1. 4.15 Halogenation of Alkanes RH + X 2  RX + HX

2. explosive for F 2 exothermic for Cl 2 and Br 2 endothermic for I 2 Energetics

3. 4.16 Chlorination of Methane

4. carried out at high temperature (400 °C) CH 4 + Cl 2  CH 3 Cl + HCl CH 3 Cl + Cl 2  CH 2 Cl 2 + HCl CH 2 Cl 2 + Cl 2  CHCl 3 + HCl CHCl 3 + Cl 2  CCl 4 + HCl Chlorination of Methane

5. 4.17 Structure and Stability of Free Radicals

6. contain unpaired electrons...... O O:O:O:O: : Examples: O 2 NO N O :..:. Li 1s 2 2s 1 ClCl... :.. Free Radicals

7. Most free radicals in which carbon bears the unpaired electron are too unstable to be isolated. Alkyl radicals are classified as primary, secondary, or tertiary in the same way that carbocations are. C R R R. Alkyl Radicals

8. Figure 4.15 Structure of methyl radical. Methyl radical is planar, which suggests that carbon is sp 2 hybridized and that the unpaired electron is in a p orbital.

9. C R R R. The order of stability of free radicals is the same as for carbocations. Alkyl Radicals

10. C H H H. Methyl radical less stable than C H3CH3CH3CH3C H H. Ethyl radical (primary) Alkyl Radicals

11. C H H H. Methyl radical less stable than C H3CH3CH3CH3C H H. Ethyl radical (primary) C H3CH3CH3CH3C CH 3 H. Isopropyl radical (secondary) less stable than Alkyl Radicals

12. C H H H. Methyl radical less stable than C H3CH3CH3CH3C H H. Ethyl radical (primary) C H3CH3CH3CH3C CH 3 H. Isopropyl radical (secondary) less stable than C H3CH3CH3CH3C CH 3. tert-Butyl radical (tertiary) Alkyl Radicals

13. The order of stability of free radicals can be determined by measuring bond strengths. By "bond strength" we mean the energy required to break a covalent bond. A chemical bond can be broken in two different ways—heterolytically or homolytically. Alkyl Radicals

14. In a homolytic bond cleavage, the two electrons in the bond are divided equally between the two atoms. One electron goes with one atom, the second with the other atom. In a heterolytic cleavage, one atom retains both electrons. In a heterolytic cleavage, one atom retains both electrons. – + Homolytic Heterolytic

15. The species formed by a homolytic bond cleavage of a neutral molecule are free radicals. Therefore, measure energy cost of homolytic bond cleavage to gain information about stability of free radicals. The more stable the free-radical products, the weaker the bond, and the lower the bond-dissociation energy. Homolytic

16. Bond-dissociation energy measurements tell us that isopropyl radical is 13 kJ/mol more stable than propyl. 410 397 CH 3 CH 2 CH 3 CH 3 CH 2 CH 2 +H.. CH 3 CHCH 3 +H.. Measures of Free Radical Stability

17. Bond-dissociation energy measurements tell us that tert-butyl radical is 30 kJ/mol more stable than isobutyl. 410 (CH 3 ) 3 CH (CH 3 ) 2 CHCH 2 +H.. (CH 3 ) 3 C + H.. 380 Measures of Free Radical Stability

18. 4.18 Mechanism of Chlorination of Methane

19. The initiation step "gets the reaction going" by producing free radicals—chlorine atoms from chlorine molecules in this case. Initiation step is followed by propagation steps. Each propagation step consumes one free radical but generates another one. :.. Cl.. : Cl..:.... Cl..:. : Cl..... + Free-radical chain mechanism. Initiation step: Mechanism of Chlorination of Methane

20. Cl:..... H 3 C : H H : Cl:.... H3CH3CH3CH3C.+ + First propagation step: Mechanism of Chlorination of Methane

21. Cl:..... H 3 C : H H : Cl:.... H3CH3CH3CH3C.+ + Second propagation step: + + First propagation step: Cl:..... H3CH3CH3CH3C. :.. Cl.. : Cl..:.. H3CH3CH3CH3C.... : ClClClCl : Mechanism of Chlorination of Methane

22. Cl:..... H 3 C : H H : Cl:.... H3CH3CH3CH3C.+ + Second propagation step: + + ++ First propagation step: Cl:..... H3CH3CH3CH3C. :.. Cl.. : Cl..:.. H 3 C : H H : Cl:.... :.. Cl.. : Cl..:.. H3CH3CH3CH3C.... : ClClClCl : H3CH3CH3CH3C.... : ClClClCl : Mechanism of Chlorination of Methane

23. Almost all of the product is formed by repetitive cycles of the two propagation steps. Cl:..... H 3 C : H H : Cl:.... H3CH3CH3CH3C.+ + Second propagation step: + + ++ First propagation step: Cl:..... H3CH3CH3CH3C. :.. Cl.. : Cl..:.. H 3 C : H H : Cl:.... :.. Cl.. : Cl..:.. H3CH3CH3CH3C.... : ClClClCl : H3CH3CH3CH3C.... : ClClClCl :

24. stop chain reaction by consuming free radicals hardly any product is formed by termination step because concentration of free radicals at any instant is extremely low hardly any product is formed by termination step because concentration of free radicals at any instant is extremely low + H3CH3CH3CH3C. Cl:..... H3CH3CH3CH3C.... : ClClClCl : Termination Steps

25. 4.19 Halogenation of Higher Alkanes

26. CH 3 CH 3 + Cl 2 CH 3 CH 2 Cl + HCl 420°C (78%) Cl + Cl 2 + Cl 2 + HCl + HCl h (73%) can be used to prepare alkyl chlorides from alkanes in which all of the hydrogens are equivalent to one another Chlorination of Alkanes

27. Major limitation: Chlorination gives every possible monochloride derived from original carbon skeleton. Not much difference in reactivity of different hydrogens in molecule. Chlorination of Alkanes

28. CH 3 CH 2 CH 2 CH 3 Cl 2 h Chlorination of butane gives a mixture of 1-chlorobutane and 2-chlorobutane. CH 3 CH 2 CH 2 CH 2 Cl CH 3 CHCH 2 CH 3 Cl (28%) (72%) Example

29. Percentage of product that results from substitution of indicated hydrogen if every collision with chlorine atoms is productive 10% 10% 10% 10% 10% 10% 10% 10% 10% 10% 10%

30. Percentage of product that actually results from replacement of indicated hydrogen 4.6% 4.6% 4.6% 18% 18% 18% 10% 4.6% 4.6% 4.6% 18%

31. divide by 4.6 4.64.6 = 1 184.6 = 3.9 A secondary hydrogen is abstracted 3.9 times faster than a primary hydrogen by a chlorine atom. 4.6% 18% Relative rates of hydrogen atom abstraction

32. Similarly, chlorination of 2-methylbutane gives a mixture of isobutyl chloride and tert-butyl chloride H CH 3 CCH 3 CH 3 CH 3 CCH 2 Cl CH 3 H Cl CH 3 CCH 3 CH 3 h Cl 2 (63%) (37%)

33. Percentage of product that results from replacement of indicated hydrogen 7.0% 37%

34. 37 7.0 divide by 7 7 = 1 7 = 5.3 A tertiary hydrogen is abstracted 5.3 times faster than a primary hydrogen by a chlorine atom. Relative rates of hydrogen atom abstraction

35. R 3 CH>R 2 CH 2 >RCH 3 chlorination: 5 41 bromination:1640821 Chlorination of an alkane gives a mixture of every possible isomer having the same skeleton as the starting alkane. Useful for synthesis only when all hydrogens in a molecule are equivalent. Bromination is highly regioselective for substitution of tertiary hydrogens. Major synthetic application is in synthesis of tertiary alkyl bromides. Selectivity of free-radical halogenation

36. Chlorination is useful for synthesis only when all of the hydrogens in a molecule are equivalent. (64%) Cl 2 h Synthetic application of chlorination of an alkane Cl

37. Bromination is highly selective for substitution of tertiary hydrogens. Major synthetic application is in synthesis of tertiary alkyl bromides. Major synthetic application is in synthesis of tertiary alkyl bromides. (76%) Br 2 hH CH 3 CCH 2 CH 2 CH 3 CH 3 Br CH 3 CCH 2 CH 2 CH 3 CH 3 Synthetic application of bromination of an alkane