From alkanes Radical halogenation

Transcription

1 From alkanes Radical halogenation or h! Alkyl alides from Alkanes 3 form, where comes from 2 form, where comes from methane and comes from 2 break bond break bond what is the role of heating or light? how can you break a strong bond of an alkane? * multi-step, with multiple intermediates

2 Radical hain Mechanism methane does not lose via cleavage of must be pulled off in a reaction heat or light breaks into radicals ( ) pulls off to form and reacts with to form 2 = 300 or h! 2 4 =

3 Mechanism: Radical Substitution Form the radicals (initiation) 2 = 300 or h! 2 4 = Atom exchange Neutral reacts with a radical to form another radical and a neutral (propagation) Radical coupling Two radicals react together to form a neutral, but not a radical (termination) Radical chain reaction - radical substitution

4 to Exchange Every in the alkane is replaced by 2 h! % % % %

5 2 h! , 4 different types 6 methyl s 1 3 methyl s different from others 1 2 methylene s 2 1 methine 3 All 12 are replaced 4 types of 4 different products Every is Replaced

6 Relative Rate: to reacts with 1 relative rate of reaction = 1 reacts with 2 relative rate of reaction = 3.9 reacts with 3 relative rate of reaction = 5.2

7 Determining Relative % 6 1 removed by at a rate of 1 a removed by at a rate of 5.7 c d b 3 1 removed by at a rate of 1 2 h! 2 2 removed by at a rate of x 1 = 6 3 x 1 = 3 2 x 3.9 = x 5.7 = Take ratio for all different 6/22.5 = % a 3/22.5 = % b 7.8/22.5 = % c 5.7/22.5 = % d Assume 100% i.e., these are the only products

8 2-Methylbutane Gives 4 Different hloride: % products % 22% 134 a 132 c 132 d 2 h! b 2 h! % 16% 136 actual 6/22.5 = % a (133) 3/22.5 = % b (136) 7.8/22.5 = % c (134) 5.7/22.5 = % d (135) predicted

9 Free Radical Reactions of Alkyl alides There is a second type of hydrogen atom in 146, b. If a chlorine radical reacts with b by an atom transfer reaction, the products are and radical 148. The reaction of 148 with additional diatomic chlorine leads to a different product, h! 2 a a b b b a If all the different hydrogen atoms in 146 react with chlorine, two products are formed, 149 and 93.

10 Free Radical Reactions of Alkyl alides From various experiments it is known that the relative rates are 1 = 1, 2 = 3.9, and 3 = 5.2. If all hydrogen atoms react equally (have the same rate constant) then the percentage of each product is given by: Relative % = (# reacting s/total # s) x 100. For 146 there are a total of ten hydrogen atoms available to react with so the % of 149 should be = 9/10 = 0.9 and the % of 93 should be 1/10 = = 0.1. owever, if the rate for reaction of the primary versus the tertiary hydrogen atom is factored in, then b reacts 5.2 times faster than a (tertiary versus primary). The rate of reaction must be factored in with the number of atoms reacting and the relative rate is: Relative % = [(# 1 s/total # s) x 1] [(# 2 s/total # s) x 3.9] [(# 3 s/total # s) x 5.2] Since there are only primary and tertiary hydrogen atoms in 146, the relative % of 149 is 9/10 x 1 and the relative % of 93 is 1/10 x 5.2. The % of 149 is therefore = 0.9/ = 0.9/1.42 = 0.63x100 = 63%. The percentage of 93 is = 0.52/ = 0.52/1.42 = 0.37x100 = 37%. The mixture of two chlorinated products is predicted to be a ratio of 63:37, 63% of 149 and 37% of 93. h! 2 10 b a b b a a

11 2-methylpropane: bromination a Br Br a b % b Br 99.5% a a b b Why such a difference? 63.4% 36.6% Br removes at a different rate than

12 Relative Rate: to Br : 1:3.9:5.2 Br Br Br reacts with 1 relative rate of reaction = 1 Br Br Br reacts with 2 relative rate of reaction = 82 Br Br Br reacts with 3 relative rate of reaction = 1640 Br reacts with times faster than with 1

13 Free Radical Reactions of Alkyl alides Photolysis of Br 2 will generate Br, and the products are alky bromides 150 and 64. The experimentally determined rates are 1 = 1, 2 = 82, and 3 = This significant difference in rate when compared to the chlorine radical means that Br reacts with 1640 tertiary hydrogen atoms to every one primary hydrogen atoms. Knowledge of the rate of reaction of 146 with Br leads to very different relative percentages of primary bromide 150 and tertiary bromide 64. The relative percentages are 9/10 x 1 and 1/10 x Therefore, the relative % of 150 is = 0.9/ = 0.9/ = and that of 64 is 16440/ = 1640/ = or 0.55% of 150 and 99.45% of a Br Br a b Br Br b

14 Transition States: hlorination versus Bromination Br 3 easier to break than 2 and easier than 1 is weaker 3 is more stable than 2 which is more stable than 1 early transition state is influenced more by starting materials so strength of is more important smaller difference between 1 and 3, so small difference in rates 1:3.9:5.2 for late transition state is influenced more by stability of products so stability of radicals is more important larger difference between 1 and 3, so large difference in rates 1:82:1640 for Br

15 Reactions: Alkane alogenation 2, h! 15 2, h! 2, h! Br 2, h! Br

16 Allylic alogenation Me Me Br Me Me 143 Br Me allylic - more reactive because a resonance 145 stabilized allylic radical is formed because of resonance, there are 2 reactive sites and therefore 2 different products rate of removal for allylic is much faster than for other types of, including 3 2, h! Me Me Me 144 Me Br Me (MINOR) (MAJOR)

17 Free Radical Reactions of Alkyl alides N-chlorosuccinimide (158; abbreviated NS) and N-bromosuccinimide (159; abbreviated NBS). Both of these reagents are derivatives of succinimide (160), which is the imide derivative of succinic acid (1,4-butanedioic acid; see chapter 16, section 16.5). Both reagents react with light to release diatomic chlorine or bromine, which subsequently react with the light or another radical to produce chlorine or bromine radicals. Therefore, NS is a chlorine surrogate and NBS is a bromine surrogate. yclohexene (161), for example, reacts with NBS in the presence of light to give 3- bromocyclohexene (162) in 45% yield by the radical substitution mechanism. 17 O O O N N Br N 158 O 159 O 160 O NBS, h! Br 45%

18 hapter 12. Based-Induced 18 Elimination Reactions A typical Brønsted-Lowry acid base reaction is shown in reaction (1), where the base donates electrons to the proton, which is directly connected to the leaving group X. After the X bond is broken, X is the conjugate base. The reaction of with NaO is typical of reaction type (1). If the acidic proton and the leaving group are separated by two carbon atoms, as shown in (2), a related acid-base reaction is possible that gives the conjugate acid (BASE:) along with the conjugate base, X. owever, removal of the acidic proton leads to transfer of electron density through the intervening carbon groups to form an alkene (=) unit as the X group leaves. BASE: X BASE: :X (1) BASE: X BASE: :X (2)

19 To begin, you should know: The structure and nomenclature of alkyl halides. (chapter 4, section 4.6 and chapter 5, section 5.6.A) The structure and nomenclature of alkenes and alkynes. (chapter 5, sections 5.1, 5.2) Understand the fundamental Brønsted-Lowry acid base reaction of a base with an acid X. (chapter 2, sections 2.1, 2.2 and chapter 6, section 6.1) Define and recognize a base. (chapter 6, section 6.4) Understand differences in base strength. (chapter 6, sections 6.3, 6.4) Understand the role of bond polarization and dipole to generate a d X d species. (chapter 3, section 3.7) An alkyl halide has a polarized X bond, where X is negative and is positive. (chapter 3, section 3.7) Bimolecular reactions are favored when there is no water present in the reaction. (chapter 11, section 11.2) Ionization of an alkyl halide to a carbocation is competitive when water is present in the reaction. (chapter 11, sections 11.4, 11.6) Tertiary alkyl halides do not undergo S N 2 reactions due to steric hindrance in the transition state. (chapter 11, sections 11.1, 11.2) Understand that a nucleophile can donate two electrons to a positive carbon or to a carbon with a positive dipole, to form a new bond to that carbon. (chapter 6, section 6.7 and chapter 11, section 11.3) Recognize intermediates. (chapter 7, section 7.4) Understand and identify intermediates and transition states in a reaction-energy curve. (chapter 7, sections 7.3, 7.4) 19

20 To begin, you should know: Be able to recognize a transition state. (chapter 7, section 7.3) Understand the fundamentals of kinetics. (chapter 7, section 7.11) Determine the relative stability of carbocation intermediates. (chapter 10, section 10.2 and chapter 11, section 11.4.B) Understand the basics of conformation applied to both acyclic molecules and cyclic molecules of ring sizes of 3-6. (chapter 8, sections 8.1, 8.4, 8.5, 8.6) Recognize a stereogenic center and assign the absolute configuration. (chapter 9, sections 9.1, 9.3) Recognize and name diastereomers. (chapter 9, sections 9.5, 9.6) 20

21 When completed, you should know: 21 When alkyl halides are heated with a base in protic solvents, an E2 reaction occurs to give an alkene. The E2 reaction follows second order kinetics, requires a base and an acidic β- hydrogen on the halide, does not have an intermediate but rather a transition state, and gives the more substituted alkene as the major product. The E2 reaction is diastereospecific and if an enantiopure halide is used, a single stereoisomeric alkene (E or Z) will be formed. Prediction of the product in an elimination reaction for a cyclic halide requires a detailed analysis of the conformations for that ring system. Ionization to a carbocation, in the presence of a base, can give unimolecular elimination to the more substituted alkenes, the E1 reaction. When a base is tethered to a leaving group, removal of a β- hydrogen occurs via an intramolecular process and a syn rotamer to give the less substituted alkene as the major product. Thermal elimination of tetraalkylammonium hydroxides, without solvent, usually proceeds by intramolecular elimination. 1,3-elimination is known, and a typical example is decarboxylation of 1,3-diacids or β- keto acids. When vinyl halides are heated with a strong base, an E2 reaction occurs to give an alkyne.

22 Alkyl alides with a Base 22 When 2-bromo-2-methylpropane (1A) is refluxed with sodium ethoxide (the conjugate base of ethanol) in ethanol, the isolated product is an alkene, 2-methyl-2-propene (2) along with sodium bromide and ethanol. The Williamson ether synthesis is the reaction of an alkoxide with an alkyl halide to give an ether. The transition state of an S N 2 reaction is so high for a tertiary halide that the reaction does not occur. Therefore, it is not a surprise that 1A does not give an ether, but formation of an alkene indicates that this is a different reaction. The keys to understanding this reaction are the observation that ethoxide is both a nucleophile and a base, and that an alkene and ethanol (the conjugate acid of ethoxide) are the products. Br NaOEt, EtO reflux 1A 2 EtO NaBr

23 Alkyl alides with a Base 23 The conversion of the tertiary bromide 1A to alkene 2 is an acid-base reaction. The alkyl halide loses a hydrogen atom as well as a bromine to form the alkene product, so the overall transformation is called an elimination reaction (the elements of and Br are eliminated to form a π-bond). Ethoxide is a nucleophile as well as a base, but why did the reaction with 1A give an alkene product? The activation barrier for a S N 2 reaction is too high due to steric hindrance in the transition state. The polarization of the bromine atom on the α-carbon is δ, and the induced dipole on the second carbon away from the bromine is also δ, which leads to a δ hydrogen (see 1B). This small positive dipole makes the hydrogen on the β-carbon slightly acidic. Br NaOEt, EtO reflux 1A 2 EtO NaBr substitution - too sterically hindered for collision to occur Br 3! 3!"!" OEt! acid-base reaction 1B

24 Transition State but No Intermediate 24 In the conversion of 3 to 5, no intermediate has been detected under the conditions at which the reaction is normally done. If there is no intermediate, then the characteristics of the reaction are described by the transition state for the reaction. As electron density increases on the β- carbon, electron density migrates towards the δ carbon (the α-carbon in 3) to form a second bond to carbon, a π-bond. The Br bond must break, with bromide ion as the leaving group. If there is no intermediate, all of the bond making and bond breaking occurs simultaneously (this is a synchronous reaction), and the transition state for the elimination reaction is represented as 4. Note that the new π-bond forms between the α- and β- carbons (see 2-methyl-1-propene; isobutylene, 5). O acid-base reaction 3 3! Br #!!" $ 3 3 3!"!"!" # O! $ 4 Br O Br