1. Khadijah Hanim bt Abdul Rahman PTT 102: Organic Chemistry School of Bioprocess Engineering, UniMAP Week 2: 18/9/2011

2.  Ability to APPLY the chemical and physical properties of each functional group carries out theoretical reaction mechanism at molecular level. - Conformation of cyclohexane - Reaction of alkanes: - Cholorination and bromination of alkanes

3.  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.

4.  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.

5.  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.

6.  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.

7.  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.

8. 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.

9.  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.

10. 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.

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

12.  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.

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

14.  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.

15. 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.

16. 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.

17.  Bromination of alkanes has the same mechanism as chlorination

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

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

20.  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

21.  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.

22.  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

23.  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.

24.  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)

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

26.  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.

27. 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.

28.  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.

29. 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

30.  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.

31.  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