Alkanes are referred to as aliphatic compounds from a Geek word mean fats, they are with a general formula C n H 2n+2

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1 Alkanes are the simplest family of organic compounds. They are relatively unreactive compounds Their study help in the studying of the basic approach of naming of the organic compounds. The simplest member of this family and indeed is the simplest organic compound is methane, 4. Alkanes are saturated cpd using sp3 hybridized carbon with tetrahedral shape of 109.5, in methane - bond length 1.10Å and 104 Kcal - bond strength.

2 Alkanes are referred to as aliphatic compounds from a Geek word mean fats, they are with a general formula n 2n+2

3 Physical properties: Alkane sometime refer to as paraffin which from the Greek work parum affinis which means low affinity which describe the chemical behaviour of alkane which have low affinity to other substance and chemically inert to most laboratory reagents. Alkanes they do interact with each other through van der waals force therefore they have low bp and mp e.g methane pb and mp -183, there are insoluble in water but soluble in organic solvent such as l4, ether, alcohol. 4 source is the natural gas as a result of the decay of living organism, it represent 97%, can be seen as marsh gas bubbling to the surface of swamp. Methane together with other gases present in natural gas is consumed as fuel.

4 The boiling point and the melting points of alkanes shows a regular increase by the increase of the molecular weight

5 Alkane with increase branching have lower boiling point

6 Reactions Methane and other alkane they don react with very reactive substances, such as oxidation with oxygen, and halogenation with halogen and react with other few substances under appropriate condition. 1) 4 + O 2 or spark flame 2) 4 Reactivity of halogen F2 > l2 > Br2 > (I2) unreactive O O + heat of combustion U.V + l 2 3 l + l or 213 Kcal/mol + l 2 2 l 2 l l + l 2 l 4 l 3 + l l 2

7 Oxidation: When natural gas burned, methane is converted into O2 and 2O and heat. eat is the important product of this reaction, and for such reaction to start high temp as provided by a flame or a spark is needed. Once started the heat generated maintain the high temp needed and permit the reaction to continue. The quantity of heat evolved when one mole of hydrocarbon is burned to O2 and 2O is called the heat of ombustion, for methane is 213 Kcal. For butane 3223 is 688 Kcal

8 alogenation: Under the influence of ultraviolet light or at temp of methane gas and chlorine gas ( gas phase chlorination) react together to give U.V 4 + l 2 3 X + l or Methylchloride The Methane methyl and alkane = alkyl In the above reaction chlorine substituted the hydrogen in an example of a broad class of reaction known as substitution reaction. The methylchloride formed can react with another molecule of chlorine to form dichloromethane 3 l + U.V l 2 2 l + 2 or dichloromethane l

9 In the same way dichloromethane can react with another chlorine to give trichloromethane (chloroform) which can react to give tetrachloromethane (arbon tetrachloride) l4 2 l 2 + U.V l 2 l + 3 or chloroform l l 3 + U.V l l 2 l + 4 or carbontetrachloride This reaction gives mixture of products, but using excess 4 can limit the reaction to monochlorination product.

10 Like chlorination methane can undergo bromination with bromine but at slower rate due to the less reactivity of bromine compared to chlorine U.V 4 + Br 2 Br 3 Br + or bromomethane Iodine does not react with methane and fluorine is too reactive hence arrangement of halogens reactivity as follows: F2 > l2 > Br2 (>I)

11 Mechanism of chlorination of methane The mechanism of chlorination is same as that of bromination of methane as well as other alkanes. U.V + l 3 l + l 4 2 or Experimental observation about above reaction: 1) This rxns is promoted by heat or light. At room temp methane and chlorine do not react as long as there is no u.v ligjt (dark). Methane and chlorine do react at room temp if the mixture irradiated with uv light. Methane and chlorine react in the darkness if the mixture heated above 250

12 2) In the light induce rxn, several thousand of methylchloride molecules are obtained for few number of light photons that is absorbed by the system. 3) The presence of small amount of oxygen slows down the rxns for a period of time after which the reaction proceeds normally, the length of time depend upon how much oxygen is present The mechanism that consistent with these observation and hence generally accepted is: 1) l l U.V or 2l 58 kcal provided by uv or temp 2) 4 + l l + 3

13 3) 3 + l l 3 l + l Then (2),(3), (2)(3).etc The 1 st step is homolytic cleavage of the chlorine molecule to produce 2 chlorine atoms (homolytic cleavage required 58 kcal/mol and this supplied by light or heat) The chlorine has an odd or (unpaired) electron, this particle with odd electron is called a free radical. The chlorine atom is extremely reactive and to complete its octet abstract hydrogen atom from methane molecule in step 2 to produce l and a methyl free radical 3

14 In step 3 highly reactive methyl radical reacts with a chlorine molecule by abstracting a chlorine atom to give methyl chloride and an other chlorine atom is produced. The new chlorine atom attack another methane molecule and cause repetition of step 2 and so step 3. This sequential stepwise mechanism in which each step generates a reactive intermediate that cause the next step to occur is called chain rxn. The chain nature of the rxn is what account for the observation that few chlorine atoms needed to cause the formation of many thousands of methyl chloride i.e need few photons of light

15 Does the process go on for ever until all the methane present is consumed? No, union of any 2 short-lived reactive particles, although infrequency will terminate the chain. This is because reactive particles are consumed but not generate l + l l l l 3 l hain terminating steps

16 hain propagation step Summary of the mechanism: U.V 1) l l 2 l or 2) 4 + l l + 3 3) 3 + l l 3 l + l hain initiating step 4) l + l l l 5) ) 3 + l 3 l hain terminating step

17 The evidence that support the mechanism: 1) independent experiment shows that chlorine dissociate into chlorine atoms when heated to 250 or irradiated with uv light. 2) in the photochemical process many thousands of methylchloride molecules are produced for each photon of light absorbed as would expected for a chain propagated process. 3) analysis of the product mixture shows the presence of 3-3 formed by combination of two methyl radicals in the chain terminating step.

18 Inhibitors: Finally, how does the mechanism account for the fact that a small a mount of oxygen slow down the rxn for a period of time, which depends upon the amount of oxygen present, after which the reaction proceeds normally. Oxygen is has a radical structure and is able to react with a methyl radical to form a peroxide radical 3 + O O 3 O O The peroxide radical is much less reactive than methyl radical and is unable to participate in the propagation steps This confirm that when all oxygen is consumed reaction proceed normally.

19 A substance that slows down or stop a rxn even though present in small amount is called an inhibitor Inhibitors are two type: 1) Those reacting to give a neutral molecule e.g nitrogen oxide. 3 + N O 3 N O 2) Inhibitors forming more stable radicals unable to continue a radical chain e.g oxygen. 3 + O O 3 O O

20 The chlorination of methane: energy changes (heat of reaction) We saw earlier that we can calculate the overall heat of reaction from bond dissociation energy (D*) We can also calculate the heat from reaction for each step of a mechanism e.g rxn of methane and chlorine to yield 3l and l. 3 + l l 3 l + l

21 hain propagation steps Initiation l l 2 l = + 58 kcal (D* = 58 kcal) l + 3 (D* = 104 kcal) l + 3 (D* = 103 kcal) = +1 kcal 3 + l l 3 l + l (D* = 58 kcal) (D* = 84 kcal) = - 26 kcal Δ * overall = Δ * bond broken - Δ * bond formed = (58+104) (103+84) = = - 25 kcal /mol Therefore the reaction is exothermic

22 In the chain terminating steps bonds are formed, but no bonds are broken. As a result, all of the chain terminating steps are highly exothermic. Why is the chlorination reaction occurs only at a high temp or in the presence of u.v. light if it is ethothermic? This is because the chain initiation step1, without which the reaction cannot occur is endothermic and take place only at a high temp. Once the chlorine atoms are formed, the two chain propagation steps ( one endothermic by 1 Kcal and the other ethothermic by 26 Kcal occur readily many times before the chain is terminated.

23 The order of the reactivity of halogen in halogenation of alkane i.e F2 > l2 > Br2 (>I) is not due to the ease at which the halogen are broken into free radicals to initiate the rxn but is a result of the Δ* of the propagation steps. Despite that D* of F2 and are 37 and 36 but F2 is too reactive whereas I2 is unreactive.

24 hain propagation steps Initiation Br Br 2 Br = +46 kcal (D* = 46 kcal) Br + 3 (D* = 104 kcal) Br + 3 (D* = 87 kcal) = +17 kcal 3 + Br Br 3 Br + (D* = 46 kcal) (D* = 70 kcal) Br = - 24 kcal Δ * overall = Δ * bond broken - Δ * bond formed = (46+104) (87+70) = = - 7 kcal /mol. Therefore the reaction is exothermic, but less than chlorination(-25 kcal)

25 hain propagation steps Initiation F F 2 F = +37 kcal (D* = 37 kcal) F + 3 F + 3 (D* = 104 kcal) (D* = 135 kcal) = - 31 kcal 3 + F F 3 F + (D* = 37 kcal) (D* = 108 kcal) F = - 71 kcal Δ * overall = Δ * bond broken - Δ * bond formed = (37+104) ( ) = = kcal /mol. Therefore the reaction is too exothermic, may cause an explosion

26 Energy of activation: E act Let us look at the rxn: 3 + l l + 3 (D* = 104 kcal) (D* = 103 kcal) If this reaction should occur, first chlorine atom and methane molecule must collide, as a bonds can form only when the atoms are in close contact. Next, for the collision to be effective, it must provide a certain minimum amount of energy (energetic collision)

27 Formation of -l bond liberates 103 Kcal while breaking of 3- bond required 104 Kcal. It might be expected that only 1 Kcal/mol is needed for the reaction to occur. owever this not so. It is found that if the rxn to occur 4 Kcal/mol of energy must be supplied. The minimum amount of energy that must be provided by a collision for a rxn to occur is called the energy of activation ( E act ). The Eact source is the kinetic energy of the moving particles. If the collision provide less energy than this minimum quantity, then they are fruitless and original particles simply bouncing apart.

28 Finally in addition to being sufficiently energetic, the collision must occur when the particles are properly oriented. In general, a chemical rxn required collision of sufficient energy (Eact) and particles properly oriented. Energy of activation provision is not only required for endothermic rxn, even some exothermic rxn have an activation energy. 3 + F F + 3 = - 31 kcal (D* = 104 kcal) (D* = 135 kcal) Eact = +1.2 Kcal

29 The reason for the fact that energy (1.2 Kca/mol) must be provided by colliding particles despite the fact the rxn is ethothermic is that 3- (- ) bond of methane must be largely broken before the -F bond is completely formed. We can illustrate graphically the rxn of chlorine atom and 3- and fluorine atom and 3- as they are transformed from reactant to products by what is called energy diagram. The attack of Br. To methane is more highly endothermic therefore E act = 18 Kcal Br + 3 Br + 3 (D* = 104 kcal) (D* = 87 kcal) = +17 kcal

30 Potential energy l 3 Transition state E act = 4 Kcal l + 3 = + 1 kcal 3 + l 3 25 K cal + l l 3 l + l Progress of Rxn

31 Potential energy F 3 Transition state E act = 1.2 Kcal/mol 3 + F = -31 kcal 3 + F Progress of Rxn

32 Reaction rate A chemical reaction is the result of collision of sufficient energy and proper orientation. The rate of a rxn can be expressed as: Rate = total number fraction of collision that X of collision have sufficient energy In other word rate = fraction that have proper orientation ollision frequency X energy factor X probability factor X

33 The collision frequency depend upon concentration, presure, size and how fast the particles are moving which in turn depends on temp. E.g increasing the concentration increases collision frequency and hence increase rate of rxn The probability factor depends upon the geometry of the particles and the rxn takes place. The most important factor determining rate is the energy factor. The fraction of collision that have sufficient energy. This factor depend upon temp which we can control. A small E act rxn means lower temp is needed, whereas high E act high temp needed.

34 At the beginning of the reaction, methane and l. have the total amount of energy indicated by the reactant level at point A on the left side of the diagram. As the two molecules crowd together, their electron repel each other, causing the energy level to rise. If the collision has occurred with sufficient force and proper orientation, the reactants continue to approach each other despite the repulsion until the new -l bond starts to form and the 3- bond starts to break. At some point (B on the diagram), a structure of maximum energy is reached, a structure called the transition state. The transition state represents the highest-energy structure involved in this step of the reaction

35 The energy difference between reactants A and transition state B is called the activation energy, Eact, and is a measure of how rapidly the reaction occurs. A large activation energy results in a slow reaction because few of the reacting molecules collide with enough energy to reach the transition state. A small activation energy results in a rapid reaction because almost all reacting molecules are energetic enough to climb to the transition state. As an analogy, think about hikers climbing over a mountain pass. If the pass is a high one, the hikers need a lot of energy and overcome the barrier slowly. If the pass is low, however, the hikers need less energy and reach the top quickly.

36 Most organic reactions have activation energies in the range kcal/mol. Reactions with activation energies less than 20 kcal/mol take place at or below room temperature, while reactions with higher activation energies often require heating to give the molecules enough energy to climb the activation barrier. The overall energy change for the reaction, however, is the energy difference between initial reactants (far left) and final products (far right), as represented by F in Figure next slide. Because the energy level of the final products is lower than that of the reactants, energy is released and the reaction is favourable. If the energy level of the final products were higher than that of the reactants, energy would be absorbed and the reaction would not be favourable.

37 3 l 3 l l 3 + l 3 + l 3 l + l

38 Ethane Next in size after methane it is 25 - bond strength a little bit smaller than in methane ( kcal - 88Kcal 1.10 Å sp3-sp3 bond 1.53 Å The values of the - bond length and the - and the angle of are quite characteristic of all alkane

39 Free rotation about the - single bond The σ-bond is cylindrical symmetry in shape which is a shape that allow free rotation around - in open chain molecules. In ethane for example this rotation around - give rise to different arrangements of the hydrogens on one carbon with respect to the hydrogens on the other carbon.

40 The different arrangements of atoms that result from bond rotation is called conformations. Molecules that have the different arrangement called conformers or conformational isomers. Therefore conformers are isomers but can not be isolated because are rapidly interconvert, this is because the energy barrier is too low between the different conformational isomers. hemist adopted three ways of representing conformational isomers on paper or back and white boards 1)Sawhorse representation 2) Newman projections 3) Flying wedge representation

41 Sawhorse representation view - bond from an oblique angle and indicate the specific orientation by showing all - bonds. Newman projection view - directly end-on and represent the two carbon as a circle, substituents on the front carbon represented by lines coming from the centre of the circle and substituents on the tear carbon are represented by lines coming from the edge of the circle

42 Fly wedge (3-dimensional representation): In this representation the - viewed from oblique angle and the spatial orientation is indicated as below Broken line : substituent Going behind plane of The paper eavy line: substituent coming out of the plane of the paper Normal line: substituent in plane of paper

43 Newman projections is more advantageous than the sawhorse representation as it clearly show the relationship between the substituents on the 2 carbon Experiments show that we don t really have entirely quite free rotation in ethane. There is a slight barrier (2.9 kcal/mol) to rotation, due to the fact that some conformational isomers are more stable than the other. The lowest energy and most stable conformation is called staggered and has all six hydrogens carbon bonds as far away from each other as possible when viewed in Newman projections.

44 Staggered conformation each hydrogen in the in front carbon is at the middle of the distant separating the two hydrogen of the back carbon

45 The highest energy conformer is called eclipsed and has the - bonds as close as possible-eclipsed in a Newman projection. The so many conformations of stabilities between staggered and eclipsed are called Skew conformations 1 kcal 1 kcal 1 kcal

46 eclipsed 2.9 kcal Skew Potential energy changes that accompany rotation of - bond of ethane

47 Most ethane molecules, naturally exist in the most stable, staggered conformation i.e any molecule spends most of its time (99%) in the most stable conformation for ethane is the staggered one. The energy required to rotate the - bond in ethane is called torsional energy. The relative instability of the eclipsed conformer or any of the skew conformer is due to the torsional strain (eclipsing strain). To what is torsional strain is due? It is a matter of disagreement, but most chemists now believe that the strain is due to the slight repulsion between electron clouds in the - bonds as they pass by each other at close distance in the eclipsed conformer.

48 onformation of propane 3 8 Propane is the next higher member in alkane series, also has a torsional barrier that result in slightly hindered rotation about - bonds. This barrier is found to be 3.4 kcal/mol, slightly higher than in ethane. The eclipsed conformer in propane has three interactions, two ethane-type - interaction and a third methyl- interaction 1.4 kcal Staggered propane 1 kacl Eclipsed propane 1 kcal

49 onformation of butane: 4 10 The conformational situation become more complex for larger alkane because not all staggered conformation or eclipsed has the same energy. In butane two hydrogens were been replaced by two It is similar to ethane except that 2 3 gps have replaced 2

50 The lowest energy conformational in butane is called anti-conformation is arrangement in which the methyl gps as far a part as possible are 180 away from each other. As rotation around 2 and 3 occurs an eclipsed conformation reach in which there are 2 3- interaction and one - interaction. This eclipsed conformation is more strained than the anti-conformation 1.4 kcal 1.4 kcal 1 kcal

51 2x (3) = 4.5 kcal 2x = 3.8 kcal 3.8 kcal 4.5 kcal 0.8 kcal Potential energy of butane rotation around 2-3

52 The same principles developed for n-butane apply to pentane and to all higher alkanes. The most favoured conformation for any alkane is the one in which all - bonds have staggered arrangement and in which large substituents are anti to each other 3 3 ( 3 )

53 igher alkanes: If we examine the molecular formula of methane 4, ethane 26, propane 38 and the butane 410, we see that each alkane contain one carbon and 2 hydrogen more than the previous one. A series of compds in which each member differs from the next member by a constant amount is called a homologous series and the member of the series called homologs. The alkane family forms a homologous series and the constant difference successive members being 2 It could also be seen that the general formula of alkanes is n2n+2

54 According to the general formula the next alkane after butane will have molecular formula 512. Butane exist in 2 isomers, whereas the 5 in pentane joined together in more than two different arrangement. 3 n butane Straight chain alkane Or normal chain In case of pentane: arrangement giving two isomers Isobutane Branched chain alkane n-pentane isopentane neopentane

55 Next member in alkane is hexane 614 and the different possible arrangements of the carbon atoms will increase to 5 isomers Therefore as the molecular formula of the alkane series increases, so does the number of isomers. Isomers are compounds have the same kind and numbers of atoms but of different the way atoms are arrange. compounds that have same number and kind of atoms and differ only in arrangement called constitutional isomers. constitutional isomers always are different compounds of different properties but with the same molecular formula.

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57 Nomenclature: Except for the first four members of alkane series, whose name is a common name of historical background, the name of the rest of alkane is derived from the Greek or (Latin) prefix for the particular number of carbon atoms in the alkane: pentane for five, hexane for six, heptane for seven and so on. We should certainly memorize the names of the first ten alkanes. See next table

58 Names of straight chain alkanes ommon names

59 From the list in the table every alkane will have a number of isomers except methane, ethane and butane. The isomers must have definite names and a prefixes are used to distinguish them. When alkane chain increase different alkane names and different isomers of different prefixes that difficult to remember, this why name of a system to name alkanes is of high importance.

60 Alkyl groups: If you imagine a hydrogen atom is removed from alkane structure, the partial structure remain is called alkyl group. Alkyl group are not stable themselves but they are a part of a large compound. The alkyl group named by replacing the ane of alkane with yl. Methane generate Methyl, Ethane generate Ethyl, butane generate butyl, usually indicated by R. methyl ethyl butyl

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62 IUPA names of alkanes: IUPA (The International Union of Pure and Applied hemistry has devised a system of nomenclature that could be used for naming all compound including alkanes, no matter how complex thes are if certain rules are followed. The IUPA rules for naming alkanes are: 1) Find the parent hydrocarbon: a) Find the longest continuous chain present in the name that chain as the parent name. The longest chain may not always be clear from the manner of writing, you may have to turn corners.

63 b) if the two different chains of equal length are present longest chain are equal choose the chain of more substitution.

64 2) Number the atoms in the main chain: a) Begin at the end near to the first branch, number each carbon in the parent chain. b) If there is branching at equal distances from both end begin from the end near to the second branching.

65 3) Identify and number the substituents: a)substituents are given the numbers of carbon to which they are attached and named in alphabetical order, use hyphens to separate substn and comma for numbers Named as heptane substituted 3 3 At 2 with methyl and 4 with Ethyl so name is ethyl-2-methylheptane 3 b) If there is more than one of the same alkyl group, then di, tri or tetra use with the indication of their position in the chain by numbers ethyl-2,4-dimethylhexane 3

66 ,2,4-trimethylpentane There are additional rules, but these given are sufficient for naming the compounds we are likely to encounter. 4) If the alkane contain a complex substituent, brackets are used to enclose this substituent (1,2-dimethylpropyl)-3-methyl)nonane

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68 lasses of carbon atoms and ydrogen: Depending upon the number of carbon atoms to which each of the carbon atoms of an alkane is attached these carbon can classified into: a) Primary 1 carbon atom is attached to only one carbon. 3 b) Secondary (2 ): carbon atom is attached to 2 other carbon

69 c) Tertiary carbon(3 ): carbon atom is attached to 3 other carbons Each hydrogen is similarly classified, being given the designation of the carbon atom to which is attached

70 Preparation of alkanes: (synthesis) From petroleum we get mixture of alkane that are difficult to separate but still pure enough to be used as fuels In our laboratory work we often have the need for pure sample of a particular alkane for these purposes, the chemical preparation or synthesis of that particular alkane is the most reliable way of obtaining it. Several methods are available: 1. ydrogenation of alkenes: alkenes add hydrogen in the presence of metal catalysts to produce alkanes Pt or Ni or Pd

71 2) ydrolysis of Grignard reagents: Grignard reagent is a reagent result from the reaction of magnesium metal with alkyl halide in dry ether it is organomagnesium comound it is member of compounds called organometallic because carbon is bonded to metal. R X + Mg Alkyl halide dry ether R Mg X (X = l, Br or I) Girgnard reagent Because magnesium is less electronegative than carbon, therefore electron pair in the bond are strongly pulled towards carbon making the bond a highly polar one R Mg X

72 The R gp although not a full carbanion ( carbon carrying a full negative charge), yet has a considerable carbanion character. Be considered as a magnesium salt of a alkane acid. R R Mg X Because of their nucleophilic strong basic character, Gignard R react readily with cpds having an acidic hydrogen i.e attached to O, N such as in -O-, -N2, R-O- and R-OO- to yield hydrocarbon

73 O R Mg X + O R + Mg X salt of weak acid stronger acid weaker acid N 2 R Mg X N 2 R + Mg X OR R Mg X + R O R + Mg The overall sequence is a useful synthesis method for alkanes from alkyl halides. X Mg dry ether 3 Mg X O l 3 3 3

74 oupling of alkyl halides with organomettallic compounds: Is method required the formation of - bond between two alkyl group, most directly by the coupling together of 2 alkyl gps to make higher alkane. A highly versatile method of doing this is coupling reaction between Lithium dialkyl copper and alkyl halide R R uli + R' X R R' + R u + LiX

75 This method was developed independently by Professor orey (arvard Univ) and Professor ouse (Massachusetts Institute of technology). Lithium dialkyl cuprate is prepared in this manner ether R X + 2 Li R Li + LiX 2 R Li + ux uli + cuprous halide R R (same as G R) LiX The overall synthesis gives excellent yields when the alkyl halide R-X is 1ry (R-2-X) and the organic gps in the organolithium may be 1, 2 or 3

76 3 Br + 2 Li dry ether 3 Li + LiBr 3 3 dry ether 3 3 Li + ui u Li 3 dry ether u Li I Li dry ether Br Li 2 3 Li 2 3 ui u Li

77 u Li Br dry ether

78 Reactions of alkanes: alogenatin of Alkanes having more than one class of : Alkanes higher than methane can also be halogenated under the same condition. owever, because of the presence of different hydrogens a mixture of isomers is obtained u.v or l 2 3 u.v or 3 + l l + l l 1-chloropropane 45% n-propylchloride (same) l 2-chloropropane 55% isopropylchloride

79 3 2 Butane 2-chlorobutane 72% 2 butylchloride u.v or 3 + l l l 1-chlorobutane 28% n-butyl chloride Iso-butane l 2 u.v or l Tertiary butyl chloride 36% l Isobutylchloride 64%

80 3 3 Bromination gives the corresponding bromides but at different proportions. u.v or Br u.v or 3 + Br Br 97% Br 98% % 2 3% 2 Br 2 Br

81 3 3 u.v or + Br Br % 3 3 Although both chlorination and bromination yield mixture of isomers, thev relative amount of the isomers differ markedly, depending upon halogen used. hlorination gives mixture in which no isomer greatly predominate. By contrast bromination gives mixture of isomers in which one predominate to such an extent as to be almost the only product, making up to 97 99% of the total mixture. This is due to the fact bromine being less reactive than chlorine therefore highly selective, it replace that are easily abstracted. traces 2 Br

82 It could be seen that halogenation is not a good method for making pure alkyl halides as a mixture of products results. Industrially, however, the method can be used, because the products can be used as solvent and a mixture is as suitable and cheaper than a pure compound.

83 Orientation of halogenation: Where is a molecule halogenation going to occur? In the reaction of propane with chlorine the relative amounts of the products depends upon the relative rates of the formation of the corresponding free radicals l 2 2l Abstract l n-propyl free radical Abstract 2 l Isopropyl free radical l l 3 l

84 The amount of each product formed will depend upon the rate of formation of the radical i.e upon abstraction of the. The easier to abstract will yield the radical that gives the predominating product. First of all the collision frequency of the 2 reactions is the same since it involve collision between the l. and propane. Next, there is probability factor. If n-propyl radical to be formed there are 6 hydrogens to be abstract from the (1 ), while there are only 2 hydrogens (2 ) to abstract from, to form the isopropyl radical. ence we expect the ratio of n-propyl chloride and isopropyl chloride formed to be 6:2 or 3:1. owever, we have seen that the 2 chlorides are formed in roughly equal amounts (55:45) which means abstraction of 2 is three times as easy as abstraction of 1. Taking probability factor in consideration Eact for abstraction of 2 is less than 1.

85 hlorination of isobutane presents a similar problem l Abstract 1 Abstract In this case abstraction of one of the 9 primary hydrogens leads to the formation of isobutyl chloride, while abstraction of the single 3ry hydrogen will give 3-butylchloride. The probability factor therefore favor the formation of isobutyl chloride by ratio of 9:1. Experiment shows the ratio roughlly is 2:1 (64:36) 2 2 -butyl free radical butyl free radical l l 3 l l

86 Again the reason for this may be that Eact for abstraction of 3 is less than Eact for abstraction of 1 and infact even less than Eact for abstraction of 2. Study of chlorination of a great number of alkanes has shown that these are typical results. Ease of abstraction of hydrogen could therefore be arranged as 3 > 2 > 1 (the arrangement is in this order exactly as sequence of bond dissociation energy) Ease of formation of free radical follows 3 > 2 > 1 >. 3. As the ease of formation sequence as above then sequence of stability is follows same sequence, therefore the more stable free radical is the more easily formed.