Cis- and Trans- Revisited: The Sequence Rules and the E- and Zconvention for describing isomerism about the double bond in structurally complex alkenes: CH3 H H H C C C C CH3 trans-2-butene CH3 CH3CH2 CH2 H H CH3 CH3 cis-2-butene H C C C C CH3 cis- or trans- ? CH3CH2 CH2 CH3 CH3 cis- or trans- ? In order to describe the structure of complex alkenes the Cahn-IngoldPrelog Rules or Sequence Rules (already used to assign R and S configurations to chiral molecules with stereogenic centres) are used to assign a priority to each of the substituents on the alkene carbons. Be clear that use of the Sequence Rules for describing alkene structure has nothing whatever to do with chirality and/or R or S configurations. When the substituents of highest priority on each alkene carbon are both on the same side of the double bond the alkene is designated Z(German: 'zusammen' = together) and when they are both on opposite sides of the double bond the alkene is designated E- (German: 'entgegen' = opposite). Cl C C H CH3 Cl Br C Br H CH3 E- H2N C C H Z- CH3 OCH3 H 2N C C C H CH2 H C H Z- Cl C CH2CH3 C H C H C Z- CH2CH2 CH3 Br CH2CH2 Br H E- CH2CH3 CH2 H EBr OCH3 CH3 E- Cl C C C CH2CH2 Br CH2CH2 CH3 Z- H2C CH3 CH3 CH C C CH3 Z- C CH3 H2C H CH3 CH3 CH C C H C CH3 CH3 E- Reading assignment: Work through the practice problem on p. 200 of McMurry (5th Ed.) and problems 6.9 to 6.11 on p. 201. Industrial Preparation of Alkenes: Thermal 'cracking' of high molecular weight alkanes: CnH2n+2 ca. 900 °C H2 + CH4 + H2C=CH2 + CH3CH=CH2 n = 2-8 + CH3 CH2CH=CH2 Laboratory Preparation of Alkenes: There are four standard laboratory methods for preparing alkenes from non-alkene starting materials: (1) From alcohols (dehydration, 1,2- or -elimination of H2O). (2) From alkyl halides and related compounds of the type RX (1,2- or -elimination of HX). (3) From aldehydes or ketones via the Wittig Reaction (This reaction will be covered in detail in module CM3001) (4) From alkynes. (1) Alkenes from alcohols (dehydration, 1,2- or elimination of H2O) N.B. OH - is a poor leaving group (good nucleophile). OH H3PO4 Heat H H + H2 O H3PO4 + OH2 H2O is a better leaving group than OH- (i.e. a weaker nucleophile). H H + [H2 PO4]- - H2 O O + H H - O P OH OH Unsymmetrical alcohols may yield more than one product: H CH3 H OH H H CH3 + H3 PO4 Heat - H2 O H H - H+ + Zaitsev (or Saytzeff) rule: the most substituted (i.e. most stable) alkene is favoured. (2) Alkenes from alkyl halides and related compounds of the type RX (1,2- or -elimination of HX). H C C X 1 2 Base - HX C C + B:H+ + X - X = Cl, Br, I, OTs (tosylate) but X ° F B = OH-, OR- (alkoxide) etc. In the reaction illustrated above an alkene is formed by removal of H and X are from an alkyl halide, i.e. a small molecule, HX, is eliminated and the reaction is referred to as an elimination: Some important points: This reaction illustrated above is called a 1,2- or -elimination to indicate that the groups being eliminated are located on adjacent atoms in the starting material as compared to a 1,1- or -elimination where both are located on the same carbon atom This, in itself, tells you nothing about the actual mechanism of the elimination process. Notice that the requirements for a -elimination reaction are: (i) A base (ii) A leaving group in the substrate such as halide, tosylate etc. and ... (iii) A C-H bond - to the leaving group. We have already seen that most strong bases are also good nucleophiles - and vice versa. For this reason nucleophilic substitution and -elimination reactions can often compete with one another if the substrate organic halide has a C- H group - to the leaving group. Alkene formation via elimination of HX from alkyl halides can follow two different mechanistic pathways identified as E2 and E1. Here is the E2 mechanism: B: H B R C H R C R R C R C R R X R X Transition State R R C R + B-H + X: C R E because it is an 'elimination' - a small molecule (here HX) is lost or 'eliminated' from one of the reactants in order to form the product. E2 because the reaction is bimolecular, i.e. two species - the base and the organic halide ot tosylate - come together to form the transition state. Therefore the reaction follows a second-order rate law: Rate = k x [RX] x [Base] The E2 reaction proceeds - like SN2 - in a single step with no intermediates. The transition state is illustrated in greater detail below: Notice that the transition state prefers a geometry in which: (i) The base, the proton being removed, the two carbon atoms and the leaving group all lie in the same plane and ... (ii) The proton being removed and the departing leaving group lie on opposite sides of the connecting C-C bond. This preferred geometry of the reacting H-C-C-X system is called anti coplanar or anti periplanar geometry. As the proton and leaving group depart, the hybridisation at carbon changes from sp3 to sp2. The sp3 orbitals of the C-H and C-X bonds must rehybridise to p and be in the same plane to overlap in a sideways-on fashion and form the component of the double bond. This developing -overlap is most efficient if the C-H and C-X bonds - and hence the developing p-orbitals - are already in the same plane in the transition state. A syn periplanar conformation also produces the desired coplanar orientation of the developing p-orbitals but requires an eclipsed relationship between the substituents on the adjacent carbon atoms and is therefore much less favourable than the anti periplanar conformation as shown by the following Newman projections: H HX • • X anti periplanar staggered - lower energy syn periplanar eclipsed - higher energy The syn and anti periplanar geometries produce different stereochemical outcomes in the E2 reaction and this can be used to show that the anti periplanar conformation is preferred: H Br C CH3 Ph C H H CH3 CH3 Ph H C C Br anti staggered syn eclipsed Base Base CH3 C CH3 C Ph CH3 CH3 H Ph H C E- C CH3 Z- E2 reactions are favoured by the same factors that favour SN2 substitution - 1° or 2° C-X, strong base (= strong nucleophile) - so that the two different process often compete under appropriate conditions: Br + Na+ OEt EtOH 53 °C OEt + E2 SN 2 12% 88% As we have seen, strong bases are normally good nucleophiles. However a strong base which is sterically demanding is likely to be a