Lthough This Chapter is Concerned With The

8/14/2019 Lthough This Chapter is Concerned With The

1/41

4ALKANES

A though this c hapter is concerne d with the chemistry of only one class ofcompounds, saturated hydrocarbons or alkanes, several fundamental prin-ciples ar e developed tha t w e shall use extensively in later chapters . T h e studyof so m e of these principles has been assoc iated traditionally m ore with physicalchemistry than with organic chemistry. W e include them here, at the beginningof our discussion of organic reactions, because they provide a sound basis forunderstanding the key qu estions concerning the practical use of organic reac-tions. Is the equilibrium point of a given reaction far enough toward the desiredprodu cts t o b e useful? C an conditions be found in which the reaction will tak eplace at a pract ical rate? How can unwanted side react ions be suppressed?Initially, we will be concerned with the physical properties of alkanesand how these properties can be correlated by the important concept of ho-mology. This will be followed by a brief survey of the occurrence and usesof hydrocarbons, with special reference to the petroleum industry. Chemicalreactions of a lkane s then will be discussed, with special emphasis on com bus-tion and substitution reactions. Th ese reactions are em ployed to il lustrate howwe c an predict and us e energy changes-particularly AH, the heat evolved o rabsorb ed by a reacting system , which often can be estim ated from bond ener-gies. Then we consider some of the problems involved in predicting reactionrates in the context of a specific reaction, the chlorination of methane. Theexample is com plex, but i t has the virtue that we are able to break the overallreaction into quite simple steps.Before proceeding further, i t will be well to reiterate what a n alkane is,lest you be confused as to the difference between alkane s and alkenes. Alkanesare compounds of carbon and hydrogen only, without double bonds, triple


2/41

4 Alkanesbonds, or rings. They all conform to the general formula C,,H,,,+, and some-times are called parafin hydrocarbons, open-chain saturated hydrocarbons,or acyclic hydrocarbons. The nomenclature of alkanes has been discussed inChapter 3, and you may find it well to review Section 3- 1 before proceeding.

4-1 PHYSICAL PROPERTIES OF ALKANES.THE CONCEPT OF HOMOLOGYThe series of straight-chain alkanes, in which n is the number of carbons inthe chain, shows a remarkably smooth gradation of physical properties (seeTable 4-1 and Figure 4-1). As n increases, each additional CH, group con-tributes a fairly constant increment to the boiling point and density, and to alesser extent to the melting point. This makes it possible to estimate the proper-ties of an unknown member of the series from those of its neighbors. Forexample, the boiling points of hexane and heptane are 69" and 98", respectively.Thus a difference in structure of one CH, group for these compounds makes adifference in boiling point of 29"; we would predict the boiling point of the nexthigher member, octane, to be 98" + 29" = 127", which is close to the actualboiling point of 126".

Table 4-1Physical Propert ies of Alkanes, CH,(CH,),-,H

Bp , "C MP , Den sity at 20,n Name (760 mm) "C dZ0,g ml- l

methaneethanepropanebutanepentanehexaneheptaneoctanenonanedecaneundecanedodecanepentadecaneeicosanetr iacontane

"At the bo i l ing point. bUn der pressure. "For the supe rcooled l iqu id.


3/41

4-1 Physical Properties of Alkanes. The Concept of Homology

Figure 4-1 Dependence on n of melting points, boiling points, and den-sities (d2O)of continuous-chain alkanes, CH,(CH,),-,H

M embers of a group of compound s, such as the alkanes, that have simi-lar chemical structures and graded physical properties, and which differ fromone another by the number of atoms in the structural backbone, are said toconstitute a homologous ser ies . When used to forecast the propert ies of un-known mem bers of th e series, the concept of h o m o l o g y wo rks m ost satisfac-torily for the higher-molecular-weight members because the introduction ofadditional CH, groups makes a smaller relative change in the overall com-position of such molecules. This is better seen from Figu re 4-2, which shows

Figure 4-2 Dependence of A T (difference in boiling and melting pointsbetween consecutive members of the series of continuous-chain alkanes)on n (number of carbon atoms)


4/41

4 Alkaneshow AT, the differences in boiling points and m elting po ints betw een co nse cu-tive members of the homologous series of continuous-chain alkanes, changeswith the number of carbons, n.Branched-chain alkanes do not exhibit the same smooth gradation ofphysical properties as do the continuous-chain alkanes. Usually there is toogreat a v ariation in m olecular structure fo r regularities to b e apparen t. Never-theless, in any one set of isomeric hydrocarbons, volatility increases withincreased branching. T his ca n be s een from the d ata in Ta ble 4-2, which liststhe physical properties of the five hexane isomers. The most striking featureof the data is the 19" difference between the boiling points of hexane and2,2-dimethylbutane.

Exercise 4-1 Use the d ata of Tables 4-1 and 4-2 to estimate the bo il ing points oftetradecane, heptade cane, 2-methyl hexane, and 2,2-dimethylpentane.

Table 4-2Phys ical Properties of Hexane Isomers

Isomer StructureDensity

BP, MP, at 20 dzO,"C "C g ml-'

hexane CH,(CH2),CH3 68.7 -94 0.659

CH3 CH,I I2,3-dimethyl butane CH3CH-CHCH, 58.0 -129 0.661CH3I2,2-dimethyl butane CH3CCH,CH3 49.7 -98 0.649IC H3


5/41

4-2 Chemical Reactions of Alkanes. Combustion of Alkanes 73Homology hardly can be overestimated as a practical aid for the or-ganic chemist to cope with the large numbers of compounds with which heworks. In the simplest approximation, the members of a homologous seriesare assum ed to h ave essentially the same propert ies, except for increases inboiling point and melting point as show n in Figure 4-1 for alkanes. Thi s gen-

erally will be true, except when the number of carbons is small and when thehydrocarbon chain has polar substituents. T o explain briefly, cons ider com-80 80pounds such a s alcohols, R O H , which have polar 0--- H groups. A s we indi-cated in Section 1-3, polarity causes m olecules to as sociate with one anoth er,which decreases their volatility, raises melting points, increases solubilityin polar l iquids, and decreases solubility in nonpolar l iquids. This explainswhy methanol , CH,OH, is much less volat i le and much more water-solublethan methane, CH ,. Bu t w e find that th e water-solubility of alcoho ls falls off

rapidly with t he length of the c arbon chain, certainly faster than e xpec ted fora simple homologous series effect. W hereas methanol , C H 3 0 H , and ethanol ,C H 3 C H 2 0 H ,ary completely so luble in wate r, bu tano l, C H 3C H 2C H 2 C H 20 H ,is only slightly soluble. This illustrates the conflicting properties conferredon molecules by polar groups compared to nonpolar hydrocarbon groups, andpoints up th at large change s in physical properties ca n be expe cted in the earlypart of a homologous series until the hydrocarbon chain is sufficiently long,usually six or m ore carbons, s o that th e hydrocarbon parts dom inate over thepolar parts of the molecules.

Exercise 4-2 Write detai led structures and predict which compound in each pairwould have (1) the lower boi l ing point and (2 ) the higher water solubil i ty.a. H2NCH2CH2N H2, 3CCH2CH2CH, d. CH3C02H,H C 0 2 C H 3b. CH30CH3 ,CH3CH20H e- CH3(CH2),CO2H, CH3(CH2),C02Hc. CH3CH2CH2CH20H,CH3),COH

4-2 CHEMICAL REACTIONS OF ALKANES.COMBUSTION OF ALKANESA s a class, alkanes general ly are unreact ive. T he names saturated hydrocarbon,o r "paraffin," wh ich literally m ean s "not enoug h affinity" [L. par ( um) , notenough, + a f i n s , affinity], arise because their chemical "affinity" for mostcommon reagents may b e regarded a s "saturated" o r satisfied. Th us none ofthe C-H or C-C bonds in a typical saturated hydrocarbon, for example ethane,are at tacked a t ordinary temperatures by a strong acid, such a s sulfuric acid( H 2 S 0 4 ) ,or by an oxidizing agent, such a s bromine (in the dark), oxygen, orpotassium permanganate (KMnO,). Under ordinary condit ions, ethane is


6/41

4 Alkaness imilarly s table to reducing agents su ch as hydrogen, ev en in the presenc e ofcatalysts such as platinum, palladium, o r nickel .How ever , a ll sa tura ted hy drocarbons a re a t tacked by oxygen a t e le -vated te mp erature s and, if oxygen is in excess , com plete combu stion to carbondioxide and water occurs . Vast quanti t ies of hydrocarbons from petroleumar e util ized a s fuels for th e production of heat and power by combu stion, a l-though i t is becoming quite clear that few of the nations of the world ar e goingto c ontinu e to satisfy their ne eds (or desires) for energy thlough use of petro-leum the way i t has been possible in the past .

Petroleums differ considerably in composition depending on their source. How-ever, a representative petroleum1 on distillation yields the following fractions:

1. Gas fraction, boiling point up to 40, contains normal and branched al-kanes from C, to C,. Natural gas is mainly methane and ethane. "Bottled"gas (liquefied petroleum gas) is mainly propane and butane.

2. Gasoline, boiling point from 40" to 180", contains mostly hydrocarbonsfrom C, to C,,. Over 100 compounds have been identified in gasoline, and theseinclude continuous-chain and branched alkanes, cycloalkanes, and alkylben-zenes (arenes). The branched alkanes make better gasoline than their con-tinuous-chain isomers because they give less "knock" in high-compressiongasoline engines.

3. Kerosine, boiling point 180" to 230, contains hydrocarbons from C,, toC,,. Much of this fraction is utilized as jet engine fuels or is "cracked" tosimpler alkanes (and alkenes).

4. Light gas oil, boiling point 230" to 305", C13 to C,,, is utilized as dieseland furnace fuels.5. Heavy gas oil and light l~lbricatirzg istillate, boiling point 305" to 405",C18 to C25.6. L~lbricants,boiling point 405" to 515", C2, to C,,, familiarly encounteredas paraffin wax and petroleum jelly (Vaseline).

7. The distillation residues are known as asphalts.The way in which petroleum is refined and the uses for it depend very much

on supply and demand, which always are changing. However, the situation forthe United States in 1974 is summarized in Figure 4-3, which shows roughlyhow much of one barrel of oil (160 liters) is used for specific purposes.

In the past three decades, petroleum technology has outpaced coal tech-nology, and we now are reliant on petroleum as the major source of fuels andchemicals. Faced with dwindling oil reserves, however, it is inevitable that coalagain will become a major source of raw materials. When coal is heated athigh temperatures in the absence of air, it carbonizes to coke and gives off agaseous mixture of compounds. Some of these gases condense to a black vis-cous oil (coal tar), others produce an aqueous condensate called ammoniacalliquors, and some remain gaseous (coal gas). The residue is coke, which isused both as a fuel and as a source of carbon for the production of steel. Themajor component in coal gas is methane. Coal tar is an incredible mixture ofcompounds, mostly hydrocarbons, a substantial number of which are arenes.Coal and coal tar can be utilized to produce alkanes, but the technology involved

,See F. D. Rossini, "Hydrocarbons in Petroleum," J. Chem. Educ. 37, 554 (1960).


7/41

4-2 Chemical Reactions of Alkanes. Combustion of Alkanes 75is m ore com plex and costly than petroleum ref ining. I t seem s inevitable thatthe co st of hydr ocarb on fuel will continue to r ise as supply problems becom emo re difficult . An d the re is yet no answ er to what will happ en w hen the world'sl imited quanti t ies of petroleum and coal are exhausted.

- 8,500,000 bar1a day

Figure 4-3 Sources and uses of petroleum in the United States in 1976


8/41

4 Alkanes

4-3 COMBUSTION. HEATS OF REACTION. BOND ENERGIESAll hydrocarbons are attacked by oxygen at elevated temperatures and, ifoxygen is in excess, complete combustion occurs to carbon dioxide and water:

CH, + 20, -+ CO, + 2H2OThe heat evolved in this process- the heat of the combustion reaction,AH- is a measure of the amount of energy stored in the C-C and C-H bondsof the hydrocarbon compared to the energy stored in the products, carbondioxide and water. It can be measured experimentally with considerable accu-racy and generally is reported as AH0 the amount of heat (in kilocalorie^)^liberated on complete combustion of one mole of hydrocarbon when the reac-tants and the products are in standard states, and at the same temperature,usually 25.3 Not all chemical reactions that occur spontaneously liberate heat-some actually absorb heat. By convention, AH0 s given a negative sign whenheat is evolved (exothermic reaction) and a positive sign when heat is absorbed(endothermic reaction). The heat evolved or absorbed also is called the entkalpychange.

For the combustion of 1 mole of methane at 25", we find by experiment(corrected from constant volume to constant pressure, if necessary) that thereaction is exothermic by 212.8 kcal. This statement can be expressed asfollows:CH4(g) + 20, (g) -+ CO, (g) + 2H 20 1) AH0= -212.8 kcalThe symbol (g) denotes that the reactants and products are in the gaseousstate except for the water, which is liquid (I). If we wish to have AH0 withgaseous water H,O (g) as the product we have to make a correction for theheat of vaporization of water (10.5 kcal mole-I at 25"):CH, (g) + 20, (g) -+ CO, (g) + 2H20 1) AH0= -212.8 kcal

2H2O (1 )- 3 2 0 g) AH0=4-2 x 10.5 kcalCH4(g)+ 20, (g) -+ C02(g)+ 2H,O (g) AH0=- 91.8 kcal21n this book we use kilocalories in place of the presently recommended (SI) joules forunits of energy. As of the date of writing, it is not clear just how general the use of thejoule will become among chemists. To convert calories to joules (or kcal to kJ), mul-tiply by 4.184.3 Y o ~ ay wonder how a reaction, such as combustion of methane, can occur a,t 25".The fact is that the reaction can be carried out at any desired temperature. The impor-tant thing is that the AH0 value we are talking about here is the heat liberated or ah-sorbed when you start with the reactants at 25" and finish with the products at 25".As long as AH0is defined this way, it does not matter at what temperature the reactionactually occurs. Standard states for gases are 1 atm partial pressure. Standard statesfor liquids or solids usually are the pure liquid or solid at 1 atm external pressure.


9/41

4-3 Combustion. Heats of Reaction. Bond EnergiesT h e task of measuring the h eats of all chemical reac tions is a formidableone and about as practical a s counting grains of sand on the beach. H owe ver,it is of practical interest to be able to estimate heats of reaction, an d this c anbe d one quite simply with the aid of bond energies. T h e nec essar y bondenergies are given in Table 4-3, and it is important to notice that they apply

only to complete dissociat ion of gaseous substances t o gaseous at om s at 25OC.Also, they do not apply, without suitable corrections, to many compounds,such as benzene, that have more than one double bond. This l imitat ion willbe discussed in Chapters 6 and 21.T o calculate AH0for the combustion of one mole of methane, first webreak bonds as follows, using 98.7 kcal mole-l for the ene rgy of each of the

Table 4-3Bond Energies (kcal mole-' at 25C)a

Diatomic Molecules

H-H 104.2 F-F 37.5 H-F 135.9O=O 1 18.9 CI-CI 58.1 H-CI 103.1N=N 226.8 Br-Br 46.4 H-Br 87.4C=Ob 257.3 1- 1 36.5 H-l 71.4

Polyatomic Molecules

C-HN-H0 - HS-HP-HN-NN=N0-0S-SN -0N=O

C-CC=CC-CC-NC=NC--NC -0C=Odc=oeC=Ofc=og

C-FC-CIC-BrC-lC-Sc=scN-FN-CI0 -F0-CI0- Br

"The b ond energies for diatomic molecu les in this table are from the extensive and up-to-datecompi lat ion of J. A. Kerr, M. J. Parsonage, and A. F. Trotman-Dickenson in the Handbook ofChemistry and Physics, 55th ed., CRC Press, 1975, pp. F-204 to F-208; those for polyatomicmolecules are from L. Pauling, The Nature of the Chemical Bond, 3rd ed., Cornell UniversityPress , Ithaca, N.Y., 1960.bCarbon mo nox ide. "For carbo n disulf ide . dFor carbo n dioxid e. "For formaldehyde. fOtheraldehydes., gKetones.


10/41

4 AlkanesC-H bonds,

WH : c : H ( ~ ) - . c - ( ~ ) + ~ H . ( ~ ) A H0 =+ 4 x9 8 .7 k ca l

W =+394.8 kcaland then 118.9 kcal molep1 for the energy of the double bond in oxygen:2 : G : : o : (g )-+ 4 : 6. (g) AH0=+2 x 118.9 kcal

=+237.8 kcalThen we make bonds, using 192 kcal molep1 for each O=C bond in carbondioxide,b . ( g ) + 2 : 0 ( g ) - - - + : ~ : : C : : o : ( g ) AH0=-2X192.0kcal

= -384.0 kcaland 110.6 kcal mole-I for each of the H - 0 bonds in water:2 : ~ ( g ) + 4 H . ( g ) + 2 ~ : ~ : H ( g ) AH( )= -4 ~1 1 0 .6 k c a l

= -442.4 kcalThe net sum of these AN0 values is 394.8 + 237.8 - 84.0 - 42.4 = 193.8kcal, which is reasonably close to the value of - 9 1.8 kcal for the heat ofcombustion of one mole of methane determined experimentally.

The same type of procedure can be used to estimate AH0 values formany other kinds of reactions of organic compounds in the vapor phase at25". Moreover, if appropriate heats of vaporization are available, it is straight-forward to compute AH0 for vapor-phase reactions of substances which arenormally liquids or solids at 25". The special problems that arise when solu-tions and ionic substances are involved are considered in Chapters 8 and 11.

It is important LO recognize that the bond energies listed in Table 4-3for all molecules other than diatomic molecules are average values. That theC-H bond energy is stated to be 98.7 kcal does not mean that, if the hydro-gens of methane were detached one by one, 98.7 kcal would have to be putin at each step. Actually, the experimental evidence is in accord with quitedifferent energies for the separate dissociation steps:

CM4 --+ CM,,. + H - AN0= 104 kcalCH:< ----+ CH,a + 11- AN0= 111 kcal

-CH,. -+ .CW +IJ. AH0= 101 kcal.C H + c . (g) +H e AH0= 81 kcalCH4 --+ C . (g ) + 4 He AH0= 397 kcal at 25"

The moral is that we should try to avoid using the bond energies in Table 4-3as a measure of AH0 for the dissociation of just one bond in a polyatomic mole-cule. For this we need what are called bond-dissociation energies, some ofwhich are given in Table 4-6. The values given have been selected to emphasize


11/41

4-3 Com bustion. Heats of Reaction. Bond Energies 99how s tructu re influences bond energy. Th us , C-H bond energies in alkanesdecrease in the order primary > secondary > tertiary ; likewise, C-H bondsdecrease in strength along the series C=C-H > C=C-W > C-C-H.How accura te a re AN0 values calculated from bond energies? Gen-erally quite good, provided nonbonded interactions between the atoms aresmall and the bond angles and distances are close to the normal values (seeSection 2-2B). A few examples of calculated and experimental heats of com-bustion of so me hydro carbon s are given in Tab le 4-4. Neg ative discrepanciesrepresent heats of combustion smaller than expected from the average bondenergies and positive values correspond to larger than expected heats ofcombustion.

Table 4-4Calculated and Experimental Heats of Combustion of GaseousHydrocarbons at 25C

AH0 of combustion, AH0 of combustion,calculate d from experimentalbon d energies, values," Discrepan cy,Hydro carbon kcal mole-' kcal mole-' kcal mole-'

CH, CH,I ICH3-C-C-CH, -1 221.8I ICH, CH,

7::H, - H,/C &

' CH, CH,I I

"Based on the in divid ual heats of formation com piled by D. R. Stull, E. F.Westrum, Jr., and G. C. Sinke, The Ch em ical Thermodynamics of Orga nicCompounds, John Wiley and Sons, Inc., New York, 1969.


12/41

4 AlkanesComparing isomers in Table 4-4, we see that 2-methylpropane and

2,2,3,3-tetramethylbutane ive off less heat when burned than do butane andoctane, and this is a rather general characteristic result of chain branching.Cyclopropane has a AH0 of combustion 27.7 kcal molev1 greater thanexpe cted from bond energies, and this clearly is associated with the abnormalC-C-C bond angles in the ring. T he se ma tters will be discussed in detail inChapte r 12. F o r cyclohexane, w hich has normal bond angles, the heat of com-bustion is close to the calculated value.

Exercise 4-3 The heat of combu stion of 1 mole of l iquid decane to give carbondioxide and l iquid water is 1620.1 kcal. The heat of vaporization of decane at 25"is 11.7 kca l mole-' . Calculate the heat of combu stion that would be observed for al lthe participants in the vapor phase.Exercise 4-4 Kilogram for ki logram, wou ld the combustion of gaseous methane orof l iqu id decane (to CO, and l iqu id water) give more heat?Exercise 4-5 Use the bond -energy table (4-3) to calculate A H 0 for the followingreactions in the vapor phase at 25":a. CH,CH,CH, + 50, - C 0 2 + 4 H 2 0b. CH, +go,- O + 2 H 2 06. CO + 3H2--+ CH, + H 2 0Exercise 4-6 Calculate A H 0 for C(s)- (g) from the heat of combustion of 1gram-atom of carbon to CO, as 94.05 kcal, an d the bo nd ene rgies in Table 4-3.Exercise 4-7 The dissociation HO-H - O. + H. or gaseous water at 25" hasA H 0 equ al to +119.9 kcal. What is AH0 for dissociation of the 0- H bond of HO .?Exercise 4-8 Methane reacts slowly with b rom ine atoms and it has been estab-l ishe d that A H 0 for the follow ing reac tion is 1 7 kcal pe r m ole of CH,:CH, + B r . -+ CH,. + HBr A H 0= 3-17 kca la. Ca lculate the C-H bo nd strength of CH, from this result and any other req uiredbond energies you choose to employ.b. The heat of the following reaction in the vapor state is 192 kcal per mole of CH,:CH, + 20, -+ CO, + 2 H 2 0 ( g ) A H 0= 192 kcalFrom AH0 and any other required bond energies in Table 4-3, compute a secondC-H bond-e nergy value for methane.c. Con sider whether the two C-H bond-e nergy values obtained in Parts a and bshould be the same in theory and experiment, provided that the experimental erroris small.


13/41

4-4 Halogenation of Alkanes. Energies and Rates of Reactions4-4 HALOGENATION OF ALKANES.ENERGIES AND RATES OF REACTIONST h e econ om ies of the highly industrialized nations of the world are based inlarge part on energy and chemicals produced from petroleum. Although themost important and versat i le intermediates for conversion of petroleum tochemicals are compounds with double or t r iple bonds, i t also is possible toprepare many valuable substances by substitution reactions of alkanes. Insuch subst i tut ions, a hydrogen is removed from a carbon chain and anotheratom o r group of atoms becomes a t tached in its place.A simple example of a su bstitution reaction is the forma tion of chloro-methane f rom m ethane a nd ch lor ine:

The equat ion for the react ion is s imple, the ingredients are cheap, and theproduct is useful . H ow eve r, if w e want to decide in adv anc e whe ther such areact ion is actual ly feasible, we have to know m ore. Part icularly , we have toknow whether the reaction proceeds in the direction it is written and, if so,whether conditions can be found under which it proc eed s at a convenient rate.Obviously, if one w ere t o mix m etha ne and chlorine and find that, at most, only1% conversion to the desired product occurred and that the 1% conversioncould be achieved only after a day or so of strong heating, this reaction wouldbe both too unfavorable and too slow for an industrial process.O ne way of visualizing the problems involved is with energy diagrams,which sh ow the energy in terms of som e arbi trary reaction coordinate that is ame asur e of progre ss between the initial and final states (F igure 4-4). Diag ram ssuch a s Figu re 4-4 m ay n ot b e familiar to you, and a mechanical analogy maybe helpful to provide better understanding of the very important ideas in-volved. Consider a two-level box containing a number of tennis balls. Ananalog to an energetically favorable reaction w ould be to ha ve all of the balls

energet ical ly favorable react ion energet ical ly unfavorable react ion( reactants ,barrier

products#'react ion coordinate

ducts

L react ion coordinateFigure 4-4 Schema tic energy diag ram s for react ions that are ener-get ical ly favorable and unfavorable when proceeding from lef t to r ightalong the react ion coordinate


14/41

82 4 Alkaneson the upper level where any disturbance would cause them to roll down tothe lower level under the influence of gravity, thereby losing energy.

u p p e rleve l +energet ical lyf a v o r a b l e ~ r o c e s s 1 I

If the u ppe r level is modified and a low fe nce ad ded to hold th e balls in place,i t will be just as energetically favorable a s when the fen ce is not there for theballs to b e at th e lower level. T h e difference is that the process will not occurspontaneously without som e major disturbance. W e can say there is an energybarrier to occurren ce of the favorable process.

energet ical lyfavorable________,but not spontaneousprocess

The situation here has a parallel in the left side of Figure 4-4 where we showan energy barrier to the spontaneous conversion of reactants to products foran energetically favorable chemical reaction.N ow , if we shake the box hard enough, the balls on the upper level canacquire enough energy to bo unce ove r the barrier and drop to the lower level.T h e balls then ca n be said to acquire enough activation energy to surmou nt thebarrier. A t the molecular level, the activation energy must be acquired eitherby collisions betwe en m olecules a s the result of their thermal m otions, or fromsome external agency, to permit the reactants to get over the barrier and betransformed into products. W e shortly will discuss this mo re, but first we wishto i l lustrate another important concept with our mechanical analogy, that ofequilibrium and equilibration.With gentle shaking of our two-level box, all of the balls on the upperlevel are expected to wind up on the lower level. There will not be enoughactivation to have them go from the lower to the upper level. In this circum-stance, we can say that the balls are not equil ibrated between the lower andupper levels. However, if we shake the box vigoro~tsly nd continuously, nomatter whe ther we start with all of the balls on the lower or the upp er level,an eq~lilibriurnwill be set up with, on the average, most of the balls in theenergetically mo re favo rable lower level, but som e in the up per level as well.

vigorous'shaking


15/41

4-4A The Question of the Equilibrium Constant 83To maintain a constant average fraction of the balls at each level with

vigorous and continued shaking, the rate at which balls go from the upper tothe lower level must be equal to the rate that they go in the opposite direction.The balls now will be equilibrated between the two levels. At equilibrium, thefraction of the balls on each of the two levels is wholly independent of theheight of the barrier, just as long as the activation (shaking) is sufficient topermit the balls to go bo th ways.

The diagrams of Figure 4-4 are to be interpreted in the same generalway. If thermal agitation of the molecules is sufficient, then equilibrium canbe expected to be established between the reactants and the products, whetherthe overall reaction is energetically favorable (left side of Figure 4-4) or ener-getically unfavorable (right side of Figure 4-4). But as with our analogy, whenequilibrium is established we expect the major portion of the molecules to bein the more favorable energy state.

What happens when =ethane is mixed with chlorine? N o measurablereaction occurs when the gases are mixed and kept in the dark at room tem-perature. Clearly, either the reaction is energetically unfavorable or the energybarrier is high. The answer as to which becomes clear when the mixture isheated to temperatures in excess of 300" or when exposed to strong violet orultraviolet light, whereby a rapid or even explosive reaction takes place.Therefore the reaction is energetically favorable, but the activation energy\is greater than can be attained by thermal agitation alone at room temperature.Heat or light therefore must initiate a pathway for the reactants to be con-verted to products that has a low barrier or activation energy.

Could we have predicted the results of this experiment ahead of time?First, we must recognize that there really are several questions here. Couldwe have decided whether the reaction was energetically favorable? That thedark reaction would be slow at room temperature? That light would cause thereaction to be fast? We consider these and some related questions in detailbecause they are important questions and the answers to them are relevantin one way or another to the study of all reactions in organic chemistry.4-4A The Question of the Equi l ibr ium ConstantPresumably, methane could react with chlorine to give chloromethane andhydrogen chloride, or chloromethane could react with hydrogen chloride togive methane and chlorine. If conditions were found for which both reactiansproceeded at a finite rate, equilibrium finally would be established when therates of the reactions in each direction became equal:CH4+ C12c-L CH3C1+ HClAt equilibrium, the relationship among the amounts of reactants and productsis given by the equilibrium constant expression

in which K,, is the equilibrium constant.


16/41

4 AlkanesThe quantities within the brackets of Equation 4-1 denote either con-centrations for liquid reactants or partial pressures for gaseous substances.If the equilibrium constant K,, is greater than 1 , then o n mixing equal volume sof each of the participant subs tanc es (all are gases above -24"), reaction toth e right will be initially faster than reaction to the left, until equilibrium is

established; at this point there will be more chloromethane and hydrogenchloride present than methane and chlorine. However, if the equilibrium con-stant were le s s than 1 , the reaction initially would proceed faster to the le f tand, at equilibrium, there would be more methane and chlorine present thanchloromethane and hydrogen ~ h l o r i d e . ~or methane chlorination, we knowfrom experiment th at the reaction go es to the right and that K,, is much grea terthan unity. Naturally, it would be helpful in planning other organic prepara-tions to be able to estimate K,, in advance.

the reaction CH, + CI, --+H,CI + HC I when CH, an d CI, are mixe d, ea ch at oneatmosphere pressure. Assume that K,, = loT8.

It is a comm on experience to associate chemical reactions with equilib-rium constan ts greater than one with the evolution of heat, in other words,with negative AH 0 values. T he re are, in fact, many striking examples. Forma-tion of chloromethane and hydrogen chloride from methane and chlorine hasa K,, of 1018 and AH0 of -24 kcal per mole of CH ,CI formed a t 25". Combus-tion of hydrogen with oxygen to give water has a K,, of 1040 and AH 0= -57kcal per mole of water formed at 2.5". I-Iowever, this correlation between K,,and AH 0 is neither universal n or rigorous. R eactions a re known that ab sorbheat (are endothermic) and yet have K,, > 1. Ot her reactions have large AH 0values and equilibrium constan ts much less than 1.The problem is that the energy change that correlates with K,, is notA H 0 but AGO (the so-called "Gibbs standard f ree en erg^")^, and if we knowAGO, we can calculate Keqby the equationAGO = -2.303RT log K,, (4-2)41f calculations based on chemical equilibrium constants are unfamiliar to you, wesuggest you study one of the general chemistry texts listed for supplemental readingat the end of Chapter 1.5Many books and references use A FOnstead of AGO. The difference between standardfree energy AGO and the free energy AG is that AGO s defined as the value of the freeenergy when all of the participants are in standard states. The free energy for AG for areaction A + B + . --+ X + Y + . . . is equal to AGO-2.303RT log [Xl[YI . . .[Al[Bl . .where the products, [X I, [Y] . . a , and the reactants, [A], [B] . . .,do not have tobe in standard states. We shall use only AGO in this book.


17/41

4-4B Entropy and Molecular Disorderin which R is the gas constant and T is the absolute temperature in degreesKelvin. Fo r our calculation s, we shall use R a s 1.987 cal deg-l mole-l and youshould not forget to convert AGO to cal.Tables of AGO values for formation of particu lar comp oun ds (at varioustemperatures and states) from the elements are available in handbook s and th eliterature. With thes e, we can calculate equilibrium con stants q uite accurately.F or exam hle, handbooks give the following data, which ar e useful for m ethanechlorination:

C ( s >+ 2H2(g) CH, (g) AGO = 12.1 kcal (25")C ( s ) + % H 2g ) + C12 g) --+CH3C1(g) AGO =- 4.0 kcal (25")

8H2(g>+ +Cl2(g) --+ I-3[Cl(g) AGO = 22.8 kcal (25")Combining thes e with p rope r regard for sign gives

C H , ( g ) ---+ C ( s ) f 2H2(g) AGO = +12.1 kcalC ( s >+ 8 H 2 ( g ) + +C12(g)--+ C H, C l (g ) AGO = -14 kcal

i H 2 ( g ) f i C 1 2 ( g )--+ H C l ( g ) AGO= 22.8 kcalC H , ( g ) + C12 ( g ) --+ C H , C l ( g ) + H C l ( g ) AGO = 24.7 kcal (25")

and log K,, = -(-24.7 x 1000)/(2.303 x 1.987 x 298.2), so K,, = 1.3 X 1018.Unfortunately, insufficient AGO values for formation reactions are availableto make this a widely applicable method of calculating K,, values.The situation is not wholly hopeless, because there is a relationshipbetween AGO and AN0 that also involves T and another quantity, AS0, thestandard entropy change of the p rocess:

This equation shows that AGO and AH0are equal when AS0is zero. Th erefo rethe sign and magnitude of TASOdetermine how well K,, correlates with AH0.N ow , we have to give attention to whether w e can estimate TASO alues wellenough to decide whether the AH0 of a given reaction (calculated from bondenergies o r other information) will give a good o r poor measure of AGO.

4-4B Entropy and M olecular DisorderT o decide whether we need to worry about AS0 with regard to any particularreaction, we have to h ave some idea what physical meaning entropy has. T obe very detailed about this subject is beyond the scope of this book, but youshould try to understand the physical basis of entropy , because if you d o, thenyou will be able to predict a t least qualitatively w heth er AH0 will be a bou t thesam e o r very different from AGO. Essentially, the en tropy of a chemical systemis a measure of its molecular disorder or randomness. Other things being thesame, the more random the system is, the m ore favorable the system is.


18/41

4 AlkanesDifferent kinds of molecules have different degrees of translational,vibrational, and rotational freedom and, hence, different average degrees ofmolecular disorder o r randomn ess. N ow , if for a chemical reaction the degreeof molecular disorder is different for the pro ducts than fo r the reactants, therewill be a change in entropy an d AS0 f 0.A spectacu lar ex am ple of the effect of molecular disorder in contributingto the difference between AH0 and AGO is afforded by the form ation of liquidnon ane, C,H,,, from solid car bo n and hyd rogen gas at 25":

9 C ( s ) 4- 10H2 g ) -+ C9H2,(1) A H 0= 54.7 kcalAGO = 5.9 kcalEquations 4-2 and 4-3 can b e rearranged to calculate ASo and Keq from AN0and AGO:

K = 10-AG0/(2.303 T )= 10-5.900/(2.303X 1.987 X 298.2) = 4 7 x 10-5eqTh ese AH 0, AS0, and Keq values can be compared to those for EI, -t l 1 2 0 ,---+ H,O , fo r which A H0 is -57 kcal, AS0 is 8.6 e.u., and Keq s 1040.Obviously,there is something unfavorable abo ut the entropy change from carbon andhydrogen to no nane. T he important thing is that there is a great diference inthe con straints on the atom s on each side of the equation. In particular, hydro-gen molecules in th e gaseous state ha ve great translational freedom and a highdegree of disorder, the gre ater part of which is lost when th e hydrogen atom sbecom e attached to a chain of carbo ns. Th is makes for a large negative AS0,which corresponds to a decrease in K,,. The differences in constraints of thecarbons are less important. Solid carbo n has an o rdered , rigid structure withlittle freedom of motion of the individual carbon atoms. These carbons areless constrained in nonan e, an d this would tend to m ake AS0 more positive andAGO more negative, corresponding to an increase in Keq (see Equations 4-2and 4-3). H ow ever, this is a sm all effect on AS0 comp ared to the enormou sdifference in the degree of disorder of hydrogen between hydrogen gas andhydrogen bound to carbon in nonane.Negative entropy effects usually are observed in ring-closure reactionssuch as the formation of cyclohexane from I-hexene, which occur with sub-stantial loss of rotational freedom (disorder) about the C-C bonds:

1 hexene cyclohexane6The entropy unit e.u. has the dimensions cal per degree or cal deg-l.


19/41

4-48 Entropy and Molecular Disorder 87There is an even greater loss in entropy on forming cyclohexane fromethene because substantially more freedom is lost in orienting three ethenemolecules to form a ring:

ethene (3 moles)F o r simple reactions, with the sam e numb er of molecules on eac h sideof the equation, with no ring formation or other unusual changes in the con-straints between the products and reactants, A S 0 usually is relatively small.

In general , for such processes, we know from experience th at Keq usually isgreater than 1 if A H 0 is mo re negative than -1 5 kcal an d usually is less than 1for A H 0 more positive than +15 kcal. W e can use this as a "rule of thumb" topredict whe ther Keq should b e greater or less than unity for vapor-phase reac-tions involving simple molecules. Some idea of the degree of success to beexpected from this rule may be inferred from the exa mples in Tab le 4-5, whichalso contains a further comparison of some experimental A H 0 values withthose calculated from bond energies.Table 4-5Comparison of Calculated and Experimental AH0Values andEq uil ibriu m Constants for Some Simp le ReactionsarbReaction Expt. AH 0 Calc. AH 0 K e qCH, + CI,- H,CI + HCIC(s) + 2H, ---+ CH,3C(s) + 4H2+ 3H8 propane)6C(s) + 7H2--+ 6H14 hexane)9C(s) + 1OH, ---+ C,H, (n on ane )I -hexene --+ cyclohexane$H, + $CI,- CI:H +$N, -+NH,CO + H,O - O, + H,C H 3 0 H + HCI --+ CH,CI + H,OaAIl participants in the gaseo us state at 298.2"K unless otherwise stated, and at constant pressure.*Thermochemical values have been taken mostly from Selected Values of Chemical Thermo-dynamic Properties by F. D. Rossini, et al. (Circular of the National Bureau of Standards 500,Washington, D.C., 1952) and Selecte d Values of Ph ysic al a nd Thermodynam ic Properties ofHydrocarbons and Related Compounds by F. D. Rossini, et al. (American Petroleum InstituteResearch P roject 44, Carne gie Press, Pittsburgh, 1953)."Calculated using AH0= 171.3 kcal per m ole for C(s) ----+ C(g), at 25".dNot an independent value because the bond energy for product was calculated from the ex-perimental AH0.


20/41

4 lkanes

Exercise 4-10 a. Calculate AHo from bond energies for the conversion of 1-hexeneto cyclohexane at 25" and from this, with A S 0 as -20.7 e.u. per mole, calcu late theequi l ibr ium constant Keq from Equation 4-2. For comparison, calculate the equil ib-r ium constant that would be expected i f the degrees of disorder of the reactantsan d the products were equ al ( i .e.,A S 0= 0). b. How large can A S 0 be at 25" for a reac-tion be fore our & I 5 kcal rule starts to g ive incorrect answers?Exercise 4-11 Knowing that the equi l ibr ium constant Keq for formation of nonanefrom sol id carbon and hydrogen gas is 4.7 X explain why l iquid nonane doesnot forthwith de comp ose into i ts elements.

Suppose AGO is positive, what hope do we have of obtaining a usefulconversion to a desired product? There is no simple straightforward andgeneral answer to this question. When the reaction is reversible the classicprocedure of removing one or more of the products to prevent equilibriumfrom being established has many applications in organic chemistry, as will beseen later. When this approach is inapplicable, a change in reagents is nec-essary. Thus, iodine does not give a useful conversion with 2,2-dimethylpro-pane, 1, to give 1-iodo-2,2-dimethylpropane, , because the position ofequilibrium is too far to the left (K, , = lop5)

Alternative routes with favorable AGO values are required. Development ofwa ys t o m ake indirectly, by efficient proce sses, w hat c anno t be made directlyis one of the most interesting and challenging activities of organic chemists.

4-4C Why Do Methane and Chlorine Failto React in the Dark at 25"?T o reach an unde rstanding of why methane and chlorine do not react in thedark, we must consider the details of how the reaction occu rs-that is, thereaction mechanism . T h e simplest mechan ism would b e for a chlorine moleculeto collide with a methane molecule in such a way as to have chloromethaneand hydrogen chloride formed directly as the result of a concerted breakingof the Cl-Cl and C-H bond s and making of the C-Cl and H-Cl bonds (seeFigure 4-5). T h e failure to react indicates that there m ust be an energy barriertoo high for this mechanism to operate. Why should this be so?


21/41

4-4C Why Do Methane and Chlorine Fail to React in the Dark at 25"?

@ CH,CI + HCIFigure 4-5 Four-center col l ision of chlorine with methane as visual izedwith bal l -and-st ick models

Fir st, this m echan ism .involves a very precisely oriented "four-center"collision between chlorine and methane that would have a low probability ofoccurrence (i .e. , a large decrease in entropy because a precise orientationmean s high mo lecular ordering). S econ d, i t requires pushing a chlorine mole-cule sufficiently dee ply into a 'methane molecule so on e of the c hlorine atom scomes close enough to the carbon to form a bond and yield chloromethane.Generally, to bring nonbonded atoms to near-bonding distances (1.2A to 1.8 A) requires a large expenditure of energy, as can be seen in Figure4-6. Interatom ic repulsive fo rces increase rapidly at sh ort distances, and push-ing a chlorine molecule into a methane molecule to attain distances similarto the 1.77-A carbon-chlorine bond distan ce in chloro me thane would requirea con siderable amo unt of comp ression (see Figure 4-7). Valuable information

kca l

Figure 4-6 Graph of the potent ial energy of pairs of neon atoms as afunction of the internuclear distance. The energy values are per mole ofneon atoms.


22/41

4 Alkanes

Figure 4-7 Models showing the deg ree of atomic com pression requiredto b r ing a chlo r ine molecu le to wi thin bon ding distanc e of carbon andhydrogen of methaneabout interatomic repulsions can be obtained with space-filling models of theC P K ty pe (Section 2-2), which have radii scaled to correspond to actual atomicinterference radii, that is, the interatomic distance at the point where curvesof the type of Figure 4-6 start to rise steeply. With such models, the degreeof atomic com pression required to bring th e nonbonded atom s to within near-bonding distance is more evident than with ball-and-stick models. It may benoted that four-center reactions of the type postulated in Figure 4-5 are en-countered only rarely.

If the concerted fou r-center mechanism for formation of chloromethaneand hydrogen chloride from chlorine and methane is discarded, all the re-maining possibilities are stepwise reaction nzechanisms. A slow stepwisereaction is dynamically analogous to the flow of sand through a successionof funnels with different stem diameters. The funnel with the smallest stemwill b e the m ost impo rtant bottleneck and , if i ts stem diameter is much smallerthan the others, i t alone will determine the flow rate. Generally, a multistepchemical reaction will have a slow rate-determining step (analogous to thefunnel with th e small stem) an d othe r relatively fast steps, which may occureither before or after the slow step.A possible set of ste ps for the ch lorination of methane follows:

slow ..( 1 ) C1, - : C I Sslow( 2 ) C H , -----+ CH, , . + Hafast(3 ) :CI.+ CH, , . ----+ CH,Clfast(4 ) : C I - +H ----+ HCI

Reactions 1 and 2 involve dissociation of chlorine into chlorine ato m s and thebreaking of a C-H bond of methane to give a methyl radical and a hydrog enatom. T h e methyl radical, l ike chlorine and hy drogen atoms, ha s one electronnot involved in bonding. Atoms and radicals usually are highly reactive, so


23/41

4-4D Why Does Light Induce the Chlorination of Methane? 91formation of chloromethane and hydrogen chloride should proceed readilyby R eactions 3 and 4. T he c rux then will be whether S teps 1 and 2 are reason-able under th e reaction conditions.In the absence of some external stimulus, only collisions d ue to the usualthermal motions of the molecules can provide the energy needed to breakthe bonds. A t tem pera tures ,belpw 100, it is very rare indeed that thermalagitation alone ca n supply sufficient energy to break an y significant num ber ofbonds stronger than 30 to 35 kcal mole-l.T h e Cl-CI bond energy from T ab le 4-3 is 58.1 kcal, which is much toogreat to allow bond breaking from therm al a gitation a t 25" in acco rd with Rea c-tion 1. F o r R eaction 2 it is not advisable to use th e 98.7 kcal C-H bond energyfrom Table 4-3 because this is one fourth of the energy required to break allfour C-H bon ds ( see Section 4-3). M ore specific bond-dissociation energiesare given in Table 4-6, and it will be seen that to break one C-H bond ofmethane requires 104 kcal at 25", which again is too much to be gained bythermal agitation. Therefore we can conclude that Reactions 1-4 can not bean important m echanism for chlorination of methane at room tem perature.One might ask whether dissociation into ions would provide viablemechanisms for methane chlorination. Part of the answer certainly is: Notin the vapor phase, as the following thermochemical data show:

CI, - c1@+ : e l : @. . AH0 = 270 kcalC H , -+ CH,:@ + H@ AHo= 400 kcalIonic dissociariorl simply does not occ ur at ordinarily accessible tem peraturesby collisions i:n m olecules in th e vapor state . W hat is neede d fo r form a-tion of ions is eithe r a highly energe tic extern al stimulus, such as bom bard m entwith fast-moving electrons, or an ionizing solvent that will assist ionization.Both of these processes will be discussed later. The point here is that ionicdissociation is not a viable step for the vapor-phase chlorination of methane.

4-4D Why Does L ight Indu ce the C hlorination of Methane?First, we should make clear that the light does more than provide energymerely to lift the m olecules of methane and chlorine over the barrier of Figure4-4. This is evident from the fact that very little light is needed, far less thanone light photon per molecule of chloromethane produced. The light couldactivate either methane or chlorine, or both. However, methane is colorlessand chlorine is yellow-green. This indicates tha t chlorine, not m ethane, inter-acts with visible light. A photon of near-ultraviolet light, such as is absorbedby chlorine gas, provides mo re than enough energy to split the m olecule intotwo chlorine atom s:


24/41

92 4 AlkanesTable 4-6Bond -Dissociation Energies at 25"

CompoundBond energy(kcal mole-')aBond energy(kcal mole-I)a Compound

/cQCH, CH+H\ /CH2

"These values are mostly from the compilations of K. W. Egger and A. T. Cocks, Helv. chim. Acta56, 1516 (1973), and J. A. Kerr, M. J. Parsonage, and A. F. Trotman-Dickenson, Handbook ofChemistry a nd Physics, 55th ed. , CRC Publishing Co., 1975, F-213 to F-216.


25/41

4-4D Why Does Light Induce the Chlorination of Methane?Once produced, a chlorine atom can remove a hydrogen atom from a

methane molecule and form a methyl radial and a hydrogen chloride mole-cule. The bond-dissociation energies of CH, (104 kcal) and HCI (103.1 kcal)suggest that this reaction is endothermic by about 1 kcal:

The attack of a chlorine atom on a methane hydrogen is not expected to requirea precisely oriented collision. Moreover, the interatomic repulsions should beconsiderably smaller than in the four-center mechanism discussed previouslyfor the reaction of molecular chlorine with methane because only two centershave to come close together (Figure 4-8). The methyl radical resulting fromthe attack of atomic chlorine on a hydrogen of methane then can remove achlorine atom from molecular chlorine and form chloromethane and a newchlorine atom:

CH,. + Cl, ---+ CH,CI + : 1 . AH0= 26 kcalUse of bond-dissociation energies gives a calculated AHo of -26 kcal for

this reaction, which is certainly large enough, by our rule of thumb, to predictthat K,, will be greater than 1. Attack of a methyl radical on molecular chlorineis expected to require a somewhat more oriented collision than for a chlorineatom reacting with methane (the chlorine molecule probably should be end-wise, not sidewise, to the radical) but the interatomic repulsion probably shouldnot b- mlich different.

-,e net result of CH, + Cl. ---+CH,. + HCl and CH,. + C1, -+CH3C1+ Cl. is formation of chloromethane arid hydrogen chloride from meth-ane and chlorine. Notice that the chlorine atom consumed in the first step isreplaced by another one in the second step. This kind of sequence of reactionsis called a chain reaction because, in principle, one atom can induce the reac-tion of an infinite number of molecules through operation of a "chain" or cycleof reactions. In our example, chlorine atoms formed by the action of light on

Figure 4-8 Mo dels show ing the degre e of atomic compression requiredto bring a chlorine atom to within b ondin g distance of a methane hydro-gen. Compare with Figure 4-7.


26/41

4 AlkanesC1, can induce th e chlorination of meth ane by the chain-propagating steps:CH4 + C1.- C l + C H 3 *

In practice, chain reactions are limited by so-called termination pro-cesses. In our example, chlorine atoms or methyl radicals are destroyed byreacting with one another, as shown in the following equations:

Ch ain reactions may be considered to involve three phases. Firs t, chaininitiation must occur, which for methane chlorination is activation and con-version of chlorine molecules to chlorine atoms by light. Second, chain-propagation steps convert reactants to products with no net consumption ofato m s or radicals. T h e propagation reactions oc cur in competition with chain-terminating steps, which result in destruc tion of atoms o r radicals. Puttingeverything together, w e can write:

chain in i t iat ionC H 4 + : C I -. -----+H, , . + H C I 1 chain propagationCH, + CI, -----+ CH,Cl + : e l ..

Ichain terminat ion

Exercise 4-12 A possible mechanism for the react ion of chlorine with methanewould be to have col l isions by which a chlorine molecule removes a hydrogen ac-cord ing to the fol lowing scheme:

slowCH3 : H + :C I :C I :. .. ---,H,. + H:CI: + :C IfastCH,. + :CI.- H,: CI:Use appropriate bond energies to assess the likelihood of this reaction mechanism.What about the possibi l i ty of a similar mechanism with elemental f luorine andmethane?


27/41

4-4D Why Does Light Induce th e Chlorination of Methane?Exercise 4-13 Calculate A H 0 for each of the propagation steps of methane chlo-r ination by a m echanism of the type

l ightCI, ------+ 2CI initiationCI + + CH, ----+ H,CI + H . propagationH + CI, -----+ Cl + CI -Compare the relative energetic feasibil i t ies of these chain-propagation steps withthose of other possible mechanisms.

The chain-termination reactions are expected to be exceedingly fastbecause atoms and radicals have electrons in unfilled shells that normally arebonding. As a result, bond formation can begin as soon as the atoms or radicalsapproach one another closely, without need for other bonds to begin to break.The evidence is strong that bond-forming reactions between atoms and radicalsusually are diBusion-controlled, that there is almost no barrier or activationenergy required, and the rates of combination are simply the rates at whichencounters between radicals or atoms occur.If the rates of combination of radicals or atoms are so fast, you mightwell wonder how chain propagation ever could compete. Of course, competi-tion will be possible if the propagation reactions themselves are fast, butanother important consideration is the fact that the atom or radical concen-trations are very low. Suppose that the concentration of Cl . is 10-llM and theCH, concentration 1M. The probability of encounters between two Cl . atomswill be proportional to 10-l1 x 10-11, and between CH, and C1. atoms it willbe 10-l1 x 1. Thus, other things being the same, CH, + Cl.- H,. +HCl(propagation) would be favored over 2C1. ---+ C1, (termination) by a factorof 10ll. Under favorable conditions, the methane-chlorination chain may gothrough 100 to 10,000 cycles before termination occurs by radical or atomcombination. Consequently the efficiency (or quantum yield) of the reactionis very high in terms of the amount of chlorination that occurs relative to theamount of the light absorbed.

The overall rates of chain reactions usually are slowed very much bysubstances that can combine with atoms or radicals and convert them intospecies incapable of participating in the chain-propagation steps. Such sub-stances are called radical traps, or inhibitors. Oxygen acts as an inhibitor inthe chlorination of methane by rapidly combining with a methyl radical toform the comparatively stable (less reactive) peroxymethyl radical, CH,OO -.This effectively terminates the chain:CH, . + 0, --+ H,OO. inhibition


28/41

4 Alkanes

4-4E Can We Predict Whether Rea ctions W ill Be Fast or Slow?To a considerable degree, we can predict relative reactivities, provided weuse comm on sense to limit our efforts to reasonable situations. In the precedingsection, we argued that reactions in which atom s or radicals combine can wellbe ex pected t o be extremely fast because e ach entity has a potentially bondingelectron in an ou ter unfilled shell, and bringing th ese togethe r to form a bo nddoes not require that other bonds b e broken:

fastC1- + .C1- 1:ClFo r the react ion C H , +C1. ---+CH,. +H C l, the methane hydrogen andcarbon valence shells are filled and, as C1- approaches, it can combine with ahyd rogen only if a C-H bond is broke n. Th is kind of process is associated witha barrier but is v ery different from a nonreactive encounter, such as two neonatom s coming together (see F igure 4-6). A s C H , and C1- get closer together,the new bond starts to form and the old bond starts to break. A t the top ofthe b arrier, the hydrogen will be bon ded pa rtly to chlorine and partly to carb on,[cl---------- --------- CH,], and this we cal l the activated complex or transitionstate. The concept of the transition state is an important one, which we willuse repeatedly later in connection with many other kinds of reactions. Thevalue of the concept l ies in the fac t that the reacting s ystem , when it reachesthe to p of the barrier, can b e thoug ht of as a chemical entity with a particular,even if not a well-defined, structure and definite thermodynamic properties.The difference between the average energy of the reactants and theenergy of the transition state is called the activation energy (Figu re 4-4). W eex pec t this energy to be smaller (lower barrier) if a weak b ond is being brok enand a strong bond is being made. Th e perceptive reader will notice that we ar esuggesting a parallel between reaction rate and AH0 because AH0 dependson the difference in the strengths of the bonds broken and formed . Ye t pre-viously (Section 4-4A), we pointed out that the energy barrier for a reactionneed bear no relationship to how energetically feasible the reaction is, andthis is indeed true for complex reactions involving many steps. But our intui-tive parallel between rate and AH0usually works quite well for the rates ofindividual steps. This is borne out by experimental data on rates of removalof a hy drogen atom from methane by atom s or radicals (X- ) ,such as F - ,Cl. ,B r . , HO . H,N., which generally parallel the strength of the new bon d form ed:

Similarly, if we look at the H-C bond-dissociation energies of the hydro-carb on s show n in Tab le 4-6, we would infer that C1. would rem ove a hydrogenmost rapidly from the carbon forming the w eakest C -H bond a nd, again, thisis very much in accord with experience. For example, the chlorination ofme thylben zene (toluene) in sun light leads to the su bstitution of a methyl hydro-


29/41

4-4E Can We Predic t Whether React ions Wil l Be Fast or Slow? 97gen rather than a ring hydrogen for the reason that the methyl C-H bonds areweaker and are attacked more rapidly than the ring C-H bonds. This can beseen explicitly in the AH0 values for the chain-propagation steps calculatedfrom the bond-dissociation energies of Table 4-6.Methyl substitution (observed).

0 C H 3 + C l C H HCI AH0=- 8 kcal0 C H 2 + C12 --+ C H 2 C I CI + AH0= 1 kcalRing substitution (not observed):

H AH0=+7 kcal

The AH0of ring-hydrogen abstraction is unfavorable by +7 kcal becauseof the high C-H bond energy (110 kcal). Thus this step is not observed. It istoo slow in comparison with the more favorable reaction at the methyl groupeven though the second propagation step is energetically favorable by -37kcal and presumably would occur very rapidly. Use of bond-dissociationenergies to predict relative reaction rates becomes much less valid when wetry to compare different kinds of reactions. To illustrate, ethane might reactwith F. to give fluoromethane or hydrogen fluoride:CH,-CH3 + F. CH,. + CH3F AH0= 20 kcalCH,-CH, + F. -+ CH3CH2.+ H F AH0= -38 kcalIt is not a good idea to try to predict the relative rates of these two reactionson the basis of their overall AH0values because the nature of the bonds madeand broken is too different.

princ ipal p rod uct to b e expected from the vapor-phase, l ght- induced monochlor ina-tion of 1, l -d imethy lcyc lopropane.


30/41

4 Alkanes4-4F How Should We Go abou t Formulatinga Reaction Mechanism?

Faced with proposing a mechanism for a reaction that involves overall makingor breaking of more than two bonds, the beginner almost invariably tries toconcoct a process wherein, with a single step, all of the right bonds break andall of the right bonds form. Such mechanisms, called concerted mechanisms,have three disadvantages. First, they are almost impossible to prove correct.Second, prediction of the relative rates of reactions involving concerted mech-anisms is especially difficult. Third, concerted mechanisms have a certainsterility in that one has no control over what happens while they are takingplace, except an overall control of rate by regulating concentrations, tempera-ture, pressure, choice of solvents, and so on.

To illustrate, suppose that methane chlorination appeared to proceed by wayof a one-step concerted mechanism:

At the instant of reaction, the reactant molecules in effect would disappearinto a dark closet and later emerge as product molecules. There is no way toprove experimentally that all of the bonds were made and formed simulta-neously. All one could do would be to use the most searching possible teststo probe for the existence of discrete steps. If these tests fail, the reaction stillwould not be proved concerted because other, still more searching tests mightbe developed later that would give a different answer. The fact is, once youaccept that a particular reaction is concerted, you, in effect, accept the proposi-tion that further work on its mechanism is futile, no matter how important youmight feel that other studies would be regarding the factors affecting the reac-tion rate.

The experienced practitioner in reaction mechanisms accepts a conceitedmechanism for a reaction involving the breaking and making of more than twobonds as a last resort. He first will try to analyze the overall transformationin terms of discrete steps that are individually simple enough surely to be con-certed and that also involve energetically reasonable intermediates.

Such an analysis of a reaction in terms of discrete mechanistic steps offersmany possibilities for experimental studies, especially in development of pro-cedures for detecting the existence, even if highly transitory, of the proposedintermediates. We shall give many examples of the fruitfulness of this kind ofapproach in subsequent discussions.

4-5 PRACTICAL HALOGENATIONS.PROBLEMS OF SELECTIVITYGiven the knowledge that a particular reaction will proceed at a suitable rate,a host of practical considerations are necessary for satisfactory operation.


31/41

4-5 Practical H alogenations. Problems of Selectivi ty 99Th ese considerations include interference by possible side reactions th at giveproducts other than those desired, the ease of separation of the desired prod-ucts from the reaction mixture, and costs of materials, apparatus, and labor.We shall consider these problems in connection with the important syntheticreactions discussed in this book.The chlorination of saturated hydrocarbons can be induced by light,but also can be carried out at temperatures of about 300" n the dark. Undersuch circumstances the mechanism is similar to that of light-induced chlori-nation, e xcep t that the chlorine atoms are formed by thermal dissociation ofchlorine molecules. Solid carbon surfaces catalyze thermal chlorination, pos-sibly by aiding in the cleavage of the chlorine molecules.Direct monohalogenation of saturated hydrocarbons works satisfac-torily only with chlorine and bromine. For the general reaction

I I-C-H + X, ---+ -C-X + H-X X = F , C l , Br, Ithe calculated AH0 value is negative and very large for fluorine, negative andmoderate for chlorine and bromine, and positive for iodine (see Table 4-7).With fluorine, the reaction evolves so much heat that it may be difficult tocontrol, and pro ducts from cleavage of carbon-carbon as well as of carbon-hydrogen bonds may be obtained. The only successful, direct fluorinationprocedure for hydrocarbons involves diffusion of minute amounts of fluorinemixed with helium into liquid or solid hydrocarbons at low temperatures,typically -78" (D ry I c e temperature). A s fluorination proceeds, the concen-tration of fluorine can be increased. The process is best suited for preparationof completely fluorinated compounds, and it has been possible to obtain inthis way amounts of (CF,) C and (CF, ) ,C-C (CF,), from 2,2-dimethyl-propane and 2,2,3,3-tetramethylbutaneorresponding to 10- 15% yields basedon the fluorine used.Bromine generally is much less reactive toward hydrocarbons thanchlorine is, both at high temperatures and with activation by light. Nonethe-less, it usually is possible to brominate saturated hydrocarbons successfully.Iodine is unreactive.Tablle 4-7Calculated Heats of Reaction for Halogenation of Hydrocarbons

X A H 0 (kcal mole- ')"

"Calculated from the bond energies of Table 4-3.


32/41

4 AlkanesThe chlorination of methane does not have to stop with the formation

of chloromethane (methyl chloride). It is usual when chlorinating methane toobtain some of the higher chlorination products: dichloromethane (methylenechloride), trichloromethane (chloroform), and tetrachloromethane (carbontetrachloride):CH,- H,Cl -+CH2CL2--+ CHCI,- Cl,chloro- dichloro- trichloro- tetrachloro-methane methane methane methaneIn practice, one can control the degree of substitution to a considerable extentby controlling the methane-chlorine ratio. For example, for monochlorinationto predominate, a high methane-chlorine ratio is necessary such that the chlo-rine atoms react with CH, and not with CH,Cl.

4-5A Selectivity in Alkane Halog ena tionFor propane and higher hydrocarbons for which more than one monosubstitu-tion product is generally possible, difficult separation problems may arisewhen a particular product is desired. For example, the chlorination of 2-methyl-butane 3 at 300" gives all four possible monosubstitution products, 4, 5, 6,and 7:

2-methyl butane3

1-chloro-2-m ethyl butane(30%)

1-chloro-3-methyl butane 3-chloro-2-methylbutane(15%) (33%)5 6


33/41

4-5A Selectivity in Alkane Halogenation l!4On a purely statistical basis, we may expect the ratio of products from 3 tocorrelate with the number of available hydrogens at the various positionsof substitution. That is, 4, 5, 6, and 7 would be formed in the ratio 6:3:2:1(50%:25%:17%:8%). However, as can be seen from Table 4-6, the strengths ofhydrogen bonds to primary, secondary, and tertiary carbons are not the sameand, -from the argument given in Section 4-4E we would expect the weakerC-H bonds to be preferentially attacked by C1.. The proportion of 7 formedis about three times that expected on a statistical basis which is in accord withour expectation that the tertiary C-H bond of 2-methylbutane should be theweakest of the C-H bonds. (See Table 4-6.)

Exercise 4-15" Use the data given above for the p ercenta ges of the monoc hloridesformed in the vapor-phase chlorination of 2-methylbutane at 300" and take into ac-count the statist ical factors for the different numbers and kinds of hydrogens in an-swer ing the fol low ing:a. From the ra tio of 1-chloro-2-methyl butane to 1-chloro-3-methyl butane formed, whatcan you say about the C-H bo nd strengths at the CH, carbons?b. Calculate the ratio of rates of attack of CI. on the individual hydrogens attachedto p rimary (C1 and C4), secondary (C3), and tert iary (C2) carbons of 2-methylbutane.Check these ra tios by show ing they are consistent with the co mp osit ion of the overal lchlor inat ion product.c. Us e your re lative rate ratios from P art b to calc ulate th e ratios of isome rs to b e ex-pected in the thermal (300") monochlorination of (a) propane, (b) 2-methylpropane,and (c) 2,2-dimethylbutane. Show your m ethod in detai l .

The factors governing selectivity in halogenation of alkanes follow:1. The rates at which the various C-H bonds of 2-methylbutane are

broken by attack of chlorine atoms approach 1: 1: 1 as the temperature israised above 300". At higher temperatures both chlorine atoms and hydrocar-bons become more reactive because of increases in their thermal energies.Ultimately, temperatures are attained where a chlorine atom essentially re-moves the first hydrogen with which it collides regardless of position on thehydrocarbon chain. In such circumstances, the composition of monochlorina-tion products will correspond to that expected from simple statistics.

2. Bromine atoms are far more selective than chlorine atoms. This isI I

not unexpected because -C-H 4- Bra -+ -C. d- HBr is endothermic,I Iwhereas corresponding reactions with a chlorine atom usually are exothermic(data from Table 4-6). Bromine removes only those hydrogens that are rela-tively weakly bonded to a carbon atom. As predicted, attack of Br. on 2-methyl-butane leads mostly to 2-bromo-2-methylbutane, some secondary bromide,


34/41

4 Alkanesand essentially no primary bromides:

CH:, CH:, CH:,lightCH,,CHCH,CH,, + Br, I> CH,,CCH,CH,, + CH,CHCHBrCH,I3. The selectivity of chlorination reactions carried on in solution isincreased m arkedly in the pre senc e of benzene o r alkyl-substituted benzenesbecause benzene and oth er arene s form loose complexes with chlorine atom s.This substantially cuts down chlorine-atom reactivity, thereby making thechlorine atom s behave more like bromine atom s.

Exercise 4-16 a. Write equations to show reasonable radical-chain ini t iation,propagation, and termination steps in the monobromination of 2-methylbutane shownabove. Explain clearly why the products of chain termination are obtained in traceamounts only.b. Use bond energies of Tables 4-3 and 4-6 and bond-dissociation energies of63 kca l for tertiary C-Br and 68 k cal for secondary C-Br bon ds to estimate AH0 fo reach of the p ropaga tion steps lea ding to the two observed produ cts. Wh ich propag a-tion step in the formation of 2-bromo-2-methylbutane is exp ecte d to b e the slow ste p?c. Calculate the relative rates of attack of bromine atoms at the tertiary C-H versusthe secondary C-H bond s from the prod uct compo sit ion in the bromination of 2-meth-ylbutane. Are the relative rates qualitatively consistent with what you would expectbased on the AH0 data?

4-5B Chemical Ini t iat ion of Ra dical-C hain Subst itutionIt is possible to achieve chlorination of alkanes using sulfuryl chloride (SO,C1,,bp 69") in place of chlorine:

I I-C-H + S02C12 -C-CI + SO, + HCII IThe reaction has a radical-chain mechanism and the chains can be initiated bylight or by chemicals, usually peroxides, ROOR. Chemical initiation requiresan initiator with a weak bond that dissociates at temperatures between 40-80".Peroxides are good examples. The 0-0 bond is very weak (30-50 kcal) andon heating dissociates to alkoxyl radicals, RO., which are reactive enough togenerate the chain-propagating radicals from the reactants. The exact sequence


35/41

4-5B Chemical Ini t iat ion of Rad ical-Chain Subst itut ion 103

of chemical initiation is not always known, but a plausible route in the presentcase would have RO. abstract hydrogen from the alkane:

I I-C-H + R O -- C. + ROH AH 0= -11.9 kcal mole-II IThe propagation steps that would follow are

I I--C-H + :CI. * C- + HCII

..1

Chlorination with sulfuryl chloride of alkanes with more than one kind of hydro-gen gives a mixture of alkyl chlorides resembling that obtained with chlorineitself. However, in some circumstances the mixture of chlorides is not the samemixture obtained with chlorine itself and when this is true, the hydrogen-abstraction step probably involves -S02C1 rather than C1.. The alternativepropagation steps then are

I I-C-H + CI-so2 --+ --C + HCI + SO2I IDifferent product ratios are expected from C1. and ClSO2. for the same rea-

son that C1. and Br. lead to different product ratios (Section 4-5A). Other re-agents that sometimes are useful halogenating agents in radical-chain reactionsinclude

tert-butyl hypochlorite brornotrichlorornethane N-brornosuccinimide

How these substances are employed is illustrated by Exercises 4-17 through4-19 and 4-36.


36/41

104 4 AI kanes

Exercise 4-17* The peroxide-induced bromination of methylbenzene with bromo-tr ichlorometha ne gives bromome thyl benzene and trichloromethane:

methyl benzene bromomethyl benzene

Write init iat ion, propagation, and termination steps for this radical-chain react ion.Est imate a AH0 for the overall react ion using the bond-dissociat ion energies ofTable 4-6. Would you expe ct bromotrichloromethane to be a select ive or nonselect ivebrominat ing agent? Explain.Exercise 4-18* tert-Butyl hypo chlorite is a useful chlorinat ing ag ent. On irradia-t ion, or with ch em ical init iators, this reagent with methylbenzene g ives chlorom ethyl-benzene:

Write a po ssib le m echa nism for the react ion, showing the propag ation steps with(CH,),CO. as the cha in-pro pa gatin g ra dic al. Us e the bond -disso ciation energ ies ofTable 4-6 to determine whether your mechanism is energetically and kinet icallyfeasible. Assume the 0- C I bon d-dissociat ion energy of tert-butyl hypoch lorite is 61kcal mole-'.Exercise 4-19*a. N-Bromo succinimide (NBS) is an excel lent brom inat ing reagent and is used wide lyto prepare bromalkenes from alkenes (Wohl-Ziegler react ion):

The react ion is init iated with c he m ical init iators (peroxide s) and is as select ive asbromination with mo lecular b romine . Write plau sible p ropag ation steps (three of them)for this react ion, given the fact that the actual brominating agent appears to be mo-lecular bromine that is generated from NBS by HBr.b. What prod ucts wo uld you expec t to be formed on brom ination of 2-methylbutanewith N-bromosuccinimide?


37/41

4-6 Nitration of Alkanes4-6 NITRATION OF ALKANESAnother reaction of commercial importance is the nitration of alkanes to givenitroparaffins. Such reactions usually are carried out in the vapor phase atelevated temperatures using nitric acid (HNO,) or nitrogen tetroxide (N,O,) asthe nitrating agent:

All available evidence points to a radical mechanism for nitration, but manyaspects of the reaction are not fully understood. Mixtures are obtained; ni-tration of propane gives not only 1- and 2-nitropropanes but nitroethane andnitromethane:

CH3CH2N02 CH3N02nitroethane (10%) nitromethane (25%)

In commercial practice, the yield and product distribution in nitrationof alkanes is controlled as far as possible by the judicious addition of catalysts(e.g., oxygen and halogens), which are believed to raise the concentration ofalkyl radicals. The products are separated from the mixtures by fractionaldistillation.

Additional Reading

S. W. Benson, "Bond Energies," J. Chem. Edu c. 42, 502 (1965).E. S. Huyser, Free -Rad ical Chain Re actions, Wiley-lnterscience , New York, 1970.A very useful procedure for calculating heats of formation and entropies of organicmo lecules is availab le that sums contributions of spe cific group s rather than summ ingbo nd ene rgies. The tables required are rather large, but the answ ers are more precisebecause second-order effects associated with particular groups are taken into ac-count. See "Additivity Rules for Estimation of Thermochemical Properties," by S. W.Benson, F. R. Cruicksh ank, D. M. Golde n, G. R. Haugen, H. E. O'Neal, A. S. Rodgers,R. Shaw, and R. Walsh, Ch em ical Reviews 69, 279 (1969).D. R. Stull, E. F. Westrum, Jr., and G. C. Sinke, The C he mic al Thermodynam ics of Or-ganic Compounds, John Wiley and Sons, Inc., New York, 1969. Summarizes A H 0 offormation of some 4500 organic compounds and AGO values for about half that many.Include s many valuab le de script ions of the use of thermodyna mic da ta in the plann ingof industrial processes.


38/41


39/41

Supplementary Exercises4-25 Explain why there is an increas ingly poor correlation between AH0 and theequil ibr ium constant K,, for the formation of methane, propane, hexane, and nonanefrom solid ca rbon and hydrogen ga s (Table 4-5).4-26 The AH0 values for formation of cyclohexan e from I-h ex en e and of hydrogenchloride from hydroge n and c hlorine dif fer by less than 3 kcal mole-' but the respec-tive equilibrium constants are different by a factor of 107.Explain.4-27" The en tropy ch ang e AS0 for the formation of chloroethane by chlorination ofethane is t-0.5 e.u., an d for the formation of chloroethane by com bina tion of hydroge nchlo ride with ethene AS0 is -31 e.u. Explain.CH3-CH, + CI,4H3CH2CI-tHCI AS0=+0.5 .u.

CH2=CH2 + HCI- H3CH2CI AS0= -31 e.u.4-28 Investigate the en ergies (A H 0) of possible chain m echan isms for the light-induced monobromination of methane and compare with those for chlorination. Whatare the prospects for iodination of methane?4-29 The heat of com bus tion of cyclopropane, (CH2)3, o give carbo n dio xid e andliqu id water is 499.8 kcal mole-'. Show how this value, assum ing normal C-H bon dstrengths, can be use d to calcu late the average C-C bon d energy of cyclo prop ane .4-30 Write a me cha nism analogous to that usually written for methane chlorina tionthat would lead to production of hexachloroethane as in Exercise 4-20a. (This reac-tion is used for commercial production of hexachloroethane.)4-31 With reference to the d ata in Table 4-6, draw the structure(s) of the majo r or-ganic produc t(s) to be exp ected from hydrogen abstract ion in the fol lowing reactions:a. (CH,),CH + Br, l ightb. O C H , SOCI, light

) Br, l ight4-32 Use the data in Tab le 4-6 to pred ict the prod ucts of the follow ing reactions.Indica te any am bigu ities that you encounter as the result of insufficient data.

C. CH3SH+ CH,CH,. ---+heatd. H 2 0 2-


40/41

108 4 Alkanes

4-33" The oxidat ion of hydroca rbons by atmos pheric oxygen to give hydroperoxide sis ca l led au tox ida t ion :

It is a detrimental reaction be cau se it leads to the deterioration of orga nic com pou ndsexpos ed to air (e.g., rubber crac king). Furthermore, the produ ct, ROOH, in comm onwith virtually al l organic compounds with -0-0- bonds, has the potential ofunde rgoing rap id decomp osit ion on heating, which in fact may occur with exp losiveviolence.The m echa nism of autoxidat ion is a radical-chain process that is init iated b yformation of a hydrocarbon radical, R .a . Write the prop agation steps for this react ion, using Re or ROO - as the chain-propagat ing radical . How do you expect that ant ioxidants added to mater ials suchas rubber ac t to he lp protect them from autoxidat ion (see Section 4-4D).b. Use the da ta of Table 4-6 to de termine the m ost favorable p roducts of autoxida-tion of cyclohexene and methylbenzene (C,H,CH,).c. I t is extremely hazardous to store some organic chemicals for long periods oft ime in un sealed containers expo sed to air and l igh t. Aldehyd es and ethers are par-t icularly dangerous chemicals to store in this way. Explain why this should be so.4-34* The first step in preparing the very useful elastomer Hypalon involves treatinga mixture of long- chain alkan es, H(CH2),H, where n = 50-200, with sulfuryl chloride(S02C12) n the presen ce of substances that can init iate rad ical-chain chlorinat ion,as des cribe d in Section 4-5B. The prod uct molecules co ntain many C-CI bond s anda few C-SO,-CI bon ds, the latter of wh ich are subse que ntly used in a curing step toimprove the physical propert ies. How can the chain mechanism for chlorinat ion withS02CI, be mo dif ied to accou nt for the formation of C-SO2-CI bond s?4-35" Explain why the pro duc t distr ibut ion in the chlorinat ion of propa ne by sulfurylchloride is expected to dif fer according to whether the hydrogen-abstract ion stepis acc om plis he d by CI. or .SO,CI,

4-36* ert-Butyl hypobromite is a radical brominating agent that is similar to tert-butyl h ypoch lorite (Exercise 4-18"), but i t is less easily prepared than the h ypoch lorite.A good substitute, provided radical bromination is possible, is a mixture of BrCCI,an d (CH,),COCI. Thus, brom ina tion of cyc loh ex en e resu lts if a hi gh ratio of brom o-trichloromethane to hypochlorite is used.

Sug gest how this react ion is init iated an d p ropaga ted, and exp lain why it is necessaryto have an excess of bromotrichloromethane.


41/41

Supplementary Exercises d 094-37* Use the data of Table 4-6 and t in-hydrogen and tin-chlorine bon d energiesof 80 kcal and 120 kca l, respectively, to determine the overall fea sibi l i ty of the follow-ing reaction:

a. Assume the react ion proceeds by a radical-chain mechanism and work out ener-get ically feasible init iat ion and propagation steps.b. Draw energy diagram s l ike those shown in Figure 4-4 that correspon d to ea ch ofthe propagation steps. Indicate clearly on your diagrams which step would be ex-pected to have the highest a ct ivat ion energy (that is, be the slower step), which pointon your curves corresponds to the transition state, and which energy differencescorre spo nd to the energy cha nge (AGO) in that step of the re action (assum e TASO=0).

Lthough This Chapter is Concerned With The

Documents

Transcript of Lthough This Chapter is Concerned With The