are ambiguous when analogies are not obvious
81. Cl Br C C H F Example What is needed:1) systematic body of rules for ranking substituents 2) new set of stereochemical symbols other than cis and trans
82. C C The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side higher lower
83. C C The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side lower higher
84. higher higher C C C C lower lower Zusammen The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side higher lower higher lower Entgegen
85. C C C C The E-Z Notational System Question: How are substituents ranked? Answer: They are ranked in order of decreasing atomic number. higher lower higher higher higher lower lower lower Entgegen Zusammen
86. The Cahn-Ingold-Prelog (CIP) System The system that we use was devised by R. S. Cahn Sir Christopher Ingold Vladimir Prelog Their rules for ranking groups were devised in connection with a different kind of stereochemistry—one that we will discuss later—but have been adapted to alkene stereochemistry.
87. higher higher Br Cl C C F H lower lower Table CIP Rules (1) Higher atomic number outranks lower atomic number Br > F Cl > H
88. higher higher Br Cl C C F H lower lower Table CIP Rules (1) Higher atomic number outranks lower atomic number Br > F Cl > H (Z )-1-Bromo-2-chloro-1-fluoroethene
89. —C(H,H,H) —C(C,H,H) Table CIP Rules (2) When two atoms are identical, compare the atoms attached to them on the basis of their atomic numbers. Precedence is established at the first point of difference. —CH2CH3 outranks —CH3
90. Table CIP Rules (3) Work outward from the point of attachment, comparing all the atoms attached to a particular atom before proceeding further along the chain. —CH(CH3)2 outranks —CH2CH2OH —C(C,H,H) —C(C,C,H)
91. Table CIP Rules (4) Evaluate substituents one by one. Don't add atomic numbers within groups. —CH2OH outranks —C(CH3)3 —C(O,H,H) —C(C,C,C)
92. Table CIP Rules (5) An atom that is multiply bonded to another atom is considered to be replicated as a substituent on that atom. —CH=O outranks —CH2OH —C(O,O,H) —C(O,H,H)
93. Table CIP Rules A table of commonly encountered substituents ranked according to precedence is given on the inside back cover of the text.
95. H H C C H H H3C H C C H H = 0.3 D Dipole moments What is direction of dipole moment? Does a methyl group donate electrons to the double bond, or does it withdraw them? = 0 D
96. = 1.4 D H H C C H H Cl H C C H H H3C H C C H H = 0.3 D Dipole moments Chlorine is electronegative and attracts electrons. = 0 D
97. = 1.4 D H H C C Cl H H3C H C C Cl H H3C H = 1.7 D C C H H = 0.3 D Dipole moments Dipole moment of 1-chloropropene is equal to the sum of the dipole moments of vinyl chloride and propene.
98. = 1.4 D H H C C Cl H H3C H C C Cl H H3C H C C H H = 0.3 D Dipole moments Therefore, a methyl group donates electrons to the double bond. = 1.7 D
99. R—C+ is more stable than H—C+ • R—C is more stable than • H—C R—C is more stable than H—C Alkyl groups stabilize sp2 hybridizedcarbon by releasing electrons
101. H R C C H H H R C C C C C C R' H disubstituted Double bonds are classified according tothe number of carbons attached to them. monosubstituted R' H R R R' H H H disubstituted disubstituted
102. R" R" R R C C C C R' H R' R"' trisubstituted tetrasubstituted Double bonds are classified according tothe number of carbons attached to them.
103. Substituent Effects on Alkene Stability Electronic disubstituted alkenes are more stable than monosubstituted alkenes Steric trans alkenes are more stable than cis alkenes
104. Figure Heats of combustion of C4H8isomers. 2717 kJ/mol + 6O2 2710 kJ/mol 2707 kJ/mol 2700 kJ/mol 4CO2 + 8H2O
105. Substituent Effects on Alkene Stability Electronic alkyl groups stabilize double bonds more than H more highly substituted double bonds are morestable than less highly substituted ones.
106. H3C CH3 C C H3C CH3 Problem Give the structure or make a molecular model of the most stable C6H12 alkene.
107. Substituent Effects on Alkene Stability Steric trans alkenes are more stable than cis alkenes cis alkenes are destabilized by van der Waalsstrain
108. van der Waals straindue to crowding ofcis-methyl groups Figure cis and trans-2-Butene cis-2-butene trans-2-butene
109. Figure cis and trans-2-Butene van der Waals straindue to crowding ofcis-methyl groups cis-2-butene trans-2-butene
110. H3C CH3 CH3 H3C CH3 H3C C C C C H H van der Waals Strain Steric effect causes a large difference in stabilitybetween cis and trans-(CH3)3CCH=CHC(CH3)3 cis is 44 kJ/mol less stable than trans
112. Cycloalkenes Cyclopropene and cyclobutene have angle strain. Larger cycloalkenes, such as cyclopenteneand cyclohexene, can incorporate a double bond into the ring with little or no angle strain.
113. H H H H Stereoisomeric cycloalkenes cis-cyclooctene and trans-cycloocteneare stereoisomers cis-cyclooctene is 39 kJ/ mol more stablethan trans-cyclooctene cis-Cyclooctene trans-Cyclooctene
114. H H Stereoisomeric cycloalkenes trans-cyclooctene is smallest trans-cycloalkene that is stable at room temperature cis stereoisomer is more stable than trans through C11 cycloalkenes trans-Cyclooctene
115. Stereoisomeric cycloalkenes cis and trans-cyclododeceneare approximately equal instability trans-Cyclododecene cis-Cyclododecene When there are more than 12 carbons in thering, trans-cycloalkenes are more stable than cis.The ring is large enough so the cycloalkene behaves much like a noncyclic one.
117. 120 pm H C C H 106 pm 106 pm 121 pm C CH3 C H 146 pm 106 pm linear geometry for acetylene Structure
118. C C Cycloalkynes Cyclononyne is the smallest cycloalkyne stable enough to be stored at room temperaturefor a reasonable length of time. Cyclooctyne polymerizeson standing.
131. H C C Acidity of Acetyleneand Terminal Alkynes
132. H2C CH2 Acidity of Hydrocarbons In general, hydrocarbons are exceedingly weak acids Compound pKa HF 3.2 H2O 15.7 NH3 36 45 CH4 60
133. HC CH H2C CH2 Acetylene Acetylene is a weak acid, but not nearlyas weak as alkanes or alkenes. Compound pKa HF 3.2 H2O 15.7 NH3 36 45 CH4 60 26
134. pKa = 60 – sp3 : C H++ H C H sp2 : pKa = 45 H++ C C C C – pKa = 26 – sp : H C C C C H++ Carbon: Hybridization and Electronegativity Electrons in an orbital with more s character are closer to thenucleus and more strongly held.
135. NaC CH + + NaOH NaC H2O CH HC CH Sodium Acetylide Objective: Prepare a solution containing sodium acetylideWill treatment of acetylene with NaOH be effective?
136. – .. .. – stronger acidpKa = 15.7 weaker acidpKa = 26 + CH C : + CH C H : H HO HO .. .. Sodium Acetylide No. Hydroxide is not a strong enough base to deprotonate acetylene. In acid-base reactions, the equilibrium lies tothe side of the weaker acid.
137. + + NaNH2 NaC NH3 CH HC CH – .. .. – + : : + H2N H H2N CH C CH C H weaker acidpKa = 36 stronger acidpKa = 26 Sodium Acetylide Solution: Use a stronger base. Sodium amideis a stronger base than sodium hydroxide. Ammonia is a weaker acid than acetylene.The position of equilibrium lies to the right.
