Alkyl Halides

1. Alkyl Halides

2. Boiling Points The size of –Br and –CH3 about the same but bromo compounds boil higher due to greater polarizability; more dispersion forces.

3. Reaction of elemental halogen and alkanes yielding haloalkanes H3C – H + X – X  H3C – 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. Bond Dissociation Energies. Energy Required to Break Bond. Chlorination: DH = 439 + 247 -351 – 431 = -96 kJ exothermic Bromination: DH = 439 +192 – 301 – 368 = -38 kJ less exothermic

4. 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….

5. Regioselectivity 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.

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

7. Predict product mixture For chlorination of 2,2,4,4-tetramethylpentane predict the product mixture. Cl2 + UV light 18 x 1 = 18 2 x 4 = 8 Number of hydrogens leading to this product Expected ratio 9 : 4 Expected fraction 9/13 4/13 Reactivity of those hydrogens

8. Chain Reaction Mechanism 1. Cl – Cl 2 Cl. 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 Weak Cl-Cl bond may be broken by heat or light. Initiation Heat or light Chlorine atom. Highly reactive, only seven electrons in valence shell 2. R – H + Cl. R. + H - Cl Chain steps. Alkyl radical, only seven electrons around the C, highly reactive alkyl radical. 3. R. + Cl – Cl R – Cl + Cl. Repeat 2, 3, 2, 3,…. Hydrogen to be abstracted. Trade bonds: R-H for H-Cl Regenerates the Cl atom used in step 2 Termination steps. Trade bonds: weak Cl-Cl for a stronger C-Cl

9. Energetics of the Chain Steps Chlorination of ethane Step 2, abstraction of the H, controls the regioselectivityof 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

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

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

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

13. Now Step 3: Stereochemistry Mirror objects. If a chiral carbon has been produced we get both configurations. Alkyl radical, sp2 hybridization, one electron in the p orbital. • Step 3 has these characteristics • Determines stereochemistry • Is exothermic • Is fast, not rate determining

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

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

16. Allylic Systems CH3CH2 – H: 411 kJ/mol (101 kCal/mol) Vinyl C – H bonds, difficult to break. Allylic C – H bonds, weak easily broken. Removal of H produces the allylic radical. Now the allylic radical…

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

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

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

20. We have a Problem: seem to have two possible reactions for an alkene with Br2. 1. Addition to the double bond yielding a dibromide. And/or 2. Substitution at allylic position.

21. Competing Reactions: Addition vs Substitution Substitution Addition Produced from Br2 at high temperature Br2 Br., not Br2 Alkene reacts directly with Br2 Happens at low Br atom concentrations. Low temperature keeps Br concentration low and thus favors addition. Alkene reacts preferentially with Br atoms if present. Favored by high Br atom concentrations. High temperature favors Br2 2Br and thus substitution.

22. Convenient allylic bromination For allylic substitution to occur we need both bromine atoms and Br2 Br: R-H + Br. R. + H-Br Br2: R. + Br2  R-Br + Br. Both Br and Br2 can be supplied from Br2 at high temperature or from NBS (N-bromosuccinimide). NBS

23. 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.

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

25. An interesting competition is occurring. Consider the allylic radical… 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….

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

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

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

29. Sample Problem 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. 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 * Rmethane = 0.30 mol H * Rethane Rethane/Rmethane = 0.4/0.3 = 1.3

30. 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.

31. If atomic orbitals overlap with each other they are bonding, nonbonding or antibonding Anti-bonding, destabilizing. Higher 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 If atoms are directly attached to each other the interactions is strongly bonding or antibonding. Bonding, stabilizing the system. Lower energy.

32. 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. Example: Allylic radical Two antibonding interactions. Only one weak, antibonding (non-bonding) interaction. All bonding interactions.

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

34. 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. 2 antibonding interactions 1 weak antibonding Interaction, “non-bonding” Antibonding interactions are symmetrically placed. 0 antibonding interactions This would be wrong.

35. Another example: hexa-1,3,5-triene Three pi bonds, six pi electrons. Each atom is sp2 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.