1. Puan Rozaini Abdullah School of Bioprocess Engineering

2. CHAPTER 2  ALKANES - Structure, nomenclature and physical properties of alkanes and cycloalkanes: Define and identify the principle in naming and physical properties of alkanes. - Conformational analysis: Discuss and interpret the conformational analysis or rotation about carbon-carbon single bonds in molecules such as ethane and butane.

3.  Alkanes:hydrocarbons containing only single bonds  Straight-chain alkanes: the carbons are in a continuous chain with no branches.  General formula: C n H 2n+2

4. Nomenclature of Alkyl Substituent  Removing a hydrogen from an alkane results in an alkyl substituent

5. Nomenclature of Alkanes  The systematic name of an alkane is obtained using the following rules: 1. Determine the number of carbons in the longest continuous chain. - This chain is called the parent hydrocarbon. - The number of carbons in the parent hydrocarbon becomes the alkane’s “last name”.

6. Same but different!

7.  2. Number the chain so that the substituent gets the lowest number - Name of any alkyl substituent that hangs off the parent hydrocarbon is placed in front + the no to designate the carbon to which the alkyl substituent is attached.

8. 3. Number the substituents to yield the lowest possible number in the number of the compound. - If more than 1 substituent is attached to parent hydrocarbon, the chain is numbered in direction that will produce the lowest possible numbers. - Substituents listed in alphabetical order

9. Note: Prefixes are ignored in alphabetizing substituent groups except for iso and cyclo- not ignored  If 2 or more substituents are same: prefixes di, tri and tetra are used- indicating the no. of substituents of the compound. 2,4-dimethylhexane 5-ethyl-2,5-dimethylheptane 3,3,6-triethyl-7-methyldecane5-isopropyl-2-ethyloctane

10. 4. Assign the lowest possible numbers to all of the substituents

11. 5. If the same substituent numbers are obtained in both directions, the first group cited receives the lower number 2-bromo-3-chlorobutane Not 3-bromo-2-chlorobutane 3-ethyl-5-methylheptane Not 5-ethyl-3-methylheptane

12. 6. If a compound has two or more chains of the same length, the parent hydrocarbon is the chain with the greatest number of substituents

13. 7. Certain common nomenclatures are used in the IUPAC system- systematic substituent names are preferred. - Systematic substituent names obtained by numbering the alkyl substituent starting at the carbon attached to the parent hydrocarbon. Carbon attached to parent hydrocarbon is always no 1 carbon. Note: if prefix such as di is part of a branch name, it is included in the alphabetization

14.  Some constituents have only a systematic name.

15. Examples:  Draw a structure for: - 2,3-dimethylhexane  Give systematic name: - Answer: 3-ethyl-6-methyl-5-propylnonane

16.  Provide an acceptable name for the alkane shown below. Answer: 2, 5-dimethylheptane

17. Nomenclature of Cycloalkanes- skeletal structures  Cycloalkanes: alkanes with their carbon atoms arranged in a ring  Has 2 fewer hydrogens than acyclic alkane  General molecular formula: C n H 2n  Cycloalkanes are named by adding prefix ‘cyclo’ to the alkane name.

18. Rules for naming cycloalkanes  Resemble the rules for naming acyclic alkanes. 1. No number is needed for a single substituent on a ring - The ring is parent hydrocarbon unless the substituent has more carbon than the ring

19. 2. Name the two substituents in alphabetical order - The no 1 position is given to the substituent listed 1 st.

20. 3. If there are more than two substituents, they are cited in alphabetical order - Substituent given the no 1 position is the one results in a 2 nd substituent getting as low no as possible.

21. Exercises:  Provide the systematic name of the compound shown.  Give the systematic name: Answer: 4-butyl-1,2- dimethylcyclohexane Answer: 4-butyl-1-ethyl-2-methylcycloheptane

22. The Physical Properties Of Alkanes  Boiling points: - Boiling point (bp) of a compound is the temperature at which the liquid form becomes a gas (vaporizes). - To vaporize- must overcome the forces that hold the individual molecules close to each other in the liquid - If the molecules are held by strong forces- a lot of energy will be needed to pull the molecules away from each other- high boiling point. - In contrast for molecules that being held by weak forces- low boiling point.

23.  Attractive forces between alkanes- relatively weak- van der waals forces. van der Waals force Dipole–dipole interaction Hydrogen bonds

24. Van der Waals Forces The boiling point of a compound increases with the increase in van der Waals force Magnitude of the van der waals forces that hold alkane together depends on- area of contact between molecules. The greater the area of contact- the stronger the van der waals forces- more energy to overcome.

25. Melting Points  Melting point (mp) of a compound is the temperature at which its solid form is converted into liquid.  mp of alkanes- as molecular weight increase, mp increases.  In addition to intermolecular interactions mentioned, mp is influenced by the type of packing- arrangement of the molecules.

26. Alkanes with odd no of carbons- pack less tightly The end molecules are facing and repelling the methyl group on the end of the other- increasing the average distance between the chains- weaker interactions.

27. Solubility  Polar compounds dissolve in polar solvents and vice versa. “like dissolves like”.  Reason ‘polar dissolves polar’- polar solvent, ie water has partial charges that can interact with partial charges on polar compound.  Clustering of the solvent molecules around the polar molecules separates them from each other- dissolve  The interaction between solvent and solute molecules- solvation

28.  Nonpolar compounds- no net charge.  Dissolves in nonpolar solvents- van der waals interactions between solvent molecules and solutes molecules

29. Exercises Which of the following has the lowest boiling point? A)CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 B) C) D) E) CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3

30.  Explain why has a lower boiling point than CH 3 CH 2 CH 2 CH 3.  Answer: CH 3 CH 2 CH 2 CH has greater van der Waals forces because it has a greater contact area than isobutane. Therefore, the boiling point of CH 3 CH 2 CH 2 CH is higher.

31. Conformations of Alkanes: Rotation about Carbon–Carbon Bonds  Conformations: different spatial arrangements of the atoms that result from rotation about a single bond.  Conformational isomers/conformers- molecules with different conformations.  The conformers produced by rotation about carbon- carbon bond ethane: staggered conformer & eclipsed conformer.

32. Staggered conformer- more stable and lower in energy than eclipse conformer. Lower energy- low intramolecular repulsions between hydrogens due to large distances.

33. Different Conformations of Ethane The above figure shows the potential energies of all conformers of ethane obtained in one complete 360 0 rotation about the C-C bond. Noticed that staggered conformers are at energy minima, whereas eclipsed conformers are at energy maxima.

34. Conformations of n-butane  Butane has 3 C-C single bonds  Rotation can occur about each of them.  The Newman projections below show the staggered and eclipsed conformers that result from rotation about the C1- C2 bond:

35.  The staggered conformers resulting from rotation about the C1-C2 bond in butane- have the same energy  The staggered conformers resulting from rotation about the C2-C3 bond- do not have the same energy. The staggered and eclipsed conformers from rotation about the C2-C3 bond in butane are:

36. Of the staggered conformers- D is more stable than B and F- the 2 methyl groups are far apart from each other. Most stable conformers- anti conformer (anti= opposite) The other 2 conformers, B and F- gauche conformers (gauche=left) ConformersCH 3 substituentsEnergy Anti conformer (D)Opposite of each other Lower energy Gauche conformer (B&F) Adjacent to each other Higher energy

38.  Anti and gauche conformers do not have the same energy  Steric strain  Steric strain: strain experienced by a molecule when atoms or groups are close enough to one another for their electron clouds to repel each other. Thus possesses additional energy.

39. Eclipsed conformers resulting from rotation about the C2-C3 bond in butane- different energies. The eclipsed conformer A is less stable than C and E.

40. CONFORMERS OF CYCLOHEXANE  Cyclic compound most commonly found contain six-membered rings- carbon rings of that size can exist in conformation- chair conformer  Bond angles in chair conformer are 111 o - close to tetrahedral bond angle 109.5 o and all adjacent bonds are staggered.

41. Drawing the chair conformer  Draw parallel line of the same length, slanted upward and beginning at the same height  Connect the top of the lines with a V whose left side is slightly longer than its right side. Connect the bottoms of the lines with an inverted V. This completes the framework of the six-membered ring.

42.  Each carbon has an axial bond and equatorial bond. The axial bonds are vertical and alternate above and below the ring  The equatorial bonds point outward from the ring. Bond angles are greater than 90o, the equatorial bonds are slant. -If axial bond points up, equatorial bond on the same C-downward slant. If axial bond points down- equatorial bond on upward slant.

43.  Note: the lower bonds of the ring are in front and the upper bonds of the ring are in back.  Cyclohexane rapidly interconverts between 2 stable chair conformers- ease of rotation about its C-C bonds.  intercoversions: ring flip.  2 chair conformers interconvert: bonds that are equatorial in a chair conformer become axial in the other chair conformer and vice versa.

44.  Also exist as a boat conformer.  Boat conformer- free of angle strain  Boat conformer less stable than chair conformer- some bonds are eclipsed.  Boat conformer further destabilize by close proximity of flagpole hydrogen- causes steric strain.  Refer to Figure 2.9 in the textbook, pg: 105.

45. The above figure showed conformers that cyclohexane assumes when interconverting from one chair to the other. Converting boat conformer to chair conformer- one of the two topmost carbons of the boat conformer must be pulled down so that it becomes the bottommost carbon of the chair conformer. When carbon is pulled down a little- the twist-boat conformer obtained. Twist-boat conformer-more stable than boat conformer- the flagpole hydrogens have moved away from each other – relieved some steric strain when the C is pulled down to the point where it is in the same plane as the sides of the boat- very unstable half-chair conformer obtained.

46. Conformers Of Monosubstituted Cyclohexanes  Monosubstituted cyclohexanes do not have 2 equivalent chair conformers like cyclohexanes.  Methyl substituent is in equatorial position in 1 conformer and in equatorial conformer in the other.  Chair conformer in equatorial position- the most stable- substituent has more room, thus less steric interactions.

47. Axial position less stable- three axial bonds on the same side of the ring are parallel to each other, any axial substituent will be relatively close to axial substituents on the other 2 carbons. Because the interacting H/substituents are in 1,3-positions- steric interactions are called 1,3-diaxial interactions. A substituent has more room if it is in equatorial position than if it is in axial position.

48.  The larger the substituent on a cyclohexane ring, the more the equatorial substituted conformer will be favored K eq = [equatorial conformer]/[axial conformer]

49. Reaction Of Alkanes  Alkanes have only σ bonds.  The electrons in C-H and C-C σ bonds- shared equally by the bonding atoms. Thus, none atoms in an alkane has significant charge.  Neither nucleophiles nor electrophiles- thus neither nucleophiles/electrophiles are attracted to them.  Alkanes = relatively unreactive compounds.

50. Chlorination And Bromination  Alkanes react with Cl 2 and Br 2 to form alkyl chlorides or alkyl bromides.  Halogenation reactions- take place at high temp or in the presence of light (hv).  The only reactions that alkanes involved (w/out metal catalyst) with exception of combustion (burning).

51.  When a bond breaks so that both of its electrons stay with 1 of the atoms- process called heterolytic bond cleavage/heterolysis  When a bond breaks so that each of the atoms retains one of the bonding electrons- homolytic bond cleavage/homolysis.

52. Mechanism For Monochlorination Of Methane In the initiation step (creating the radicals): Energy (heat/light) required to break the Cl-Cl bond, homolytically. (Note: a radical is a species containing an atom with unpaired electron- highly reactive- only acquire an electron to complete its octet). In the propagation step: 1. Cl radical formed in previous step removes a H atom from methane (CH 4 )- forming HCl and a methyl radical. 2. The methyl radical removes a Cl atom from Cl 2 (more starting materials is used) forming chloromethane and another Cl radical which can remove a H atom from another molecule of CH 4.

53. Step 2 & 3- propagation steps- the radical created in the 1 st propagation step reacts in the 2 nd propagation step to produce the radical that participates in the 1 st propagation step. -These 2 propagation steps are repeated -The 1 st propagation step is the rate-determining step of the overall reaction. termination step: any 2 radicals in reaction mixture can combine to form a molecule in which all the electrons are paired. - Termination step- helps bring the reaction to an end by decreasing the no of radicals available to propagate the reaction.

54. Mechanism For The Monobromination Of Ethane  Bromination of alkanes has the same mechanism as chlorination

55. exercise  Write equations showing the initiation, propagation and termination steps for the monochlorination of ethane.

56. Radical stability  Radicals are stabilize by electron-donating alkyl-groups  Relative stabilities follow the order of primary, secondary and tertiary alkyl radicals.

57.  Alkyl groups stabilize radicals by hyperconjugation- delocalization of electrons  Stabilization results from overlap between a filled C-H or C-C σ bond and a p orbital that contains an electron: a three- electron system.  In the two-electron system- both electrons are in bonding molecular orbital (MO).  In three-electron system-1 of the electrons has to go into an antibonding MO. Radical: One of the electrons is in the antibonding orbital

58.  Overall, the three-electron system is stabilizing because there are more electron in the bonding MO than in the antibonding MO  But it is not as stabilizing as the 2 electron system which does not have an electron in the antibonding MO.  Consequently, a two-electron system stabilizes 5-10 times better than the three- electron system.

59. The distribution of products depends on probability and reactivity  2 different alkyl halides are obtained from monochlorination of butane.  Substitution of H bonded to terminal Cs produces 1- chlorobutane.  Substitution of H bonded to internal Cs produces 2- chlorobutane.  The expected distribution of products is 60% 1-chlorobutane and 40% of 2-chlorobutane

60.  Probability of Cl radical colliding with primary H is more than colliding with internal H.  Experimentally, 29% is 1-chlorobutane and 71% is 2-chlorobutane. Therefore, probability alone does not explain the product formation.  Because, it is easier to remove H atom from 2 o C than to remove H atom from primary C.  2 o radical is more stable than 1 o radical. The more stable the radical, the easier it is formed.

61.  To determine the relative amounts of different products obtained from radical chlorination of alkane, both probability and reactivity must be taken into account.  Relative amount of 1-chlorobutane: no of Hs x relative reactivity = 6 x 1.0 = 6.0  Relative amount of 2-chlorobutane: no of Hs x relative reactivity = 4 x 3.8 = 15  The % yield of each alkyl chloride is by dividing the relative amount by the sum of relative amount (6.0 + 15.0 = 21)

62.  % yield of 1-chlorobutane 6.0/21 = 21%  % yield of 2-chlorobutane 15.0/21 = 71%

63. The Reactivity-selectivity Principle  The relative rates of radical formation by a bromine radical are different from the relative rates of radical formation by a chlorine radical.  Radical bromination is more selective than radical chlorination: At 125 o C, a bromine radical removes a hydrogen atom from a tertiary carbon 1600 times faster than from a primary C and removes a H atom from a secondary C 82 times faster than from a primary C.

64. When a bromine radical removes a H atom, the differences in reactivity are so great that the reactivity factor is more important than the probability factor E.g. radical bromination of butane gives a 98% yield of 2- bromobutane compared with the 71% yield of 2-chlorobutane obtained when butane is chlorinated. Bromination is more highly selective than chlorination.

65. Why are the relative rates of radical formation is different?  Relative rates of radical formation is different when a Br radical rather than Cl radical is used as the H-removing reagent is due to: - Bromination is a much slower reaction than chlorination - The activation energy for removing a H atom by Br radical is 4.5 times greater than that for removing a H atom by a Cl radical.

66. Radical stability not important Radical stability important Reaction diagrams for the formation of primary, secondary and tertiary radicals by Cl radical and Br radical using calculated ΔH o (enthalphy) values and activation energies. Chlorination reaction to form primary, secondary and tertiary radical- exothermic, the transition states resemble the reactants. Reactants- all have approximately the same energy so only have a small difference in the activation energies for removal of H atom from primary, secondary and tertiary carbon. Bromination- endothermic, the transition states resemble products. There is significant difference between the activation energies Thus, Cl radical makes primary, secondary and tertiary radicals with equal ease Br radicals preference for forming the easiest to form tertiary radical

67.  Since Br radical is relatively unreactive- it is highly selective of which H atom it removes  The more reactive Cl radical is consider less selective  The reactivity-selectivity principle: the greater the reactivity of a species, the less selective it will be.  The fluorine radical is the most reactive halogen radical- reacts violently with alkanes  Iodine is the least reactive of the halogen radicals. It unable to remove H atom from an alkane.

68. Exercise  What is the major monochlorination product of the following reaction? Disregard stereoisomers.  What would be the % yield of the major product? CH 3 hv CH 3 CHCH 2 CH 3 + Cl 2