ALKENES Alkene Nomenclature Alkenes Alkenes are hydrocarbons that contain a carbon-carbon double bond also called "olefins" characterized by molecular formula CnH2n said to be "unsaturated" Alkene Nomenclature H2C CH2 Ethene or Ethylene (both are acceptable IUPAC names) H2C CHCH3 Propene (Propylene is sometimes used but is not an acceptable IUPAC name) Alkene Nomenclature H2C CHCH2CH3 1-Butene 1) Find the longest continuous chain that includes the double bond. 2) Replace the -ane ending of the unbranched alkane having the same number of carbons by -ene. 3) Number the chain in the direction that gives the lowest number to the doubly bonded carbon. Alkene Nomenclature H2C CHCHCH2Br CH3 4) If a substituent is present, identify its position by number. The double bond takes precedence over alkyl groups and halogens when the chain is numbered. The compound shown above is 4-bromo-3-methyl-1-butene. Alkene Nomenclature H2C CHCHCH2OH CH3 4) If a substituent is present, identify its position by number. Hydroxyl groups take precedence over the double bond when the chain is numbered. The compound shown above is 2-methyl-3-buten-1-ol. Alkenyl groups methylene H2C vinyl H2C CH allyl H2C CHCH2 isopropenyl H2C CCH3 Cycloalkene Nomenclature Cyclohexene 1) Replace the -ane ending of the cycloalkane having the same number of carbons by -ene. Cycloalkene Nomenclature CH3 CH2CH3 1) Replace the -ane ending of the cycloalkane having the same number of carbons by -ene. 2) Number through the double bond in the direction that gives the lower number to the first-appearing substituent. Cycloalkene Nomenclature CH3 6-Ethyl-1-methylcyclohexene CH2CH3 1) Replace the -ane ending of the cycloalkane having the same number of carbons by -ene. 2) Number through the double bond in the direction that gives the lower number to the first-appearing substituent. Structure and Bonding in Alkenes Structure of Ethylene bond angles: H-C-H = 117° H-C-C = 121° bond distances: C—H = 110 pm C=C = 134 pm planar Bonding in Ethylene • Framework of bonds • Each carbon is sp2 hybridized Bonding in Ethylene • Each carbon has a half-filled p orbital Bonding in Ethylene • Side-by-side overlap of halffilled p orbitals gives a bond Isomerism in Alkenes Isomers Isomers are different compounds that have the same molecular formula. Isomers Constitutional isomers Stereoisomers Isomers Constitutional isomers different connectivity Stereoisomers same connectivity; different arrangement of atoms in space Isomers Constitutional isomers Stereoisomers consider the isomeric alkenes of molecular formula C4H8 H CH2CH3 C H 1-Butene H3C H H C C H C H3C H 2-Methylpropene CH3 C H3C H H3C C C H cis-2-Butene H C CH3 trans-2-Butene H CH2CH3 C H 1-Butene H3C H C C H H3C C H 2-Methylpropene CH3 C H H3C C Constitutional isomers H cis-2-Butene H CH2CH3 C H3C C C H H H 1-Butene C H3C H 2-Methylpropene H H3C C Constitutional isomers H C CH3 trans-2-Butene Stereoisomers H3C CH3 C H H H3C C C H cis-2-Butene H C CH3 trans-2-Butene Stereochemical Notation cis (identical or analogous substituents on same side) trans (identical or analogous substitutents on opposite sides) Figure 5.2 Interconversion of stereoisomeric alkenes does not normally occur. Requires that component of double bond be broken. cis trans Figure 5.2 cis trans Naming Stereoisomeric Alkenes by the E-Z Notational System Stereochemical Notation CH2(CH2)6CO2H CH3(CH2)6CH2 C H C Oleic acid 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 Cl Example Br C H C F What is needed: 1) 2) systematic body of rules for ranking substituents new set of stereochemical symbols other than cis and trans The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side higher C lower C The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side lower C C higher The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side higher C lower lower C higher Entgegen The E-Z Notational System E : higher ranked substituents on opposite sides Z : higher ranked substituents on same side higher C lower lower C higher Entgegen higher C lower higher C lower Zusammen The E-Z Notational System Question: How are substituents ranked? Answer: higher C lower They are ranked in order of decreasing atomic number. lower C higher Entgegen higher C lower higher C lower Zusammen 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 in Chapter 7—but have been adapted to alkene stereochemistry. Table 5.1 CIP Rules (1) Higher atomic number outranks lower atomic number Br > F Cl > H higher Br C lower F Cl higher H lower C Table 5.1 CIP Rules (1) Higher atomic number outranks lower atomic number Br > F Cl > H higher Br C lower F Cl higher H lower C (Z )-1-Bromo-2-chloro-1-fluoroethene Table 5.1 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 —C(C,H,H) —C(H,H,H) Table 5.1 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,C,H) —C(C,H,H) Table 5.1 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) Table 5.1 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) (A table of commonly encountered substituents ranked according to precedence is given on the inside back cover of the text.) Physical Properties of Alkenes Dipole moments H What is direction of dipole moment? Does a methyl group donate electrons to the double bond, or does it withdraw them? H C C H H =0D H3C H C H C H = 0.3 D Dipole moments = 1.4 D H H C H H C C H Cl C H H =0D H3C H C H C H = 0.3 D Chlorine is electronegative and attracts electrons. Dipole moments = 1.4 D Dipole moment H of 1C chloropropene is equal to the H sum of the dipole moments of vinyl chloride H C 3 and propene. C H H C Cl H3C H C C H H C H = 0.3 D = 1.7 D Cl Dipole moments = 1.4 D H H C Therefore, a methyl group donates electrons to the double bond. C H Cl H3C H C C Cl H H3C H C H C H = 0.3 D = 1.7 D Alkyl groups stabilize sp2 hybridized carbon by releasing electrons R—C+ is more stable than H—C+ R—C . is more stable than H—C . R—C is more stable than H—C Relative Stabilities of Alkenes Double bonds are classified according to the number of carbons attached to them. H R C C H H H R R' R C R' monosubstituted C C H disubstituted H H R C C H disubstituted H C R' disubstituted Double bonds are classified according to the number of carbons attached to them. R" R C R' R" R C C H trisubstituted R' C R"' tetrasubstituted Substituent effects on alkene stability Electronic disubstituted alkenes are more stable than monosubstituted alkenes Steric trans alkenes are more stable than cis alkenes Fig. 5.4 Heats of combustion of C4H8 isomers. 2717 kJ/mol + 6O2 2710 kJ/mol 2707 kJ/mol 2700 kJ/mol 4CO2 + 8H2O Substituent effects on alkene stability Electronic alkyl groups stabilize double bonds more than H more highly substituted double bonds are more stable than less highly substituted ones. Problem 5.8 Give the structure or make a molecular model of the most stable C6H12 alkene. C C Problem 5.8 Give the structure or make a molecular model of the most stable C6H12 alkene. H3C CH3 C H3C C CH3 Substituent effects on alkene stability Steric effects trans alkenes are more stable than cis alkenes cis alkenes are destabilized by van der Waals strain van der Waals strain due to crowding of cis-methyl groups cis-2-butene Figure 5.5 cis and trans-2-Butene trans-2-butene Fig. 5.5 cis and trans-2-butene van der Waals strain due to crowding of cis-methyl groups cis-2-butene trans-2-butene Van der Waals Strain Steric effect causes a large difference in stability between cis and trans-(CH3)3CCH=CHC(CH3)3 cis is 44 kJ/mol less stable than trans CH3 H3C H3C H3C C C C H CH3 C H CH3 Cycloalkenes Cycloalkenes Cyclopropene and cyclobutene have angle strain. Larger cycloalkenes, such as cyclopentene and cyclohexene, can incorporate a double bond into the ring with little or no angle strain. Stereoisomeric cycloalkenes cis-cyclooctene and trans-cyclooctene are stereoisomers cis-cyclooctene is 39 kJ/ mol more stable than trans-cyclooctene H H cis-Cyclooctene H H trans-Cyclooctene Stereoisomeric cycloalkenes trans-cyclooctene is smallest trans-cycloalkene that is stable at room temperature cis stereoisomer is more stable than trans through C11 cycloalkenes cis and trans-cyclododecene are approximately equal in stability H H trans-Cyclooctene Stereoisomeric cycloalkenes trans-cyclooctene is smallest trans-cycloalkene that is stable at room temperature cis stereoisomer is more stable than trans through C11 cycloalkenes cis and trans-cyclododecene are approximately equal in stability cis-Cyclododecene trans-Cyclododecene Stereoisomeric cycloalkenes trans-cyclooctene is smallest trans-cycloalkene that is stable at room temperature cis stereoisomer is more stable than trans through C11 cycloalkenes cis and trans-cyclododecene are approximately equal in stability When there are more than 12 carbons in the ring, trans-cycloalkenes are more stable than cis. The ring is large enough so the cycloalkene behaves much like a noncyclic one. Preparation of Alkenes: Elimination Reactions -Elimination Reactions Overview •dehydrogenation of alkanes: H; Y = H •dehydration of alcohols: H; Y = OH •dehydrohalogenation of alkyl halides: H; Y = Br, etc. H C C Y C C + H Y Dehydrogenation • limited to industrial syntheses of ethylene, propene, 1,3-butadiene, and styrene • important economically, but rarely used in laboratory-scale syntheses CH3CH3 CH3CH2CH3 750°C 750°C H2C CH2 + H2 H2C CHCH3 + H2 Dehydration of Alcohols Dehydration of Alcohols CH3CH2OH OH H2SO4 160°C H2C CH2 + H2O + H2O H2SO4 140°C (79-87%) CH3 H3C C CH3 OH H2SO4 heat H3C C H3C CH2 (82%) + H2O R' Relative Reactivity R C OH tertiary: most reactive R" R' R C OH H H R C H OH primary: least reactive Regioselectivity in Alcohol Dehydration: The Zaitsev Rule Regioselectivity H2SO4 HO + 80°C 10 % • 90 % A reaction that can proceed in more than one direction, but in which one direction predominates, is said to be regioselective. Regioselectivity CH3 CH3 OH H3PO4 CH3 + heat 84 % • 16 % A reaction that can proceed in more than one direction, but in which one direction predominates, is said to be regioselective. The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having the fewest hydrogens. R R OH C C H CH3 CH2R three protons on this carbon The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having the fewest hydrogens. R R OH C C H CH3 CH2R two protons on this carbon The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having the fewest hydrogens. R R OH C C H CH3 CH2R only one proton on this carbon The Zaitsev Rule • When elimination can occur in more than one direction, the principal alkene is the one formed by loss of H from the carbon having the fewest hydrogens. R R OH C C H CH3 R CH2R C CH2R R C CH3 only one proton on this carbon The Zaitsev Rule Zaitsev Rule states that the elimination reaction yields the more highly substituted alkene as the major product. The more stable alkene product predominates. Stereoselectivity in Alcohol Dehydration Stereoselectivity H2SO4 + heat OH (25%) (75%) • A stereoselective reaction is one in which a single starting material can yield two or more stereoisomeric products, but gives one of them in greater amounts than any other. The Mechanism of the Acid-Catalyzed Dehydration of Alcohols A connecting point... • The dehydration of alcohols and the reaction of alcohols with hydrogen halides share the following common features: • 1) Both reactions are promoted by acids • 2) The relative reactivity decreases in the order tertiary > secondary > primary These similarities suggest that carbocations are intermediates in the acid-catalyzed dehydration of alcohols, just as they are in the reaction of alcohols with hydrogen halides. Dehydration of tert-Butyl Alcohol CH3 H3C C CH3 OH H2SO4 H3C C heat CH2 + H2O H3C •first two steps of mechanism are identical to those for the reaction of tert-butyl alcohol with hydrogen halides Mechanism Step 1: Proton transfer to tert-butyl alcohol H .. + (CH3)3C O : + H O .. H H fast, bimolecular H + (CH3)3C O : H + H tert-Butyloxonium ion :O: H Mechanism Step 2: Dissociation of tert-butyloxonium ion to carbocation H + (CH3)3C O : H slow, unimolecular H + (CH3)3C + tert-Butyl cation :O: H Mechanism Step 3: Deprotonation of tert-butyl cation. H H3C +C H + :O: H CH2 H3C fast, bimolecular H H3C C H3C CH2 + H + O: H Carbocations are intermediates in the acid-catalyzed dehydration of tertiary and secondary alcohols Carbocations can: •react with nucleophiles •lose a -proton to form an alkene (Called an E1 mechanism) Dehydration of primary alcohols CH3CH2OH H2SO4 160°C H2C CH2 + H2O •A different mechanism from 3 o or 2 o alcohols •avoids carbocation because primary carbocations are too unstable •oxonium ion loses water and a proton in a bimolecular step Mechanism Step 1: Proton transfer from acid to ethanol H .. CH3CH2 O : + H O .. H H fast, bimolecular H + CH3CH2 O : H Ethyloxonium ion H + :O: H Mechanism Step 2: Oxonium ion loses both a proton and a water molecule in the same step. H H + : O : + H CH2 CH2 O : H H slow, bimolecular H + :O H H H + H2C CH2 + :O: H Mechanism Step 2: H + :O H Oxonium ion loses both a proton and a water molecule in the same step. H H + : O : + H CH2 CH2 O : H H Because rate-determining step is bimolecular, thisbimolecular slow, is called the E2 mechanism. H H + H2C CH2 + :O: H Rearrangements in Alcohol Dehydration Sometimes the alkene product does not have the same carbon skeleton as the starting alcohol. Example OH H3PO4, heat + 3% + 64% 33% Rearrangement involves alkyl group migration CH3 CH3 C CHCH3 + CH3 3% • carbocation can lose a proton as shown • or it can undergo a methyl migration • CH3 group migrates with its pair of electrons to adjacent positively charged carbon Rearrangement involves alkyl group migration CH3 CH3 CH3 C CHCH3 + 97% CH3 + C CHCH3 CH3 CH3 3% • tertiary carbocation; more stable Rearrangement involves alkyl group migration CH3 CH3 CH3 C CHCH3 + 97% CH3 + C CH3 CH3 3% CHCH3 Another rearrangement CH3CH2CH2CH2OH H3PO4, heat CH3CH2CH 12% CH2 + CH3CH CHCH3 mixture of cis (32%) and trans-2-butene (56%) Rearrangement involves hydride shift CH3CH2CH2CH2 H + O: H CH3CH2CH CH2 oxonium ion can lose water and a proton (from C-2) to give1-butene doesn't give a carbocation directly because primary carbocations are too unstable Rearrangement involves hydride shift CH3CH2CH2CH2 H + O: H CH3CH2CH CH2 CH3CH2CHCH3 + hydrogen migrates with its pair of electrons from C-2 to C-1 as water is lost carbocation formed by hydride shift is secondary Hydride shift H CH3CH2CHCH2 + O: H H + CH3CH2CHCH2 + H H : O: H Rearrangement involves hydride shift CH3CH2CH2CH2 H + O: CH3CH2CHCH3 + H CH3CH2CH CH2 CH3CH CHCH3 mixture of cis and trans-2-butene Carbocations can... •react with nucleophiles •lose a proton from the -carbon to form an alkene •rearrange (less stable to more stable) (alkyl shift or hydride shift) Dehydrohalogenation of Alkyl Halides -Elimination Reactions Overview •dehydrogenation of alkanes: H; Y = H •dehydration of alcohols: H; Y = OH •dehydrohalogenation of alkyl halides: H; Y = Br, etc. H C C Y C C + H Y -Elimination Reactions Overview •dehydrogenation of alkanes: industrial process; not regioselective •dehydration of alcohols: acid-catalyzed •dehydrohalogenation of alkyl halides: consumes base H C C Y C C + H Y Dehydrohalogenation A useful method for the preparation of alkenes Cl NaOCH2CH3 ethanol, 55°C (100 %) likewise, NaOCH3 in methanol, or KOH in ethanol Dehydrohalogenation When the alkyl halide is primary, potassium tert-butoxide in dimethyl sulfoxide is the base/solvent system that is normally used. CH3(CH2)15CH2CH2Cl KOC(CH3)3 dimethyl sulfoxide CH3(CH2)15CH (86%) CH2 Regioselectivity KOCH2CH3 Br + ethanol, 70°C 29 % 71 % follows Zaitsev's rule: more highly substituted double bond predominates Stereoselectivity KOCH2CH3 ethanol Br + (23%) (77%) •more stable configuration of double bond predominates Stereoselectivity Br KOCH2CH3 ethanol + (85%) •more stable configuration of double bond predominates (15%) Mechanism of the Dehydrohalogenation of Alkyl Halides: The E2 Mechanism Facts • (1) Dehydrohalogenation of alkyl halides exhibits second-order kinetics first order in alkyl halide first order in base rate = k[alkyl halide][base] implies that rate-determining step involves both base and alkyl halide; i.e., it is bimolecular (second-order) Facts • (2) Rate of elimination depends on halogen weaker C—X bond; faster rate rate: RI > RBr > RCl > RF implies that carbon-halogen bond breaks in the rate-determining step The E2 Mechanism •concerted (one-step) bimolecular process •single transition state C—H bond breaks component of double bond forms C—X bond breaks The E2 Mechanism R .. – O .. : H C C : X: .. Reactants The E2 Mechanism R .. – O .. : H C C : X: .. Reactants The E2 Mechanism R – .. O .. Transition state H C C – : X: .. The E2 Mechanism R .. O .. H C C .. – : X: .. Products Anti Elimination in E2 Reactions Stereoelectronic Effects Isotope Effects Stereoelectronic effect Br KOC(CH3)3 (CH3)3COH (CH3)3C cis-1-Bromo-4-tertbutylcyclohexane (CH3)3C Stereoelectronic effect (CH3)3C trans-1-Bromo-4-tertbutylcyclohexane Br (CH3)3C KOC(CH3)3 (CH3)3COH Stereoelectronic effect cis Br KOC(CH3)3 (CH3)3COH (CH3)3C Rate constant for dehydrohalogenation of cis is 500 times greater than that of trans (CH3)3C Br (CH3)3C trans KOC(CH3)3 (CH3)3COH Stereoelectronic effect cis Br KOC(CH3)3 (CH3)3COH (CH3)3C H H (CH3)3C H that is removed by base must be anti periplanar to Br Two anti periplanar H atoms in cis stereoisomer Stereoelectronic effect trans H Br H (CH3)3C KOC(CH3)3 (CH3)3COH H H (CH3)3C H that is removed by base must be anti periplanar to Br No anti periplanar H atoms in trans stereoisomer; all vicinal H atoms are gauche to Br Stereoelectronic effect cis more reactive trans less reactive Stereoelectronic effect An effect on reactivity that has its origin in the spatial arrangement of orbitals or bonds is called a stereoelectronic effect. The preference for an anti periplanar arrangement of H and Br in the transition state for E2 dehydrohalogenation is an example of a stereoelectronic effect. Isotope effect Deuterium,D, is a heavy isotope of hydrogen but will undergo the same reactions. But the C-D bond is stronger so a reaction where the rate involves breaking a C-H (C-D) bond, the deuterated sample will have a reaction rate 3-8 times slower. In other words comparing the two rates, i.e. kH/kD = 3-8 WHEN the rate determining step involves breaking the C-H bond. A Different Mechanism for Alkyl Halide Elimination: The E1 Mechanism Example CH3 CH3 CH2CH3 C Br Ethanol, heat H3C CH3 H2C + C CH2CH3 (25%) H C C CH3 H3C (75%) The E1 Mechanism 1. Alkyl halides can undergo elimination in absence of base. 2. Carbocation is intermediate 3. Rate-determining step is unimolecular ionization of alkyl halide. 4. Generally with tertiary halide, base is weak and at low concentration. CH3 Step 1 CH3 CH2CH3 C : Br: .. slow, unimolecular CH3 C CH3 + CH2CH3 .. – : Br : .. CH3 Step 2 CH3 C + CH2CH3 – H+ CH3 CH2 + C CH3 CH2CH3 C CH3 CHCH3