Alkenes Bonding, Nomenclature, Properties Structure chapter 5 Alkenes Bonding, Nomenclature, Properties Z E Structure Hydrogen Deficiency Nomenclature Physical Properties Naturally Occurring Alkenes/Terpenes Lecture 10, February 25, 2003 b

R or S? R or S?

Which is 2S,3R-propoxyphen & 2R,3S-propoxyphen? antitussive NovraD | DarvoN analgesic

Unsaturated Hydrocarbons 1. Alkene: contains a carbon-carbon double bond ethene 2. Alkyne: contains a carbon-carbon triple bond (Ch 7) ethyne (acetylene) 3. Arene: benzene and its derivatives (Ch 21-22) phenyl. Ph

Structure of Alkenes -bond Double Bond: 1 -bond formed by overlap of 2 sp2 hybrid orbitals 1 -bond formed by overlap of 2 parallel 2p orbitals -bond perpendicular to the plane -bonded atoms all in a plane  trigonal - bond angles ~ 120° H C

Structure of Alkenes No rotation about a C=C bond - why? Rotation requires breaking  bond ~63 kcal/mol

Nomenclature IUPAC: Root name - longest continuous chain containing the olefin. Number chain - olefin lowest number #- alkene indicates C=C in chain Number of C’s in C=C chain Start 8 sept 25, 2001 1 2 3 4 5 6 2-ethyl-4-methyl-1-pentene 2-hexene

IUPAC names? (7S)-7-bromo-4-(2-iodoethyl)-6-methyl-3-octene 7-bromo-4-(2-iodoethyl)-6-methyl-3-octene

Number of C’s in chain with an C=C Nomenclature IUPAC: Root name - longest continuous chain containing the olefin #-cycloalkene #- alkene Number chain - olefin lowest number #- 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

4-bromo-4-methylcycloheptene 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

The Cis,Trans System recall cis/trans isomers Cl C Cl H cis-1,2-dichloroethene trans-1,2-dichloroethene trans generally more stable than cis - dipoles and 1,2 interactions

The Cis,Trans System trans-3-hexene cis-3,4-dimethyl-2-pentene Configuration is determined by the orientation of atoms of the main chain trans-3-hexene cis-3,4-dimethyl-2-pentene

Cycloalkenes- 3 to 7 “cis” olefins rings not large enough to accommodate trans double bonds C8 limited stability as trans

E,Z Configuration E Z uses priority rules (Chapter 3) higher priority groups - same side, Z higher priority - opposite sides, E Z E

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. C H C H

The E,Z System Example: name each alkene and specify its configuration by the E,Z system

?(trans)-longest chain? IUPAC names? Stereochemistry? ?(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

priority goes to the next point of difference If same atoms, priority goes to the next point of difference (E)- (4E)- 9-bromo-5-(2-methylpropyl)-4-nonene

number so enes (enyne, ynes) have the lowest possible number 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 5 1 2 3 4 6 7 1,4-cycloheptadiene

Dienes, Trienes, and Polyenes alkenes with n double bonds (that can be cis, trans) 2n stereoisomers are possible example 2,4-heptadiene

Dienes - E or Z per olefin with number ( 2Z, 4S, 6E)- 1,6-dichloro-2,4-dimethyl-2,6-decadiene ( 2Z, 4S, 6E)- to 6.1

Physical Properties Alkenes are nonpolar compounds attractive forces between molecules are dispersion forces The physical properties of alkenes are similar to those of alkanes to 6.1

Common Names used lab & lecture not on tests

Index of Hydrogen Deficiency Index of hydrogen deficiency (IHD): IHD =  (number of rings + number of  bonds) Compare Hs of alkane with Hs in a compound CnH2n+2 CnHx e.g. C6H10 C6H2(6)+2 IHD = 2 (H reference - H molecule ) = (14 - 10) 2 = 2 to 6.1

Index of Hydrogen Deficiency Other elements present F, Cl, Br, I add one H - (Group 7) C6H7ClO O No correction to unknown formula N, P subtract one H (Group 5) C6H8ClNO e.g. C6H9Cl 8 + 1 - 1 9 + 1 7 + 1 to 6.1

Index of Hydrogen Deficiency Problem: calculate the IHD for niacin, molecular formula C6H6N2O reference hydrocarbon C6H14 IHD = [14 - (6-2)]/2 = 5 end to 6.1