Chapter 7 Alkenes and Alkynes I: Properties and Synthesis. Elimination Reactions of Alkyl Halides Ch. 7 - 1 1. Introduction Alkenes ● Hydrocarbons containing C=C ● Old name: olefins CH2OH Vitamin A H3C H3C H H Cholesterol HO Ch. 7 - 2 Alkynes ● Hydrocarbons containing C≡C ● Common name: acetylenes H N O I Cl C O C O Cl F3C C Cl C Cl Efavirenz Haloprogin (antiviral, AIDS therapeutic) (antifungal, antiseptic) Ch. 7 - 3 2. The (E) - (Z) System for Designating Alkene Diastereomers Cis-Trans System ● Useful for 1,2-disubstituted alkenes ● Examples: H Br Cl Cl (1) Br vs H H H trans -1-Bromo- cis -1-Bromo- 2-chloroethene 2-chloroethene Ch. 7 - 4 ● Examples H (2) vs H H H trans -3-Hexene cis -3-Hexene Br (3) Br Br vs Br trans -1,3- cis -1,3- Dibromopropene Dibromopropene Ch. 7 - 5 (E) - (Z) System ● Difficulties encountered for trisubstituted and tetrasubstituted alkenes CH3 e.g. Cl cis or trans? Br H Cl is cis to CH3 and trans to Br Ch. 7 - 6 The Cahn-Ingold-Prelog (E) - (Z) Convention ● The system is based on the atomic number of the attached atom ● The higher the atomic number, the higher the priority Ch. 7 - 7 The Cahn-Ingold-Prelog (E) - (Z) Convention ● (E) configuration – the highest priority groups are on the opposite side of the double bond “E ” stands for “entgegen”; it means “opposite” in German ● (Z) configuration – the highest priority groups are on the same side of the double bond “Z ” stands for “zusammer”; it means “together” in German Ch. 7 - 8 ● Examples CH3 Cl 1 2 Br H On carbon 2: Priority of Br > C On carbon 1: Priority of Cl > H ⇒ highest priority groups are Br (on carbon 2) and Cl (on carbon 1) Ch. 7 - 9 ● Examples CH3 Cl Br H ⇒ (E )-2-Bromo-1-chloropropene Br Cl CH3 H ⇒ (Z )-2-Bromo-1-chloropropene Ch. 7 - 10 ● Other examples H (E )-1,2-Dichloroethene (1) Cl [or trans-1,2-Dichloroethene] 1 2 Cl H C1: Cl > H C2: Cl > H Cl 2 (Z )-1-Bromo-1,2-dichloroethene (2) 1 Cl Br C1: Br > Cl C2: Cl > H Ch. 7 - 11 ● Other examples Br 3 1 4 (3) 2 7 5 8 6 (Z )-3-Bromo-4-tert-butyl-3-octene C3: Br > C C4: tBu > nBu Ch. 7 - 12 3. Relative Stabilities of Alkenes Cis and trans alkenes do not have the same stability crowding R R H R C C C C H H R H Less stable More stable Ch. 7 - 13 3A. Heat of Reaction Pt C C + H H C C H H Heat of hydrogenation ● ∆H° ≃ -120 kJ/mol Ch. 7 - 14 + H2 7 kJ/mol + H2 ∆H° = -127 kJ/mol 5 kJ/mol + H2 ∆H° = -120 kJ/mol Enthalpy ∆H° = -115 kJ/mol ≈ ≈ ≈ Ch. 7 - 15 3B. Overall Relative Stabilities of Alkenes The greater the number of attached alkyl groups (i.e., the more highly substituted the carbon atoms of the double bond), the greater the alkene’s stability. Ch. 7 - 16 Relative Stabilities of Alkenes R R R R R H R H R R R H H H > > > > > > R R R H R H H R H H H H H H tetra- tri- di- mono- un- substituted substituted substituted substituted substituted Ch. 7 - 17 Examples of stabilities of alkenes (1) > (2) > Ch. 7 - 18 4. Cycloalkenes Cycloalkenes containing 5 carbon atoms or fewer exist only in the cis form cyclopropene cyclobutene cyclopentene Ch. 7 - 19 Trans – cyclohexene and trans – cycloheptene have a very short lifetime and have not been isolated cyclohexene Hypothetical trans - cyclohexene (too strained to exist at r.t.) Ch. 7 - 20 Trans – cyclooctene has been isolated and is chiral and exists as a pair of enantiomers cis - cyclooctene trans - cyclooctenes Ch. 7 - 21 5. Synthesis of Alkenes via Elimination Reactions Dehydrohalogenation of Alkyl Halides H H H H H base C C -HX H X H H H Dehydration of Alcohols H H H H+, heat H H C C H OH -HOH H H H Ch. 7 - 22 6. Dehydrohalogenation of Alkyl Halides The best reaction conditions to use when synthesizing an alkene by dehydrohalogenation are those that promote an E2 mechanism H E2 B: C C C C + B:H + X X Ch. 7 - 23 6A. How to Favor an E2 Mechanism Use a secondary or tertiary alkyl halide if possible. (Because steric hinderance in the substrate will inhibit substitution) When a synthesis must begin with a primary alkyl halide, use a bulky base. (Because the steric bulk of the base will inhibit substitution) Ch. 7 - 24 Use a high concentration of a strong and nonpolarizable base, such as an alkoxide. (Because a weak and polarizable base would not drive the reaction toward a bimolecular reaction, thereby allowing unimolecular processes (such as SN1 or E1 reactions) to compete. Ch. 7 - 25 Sodium ethoxide in ethanol (EtONa/EtOH) and potassium tert- butoxide in tertbutyl alcohol (t-BuOK/t- BuOH) are bases typically used to promote E2 reactions Use elevated temperature because heat generally favors elimination over substitution. (Because elimination reactions are entropically favored over substitution reactions) Ch. 7 - 26 6B. Zaitsev’s Rule Examples of dehydrohalogenations where only a single elimination product is possible EtONa (1) (79%) Br EtOH, 55oC EtONa (2) (91%) Br EtOH, 55oC t -BuOK (3) ( ) Br ( ) (85%) n t -BuOH, 40oC n Ch. 7 - 27 H Rate = k H3C C CH3 EtO Br (2nd order overall) ⇒ bimolecular ̶̶̶ Ha B Ha Hb 2-methyl-2-butene Br ̶̶̶ Hb 2-methyl-1-butene Ch. 7 - 28 When a small base is used (e.g. EtO⊖ or HO⊖) the major product will be the more highly substituted alkene (the more stable alkene) Examples: a b H H NaOEt (1) + EtOH 70oC Br 69% 31% (eliminate Ha) (eliminate Hb) Br KOEt (2) + + EtOH 51% 18% 31% 69% Ch. 7 - 29 Zaitsev’s Rule ● In elimination reactions, the more highly substituted alkene product predominates Stability of alkenes Me Me Me Me Me H C C > C C > C C Me Me Me H H Me Me Me Me H > C C > C C H H H H Ch. 7 - 30 Mechanism for an E2 Reaction Et O Et O H CH3 H CH3 α δ− H3C CH3 C C CH3 C C CH3 C C H C H C 3 β 3 δ− H CH3 H Br H Br + Et OH + Br EtO removes Partial bonds in a β ⊖proton; the transition C=C is fully C−H breaks; state: C−H and formed and new π bond C−Br bonds the other forms and Br break, new π products are begins to C−C bond forms EtOH and Br depart Ch. 7 -⊖31 δ− O Et H3C H δ− CH3CH2 C C Et O δ− H Br H H CH3 C C CH3 H3C δ− ‡ Br ∆G y H 1 g ‡ r G e ∆ 2 n E CH 3 e - e CH3 CH CH C CH + EtOH + Br r 3 2 2 F - EtO + CH3CH2 C CH3 Br CH3 - CH3CH C CH3 + EtOH + Br Reaction Coordinate Ch. 7 - 32 6C. Formation of the Less Substituted Alkene Using a Bulky Base Hofmann’s Rule ● Most elimination reactions follow Zaitsev’s rule in which the most stable alkenes are the major products. However, under some circumstances, the major elimination product is the less substituted, less stable alkene Ch. 7 - 33 ● Case 1: using a bulky base EtO CH3CH CHCH3 (80%) + (small) CH3CH2CH CH2 (20%) CH3CH2CHCH3 Br t BuO CH3CH CHCH3 (30%) + (bulky) CH3CH2CH CH2 (70%) EtO⊖ (small base) H H H H tBuO⊖ H C C C C H (bulky base) H H Br H Ch. 7 - 34 ● Case 2: with a bulky group next to the leaving halide less crowded β-H Me H Br H Me H EtO H3C C C C C H H3C C C C CH2 Me H Me H Me H Me (mainly) more crowded β-H Ch. 7 - 35 Zaitsev Rule vs. Hofmann Rule ● Examples Ha Hb (1) + Br (eliminate Ha) (eliminate Hb) NaOEt, EtOH, 70oC 69% 31% KOtBu, tBuOH, 75oC 28% 72% Ch. 7 - 36 ● Examples Hb Br Ha (2) + (eliminate Ha) (eliminate Hb) NaOEt, EtOH, 70oC 91% 9% KOtBu, tBuOH, 75oC 7% 93% Ch. 7 - 37 6D. The Stereochemistry of E2 Reactions The 5 atoms involved in the transition state of an E2 reaction (including the base) must lie in the same plane The anti coplanar conformation is the preferred transition state geometry ● The anti coplanar transition state is staggered (and therefore of lower energy), while the syn coplanar transition state is eclipsed Ch. 7 - 38 B B H H LG C C C C LG Anti coplanar Syn coplanar transition state transition state (preferred) (only with certain rigid molecules) Ch. 7 - 39 Orientation Requirement ● H and Br have to be anti periplanar (trans-coplanar) ● Examples CH3CH2 + EtO CH3CH2 Br CH3 CH3 since: Br CH3CH2 H Only H is H anti periplanar H CH3 to Br EtO Ch. 7 - 40 E2 Elimination where there are two axial β hydrogens EtO (a) 1 H3C 4 CH(CH3)2 a 3 2 EtO b H H 1-Menthene (78%) CH(CH ) 1 3 2 (more stablealkene) H3C 4 3 H H 2 H Cl 1 H C 4 CH(CH ) (b) 3 3 2 Both Ha and Hb hydrogens 3 2 are anti to the chlorine in 2-Menthene (22%) this, the more stable (less stable alkene) conformation Ch. 7 - 41 E2 elimination where the only axial β hydrogen is from a less stable Conformer H CH3 H Cl 1 CH(CH3)2 H3C 4 2 H 3 Cl H H H H H H CH(CH3)2 Menthyl chloride Menthyl chloride (more stable conformer) (less stable conformer) Elimination is not possible Elimination is possible for for this conformation this conformation because because no hydrogen is anti the green hydrogen is anti to the leaving group to the chlorine Ch.