Synthesis and Properties of Alkenes PPT(1)(1).pdf

Alkenes Properties and Synthesis  2  The (E)-(Z) System for Designating Alkene Diastereomers The Cahn-Ingold-Prelog convention is used to assign the groups of highest priority on each...

Alkenes Properties and Synthesis  2  The (E)-(Z) System for Designating Alkene Diastereomers The Cahn-Ingold-Prelog convention is used to assign the groups of highest priority on each carbon  If the group of highest priority on one carbon is on the same side as the group of highest priority on the other carbon the double bond is Z (zusammen)  If the highest priority groups are on opposite sides the alkene is E (entgegen)  3  Relative Stabilities of Alkenes Generally cis alkenes are less stable than trans alkenes because of steric hindrance  Heat of Hydrogenation The relative stabilities of alkenes can be measured using the exothermic heats of hydrogenation  Heat of Hydrogenation Exp. DH Theo. DH Compound Comment (Kcal/mol) (Kcal/mol) -30.3 N/A -28.6 N/A substitution -27.6 N/A trans -60.1 -60.6 -54.1 -57.9 conjugation Extended -80.03 -88.2 conjugation  Extended Conjugation -carotene -carotene Distyrylbiphenyl trans-trans cis-trans  6 Heats of hydrogenation of three butene isomers:  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  7  Structural Information from Molecular Formulas and the Index of Hydrogen Deficiency (IHD)  Unsaturated and Cyclic Compounds A compound with the general molecular formula CnH2n will have either a double bond or a ring A compound with general formula CnH2n-2 can have a triple bond, two double bonds, a double bond and a ring or two rings Index of Hydrogen Deficiency: the number of pairs of hydrogen atoms that must be subtracted from the molecular formula of the corresponding alkane to give the molecular formula of the compound under consideration  8 Example: A compound with molecular formula C6H12 Hydrogenation allows one to distinguish a compound with a double bond from one with a ring  Compounds Containing Halogens, Oxygen, or Nitrogen For compounds containing halogen atoms, the halogen atoms are counted as if they were hydrogen atoms Example: A compound with formula C4H6Cl2  This is equivalent to a compound with molecular formula C4H8 which has IHD=1  9 For compounds containing oxygen, the oxygen is ignored and IHD is calculated based on the rest of the formula  Example: A compound with formula C4H8O has IHD = 1 For compounds containing nitrogen, one hydrogen is subtracted for each nitrogen and the nitrogen is ignored in the calculation  Example: A compound with formula C4H9N is treated as if it has formula C4H8 and has IHD = 1  10  Elimination Reactions of Alkyl Halides  Dehydrohalogenation Used for the synthesis of alkenes  Elimination competes with substitution reaction  Strong bases such as alkoxides favor elimination 11  12  The E2 Reaction E2 reaction involves concerted removal of the proton, formation of the double bond, and departure of the leaving group Both alkyl halide and base concentrations affect rate and therefore the reaction is 2nd order 13  14  The E1 Reaction The E1 reaction competes with the SN1 reaction and likewise goes through a carbocation intermediate  15  Substitution versus Elimination  SN2 versus E2 Primary substrate  If the base is small, SN2 competes strongly because approach at carbon is unhindered Secondary substrate  Approach to carbon is sterically hindered and E2 elimination is favored  16 Tertiary substrate  Approach to carbon is extremely hindered and elimination predominates especially at high temperatures Temperature  Increasing temperature favors elimination over substitution  17 Size of the Base/Nucleophile  Large sterically hindered bases favor elimination because they cannot directly approach the carbon closely enough to react in a substitution  Potassium tert-butoxide is an extremely bulky base and is routinely used to favor E2 reaction  18  Synthesis of Alkenes via Elimination Reactions  Dehydrohalogenation Reactions by an E2 mechanism are most useful  E1 reactions can be problematic E2 reaction are favored by:  Secondary or tertiary alkyl halides  Alkoxide bases such as sodium ethoxide or potassium tert-butoxide Bulky bases such as potassium tert-butoxide should be used for E2 reactions of primary alkyl halides  19  Zaitsev’s Rule: Formation of the Most Substituted Alkene is Favored with a Small Base Some hydrogen halides can eliminate to give two different alkene products Zaitzev’s Rule: when two different alkene products are possible in an elimination, the most highly substituted (most stable) alkene will be the major product  This is true only if a small base such as ethoxide is used  20 The transition state in this E2 reaction has double bond character The trisubstituted alkene-like transition state will be most stable and have the lowest DG‡ Kinetic control of product formation: When one of two products is formed because its free energy of activation is lower and therefore the rate of its formation is higher  This reaction is said to be under kinetic control  21  Formation of the Least Substituted Alkene Using a Bulky Base Bulky bases such as potassium tert-butoxide have difficulty removing sterically hindered hydrogens and generally only react with more accessible hydrogens (e.g. primary hydrogens)  22 Stereochemistry of E2 Reactions: Orientation of Groups in the Transition State All four atoms involved must be in the same plane  Anti coplanar orientation is preferred because all atoms are staggered β - Carbon With Two Labile Hydrogens 23  24 β - Carbon With One Labile Hydrogen  25 Stereochemistry of E1 Reactions  Both the E and Z products will be formed  The major product will be the one with the bulkiest groups on opposite sides of the double bond  Both syn and anti eliminations can occur, regardless of whether the β-carbon is bonded to one or two hydrogen atoms  26 Cyclic Molecules In a cyclohexane ring, the eliminating substituents (i.e. H and X) must be diaxial to be anti coplanar  27 When cis-1-bromo-2-methylcyclohexane undergoes an E2 reaction, two products are formed. What are these two cycloalkenes? Which one is the major product? Br H H H CH3 CH3 B H Br H H H CH3 CH3 H B  28 When trans-1-bromo-2-methylcyclohexane undergoes an E2 reaction, only one product is formed. What is this product? Br H H H H CH3 B CH3  29 Neomenthyl chloride and menthyl chloride give different elimination products because of this requirement In neomenthyl chloride, the chloride is in the axial position in the most stable conformation  Two axial hydrogens anti to chlorine can eliminate; the Zaitzev product is major  30 In menthyl chloride the molecule must first change to a less stable conformer to produce an axial chloride  Elimination is slow and can yield only the least substituted (Hoffman) product from anti elimination  31  Acid Catalyzed Dehydration of Alcohols Recall that elimination is favored over substitution at higher temperatures Typical acids used in dehydration are sulfuric acid and phosphoric acid The temperature and concentration of acid required to dehydrate depends on the structure of the alcohol  Primary alcohols are most difficult to dehydrate, tertiary are the easiest Rearrangements of the carbon skeleton can occur  32  Mechanism for Dehydration of Secondary and Tertiary Alcohols: An E1 Reaction Only a catalytic amount of acid is required since it is regenerated in the final step of the reaction  33  Carbocation Stability and the Transition State Recall the stability of carbocations is:  The second step of the E1 mechanism in which the carbocation forms is rate determining The transition state for this reaction has carbocation character Tertiary alcohols react the fastest because they have the most stable tertiary carbocation-like transition state in the second step  34 The relative heights of DG‡ for the second step of E1 dehydration indicate that primary alcohols have a prohibitively large energy barrier  35  A Mechanism for Dehydration of Primary Alcohols: An E2 Reaction Primary alcohols cannot undergo E1 dehydration because of the instability of the carbocation-like transition state in the 2nd step In the E2 dehydration the first step is again protonation of the hydroxyl to yield the good leaving group water  36  Carbocation Stability and the Occurrence of Molecular Rearrangements  Rearrangements During Dehydration of Secondary Alcohols Rearrangements of carbocations occur if a more stable carbocation can be obtained Example The first two steps are to same as for any E1 dehydration  37 In the third step the less stable 2o carbocation rearranges by shift of a methyl group with its electrons (a methanide)  This is called a 1,2 shift The removal of a proton to form the alkene occurs to give the Zaitzev (most substituted) product as the major one  38 A hydride shift (migration of a hydrogen with its electrons) can also occur to yield the most stable carbocation Carbocation rearrangements can lead to formation of different ring sizes  39  Hydrogenation of Alkenes Hydrogen adds to alkenes in the presence of metal catalysts Heterogeneous catalysts: finely divided insoluble platinum, palladium or nickel catalysts Homogeneous catalysts: catalyst(typically rhodium or ruthenium based) is soluble in the reaction medium  Wilkinson’s catalyst is Rh[(C6H5)3P]3Cl This process is called a reduction or hydrogenation  An unsaturated compound becomes a saturated (with hydrogen) compound  40  Hydrogenation: The Function of the Catalyst The catalyst provides a new reaction pathway with lower DG‡ values  41 In heterogeneous catalysis the hydrogen and alkene adsorb to the catalyst surface and then a step-wise formation of C-H bonds occurs Both hydrogens add to the same face of the alkene (a syn addition)  Addition to opposite faces of the double bond is called anti addition