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Z. E. chapter 5. Alkenes. Bonding, Nomenclature, Properties. Structure Hydrogen Deficiency Nomenclature Physical Properties Naturally Occurring Alkenes/Terpenes. b. R or S?. R or S?. Which is 2S , 3R -propoxyphen & 2R , 3S -propoxyphen?. ( 2R , 3S )-.
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Z E chapter 5 Alkenes Bonding, Nomenclature, Properties Structure Hydrogen Deficiency Nomenclature Physical Properties Naturally Occurring Alkenes/Terpenes b
R or S? R or S?
Which is 2S,3R-propoxyphen & 2R,3S-propoxyphen? (2R,3S)- antitussive NovraD | DarvoN analgesic
1. Alkene: contains a carbon-carbon doublebond ethene 2.Alkyne: contains a carbon-carbon triple bond(Ch 7) ethyne(acetylene) 3.Arene: benzene and its derivatives (Ch 21-22) phenyl. Ph UnsaturatedHydrocarbons
Double Bond: 1-bond formed by overlap of 2 sp2 hybrid orbitals 1 -bondformed by overlap of 2 parallel 2p orbitals -bond -bonded atoms all in a plane trigonal - bond angles ~ 120° H C H perpendicular H C H to the plane Structure of Alkenes
No rotation about a C=C bond - why? H C H H C H Structure of Alkenes Rotation requires breaking bond ~63 kcal/mol
Number chain - olefin lowest number #- indicates C=C in chain 6 5 1 2 4 3 Nomenclature IUPAC: Root name - longest continuous chain containing the olefin. alkene Number of C’s in C=C chain 2-ethyl-4-methyl-1-pentene 2-hexene
IUPAC names? 7-bromo-4-(2-iodoethyl)-6-methyl-3-octene (7S)-7-bromo-4-(2-iodoethyl)-6-methyl-3-octene
Number chain - olefin lowest number #-cycloalkene #- #- alkene Nomenclature IUPAC: Root name - longest continuous chain containing the olefin alkene indicates C=C in chain Number of C’s in chain with an C=C Cyclic Olefin (functional group) - positions 1,2 Number around the ring to best accommodate substituents
IUPAC names? cycloheptene -bromo- -methyl 4-bromo-4-methylcycloheptene (3 ,5 )- -chloro- -fluoro- -methylcyclopent-1-ene (3S,5R)-5-chloro-1-fluoro-3-methylcyclopent-1-ene
Cl Cl Cl H C C C C H H H Cl The Cis,Trans Systemrecall cis/trans isomers cis-1,2-dichloroethene trans-1,2-dichloroethene transgenerally more stable thancis - dipoles and 1,2 interactions
Configuration is determined by the orientation of atoms of the main chain The Cis,Trans System trans-3-hexene cis-3,4-dimethyl-2-pentene
C8 limited stability as trans Cycloalkenes- 3 to 7 “cis” olefins rings not large enough to accommodate trans double bonds
uses priority rules (Chapter 3) higher priority groups - same side, Z higher priority - opposite sides, E E,Z Configuration E Z
H C C H C C H H C C C C C C H C C H C C H H E/Z - priorities of groups on ends of C=C (1) Atom assigned a priority, higher atomic number higher priority. (2) Isotopes - higher atomic mass, higher priority 1H < 2H < 3H [H < D < T] (3) If priority the same, go to the next set of atoms: -CH2-H < -CH2-O-H < -CH2-Cl (4) double (triple) bonds replaced by single bonds.
Example:name each alkene and specify its configuration by the E,Z system The E,Z System
Stereochemistry? IUPAC names? ?(trans)-longest chain? (3Z,7S)-7-bromo-4-(2-chloroethyl)-6-methyl-3-octene (7S)-7-bromo-4-(2-chloroethyl)-6-methyl-3-octene (S)-6-ethyl-1-fluoro-5,5-dimethylcyclooctene
If same atoms, priority goes to the next point of difference 9-bromo-5-(2-methylpropyl)-4-nonene (E)- (4E)-
1 2 7 3 4 6 5 Nomenclature - more than 1 unsaturated group Diene (or diyne, en-yne) longest chain (ring) with both groups: #,#-alkadiene #,#-alkadiyne #-alken-#-yne (if the same ene > yne). number so enes (enyne, ynes) have the lowest possible number 1,4-cycloheptadiene
alkenes with n double bonds (that can be cis, trans)2n stereoisomers are possible example 2,4-heptadiene Dienes, Trienes, and Polyenes
Dienes - E or Z per olefin with number 1,6-dichloro-2,4-dimethyl-2,6-decadiene ( 2Z, 4S, 6E)- ( 2Z, 4S, 6E)- to 6.1
Alkenes are nonpolar compounds attractive forces between molecules are dispersion forces The physical properties of alkenes are similar to those of alkanes Physical Properties to 6.1
Index of hydrogen deficiency (IHD): IHD = (number of rings + number ofbonds) Compare HsofalkanewithHsin a compound CnH2n+2CnHx (14 - 10) 2 = ) (H - H reference molecule IHD = 2 Index of Hydrogen Deficiency e.g. C6H10 C6H2(6)+2 = 2 to 6.1
Other elements present F, Cl, Br, I add one H - (Group 7) O No correction to unknown formula N, P subtract one H (Group 5) C6H8ClNO C6H9Cl C6H7ClO e.g. 8 + 1 - 1 7 + 1 9 + 1 Index of Hydrogen Deficiency to 6.1
Problem: calculate the IHD for niacin, molecular formula C6H6N2O reference hydrocarbon C6H14 IHD = [14 - (6-2)]/2 = 5 Index of Hydrogen Deficiency end to 6.1