Alkenes. Bonding and Structure: Carbons in the double bond of butene are sp 2 hybridized. Side on p-p orbital overlap creates a π-bond.

Transcription

1 Alkenes Bonding and Structure: Carbons in the double bond of butene are sp 2 hybridized. Side on p-p orbital overlap creates a π-bond. Angles around the carbons in the double bond are ~ 120º. Thus, all of the atoms bonded to the sp 2 hybridized carbon lie in a plane. Carbon-Carbon double bond length is ~ 1.34 Å (single bonds in alkane are ~ 1.54 Å. 3 C C C 90 C 3 3 C C 3 In order to interconvert between the isomers with the methyl groups on the same, and opposite sides, the double bond must be broken. As a result the two isomers do not interconvert at ordinary temperatures.

2 Introduction to Stereoisomers For 2 butene there are two isomers with the same connectivity, one with the methyl groups on the same side of the double bond and the other with the methyl groups on the opposite sides. C 3 3 C C 3 3 C trans-2-butene (E)-2-butene cis-2-butene (Z)-2-butene Thus, while the compounds have the same connectivity, they have their atoms fixed in different regions of space. Such isomers are called stereoisomers. The carbons are called stereo centers or stereogenic atoms. The cis case, both methyl groups are on the same side of the double bond and in the trans case, opposite sides of the double bonds. For more complicated olefins, a system based on the priorities of the groups attached to the double bond is used. This is called the E and Z priority system where E is used for the isomer with the high priority groups on the opposite side of the double bond and Z when they are on the same side more later. 2

3 Nomenclature (IUPAC) i. The same prefixes are used for alkenes as alkanes but the ane is replaced by ene. The principle chain is the longest chain containing the greatest number of double bonds. (It is not necessarily the longest chain in the molecule). Example: propyl-1-hexene not an octane (longest hydrocarbon chain) not 2-butyl-1-pentene 5 ii. The chain is numbered such that the double bond(s) receive(s) the lowest number(s). (First point of difference rule applies.) Example: heptene...correct 4-heptene...incorrect 3

4 First point of difference: Example: ,3-diethyl-4-methyl-1,3-pentadiene not 3,4-diethyl-2-methyl-2,4-pentadiene First point of difference with numbering of double bonds 2,3-diethyl-4-methyl-1,3-pentadiene position of substituents # of double bonds position of double bonds 4

5 For cycloalkenes, begin numbering at the double bond and proceed through the bond in the direction to generate the lowest number at the first point of difference (FPOD). (Thus a double bond should always be carbons 1 and 2.) One of the most common mistakes in naming cycloalkenes is to generate the lowest number sequence around the ring, disregarding this rule. Once again, the numbering must begin at the double bond and proceed through the bond in the direction to generate the lowest number sequence ,5-diethylcyclopentene not 2,3-diethylcyclopentene not 1,2-diethylcyclo-2-pentene Note # of ethyl determined by FPOD Note that the alkene is assumed to be in position 1 and thus is not stated explicitly 5

6 Some common names of molecules containing double bonds: 2 C C 2 ethylene propylene styrene isoprene Some common names of groups containing double bonds: R C C 2 R R vinyl allyl isopropenyl Examples: Br allylbromide 3-bromo-1-propene Br isopropenylbromide 2-bromo-1-propene 6

7 Nomenclature for stereoisomers: The Z (zusammen, together in German) and (E entgegen, across in German) nomenclature. Low Low igh Low C b C a C b C a igh igh Low igh Z-isomer E-isomer Z is similar to cis E is similar to trans Cis and trans typically used only when the low priority groups are hydrogens. 1) Consider each carbon of double bond separately. 2) Rank the priority of the substituent on the carbon as follows: The atom with the higher molecular weight takes top priority. If there are two isotopes of the same atom, the isotope with the higher mass takes priority. Low 3 C Low Low 3 C igh D 3 C Low C b C a C b C a C b C a Br igh Cl igh Br igh Low 3 C Cl igh Z-isomer E-isomer no stereocenter on C b 7

8 If the atom directly attached to the alkene does not distinguish, then move down the chain of the substituent assigning priorities down the chain until you reach the first point of difference to assign the priority. FPOD Low Low C 2 C 2 C 2 C 2 I Cl > C 3 C igh C 2 C 2 Cl igh FPOD Z-isomer Multiple bonds count as multiples of that same atom. X C a C b C b C a treated as C b C a X alkene X alkene X Example: igh Low 3 C C 2 O 3 C O [O,,] C b C a counts as C b C a C O O [O,O,] Low O C igh E-isomer 8

9 Calculating Degrees of Unsaturation Alkanes, (C n 2n+2 ), are fully saturated. Cycloalkanes, (C n 2n ), have two fewer hydrogens than the parent hydrocarbon since two of the hydrogens on the terminal methyl groups are lost in order to form the carbon-carbon bond that closes the ring: cyclopentane is C 5 10, while pentane is C Alkenes, have one, (C n 2n ), or more double bonds. An alkene with one double bond has two fewer hydrogens than the saturated hydrocarbon of the same length. For example, butene ( 2 C=CC 2 C 3 ) has the molecular formula C 4 8 and butane (C 3 C 2 C 2 C 3 ) has the formula C Alkynes have one, (C n 2n-2 ), or more triple bonds. An alkyne with one triple bond has four fewer hydrogens than the saturated hydrocarbon of the same length. For example, butyne (C CC 2 C 3 ) has the molecular formula C 4 6 and butane (C 3 C 2 C 2 C 3 ) has the formula C Knowing this relationship, it is possible to take a molecular formula and calculate the degree of unsaturation; that is, the total number of multiple bonds or rings in a molecule. For hydrocarbons, the process is simple: take the parent hydrocarbon and calculate the number of hydrogens using the 2n + 2 rule, every two hydrogens that are "missing" in the analysis of the unknown represents one degree of unsaturation. 9

10 For compounds containing elements other than carbon and hydrogen, degrees of unsaturation can be calculated as follows: alogen containing compounds: since a halogen is simply a replacement for a hydrogen in an organic molecule (a valence of one), you simply add the total number of halogens to the carbon-hydrogen analysis, and calculate the unsaturation number as described above. Oxygen containing compounds: since oxygen is divalent, it has no effect on the calculation for the degree of unsaturation, and can simply be ignored. Considering methanol (C 3 O) removing the oxygen yields methane (C 4 ). For a carbonyl, (i.e., acetone, C 3 COC 3 ), ignoring the oxygen gives C 3 6, two hydrogens short of (2n + 2), and one degree of unsaturation. The carbonyl is therefore equivalent to one degree of unsaturation. Nitrogen containing compounds: since nitrogen is trivalent, a nitrogen containing compound has one more hydrogen than an equivalent hydrocarbon has. Therefore you should subtract the number of nitrogens from the total number of hydrogens and calculate as described above. The unsaturation number, U is the sum of the number of multiple bonds and/or the rings in a compound. Double bond adds one, triple bond adds two, (these bonds need not be between the same type of atoms), ring adds one. U = 0.5 [2C N - ( + X)] C = # of Carbons N = # of Nitrogens = # of ydrogens X = # of alogens 10

11 Examples: C 8 8 U = 0.5 x [2(8) ( 8)] = 5 C 3 1 ring 4 double bonds 0 triple bonds 1 ring 2 double bonds 1 triple bonds C C 3 C C C C C C 2 rings 3 double bonds 0 triple bonds 0 rings 3 double bonds 1 triple bonds Examples: C 9 16 ClNO 2 U = 0.5 x [2(9) ( )] = 2 O 2 N O Cl 11

12 Relationship between Equilibrium Constant and Free Energy Gº = º-T Sº Gº is the Gibbs free energy change at equilibrium for standard states of reactants and products (for gases 1 atm, for solution [1 M]). º is the standard enthalpy of the reaction. Sº is the standard entropy of the reaction. T is the temperature in Kelvin. So, Gº is defined as the free energy change, G, of an ideal solution in which each of the reactants and products are present at 1 Molar. For such a system G= Gº. owever, at equilibrium G = 0, and concentration of all of the components need not be 1 Molar. At this point: G = Gº + RT ln K eq Where: So Gº = -RT ln K eq R is the universal gas constant x 10-3 kcal/deg mol, and K eq = [Products] eq [Reactants] eq Or Gº = -RT2.3 log K eq Gº, is the driving force of a reaction. As we can see from the equation above, it is related to equilibrium constant: logk eq = -ΔG 0 2.3RT 12

13 Reactions tend to go in the direction to create more stable compounds. REACTANTS PRODUCTS G products ΔG 0 > 0, K eq < 1 FREE ENERGY G reactants G reactants ΔG 0 < 0, K eq > 1 G products Reaction goes to the left Reaction goes to the right 13

14 Reaction Coordinate Diagram Standard Free Energy Reactants Transition state ΔG (activation energy) ΔG o Products Reaction Coordinate The height of the barrier for the transition state (ΔG ) determines the rate of the reaction. The transition state is an activated complex that is the highest energy point on a simple reaction coordinate. The transition state is unstable with respect to both the reactants and products, and, therefore, is not present at any appreciable concentration and exists for extremely short periods of time. The direction of the reaction is determined by standard free energy of the reaction (ΔG ). 14

15 Examples: E A B A B A B reaction doesn t occur reaction slow reaction fast ΔG is extremely high ΔG is high ΔG is low A B A B A B A B fast fast slow slow (reactants favored) (products favored) (reactants favored) (products favored) ΔG is low ΔG is low ΔG is high ΔG is high ΔG is positive ΔG is negative ΔG is positive ΔG is negative 15

16 More Complex Reaction Coordinate Diagrams Standard Free Energy ighest energy process is slowest step, so this is the rate determining step, RDS. Transition state (t.s.) #1 ΔG (1) ΔG (2) Intermediate Transition state (t.s.) #2 Reactants Products ΔG o Reaction Coordinate A reaction mechanism is a detailed series of steps that track relatively low energy bond breaking and making processes, which taken in sequence, provides a chemically reasonable path from the reactants to the products. Such a mechanism may typically involve the formation of intermediates, which are metastable species that are higher in energy than either the reactant or the product but lower in energy than the transition state. Intermediates may or may not have significant lifetimes. The rate determining step is the slowest step in a multistep process. This acts as a bottleneck for the progress of a reaction that involves intermediates. 16

17 ammond Postulate There is no fundamental relationship between the energy of the transition state and the stability of the reactants and products (intermediates). owever, there are trends that often occur and are worth noting. Energy Slower Reaction Faster reaction Less stable product Energy Slower Reaction Faster reaction Less stable product More stable product Reaction coordinate The reaction with the lower energy transition state (faster reaction) gives the more stable product (intermediate). More stable product Reaction coordinate The reaction with the lower energy transition state (faster reaction) gives less stable product (intermediate). In general, the situation on the left is what is observed and NOT the righthand case. Thus, typically the more stable intermediate gives the faster reaction, when there is a choice. George ammond summarized these observations in what is now known as the ammond Postulate. 17

18 The diagrams below help to illustrate the point. On the left, the reaction is endergonic and the transition state is similar in energy as the product. On the right, the reaction in exergonic and the transition state is similar in energy to the reactant. T.S. T.S. Energy Product Energy Reactant Reactant Reaction Coordinate Reaction Coordinate Product ammond Postulate: The structure of a transition state resembles that of the nearest stable species (in energy). The structure of the transition state on the left resembles that of the product and the structure of the transition state on the right resembles that of the reactant. 18

19 Solvents Although we draw chemical reactions with reagents on the left of an arrow and the products on the right, the details of how the reaction is carried out in the lab are absolutely critical for a successful outcome. Various aspects of conditions that must be considered (not in any particular order of importance) are: 1. order of addition of reagents 2. rate at which one reagent is added to the other 3. temperature at which the reagents are mixed 4. temperature at which the reaction is allowed to proceed 5. time of the reaction 6. details of stoichiometry--is one reagent used in excess 7. concentration of the reagents 8. is an equilibrium forced to completion by removing one of the products 9. adequate stirring 10. use of catalyst 11. inherent incompatibility of reagents 12. choice of solvent A chemist must consider each of these factors carefully when planning a reaction. Lack of consideration of these factors will often result in: mixtures of products, no desired product, decomposition of the desired product, or accidents -- explosions, liberation of noxious gases, etc. For now we will focus a bit on properties of solvents. In the next chapters it will become clear why choices of particular solvents are favorable for various reactions. 19

20 Choosing Solvents Choice of solvent is dictated by many things including boiling point, solubility of the reactants and products, whether the solvent is inert toward reactants and products, ability of the solvent to stabilize transition states by virtue of its polarizability, its polarity, or its ability to form hydrogen bonds. While there are many ways to classify solvents, organic chemists typically use three classifications: 1. Polar or apolar (really a continuous gradation) 2. Protic and aprotic (referring to its ability to act as a hydrogen bond donor) 3. Donor or nondonor (referring to its Lewis basity) There are two functional definitions of solvent polarity: One is simply the dipole moment of the solvent The other refers to the solvent ability to (by whatever mechanism) stabilize charge or decrease the attractive or repulsive interaction between ions This latter property goes back to our discussion of ions, dipoles and the energy of charges interacting with each other back in the chapter on acids and bases. As promised, the energy of interaction of charges rears its ugly head once again. 20

21 One simple definition of a solvent polarity in the latter sense of the word is the magnitude of a solvent s dielectric constant. The potential energy of two charges separated by some distance is given by: PE = q 1 q 2 K εr 12 The ε in the denominator is the dielectric constant of the medium between the two charges. The larger the dielectric constant, the lower the interaction energy between ions. A solvent, which lowers the interaction between charges, is said to screen or shield the charges from each other. Organic chemists (somewhat arbitrarily) define a polar solvent as one with a dielectric constant greater than 15. Solvents lower the interaction energy between ions by aligning their dipoles around the ion in such a manner as to stabilize the ion as shown below. 21

22 Also, specific interactions between the solvent and the ions can have a profound effect on a solvent's ability to screen ions. In particular, the ability of a solvent to act as either a hydrogen bond donor or acceptor plays an extremely important role in solvent stabilization. Protic (i.e., hydrogen bond donor) solvents can form hydrogen bonds with Lewis bases because they preferentially stabilize anions. O O O O O O O O O O O O O O O O O O Real molecules are more complex than arrows representing dipoles. Consider DMSO (below), its dipole moment arises from a significant contribution of the right hand, charge-separated resonance structure. owever, the positive end is sterically congested relative to the oxygen. As a result, the oxygen can have a closer approach to a positively (or partially positively) charged species than the anion. Thus, positively (or partially positively) charged species will be preferentially stabilized by solvents in which the negative end of the dipole is exposed and the positive end is sterically crowded. r r' O S 3 C C 3 dimethylsulfoxide DMSO O S 3 C C 3 22

23 Effects of Solvation on Reaction Rates ΔG ΔG ΔG ΔG Reference Solvent Solvent that stabilizes T.S. Reference Solvent Solvent that stabilizes reactants ΔΔG is negative reaction is faster ΔΔG is positive reaction is slower 23

24 Classification of Solvents Dielectric Dipole Polarizability Solvent Constant Moment α (cm 3 x ) ε µ (D) Apolar Protic C 3 COO (C 3 ) 3 CO exanol Apolar Aprotic CCl exane Diethylether Tetrahydrofuran (TF) Polar Protic Water COO Methanol Ethanol Polar Aprotic C 3 COC 3 (Acetone) (C 3 ) 2 NCO (DMF) C 3 CN (Acetonitrile) DMSO