1. Alkyl Halides

2. Boiling Points The size of –Br and –CH 3 about the same but bromo compounds boil higher due to greater polarizability; more dispersion forces.

3. Boiling Points Branching in an alkane decreases the boiling point. Likewise for haloalkanes.

4. Boiling Points Fluoroalkanes and alkanes of same MW have about the same BP. In both cases the electrons are tightly held, not very polarizable.

5. Reaction of elemental halogen and alkanes yielding haloalkanes H 3 C – H + X – X  H 3 C – X + H - X Reaction Characteristics Requires heat or light to initiate. Fluorine is explosive. Reactivity decreases fluorine to chlorine to bromine to iodine which does not react. Bonds being broken.Bonds being made. H 3 C - H439 kJ Cl - Cl247H 3 C - Cl351H – Cl431 Br - Br192H 3 C - Br301H - Br368 Bromination:  H = 439 +192 – 301 – 368 = -38 kJ less exothermic Bond Dissociation Energies. Energy Required to Break Bond. Chlorination:  H = 439 + 247 -351 – 431 = -96 kJ exothermic

6. Regioselectivity Starting Point in Analysis: Random Substitution. Assume that all hydrogens are equally likely to be replaced by X. There are 8 H in the molecule. Equally likely to be replaced if random…. Random substitution 2/8 = 25%6/8 = 75%

7. Regioselectivity SecondaryPrimary Random substitution2/8 = 25%6/8 = 75% X = Cl, experimental57%43% X = Br, experimental92%8% Now include experimental results for X = Cl, Br. Replacement of secondary H is favored over primary H. Generally order of reactivity is tertiary > secondary > primary > methyl. Bromination displays greater selectivity than does chlorination.

8. Reactivities of Hydrogens For chlorination tertiary:secondary:primary = 5:4:1 For bromination tertiary:secondary:primary = 1600:80:1 Quantitative Analysis: Deviations from random replacement quantified by assigning a Reactivity to each kind of Hydrogen (primary, secondary, or primary). Substitution at a carbon proportional to (# hydrogens at C) x (Reactivity of the H’s)

9. Predict product mixture For chlorination of 2,2,4,4-tetramethylpentane predict the product mixture. Cl 2 UV light + 18 x 1 = 182 x 4 = 8 Number of hydrogens leading to this product Reactivity of those hydrogens Expected ratio 9 : 4 Expected fraction 9/13 4/13

10. The life of the chain depends on the ongoing presence of the highly reactive Cl atoms and alkyl radicals. Eliminating these species ends chains. 4. 2 Cl. Cl – Cl 5. 2 R. R – R 6. R. + Cl. R - Cl Heat or light Chain Reaction Mechanism 1. Cl – Cl 2 Cl. 2. R – H + Cl. R. + H - Cl 3. R. + Cl – Cl R – Cl + Cl. Repeat 2, 3, 2, 3,…. Chain steps. Chlorine atom. Highly reactive, only seven electrons in valence shell Weak Cl-Cl bond may be broken by heat or light. Hydrogen to be abstracted. Trade bonds: R-H for H-Cl Alkyl radical, only seven electrons around the C, highly reactive alkyl radical. Trade bonds: weak Cl-Cl for a stronger C-Cl Regenerates the Cl atom used in step 2 Termination steps. Initiation

11. Energetics of the Chain Steps Chlorination of ethane Step 2, abstraction of the H, controls the regioselectivity of the reaction. Isothermic or slightly exothermic for Cl; endothermic for Br. Step 3, attachment of the halogen, controls the stereochemistry, which side the halogen attaches. Exothermic

12. Step 2 and Bond Dissociation Energies, breaking bonds… More highly substituted radicals are easier to make. This gives rise to regioselectivity = non-random replacement.

13. Now bromination. Compare chlorination and bromination of 2-methyl propane. Bromination is more Regioselective. Examine Step 2 only for regioselectivity. BDE, reflecting different radical stabilities Slightly exothermic Endothermic H-Cl is a more stable bond than H-Br. Step 2 is exothermic for chlorination. Endothermic for Br First chlorination. Two kinds of H.

14. Step 2 Transition State Energetics, Cl vs Br chlorination Exothermic, tertiary radical more stable Early transition states, little difference in energies of activation, rates of abstraction and regioselectivity bromination Endothermic, but tertiary radical still more stable by same amount. Late transition states, larger difference in energies of activation, rates of abstraction and regioselectivity. R … H ………. Cl R ………. H.. Br Halogenation of 2-methylpropane yields two differerent radicals, primary and tertiary.

15. Now Step 3: Stereochemistry Alkyl radical, sp 2 hybridization, one electron in the p orbital. Mirror objects. If a chiral carbon has been produced we get both configurations. Step 3 has these characteristics Determines stereochemistry Is exothermic Is fast, not rate determining

16. Simple Example: monochlorination of 2-methylbutane Observations: Optically inactive molecule (can show reflection plane) and products will be optically inactive. First, look carefully at molecule Next, organize approach, label the carbons. a a’ b c From a and a’. From b d From c From d Four optically inactive fractions if distilled. Chiral carbon

17. Example a a’ bb’ c a a’ b Diastereomers both sides used. b’ Diastereomers both sides used. meso b b’enantiomers First get stereochemical relationships between carbons. enantiomers Can get relative amounts made of each using reactivities of 1:4:5 3 x 1 1 x 5 / 2 2 x 4 / 2 Distillation would yield 5 optically inactive fractions. c c

18. Allylic Systems Allylic C – H bonds, weak easily broken. Removal of H produces the allylic radical. Vinyl C – H bonds, difficult to break. CH 3 CH 2 – H: 411 kJ/mol (101 kCal/mol) Now the allylic radical…

19. Resonance in Allylic Radical Resonance provides the stabilization. The pi system is delocalized. Odd electron located on alternate carbons, C1 and C3, not C2.

20. Allylic Substitution Allylic C-H, 372 kJ Br-Br 192 kJ Allylic C-Br 247 kJ H-Br 368 kJ  H = -247 - 386 – (-372 – 192) = -51kJ

21. Mechanism initiation Chain steps Weakest C-H bond selected, highest reactivity 372 kJ 368 kJ Termination: usual combining of radicals

22. We have a Problem: seem to have two possible reactions for an alkene with Br 2. 1. Addition to the double bond yielding a dibromide. 2. Substitution at allylic position. And/or

23. Competing Reactions: Addition vs Substitution Alkene reacts preferentially with Br atoms if present. Favored by high Br atom concentrations. High temperature favors Br 2  2Br and thus substitution. Alkene reacts directly with Br 2 Happens at low Br atom concentrations. Low temperature keeps Br concentration low and thus favors addition. Br 2 Br., not Br 2 Addition Substitution Produced from Br 2 at high temperature

24. Convenient allylic bromination For allylic substitution to occur we need both bromine atoms and Br 2 Br: R-H + Br.  R. + H-Br Br 2 : R. + Br 2  R-Br + Br. Both Br and Br 2 can be supplied from Br 2 at high temperature or from NBS (N-bromosuccinimide). NBS

25. Allylic Rearrangement Expect bromination of but-2-ene to yield 1-bromo but-2-ene by replacing allylic hydrogen. But get rearranged product as well…. Major product, more stable with subsituted double bond.

26. Mechanism of Rearrangement Two different sites of reactivity More highly substituted alkene (more stable, recall hydrogenation data) is the major product

27. The actual radical is a blending of these two structures. Secondary radicals are more stable than primary. This predicts most of the radical character at the secondary carbon, favoring this structure. But… But more highly substituted alkenes are more stable. This predicts most radical character at primary carbon favoring this structure. This appears to be the dominant factor leading to dominant product. But also…. An interesting competition is occurring. Consider the allylic radical…

28. Results of Calculation of Spin Densities in radical formed from 1-butene Blue is unpaired electron density. More at primary than secondary.

29. Anti Markovnikov addition of HBr Only with HBr, not HCl, HI

30. Mechanism of anti-Markovnikov Radical Addition of HBr Chain Steps Contrast radical and ionic addition of HBr Radical Ionic Not a Chain Process Common Concept: More stable intermediate formed, secondary radical or secondary carbocation

31. Autoxidation, reaction of allylic sites with oxygen. Double allylic Lower energy since double bonds are in conjugation with each other

32. Cont’d

33. 4. A mixture of 1.6 g of methane and 1.5 g of ethane are chlorinated for a short time. The moles of methyl chloride produced is equal to the number of moles of ethyl chloride. What is the reactivity of the hydrogens in ethane relative to those in methane? Show your work. Sample Problem Solution: Recall: The amount of product is proportional to the number of hydrogens that can produce it multiplied by their reactivity. Number of hydrogens leading to methyl chloride = 1.6g * (1 mol/16 g) * (4 mol H/1 mol methane) = 0.40 mol H Number of hydrogens leading to ethyl chloride = 1.5 g * (1 mol/30 g) * (6 mol H/ 1 mol ethane) = 0.30 mol H 0.40 mol H * R methane = 0.30 mol H * R ethane R ethane /R methane = 0.4/0.3 = 1.3

34. How do we form the orbitals of the pi system… First count up how many p orbitals contribute to the pi system. We will get the same number of pi molecular orbitals. Three overlapping p orbitals. We will get three molecular orbitals.

35. If atomic orbitals overlap with each other they are bonding, nonbonding or antibonding Anti-bonding, destabilizing. Higher Energy If atoms are directly attached to each other the interactions is strongly bonding or antibonding. Bonding, stabilizing the system. Lower energy. But now a particular, simple case: distant atomic orbitals, on atoms not directly attached to each other. Their interaction is weak and does not affect the energy of the system. Non bonding

36. Molecular orbitals are combinations of atomic orbitals. They may be bonding, antibonding or nonbonding molecular orbitals depending on how the atomic orbitals in them interact. All bonding interactions. Only one weak, antibonding (non- bonding) interaction. Two antibonding interactions. Example: Allylic radical

37. Allylic Radical: Molecular Orbital vs Resonance Note that the odd electron is located on the terminal carbons. Molecular Orbital. We have three pi electrons (two in the pi bond and the unpaired electron). Put them into the molecular orbitals. Resonance Result Again the odd, unpaired electron is only on the terminal carbon atoms.

38. But how do we construct the molecular orbitals of the pi system? How do we know what the molecular orbitals look like? Key Ideas: For our linear pi systems different molecular orbitals are formed by introducing additional antibonding interactions. Lowest energy orbital has no antibonding, next higher has one, etc. 0 antibonding interactions 1 weak antibonding Interaction, “non-bonding” 2 antibonding interactions Antibonding interactions are symmetrically placed. This would be wrong.

39. Another example: hexa-1,3,5-triene Three pi bonds, six pi electrons. Each atom is sp 2 hybridized. Have to form bonding and antibonding combinations of the atomic orbitals to get the pi molecular orbitals. Expect six molecular orbitals. # molecular orbitals = # atomic orbitals Start with all the orbitals bonding and create additional orbitals. The number of antibonding interactions increases as we generate a new higher energy molecular orbital.