1. Chapter 3: Alkenes and Alkynes © R. Spinney 2013

2. General Properties Alkenes contain double bonds and alkynes triple bonds. Both classes of compounds are hydrocarbons, containing only C and H atoms. – a double bond consists of 1  and 1  bond, – a triple bond consists of 1  and 2  bonds. They are also “unsaturated” as they contain  bonds and are also known as olefins.

3. General Properties (cont’d) They contain less H atoms that the corresponding alkane, generic chemical formulas are; C n H 2n+2 alkanes C n H 2n alkenes C n H 2n-2 alkynes Every  bond results in the loss of a pair of H atoms.

4. General Properties (cont’d) With multiple double (or triple) bonds three possible arrangements arise: cumulated, conjugated or isolated (non-conjugated). Conjugated are especially important as the  bonds can interact. CumulatedConjugatedIsolated C=C=CC=C-C=CC=C-C-C=C C=C=C=CC≡C-C≡CC≡C-C-C-C≡C

5. Bonding in Alkenes Alkenes are sp 2 hybridized Trigonal planar – bond angle ~120° 3  and 1  bond (or 2 single and 1 double) C=C double bond ~ 1.34 Å The  bond lock the geometry to planar

6. Cis – Trans Isomerism The double bond in an alkene is rigid, that is it will not rotate freely. Therefore substituents on the carbon atoms will produce geometric isomers the same as on a cycloalkane ring.

7. Cis – Trans Isomerism (cont’d) If the two non-hydrogen atoms or groups are on the same side of the double bond it is a cis- arrangement, on opposite sides a trans- arrangement, i.e. Note that they have different physical properties, this is because their dipole moments are different.

8. Cis – Trans Isomerism (cont’d) Under normal conditions the two isomers will not interconvert since you would have to break the  bond, however using light or heat it is possible to interconvert the two isomers. i.e.

9. Chemical Reactivity addition reactions The chemical reactivity of alkenes arise from the  electrons. The  bond is weaker than the  bond so these electron react first. The reagent will add across the double bond so these are addition reactions.

10. Electrophilic Addition Mechanism carbocation The basic mechanism is the same for all reagents, a two step reaction where an electrophile (E) add to the  bond in the first step creating a carbocation intermediate. A nucleophile (Nu) then adds to the carbocation in the second step.

11. Carbocations Are electron deficient and have an empty p orbital (sp 2 hybridized). Not all carbocations are equally stable so there are predictable patterns for which ones will form. Note: this is the same order of stability as carbon radicals.

12. Carbocations (cont’d) The order of carbocation stability arise from three sources. 1)Inductive electron donation 1)Inductive electron donation: the electrons in C-C  bonds will be pulled closer to the C + helping to minimize the charge. Note: this does not work for C-H bonds. 2)Hyperconjugation 2)Hyperconjugation: this is a orbital interaction between adjacent C-H bonds that can overlap the empty p orbital of the C +, this again helps to minimize the charge on the C +.

13. Carbocations (cont’d) 3)Resonance 3)Resonance: a carbocation immediately adjacent to a  system (double bond, triple bond or aromatic ring) can be stabilized by resonance. This lowers the energy by spreading the charge over more atoms, i.e.

14. Carbocations (cont’d) One final complexity for carbocations: rearrangement. Carbocations are susceptible to 1,2-hydride shifts, where an adjacent H atom (and its bonding electrons) shift to the C + to produce a more stable carbocation, i.e. This is also possible for methyl groups. These shifts are common when a tertiary, allylic or benzylic carbocation is produced.

15. Halogenation Addition of Cl 2 or Br 2 across the double bond to product 1,2-dihalide alkane. Halogen is usually dissolved in chloroform or carbon tetrachloride. Reaction is rapid at room temperatures Addition of bromine is a common chemical test for the presence of double bonds as the red colored bromine solution turns colorless when it reacts with an alkene Reaction relies on the polarizability of the halogen bonds

16. Halogenation (cont’d) In the first step the  electrons act as a nucleophile attacking the bromine displacing a bromide ion and forming the cyclic bromonium cation intermediate. In the second step the nucleophilic bromide ion attacks the side of the bromonium ion away from the bromine atom opening the highly strain cyclic structure and producing the 1,2 dibromo hydrocarbon. Note this is a trans addition since the two Br atoms add to opposite side of the double bond.

17. Symmetry and Addition Reactions regioisomer The halogenation of ethene is a very symmetry reaction as both the reagent (Br 2 ) and the substrate (ethene) are symmetric. In this situation there can only be a single product. What about asymmetric reactions? Two products are possible now, which regioisomer will form?

18. Markovnikov’s Rule Fortunately there is a simple rule of thumb to predict which product will form: Markovnikov’s rule, which states that when an unsymmetrical reagent adds to a double bond the electrophilic part of the reagent adds to the carbon with the most hydrogen atoms on it.

19. Markovnikov’s Rule (cont’d) Why? This mode of addition will always produce the most stable carbocation intermediate, i.e. consider the addition of H + to propene…

20. Hydration Addition of water across the double bond Product is an alcohol Requires an acid as a catalyst as water is not acidic enough to produce the electrophile

21. Hydration (cont’d) First step is the protonation of the alkene in a Markovnikov orientation to generate the most stable carbocation. The second step is the attack of the nucleophilic water molecule on the carbocation Finally an acid/base reaction deprotonates the alkyloxonium ion to form the alcohol.

22. Hydrohalogenation Addition of HX across a double bond to produce a Markovnikov halide alkane Reactivity: HI > HBr > HCl > HF (parallels acidity) HBr needs to be used in the dark and under an inert atmosphere to prevent a free radical addition process that produces the anti- Markovnikov product. Note: this same mechanism applies to other acids such as H 2 SO 4

23. Hydrohalogenation (cont’d) First step in the Markovnikov addition of the electrophilic acidic proton to produce the most stable carbocation intermediate. The second step is the nucleophilic attack of the halide anion on the carbocation intermediate generating the alkyl halide product.

24. Thermodynamics & Equilibrium For a generic chemical reaction: aA + bB ⇌ cC + dD We can define the where the equilibrium lies by:

25. Thermodynamics & Equilibrium The size of the equilibrium constant, K eq tells us which direction of the reaction is favored. Greater than 1, the products and the reaction proceeds from left to right. The larger K eq is the further towards the products the equilibrium lies.

26. Thermodynamics & Equilibrium The equilibrium can be shifted, this is Le Châtelier’s Principle. Adding a reactant or removing a product will help “drive” the reaction forward, i.e. Adding ethanol, or removing water will drive this reaction to the right.

27. Thermodynamics & Equilibrium So what determines if a reaction proceeds to the right or left? exothermic Spontaneous chemical processes release heat as a product. These reactions are exothermic (the enthalpy,  H < 0). Endothermic Endothermic (  H > 0) processes can be made to work, but you have to provide heat to “drive” the reaction.

28. Thermodynamics & Equilibrium Transition State Theory The equilibrium constant and enthalpy tell us which direction a reaction proceeds in, and whether it is spontaneous or not, but neither tells us how fast it will go. For that we need to use Transition State Theory. This theory states that the reactants will proceed through a high energy “Transition State” (TS) and then proceed to the products.

29. Thermodynamics & Equilibrium The TS is the point where the molecules are colliding and bonds are being broken and reformed. It is commonly represented in a “reaction coordinate diagram”.

30. Thermodynamics & Equilibrium In contrast to an intermediate, the TS has no real life time. You cannot isolate the molecules in the TS. Currently the TS state is studied primarily by computational chemical models, although new laser spectroscopy techniques have a time frame short enough to provide information on the TS.

31. Thermodynamics & Equilibrium So how does this work? We can explain the reaction of propene with a proton based on a reaction coordinate diagram. The reason the 1° carbocation is less stable than the 2° is because it has a TS higher in energy and so is harder to form.

32. Thermodynamics & Equilibrium The reaction of HBr with ethene can be explained in a similar fashion. TS 1 is high E as we are in the process of breaking the H-Br bond but have not yet formed the new C-H bond. The intermediate carbocation is lower in E as the new C-H bond has formed. TS2 is higher in E as the C-Br bond is in the process of forming. Finally the product has the lowest E so the reaction is exothermic and spontaneous providing there is sufficient E to form TS1.

33. Thermodynamics & Equilibrium Catalysts act by providing an alternate pathway for the reaction which has a lower activation barrier. This applies to enzymatic and non- enzymatic catalysts.

34. Thermodynamics & Equilibrium Thermal effects change the energy of the reactants. If the temperature is raised the reactants have a higher starting energy and hence have a low barrier to cross, the E a. Note this has no effect of the energy at the TS or the products.