1559Tch0338-5410/22/052:59Page38 EQA 3 Reactions of Alkanes:Bond-Dissociation Energies, Radical Halogenation,and Relative Reactivity how any member of class of compoundsis likely to behave under certain reaction conditions.We ship between the energ ncepts introd in Chapter 2 and rea bonds.and nothing else Therefore alkanes are essentially unreactive toward ionic or polar materials:indeed alkanes are jst about the least reactive of all compound bond to cleave is leaving one electron with each of the formerly bonded atoms C-C→C·+·CorCH■ C.+H.This kind of bond cleavage is difficult and only occurs at high em s three n (by halo m),and combustion (high tcmperature and oygen Youill noteronmpha on dis- cussion ond energies aiepy-sep.boaayh be Outline of the Chapter 3-1 Strength of Alkane Bonds:Radicals Exactly what does it take to cleave bonds in alkanes? 3-2 Alkyl Radicals and Hyperconjugation The nature of the species obtained upon alkane bond cleavage. 3-3 Convers 3.4 Chlorination of Methane:The Radical Chain Mechanism Radical substitution of chlorine for hydrogen in methane:mechanism and energetics 3-5 Other Radical Halogenations of Methane Similarities and differences
3 Reactions of Alkanes: Bond-Dissociation Energies, Radical Halogenation, and Relative Reactivity Discussing reactions of alkanes at the start of an organic chemistry course allows us to learn to work with several concepts that will be useful later. These include the idea of a general reaction mechanism that describes how any member of an entire class of compounds is likely to behave under certain reaction conditions. We also see the relationship between the energy concepts introduced in Chapter 2 and reactions that require more than one step. Alkanes do not contain any functional groups: They are made up of nonpolar COC and COH bonds, and nothing else. Therefore, alkanes are essentially unreactive toward ionic or polar materials; indeed, alkanes are just about the least reactive of all compound classes. Therefore, their chemistry is limited to processes that can lead to cleavage of nonpolar bonds. Thus, the only reasonable way for an alkane bond to cleave is homolytically, leaving one electron with each of the formerly bonded atoms: . This kind of bond cleavage is difficult and only occurs at high temperatures or in the presence of certain especially reactive species like halogen atoms. This chapter covers three major ways alkane bonds are cleaved: pyrolysis (high temperature), halogenation (by halogen atoms), and combustion (high temperature and oxygen). You will note a strong emphasis on discussions involving bond energies. This should not surprise you, because bond cleavage requires an input of energy. The mechanism presents the reaction in terms of a step-by-step, bond-by-bond analysis that is helpful for spotting trends and analogies. Outline of the Chapter 33-1 Strength of Alkane Bonds: Radicals Exactly what does it take to cleave bonds in alkanes? 33-2 Alkyl Radicals and Hyperconjugation The nature of the species obtained upon alkane bond cleavage. 33-3 Conversion of Petroleum: Pyrolysis A practical example. 33-4 Chlorination of Methane: The Radical Chain Mechanism Radical substitution of chlorine for hydrogen in methane: mechanism and energetics. 33-5 Other Radical Halogenations of Methane Similarities and differences. 33-6 Chlorination of Higher Alkanes What happens when substitution for different hydrogens in an alkane can give different products? CC C H C or C C H 38 1559T_ch03_38-54 10/22/05 2:59 Page 38
1559T_ch03_38-5410/22/052:59Pa9e39 EQA Keys to the Chapter·39 3-7 Selectivity with Other Halogens Energetic comparisons of the reactions involving F2 and Br2 3-8 Synthetic Aspects More practical considerations. E AIL e energetics associated with a chemical reaction Keys to the Chapter 3-1.Strength of Alkane Bonds:Radicals A minor but annoying point of confusion is often encountered when one discusses bond strengths.A bond's e energy input require Energy is released. cocction of these to aions ho the honded oe o in nry and B by an an the sneny that has to be put in t ra (g ule ar ting large DH lues with high-en ry species.Large DH ing of theme age of any bond in an alkane generates radicals:species with a single unpaired electron where used to be.The section i strates four such examples:methyl,.CH:ethyl,.CHCH: (tetrahedral,as in alkanes).Why should this be?A partial rease n goes back to basic elec rostatics.The shape or VSEPR?) om (re accommodated by midal shape:Repulsion betwe n the lone pair and the ele trons in the N ds is important in ca etry to be pren ing the number of no between the pairs in theCH bonds dominates.situation leading to shybridization and trigonal planar geometry.which allows the( -H bonding electro ns to spread out far away from one another. methyl radical is the least stable.Hyperconjugation is one concept often used to explain this stabilization.Phy: cally,and ele ly,the radica on can be viewed a (7 valen
Keys to the Chapter • 39 33-7 Selectivity with Other Halogens Energetic comparisons of the reactions involving F2 and Br2. 33-8 Synthetic Aspects More practical considerations. 33-9 Synthetic Chlorine Compounds and the Stratospheric Ozone Layer Halogens in the “real world.” 3-10 Combustion and the Relative Stability of Alkanes Included is a detailed introduction to the evaluation of the energetics associated with a chemical reaction. Keys to the Chapter 3-1. Strength of Alkane Bonds: Radicals A minor but annoying point of confusion is often encountered when one discusses bond strengths. A bond’s strength, or more properly, bond-dissociation energy (DH°), is defined as the energy released when a bond forms or, equivalently, the energy input required to break a bond: Aj Bj On AOB H° DH° Energy is released. AOB On Aj Bj H° DH° Energy is put in. Inspection of these two equations shows that the bonded molecule AOB is more stableOlower in energy contentOthan the separated atoms A and B by an amount equal to DH°. When the linkages in a molecule are strong (high DH°), the molecule is usually relatively low in energy content (e.g., stable). As long as you remember that DH° is the energy that has to be put in to break a bond, you won’t fall into the common trap of associating large DH° values with high-energy species. Large DH° values imply low energy, strongly bonded, stable species. The tables and figures in this section should further help you develop a comfortable understanding of the meaning of DH° values, in preparation for their use later on. 3-2. Alkyl Radicals and Hyperconjugation Homolytic cleavage of any bond in an alkane generates radicals: species with a single unpaired electron where an attached group used to be. The section illustrates four such examples: methyl, jCH3; ethyl, jCH2CH3; isopropyl, jCH(CH3)2; and tert-butyl, jC(CH3)3. Several points are made in the section. First, radical carbons are sp2 hybridized (planar), not sp3 hybridized (tetrahedral, as in alkanes). Why should this be? A partial reason goes back to basic electrostatics. The shape of a species will be that which minimizes repulsion between electrons around a central atom (remember valence shell electron pair repulsion, or VSEPR?). In ammonia, NH3, the four electron pairs around N are best accommodated by a pyramidal shape based on sp3 hybridization: Repulsion between the lone pair and the electrons in the NOH bonds is important in causing this geometry to be preferred. Reducing the number of nonbonded electrons from two to one as in methyl radical, jCH3, changes the situation. Now, electron repulsion between the pairs in the COH bonds dominates, a situation leading to sp2 hybridization and trigonal planar geometry, which allows the COH bonding electrons to spread out far away from one another. The second main point in the chapter is the stabilization of a radical center by the presence of alkyl groups attached to the radical carbon. So, tert-butyl radical is more stable than isopropyl, which is better than ethyl; methyl radical is the least stable. Hyperconjugation is one concept often used to explain this stabilization. Physically, and electrostatically, the radical carbon can be viewed as somewhat electron deficient (7 valence electrons instead of an octet). Hyperconjugation provides a means for bonds in neighboring alkyl groups to “lend” a little electron density to the radical center, thereby making it feel a little less electron-poor. In doing so, the 1559T_ch03_38-54 10/22/05 2:59 Page 39
1559rch0338-5410/22/052:59Page40 EQA 40.Chapter 3 REACTIONS OF ALKANES lem"to be diluted over a larger area.rather than being the concentrated burden of a single atom. lty of alky 3-3.Conversion of Petroleum:Pyrolysis aspects of bond cle hat.issue of eaction the mod ch a chemical transformation is carried out in order to give a desired mole Itscto the choine olcueTe pre Chloringtion of Methane:The radical chain mechanism CH,+C2→HCI+CH,C kane )On ce the functic I group is pre many more k agation,and termination)but also to the finer details re state structure for Although some o t the terminology intro ced he information that the mechanism contains is critical to an understanding of how and why reactions oc Take some time in this section to study each reaction stp. What are its energetic ve"r nismsr intended toallowon to makesesou of oranic chemisry.Give thison the time to do that for you dakins th somtoochemical reaction from the △ection=DH°(bonds broken)-DH°(bonds formed Energy inpu nergy CH,-CH+H-H一2CH-H DH°= 90 104 10 kcal mol- =[0+104]-[2(105)]=-16 kcal mol c1 cess that,although exothermic.requires very high temperatures to
40 • Chapter 3 REACTIONS OF ALKANES alkyl groups effectively take some of the electron deficiency onto themselves, spreading it out, or “delocalizing” it. Delocalization of an electron deficiency or an electron excess over more than just the atom on which it is nominally located is often an energetically favorable, stabilizing process. It effectively allows the “problem” to be diluted over a larger area, rather than being the concentrated burden of a single atom. The ability of alkyl groups to stabilize electron-deficient centers like radicals is often taken to imply that they are better donors of electrons than hydrogen atoms. Alkyl groups are therefore referred to as electron donating. 3-3. Conversion of Petroleum: Pyrolysis The practical, “real-world” aspects of bond cleavage and radical formation are explored. Pyrolysis is a process that often gives mixtures of many products, and methods have been developed (mostly within the petroleum industry) to control this reaction somewhat. We will frequently explore the issue of reaction control: the modification of conditions under which a chemical transformation is carried out in order to give a desired molecule as a major or exclusive product. 3-4. Chlorination of Methane: The Radical Chain Mechanism In this section the reaction of methane with chlorine molecules is discussed. The process CH4 Cl2 On HCl CH3Cl is important because it converts a nonfunctionalized molecule (an alkane) into a molecule containing a functional group (a haloalkane). Once the functional group is present, many more kinds of chemical reactions will become possible. This section also presents the mechanism of this reaction in full detail. Pay close attention not only to the steps of the reaction (initiation, propagation, and termination) but also to the finer details relating H°, Ea, and transition state structure for each step. Although some of the terminology introduced here is appropriate only for radical mechanisms and not for the majority of reactions to come later, the type of information that the mechanism contains is critical to an understanding of how and why organic reactions occur. Take some time in this section to study each reaction step. What are its energetic circumstances, under what conditions does it occur, what role does it play in the overall process? Try to establish a feeling for the species involved as “stable” or “unstable,” “reactive” or “unreactive,” relatively speaking. Reaction mechanisms are intended to allow one to make sense out of organic chemistry. Give this one the time to do that for you. Be sure that you understand the procedure for calculating the H° value for a chemical reaction from the DH° values of the bonds taking part in the transformation. The general formula is To illustrate with a reaction different from those in the text, let us calculate H° for the process C2H6 H2 n 2 CH4. Using the data from Tables 3-1 and 3-2 in the text, Comment: This is a “hydrocracking” process that, although exothermic, requires very high temperatures to occur (cleavage of COC bond is necessary). reaction [90 104] [2(105)] 16 kcal mol1 kcal mol1 DH 90 104 105 Note 2 methane C—H bonds CH3 CH3 H H 2 CH3 H H reaction DH (bonds broken) Energy input DH (bonds formed) Energy output H 1559T_ch03_38-54 10/22/05 2:59 Page 40
1559T_ch03_38-5410/22/052:59Page41 EQA Keys to the Chapter·41 Note on energetics △f=-27 keal mol-l Sum:CH4+er+e+C2→HC+e+CH,C+e AH=-25 kcal mol-! Rem oa part of the enthalpy change as it is defined for the stoichometriceaction.Whenem the heat of ntally,th qual toH for the frequently relat part c on processe 3-5.Other Radical Halogenations of Methane features of organic ch ct mchanism can hold for many with the other halogens.The similarities are qualitative.however.Differences in energetics are significant.and as a result the reac tion-coordinate diagrams differ in appearance in a way that will become important as we continue through the chapter 3-6.Chlorination of Higher Alkanes monly used symbols.)The weakest(3) re the most re rihes this se factors combine to produce the observed ratios of products in several representative systems. tivity with Other ignificant point is that reactivity and selectivity in radica halogenations are inversely related.Simply put,the more reactive halogens are less picky and show less pref erence for 3 VS. ve t react and very similar to one another.Fluorine thus reacts very rapidly with any CH bond in a molecule.The reactio ine have large actva es,with and bro mine is much more discriminatin (selective)in its reactions too greatly preferring 3 over 2 or 3-8.Synthetic Aspects Synthes ated with oranic chemistry.In synthes s.we strive to produce a de thetically useless.In this chapter we have of a single reactior tically.The best ones start wit
Keys to the Chapter • 41 Note on energetics: The overall enthalpy of a radical chain reaction is the sum of the H° values for only the propagation steps. If we “sum up” the species in these steps, we see that the free atoms and radicals “cancel out,” leaving only the molecular species of the overall reaction: Propagation step 1 H° 2 kcal mol1 Propagation step 2 H° 27 kcal mol1 Sum: H° 25 kcal mol1 Removing the Clj and jCH3 that appear on both sides of the equation leaves just the molecules of the overall process. What about the initiation and termination steps and their H°’s? They are separate. Their H° values are not a part of the enthalpy change as it is defined for the stoichiometric reaction. When we measure the heat of a radical reaction experimentally, the value we obtain will not be precisely equal to H° for the propagation steps alone; initiation and termination steps are occurring, too, and their H°’s will introduce an error. This deviation will usually be small, however, because initiation and termination steps occur only infrequently relative to the propagations, and because the H°’s for the endothermic initiation are for the most part canceled out by those of the exothermic termination processes. 3-5. Other Radical Halogenations of Methane One of the best features of organic chemistry is the fact that one mechanism can hold for many individual reactions. Thus, the same types of steps that occur in the chlorination of methane are followed in its reactions with the other halogens. The similarities are qualitative, however. Differences in energetics are significant, and as a result the reaction-coordinate diagrams differ in appearance in a way that will become important as we continue through the chapter. 3-6. Chlorination of Higher Alkanes The same mechanism also applies qualitatively to chlorination of other alkanes. The only difference is in the nature of the COH bonds available in the alkane to be broken. They are generally less strong than those in methane, following a DH° order of CH4 1° 2° 3°. (Note: 1° primary, 2° secondary, and 3° tertiary. These are commonly used symbols.) The weakest (3°) are the most readily broken; thus, alkanes with different types of COH bonds display a built-in selectivity of 3° 2° 1° in their reactions with chlorine. This section describes this selectivity quantitatively, illustrating how both reactivity differences and statistical factors combine to produce the observed ratios of products in several representative systems. 3-7. Selectivity with Other Halogens An extension of the previous sections. The most significant point is that reactivity and selectivity in radical halogenations are inversely related. Simply put, the more reactive halogens are less picky and show less preference for 3° vs. 2° vs. 1° COH bonds relative to less reactive halogens. The reason lies in the different activation energies associated with the COH bond-breaking step. The values for fluorine are all very small, and very similar to one another. Fluorine thus reacts very rapidly with any COH bond in a molecule. The reactions for bromine have large activation energies, with significant differences associated with the different types of COH bonds present. The result is that bromine is much, much slower than fluorine to react with any alkane, and bromine is much more discriminating (selective) in its reactions, too, greatly preferring 3° over 2° or 1° COH bonds. The contrasts between reaction-coordinate diagrams for the two halogens provide a pictorial representation of these differences. 3-8. Synthetic Aspects Synthesis is one primary function associated with organic chemistry. In synthesis, we strive to produce a desired product in good yield with high selectivity, to minimize the effort required to separate this material from side products. In particular, a reaction that gives rise to a hard-to-separate mixture of many components is synthetically useless. In this chapter we have seen a large number of possible permutations of a single reaction: alkane halogenation. Not all of the examples shown are equally useful synthetically. The best ones start with CH4 Cl CH3 Cl2 HCl CH3 CH3Cl Cl CH3 Cl2 CH3Cl Cl CH4 Cl HCl CH3 1559T_ch03_38-54 10/22/05 2:59 Page 41
1559r.ch03.38-5410/22/052:59Page42 EQA 42.Chapter 3 REACTIONS OF ALKANES analkane in which all hydrogens are chemically indistinguishable(methane.ethane.opent)bec ause they mined by the number of different types of hydrogens present and whether the desired product derives from substitution of a more reactive or a less reactive hydrogen in the molecule. ter example.there are one an carbon.a less CH: CHa-CH-CH3 Br2-CH3-C-CH3 Major product CH3-CH-CHs F2-F-CH2-CH-CHs Major product 3-10.Combustion and Relative Stability In order to obtain thermodynamic information experimentally.several methods may be used.The measure on.h the a ot a re sure of the energy content of the compound relative to that of the product molecules.COand HO.Such data ⊕ ve ent structure Solutions to Problems 13.This revious chapter.For shorthand purposes.we (a)CH:CH2CH2CHs (b)CH;CHzCH2CH2CHs 2 1 2° d”一HCH-” e CH2-CH2 14.(a Table2-4) CH.CH.CH.CH Primary (ess stable Remember:Identify radicals as 1 2.or 3 by the radical carbon.None of the other carbons matter.Two hyperconiugation pictures may be drawn for 1-methylpropyl radical.one with two
42 • Chapter 3 REACTIONS OF ALKANES an alkane in which all hydrogens are chemically indistinguishable (methane, ethane, neopentane), because they can produce only one monohalogenation product. In the case of most alkanes, synthetic utility will be determined by the number of different types of hydrogens present and whether the desired product derives from substitution of a more reactive or a less reactive hydrogen in the molecule. In isobutane, for example, there are one 3° and nine 1° hydrogens. If we desire to halogenate at the 3° center, the natural selectivity of bromine makes it the obvious halogen to choose. If we desire to halogenate a 1° carbon, a less selective, more reactive halogen would allow us to take best advantage of the statistical factor of nine possible 1° hydrogens available to be replaced in each molecule. Thus, 3-10. Combustion and Relative Stability In order to obtain thermodynamic information experimentally, several methods may be used. The measurement of equilibrium constants gives energy differences between species. Directly measuring the heat of a reaction accomplishes the same thing. When the reaction is combustion of a hydrocarbon, the result is a measure of the energy content of the compound relative to that of the product molecules, CO2 and H2O. Such data allow comparisons to be made between related compounds, which in turn reveal factors influencing the relative stabilities of different structures. Solutions to Problems 13. This problem is really a reminder of material from the previous chapter. For shorthand purposes, we use the symbols 1° primary, 2° secondary, and 3° tertiary. (a) (b) (c) (d) 14. (a) Remember: Identify radicals as 1°, 2°, or 3° by the radical carbon. None of the other carbons matter. Two hyperconjugation pictures may be drawn for 1-methylpropyl radical, one with two 1-Methylpropyl (sec-butyl; see Table 2-4) Secondary (2), more stable CH3CH2CHCH3 Butyl radical Primary (1), less stable CH3CH2CH2CH2 CH3 1 2 2 3 CH2 CH2 CH2 CH2 H C As you will see in Chapter 4, most “ring” compounds can be treated just like molecules without rings. CH3 1 1 2 3 CH3CH2CH2CH2CH3 1 2 1 CH3CH2CH2CH3 1 2 1 CH3 CH CH3 Br2 CH3 CH3 C CH3 Major product CH3 Br CH3 CH CH3 F2 F CH3 CH2 CH CH3 Major product CH3 1559T_ch03_38-54 10/22/05 2:59 Page 42
1559T_ch03_38-5410/22/052:59Page43 ⊕ EQA Solutions to Problems.43 Cbond participating (b)In naming these,remember that the radical carbon is alwaysCl(just like the"point of CH-CHz CH3-CH2 CH-CH-CH-CH CH-CH-C-CH 2-Ethylbutyl radical 1-Ethyl-1-mthypropyl radica Primary,less stable ⊕ (c)Left to right:1.2-dimethylpropyl radical,secondary.intermediate in stability:1.1-dimethylpropyl radical,tertiary,most stable;3-methylbutyl radical,primary,least stable Hyperconjugation in the 1.1-dimethylpropyl radical is the same as in 1-ethyl-1-methylpropyl [(b)above]: in your picture.an H should replace one of the end CH groups. aCH,C,CH,一CH,cH,·+·CH,c-Cond c Then there are three possible recombinations: (d)CH3 CH3CH2-CHaCH2CH3 (Reverse of first step) Propane
Solutions to Problems • 43 COH bonds overlapping with the radical p orbital, and one with a COC bond participating instead of a COH: (b) In naming these, remember that the radical carbon is always C1 (just like the “point of attachment” carbon in alkyl groups). The parent name is based on the longest carbon chain beginning at C1, and all appendages are named as substituents: (c) Left to right: 1,2-dimethylpropyl radical, secondary, intermediate in stability; 1,1-dimethylpropyl radical, tertiary, most stable; 3-methylbutyl radical, primary, least stable. Hyperconjugation in the 1,1-dimethylpropyl radical is the same as in 1-ethyl-1-methylpropyl [(b) above]; in your picture, an H should replace one of the end CH3 groups. 15. Work problems like this “mechanistically”: Proceed via general reaction steps as you have previously seen illustrated in the text until you reach stable molecules. Pyrolysis of propane starts as follows: (a) Then there are three possible recombinations: (b) (c) (d) Propane CH3 CH3CH2 CH3CH2CH3 (Reverse of first step) Butane 2 CH3CH2 CH3CH2CH2CH3 Ethane 2 CH3 CH3CH3 CH3CH2 CH3 CH3CH2 CH3 C—C bond cleavage H H H C H2C CHCH3 CH3HC and H H H H C C H2C CH3 H H H H C C H2C CH3 1559T_ch03_38-54 10/22/05 2:59 Page 43
1559T.ch03.38-5410/22/052:59Page44 EQA 44.Chapter 3 REACTIONS OF ALKANES Two possible hydrogen abstractions can occur: H aa+an一+atcw 气H Abstraction only occurs from the carbon next to a radical carbon.Methyl radical.CH.doesn't have another carbon next to its radical center.so it cannot give up a hydrogen in an abstraction.It can still acking of propane:methane,ethane.butane,and ethen 16.(a)The weakest bond in butane is the C2-C3 bond.DH=88 kcal mol-!(Table 3-2).Pyrolysis should therefore proceed as follows: ①CH,c闻间6H,CH,一2CH,C:c-Cbd deag (2)2CH,CH2·一CHCH2CH,CH Reverse of(1) Ethane (b)The weakest bonds are the three equivalent C-Cbonds.D=88 kcal molTherefore. @2a一 (6)2(CH2CH ④CH,CHCH一(CH,CH Revene of ()bi 句cH,fYam6ch一GL+cH=ca,lyop nc (6)(CHCH.HCHCHCH,-CH,CH.CH +CH:-CHCH, Propane Propene 17.The D data are readily found (Tables 3-1 and 3-4).Values in kcal mol- (a)104+38-2136)=-130 (b)104+58-2103)=-44 (c)104+46-2(87)=-24 (d104+36-2(71)=-2 (e)96.5+38-(110+136)=-1115 f)96.5+58-(85+103)=-335 (g)96.5+46-(71+87)=-15.5 h)96.5+36-(55+71)=+6.5
Two possible hydrogen abstractions can occur: (e) (f ) Abstraction only occurs from the carbon next to a radical carbon. Methyl radical, jCH3, doesn’t have another carbon next to its radical center, so it cannot give up a hydrogen in an abstraction. It can still accept a hydrogen, however [reaction (e), above]. So there are four new products formed from cracking of propane: methane, ethane, butane, and ethene (ethylene). 16. (a) The weakest bond in butane is the C2OC3 bond, DH° 88 kcal mol1 (Table 3-2). Pyrolysis should therefore proceed as follows: (1) (2) (3) (b) The weakest bonds are the three equivalent COC bonds, DH° 88 kcal mol1 . Therefore, (1) (2) (3) (4) (5) (6) 17. The DH° data are readily found (Tables 3-1 and 3-4). Values in kcal mol1 . (a) 104 38 2(136) 130 (b) 104 58 2(103) 44 (c) 104 46 2(87) 24 (d) 104 36 2(71) 2 (e) 96.5 38 (110 136) 111.5 (f ) 96.5 58 (85 103) 33.5 (g) 96.5 46 (71 87) 15.5 (h) 96.5 36 (55 71) 6.5 (CH3)2CH Propane Propene H CH2 CHCH3 CH3CH2CH3 CH2 CHCH3 CH3 Hydrogen abstractions Methane Propene H CH2 CHCH3 CH4 CH2 CHCH3 CH3 CH(CH3)2 (CH3)3CH Reverse of (1); recombinations 2,3-Dimethylbutane 2 (CH3)2CH (CH3)2CHCH(CH3)2 Ethane 2 CH3 CH3CH3 (CH3)2CH (CH CH3 3)2CH CH 3 Cleavage Hydrogen abstraction Ethane Ethene CH3CH2 H CH CH2 CH2 3CH3 CH2 CH2 2 CH3CH2 CH3CH2CH2CH3 Reverse of (1) CH3CH2 CH2CH3 2 CH3CH2 C—C bond cleavage Ethane Ethene CH3CH2 CH3CH3 CH2 CH2 H CH2 CH2 Methane Ethene CH3 CH4 CH2 CH2 H CH2 CH2 44 • Chapter 3 REACTIONS OF ALKANES 1559T_ch03_38-54 10/22/05 2:59 Page 44
1559T_ch03_38-5410/22/052:59Page45 ⊕ EQA Solufions to Problems45 18.(a)Two:I-halobutane and 2-halobutane (b)Three (see solution to Problem 19) (e)Four(see solution to Problem 19) (d)Four: CH2X X CHa 19.acH,CH,C用，C() ,and CH CH-CHCICH-CH (3-chloropentane) CH,CH.CCI(CHJ)CH.CH, (3-chloro-3-methylpentane),and CH.CH2CH(CH2Cl)CH2CH3 [3-(chloromethyl)pentane]. (b)To answer this question.you first have to count up and identify by type (1e.2.or 3)all the hydrogens whose remo ould cad to each one of the product Then multiply the numbr of cedure gives the relative amount of product corresponding to removal of those hydrogens.After you have don his Tor all the pro you convert these relative amounts into percentage yields by normalizing is formed by chlorination of any of the four secondary hydro y4)on carbons 2 and 4:its relative yield is43-Chioropentane is formed by c ield for each Relative yield of the product Sm of raiv yicds forlodyeld of tapoduc So.Yicld of 1-chloropentanc 100%x 6/(6+16+8) =100%X6/50=209 Chloro-3-mcthylpentane is fomdieldrimayyr carbons 2 its relativ 16.3Chloro-3-mcthylpentan o-3api3Sn0o四eao of the three primary hyd 1)on theme uced
Solutions to Problems • 45 18. (a) Two: 1-halobutane and 2-halobutane (b) Three (see solution to Problem 19) (c) Four (see solution to Problem 19) (d) Four: 19. (a) (i) CH3CH2CH2CH2CH2Cl (1-chloropentane), CH3CH2CH2CHClCH3 (2-chloropentane), and CH3CH2CHClCH2CH3 (3-chloropentane). (ii) CH3CH2CH(CH3)CH2CH2Cl (1-chloro-3-methylpentane), CH3CH2CH(CH3)CHClCH3 (2-chloro-3-methylpentane), CH3CH2CCl(CH3)CH2CH3 (3-chloro-3-methylpentane), and CH3CH2CH(CH2Cl)CH2CH3 [3-(chloromethyl)pentane]. (b) To answer this question, you first have to count up and identify by type (1°, 2°, or 3°) all the hydrogens whose removal could lead to each one of the products. Then multiply the number of hydrogens you counted by the relative reactivity for that type of hydrogen in a chlorination reaction at 25°C, the conditions for the reaction stated in the problem. This procedure gives you the relative amount of product corresponding to removal of those hydrogens. After you have done this for all the products, you convert these relative amounts into percentage yields by normalizing to 100% (as shown below). (i) 1-Chloropentane is formed by chlorination of any of the six primary hydrogens (each with a relative reactivity 1) on carbons 1 and 5; its relative yield is 6 1 6. 2-Chloropentane is formed by chlorination of any of the four secondary hydrogens (each with a relative reactivity 4) on carbons 2 and 4; its relative yield is 4 4 16. 3-Chloropentane is formed by chlorination of any of the two secondary hydrogens (each with a relative reactivity 4) on carbon 3; its relative yield is 2 4 8. The absolute % yield for each product is calculated as follows: 100% % yield of that product So, Yield of 1-chloropentane 100% 6/(6 16 8) 100% 6/30 20% Yield of 2-chloropentane 100% 16/30 53% Yield of 3-chloropentane 100% 8/30 27% (ii) 1-Chloro-3-methylpentane is formed by chlorination of any of the six primary hydrogens (relative reactivity 1) on carbons 1 and 5; its relative yield is 6 1 6. 2-Chloro-3- methylpentane is formed by chlorination of any of the four secondary hydrogens (relative reactivity 4) on carbons 2 and 4; its relative yield is 4 4 16. 3-Chloro-3-methylpentane is formed by chlorination of the single tertiary hydrogen (relative reactivity 5) on carbon 3; its relative yield is 1 5 5. 3-(Chloromethyl)pentane is formed by chlorination of any of the three primary hydrogens (relative reactivity 1) on the methyl group attached to carbon 3; its relative yield is 1 3 3. Relative yield of the product Sum of relative yields for all products (Halomethyl)- cyclopentane CH2X 1-Halo-1-methylcyclopentane X CH3 1-Halo-2-methylcyclopentane CH3 X 1-Halo-3-methylcyclopentane CH3 X 1559T_ch03_38-54 10/22/05 2:59 Page 45
1559p.eh0338-5410/22/052:59Pmge46 EQA 46.Chapter 3 REACTIONS OF ALKANES Yield of 2-chloro-3-methylpentane=100%x 16/30=53% oromethyl)pentane 00% (c)Propagation step 1 [values below are D(kcal mol)for bonds made or broken] CH3 CH.CH CHCH CH+Cl.HCI+CHCH.CCH.CH 1034=96.5-103=-6 Propagation step 2: CHs CHCH2CCH-CH;+Cl-Cl.+CH;CH2CCICH2CH3 58 85△r=-27 For the reaction overall.A=-33.5 kcal mol Br2-2Br Propagation: ()Br·+CHL一CH·+HBr (2②CH+Br2一CBr+Br Termination: Br.+Br 一B2 CH3·+Br·→CHB →CHCH 21.Initiation: Br2一2Br H=+46 kcal mol- Propagation: Br·+C6H6一HBr+Casr=+25 keal mol- (②)CH+B2- →C.HsBr+Br· 4°=-35 kcal mol- Overall △=-l0 kcal mol determining first propagation step in the reaction of benzene is much more endothermic than any of the
So, Yield of 1-chloro-3-methylpentane 100% 6/(6 16 5 3) 100% 6/30 20% Yield of 2-chloro-3-methylpentane 100% 16/30 53% Yield of 3-chloro-3-methylpentane 100% 5/30 17% Yield of 3-(chloromethyl)pentane 100% 3/30 10% (c) Propagation step 1 [values below are DH° (kcal mol1 ) for bonds made or broken]: Propagation step 2: For the reaction overall, H° 33.5 kcal mol1 . 20. The mechanism is qualitatively the same as that of chlorination of methane (textbook Section 3-4): Initiation: Propagation: (1) (2) Termination: 21. Initiation: Propagation: (1) (2) The overall H° is not very different from those of typical alkane COH bonds: methane, H° 6 kcal mol1 ; 1° COH, H° 10; 2° COH, H° 13.5; 3° COH, H° 15.5. However, the ratedetermining first propagation step in the reaction of benzene is much more endothermic than any of the C6H5 Br2 C6H5Br Br Overall H 35 kcal mol1 H 10 kcal mol1 Br C6H6 HBr C6H5 H 25 kcal mol1 Br2 2 Br H 46 kcal mol1 Br Br Br2 CH3 Br CH3Br CH3 CH3 CH3CH3 CH3 Br2 CH3Br Br Br CH4 CH3 HBr Br2 2 Br 58 85 H 27 CH3 CH3CH2CCH2CH3 Cl2 Cl CH3 CH3CH2CClCH2CH3 96.5 103 H 96.5 103 6.5 CH3CH2CHCH2CH3 CH3 CH3 Cl HCl CH 3CH2CCH2CH3 46 • Chapter 3 REACTIONS OF ALKANES 1559T_ch03_38-54 10/22/05 2:59 Page 46
1559T_ch03_38-5410/22/052:59Page47 EQA Solutions to Problems.47 alkane reaction kinetically with bromination reactions of typical alkanes (a)No reaction.lodination of alkanes is endothermic. (b)CH:CHFCH,+ OC及Br数epe CHs CH3 CH3 CH3 (d)CH3-C-CH2-C-CH3+Cl-CH2-CH-CH2-C-CH3 CH3 CH3 CH3CH3 CH,-CH-CH-C-CH;+CH;-CH-CH2-C-CH2-Cl CI CH 个g物a,cthe”的 CH3 CH, (e)CHs-C-CH2-C-CHs Br2 is very selective for 3C-H bonds 23.Relative yield=number of hydrogens of a given type in the starting alkane X relative reactivity relative vield of one product X 100%=%vield of that product sum of relative vields of all products Product Hydrogen etrtvity yield 之 (d)(CH).CCICH.C(CH) CICH-CH(CHCH.C(CH) d (CH).CHCH.C(CH.).CH.C 9 (e)and (e)3 substitution by Br2 is about 73%selective in (c)and 91%in (e). 24o mnthetic methods.The othe reactions y useful.The fluorination (b) might look good on paper,but use of F as a reagent in practice is very difficult
Solutions to Problems • 47 alkane reactions, due to the exceptional strength of the COH bonds in benzene. The result is that bromination of benzene by this mechanism is exceedingly difficult (very slow) and does not compete kinetically with bromination reactions of typical alkanes. 22. Unless otherwise stated, assume that no more than one halogen atom attaches to each alkane molecule. (a) No reaction. Iodination of alkanes is endothermic. (b) CH3CHFCH3 CH3CH2CH2F F2 is not very selective. (c) Bromination selects the tertiary position whenever possible. See note in answer to Problem 13(d). (d) A complex mixture is obtained. Cl2 is more selective than F2, but still prefers 3° to 1° positions by a factor of only about 5 to 1. (e) 23. Relative yield number of hydrogens of a given type in the starting alkane relative reactivity 100% % yield of that product Hydrogen Number of Relative Relative % Product type hydrogens reactivity yield yield (b) CH3CHFCH3 2° 2 1.2 2.4 29 CH3CH2CH2F 1° 6 1.2 6.2 71 (d) (CH3)2CClCH2C(CH3)3 3° 1 5.2 5.2 18 ClCH2CH(CH3)CH2C(CH3)3 1° 6 1.2 6.2 21 (CH3)2CHCHClC(CH3)3 2° 2 4.2 8.2 29 (CH3)2CHCH2C(CH3)2CH2Cl 1° 9 1.2 9.2 32 (c) and (e) 3° substitution by Br2 is about 73% selective in (c) and 91% in (e). 24. Only the bromination reactions (c) and (e) are really acceptable as synthetic methods. The other reactions, giving several products in comparable yields, are not synthetically useful. The fluorination (b) might look good on paper, but use of F2 as a reagent in practice is very difficult. relative yield of one product sum of relative yields of all products CH3 C CH2 CH3 CH3 Br CH3 Br2 C CH3 is very selective for 3 C—H bonds. CH3 C CH2 CH3 CH3 Cl CH3 Cl C CH3 Cl CH CH2 CH2 CH3 CH3 CH3 C CH3 CH3 CH CH2 CH3 CH3 CH3 C CH2 Cl CH3 CH3 CH3 CH C CH3 CH3 CH CH3 Br 1559T_ch03_38-54 10/22/05 2:59 Page 47
