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Msc Chemistry Subject Chemistry Paper No and Title Paper-9, Organic Chemistry-III (Reaction Mechanism-2) Module No and Title Module-23, Elimination Reactions: E1 and E2 mechanisms- Part-II Module Tag CHE_P9_M23 CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3. Regio-Chemical Aspects of E1 And E2 Eliminations: Orientation of the Double Bond 3.1 Regiochemistry of E1 elimination 3.2 Regiochemistry of E2 elimination 4. Stereochemical aspects of E1 and E2 Elimination Mechanisms 4.1 Stereochemistry of E1 elimination 4.2 Stereochemistry of E2 elimination 5. E2 elimination in cyclohexane rings 6. Summary CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II 1. Learning Outcomes After studying this module, you shall be able to Know what is meant by elimination reactions Perceive factors favouring elimination over substitution Learn about the types of elimination reactions Differentiate between the E1 and E2 mechanism Identify set of conditions responsible for E1 or E2 mechanism 2. Introduction The E1 and E2 eliminations present a fair degree of regioselectivity and stereo-selectivity which are governed by their mechanistic aspects. Before proceeding to a detailed discussion on these aspects, let us recall the main features of E1 and E2 eliminations as shown in table-1. E1 elimination E2 elimination Rate = k [substrate] Rate = k [substrate][Base] unimolecular Bimolecular Leaving group first departs and Concerted process; both the groups carbocation is formed from which (proton and leaving group) are removal of H takes place eliminated simultaneously. Preferred in 3o >2o >1o substrates No such preference No strong base required Stronger and bulky base in high concentration is required Strongly dependent on leaving group. Hydroxyl can never be a leaving group Hydroxyl can be converted to a better for E2 elimination leaving group via protonation by use of an acid No stereochemical requirement The leaving group must be trans to the Hydrogen involved 3. Regio-Chemical Aspects of E1 And E2 Eliminations: Orientation of the Double Bond During elimination reaction, a double bond is generated. So, one is concerned with the regiochemistry i.e., the production of structural isomers which have the double bond in different positions. Elimination reactions may be either regiospecific or regioselective. Regiospecific elimination reactions produce only one isomer of an alkene. Regioselective elimination reactions, on the other hand, produce several different isomers, but give one isomer in major quantity than the others. This, in turn, depends on whether the reaction prefers to eliminate only one particular β-hydrogen or a mixture of β-hydrogens. CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II If an elimination reaction produces the more highly substituted alkene, the reaction is said to follow the Saytzeff rule. On the other hand, if the elimination reaction forms the less substituted alkene, the reaction is said to follow the Hofmann rule. A number of factors including the nature of substrate, the leaving group, reaction conditions etc. determine what rule a given reaction may follow. 3.1 Regiochemistry of E1 elimination E1 eliminations are regioselective and generate the alkene that has the more substituents, as the major product. Thus, the E1 reactions follow Saytzeff rule. This is due to the fact that in E1 eliminations, leaving group departs to produce first a carbocation. Thus the direction of the double bond is determined almost entirely by the relative stabilities of the olefins produced via elimination. The major product is the alkene that has the more substituents, because this alkene is the more stable of the two possible products. For example, 2-bromo-3-methylbutane undergoes E1 elimination to yield two alkene products viz. 3-methyl-1-butene and 2-methyl-2-butene. The major product is 2-methyl-2-butene because there are three alkyl substituents attached to the carbon—carbon double bond. ‘More substituted alkene is more stable’ does not necessarily explain why it is the one that forms faster. For this, we need to have a look on the transition state leading to the two alkenes. In the carbocation there is hyperconjugation involving each β-hydrogen. Since the hyperconjugation structures possess some double-bond character, the interaction with hydrogen is greatest at more highly substituted carbons; that is, there will be greater weakening of C−H bonds and more double-bond character at more highly substituted carbon atoms. This structural effect in the carbocation intermediate governs the direction of elimination and leads to the preferential formation of the more highly substituted alkene. More stable transition state is lower in energy, and results in faster formation of more substituted alkene, i.e. it is the thermodynamic product. CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II 3.2 Regiochemistry of E2 elimination Most of the E2 reactions, involving good leaving groups (halides, sulfonates etc.) also follow the Saytzeff rule, i.e. lead to a more substituted alkene. This can be attributed to a similar transition state as discussed in the case of E1 elimination. Hence, halides (Br, Cl, I) give more substituted alkene, but E2 elimination of alkyl fluorides results in the least substituted alkene. When a base begins to abstract a proton from an alkyl fluoride, the fluoride ion has less tendency to leave compared to other halide ions as it is a poor leaving group. Therefore negative charge develops on the carbon that is losing the proton, giving the transition state a carbanion character (rather than an alkene character as shown previously) which is stabilized by highly electron-withdrawing fluorine and fewer electron-donating alkyl substituents. This subsequently results in the formation of least-substituted alkene. CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II Similarly, E2 reactions involving other poor leaving groups, particularly those with quaternary ammonium salts, are said to follow the Hofmann rule and give primarily the less-substituted alkene. Moreover, when the proton to be removed is in the sterically more hindered position or when the base used for E2 elimination is a bulky base, exceptions to the Saytzeff rule occur. These situations lead to a high proportion of the less substituted alkene (Hofmann orientation). However, when formation of a double bond can lead to extended conjugation with already present functional groups (C=O or C=C) or aryl ring, then irrespective of the mechanism, the conjugated product only predominates. Similarly, eliminations in bridged cyclic compounds follow the Bredt rule, i.e. irrespective of the mechanism, a double bond never goes to the bridgehead carbon in rings of smaller size. This is simply a consequence of the strain induced by a planar bridgehead carbon, and so having a double bond on a bridgehead would be too unstable. 4. Stereochemical aspects of E1 and E2 Elimination Mechanisms The E1 reaction mechanism is a two-step process that, as with the SN1 mechanism, usually loses all the stereochemical information of the substrate as the reaction proceeds. On the other hand, the E2 mechanism is a concerted mechanism. A concerted reaction usually requires that the substrate have a specific conformation that allows the orbitals of the bonds being broken to overlap the bonds being formed. It is only when the orbitals overlap, the electrons flow smoothly CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II from the breaking bonds to the forming bonds and consequently the reaction takes place. These aspects of elimination reactions are discussed as follows: 4.1 Stereochemistry of E1 eliminations Considering the geometrical isomerism of the E1 elimination product, usually the formation of E- alkene is favoured over the Z-alkene. This is due to the steric factors of the transition states leading to these possible geometries, favouring E-geometry because the substituents can get farther apart from each other. So, E-alkene forms faster. 4.2 Stereochemistry of E2 eliminations As discussed before, E2 elimination requires a stricter stereo-chemical requirement to occur. The alkene is formed by overlap of the C–H σ-bond with the C–X σ* antibonding orbital (X=leaving group). The two orbitals must lie in the same plane for best overlap, and there are two possible conformations for this viz. Hydrogen and leaving-group are either syn-periplanar, or anti- periplanar. Of these two conformations, the anti-coplanar conformation (staggered) is generally more stable. Thus, most E2 eliminations occur when the substrate is in the anti-coplanar conformation. This is also illustrated below: CHEMISTRY PAPER No. 9 : Organic Chemistry-III (Reaction Mechanism-2) Module-23, Elimination Reactions: E1 and E2 mechanisms-Part-II Syn-elimination Anti-elimination Not so efficient overlap between σ C-H Efficient overlap possible between σ C-H bonding and σ* C-LG anti-bonding orbital. bonding and σ* C-LG anti-bonding orbital. Repulsion between the attacking electron- rich base and electron-rich halogen on the No repulsion between the attacking same side of the molecule. electron-rich base and electron-rich halogen on the opposite sides of the molecule. Thus, anti-elimination is faster. However, when the substrate itself is quite rigid and only one proton is available for taking part in elimination as shown in the figure below for the two diastereomers of (1- bromopropane-1,2-diyl)dibenzene, the E2-elimination can be stereo-specific and the reaction outcome will depend upon which diastereomer has been used for the reaction.
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