1. Course outcome(CO) Understand the various type of polymer synthesis, their mechanism and kinetics. Design synthetic route to polymer synthesis based on chemical and physical properties. Utilise the various reaction conditions to affect structural and chemical modification of polymers. Decide the polymerization system for optimum output during polymer synthesis. Appreciate the various factors affecting an experimental design in polymer synthesis. Develop skill in literature review, communication (oral + written) and result presentation EBP 201 Polymer Synthesis Coursework 30% Final Exam 70%

2. POLYMER SYNTHESIS kinetic and mechanism – chain growth - step growth – 7 Specialised polymerisation techniques - Ziegler-Natta - Metallocene - 3 Copolymerisation – random, block, alternating and graft - copolymer equation and reactivity ratio - 10 Polymerisation system – bulk, solution, suspension and emulsion - 8 Specific polymerisation example – epoxide resin - polyurethane - polyimide - polysiloxane - 8 Polymerisation techniques – laboratory scale - 2 References: - R.J.Young, Introduction to Polymers, Chapman & Hall - G. Odian, Principles of Polymerisation, 3 rd Ed., John Wiley & Sons Inc - D Braun et. al.,Polymer Synthesis: Theory and Practice, 4 th Ed, Springer

3. 2 methods for polymer synthesis: i. Chain growth polymerisation ii. Step growth polymerisation i. Chain growth polymerisation - radical or ionic mechanism - requires double bond -Example: Polystyrene, PMMA ii. Step growth polymerisation -Condensation mechanism - monomers requires bifunctional. If more than cross-linking occur -Example: Thermoplastic – polyester, polyurethane, polyimide, epoxy + H 2 O

4. Kinetics of radical polymerisation (not ionic) Why study kinetics? i.Thermodynamic refer to equilibrium of a reaction but kinetic refer to rate of a reaction ii.Study the mechanism of a reaction iii.To obtain rate law of a reaction. iv.Determine the dependence of reactants, initiator, catalyst, solvent, temperature on reaction rate. Kinetic factor Thermodynamic factor Energy Reaction coordinate

5. i.Initiation - initiator breaks down to radical fragments through homolysis to give active centres. I – I 2R slow rate determining step d[R ] = 2 k i [I] dt 2 because two radicals produced assuming (i)100% efficient and (ii) no effect from monomer concentration ii. Propagation - each radical initiator will react with a new monomer M i to give new radical chain which then attack another new monomer M M i + M M i+1 -d[M] = k p [M][M 1 ] + k p [M][M 2 ] + k p [M][M 3 ] + …. dt -d[M] = k p [M] [M i ] (1) dt kiki kpkp Σ

6. M1M1 MaMa M1MaM1Ma M1MaM1Ma KpKp M2M2 KpKp MbMb M2MbM2Mb M2M2 McMc KpKp M3McM3Mc

7. iii. Termination Consider combination and disproportionation, -Each chain radical i will react with another chain radical j ie. if there is n radical chain then the number of termination reaction is n 2. - since there are two types of termination reaction, the overall rate will be double. M i+j (combination) m i + m j M i + M j (disproportionation) Thus, -d[M i ] = 2k t ( [M i ]) 2 dt where k t = (k c + k d ) kckc kdkd Σ Difficult to determine [M i ] so use steady state assumption

8. Steady state assumption: Rate of production of radicals is equal to rate of its destruction - Traditionally, this is the region where rate of radical reaction is measured. Rate of radical production = d[R ] dt Rate of radical destruction = -d[M i ] dt Therefore d[R ] = -d[M i ] dt dt 2k i [I] = 2 kt ( [Mi ]) 2 [ Σ M i ] = k i [I] 1/2 k t So no need to measure [M i ] but only need to know the concentration of initiator Conversion % time Steady state Σ Pre-steady state

9. From 1, -d[M] = k p [M] [M i ] dt = k p [M] = k p k i 1/2 [M] [I] 1/2 k t 1/2 Σ ki [I] ½ kt This is ‘rate of polymerisation’ performed under steady state ie Rate of polymerisation is proportional to the monomer concentration and square root of initiator concentration. -d[M] = k r [M] dt wherek r = k p k i 1/2 [I] 1/2 k t 1/2 The value of k r is true only if the initiator efficiency is 100% if not efficiency factor f has to be included.

10. Rate law: -d[M] = k r [M] dt Thus, the rate is first order dependence of [M]. Rearranging and integration gives ln [M] t = -k r t [M] o ln M t M o -k r t Order of reaction: Second: -d[M] = k r [M] 2 dt

11. Rate (Ms -1 ) rate [I] 1/2 [M] Methyl methacrylate + benzoyl peroxide Vinyl benzoate + AZBN Methyl methacrylate + benzoyl peroxide

12. Assumption: Initiation step occur with 1.f = 1 ie efficiency of radical production is 100% 2.No effect from monomer concentration If f ≠ 1, ie due to non-productive radicals, then f = f’ [M] Substitute into Ie. order for [M] and ½ order for [I]

13. 1.Calculate the initial rate of polymerisation of acrylonitrile in benzene using AIBN as the initiator under the following condition. 1.[acrylonitrile] = 0.2 M 2.[AIBN] = 0.05 M 3.Initiation k i = 1.5 x 10-5 s -1 4.Propagation k p = 2.0 x 10 3 dm 3 mol -1 s -1 5.Termination k t = 7.8 x 10 8 dm 3 mol -1 s -1 Ans: 1.24 x 10 -5 mol dm -3 s -1 2. What happen when [M]o increase by 4X at constant [I] concentration and [I] increase 4X at constant [M] on a.Total radical concentration at steady state b.Rate of polymerisation

14. Q1. The rate of change of ethylene monomer performed during radical polymerisation at 35 o C is shown below: Time/min[CH 2 CH 2 ]/mmHg 0306.5 10262.7 20243.2 30224.5 40207.5 50191.2 60176.8 70163.5 80151.9 90140.1 100129.4 Calculate the rate constant and half-life for this reaction. Ans: k = 7.88 min -1, t 1/2 = 88 min

15. Determine using Arrhenius plot where rate constant dependent on temperature. A = collision factor / frequency factor Ea reactant product Transition state Rate of polymerisation Depends on activation energy Ea – minimum amount of energy needed to initiate a reaction. Catalyst – decrease E a Temperature – supply kinetic energy to overcome E a ln k = ln A – Ea/RT (T in unit kelvin)

16. ln k 1/T Slope = - E a /R ln A

17. Ans: 97.0 kJmol -1, 9.06 x 10 13 s -1 k x 10 5 /s -1 T/ o C 2.460 47.520 57640 548060 Example: to determine E a and Frequency factor, A. rate = k[M] Method: Let [M] constant, then measure rate at different temperature. Plot 1 st order graph to obtain rate constant k from the gradient of the plot. Determine E a and collision frequency constant, A given R = 8.314 JK -1 mol -1

18. Change in rate of polymerisation Example for radical polymerisation: from Arrhenius relationship, k i = A i exp(-E i /RT) k p = A p exp(-E p /RT) k t = A t exp(-E t /RT) Substitute into equation Usually negative Conclusion: Rate increases as temp increases

19. Example Calculate change in rate of polymerisation of benzene with AZBN if temperature increase from 60 o C to 70 o C given: A p = 2.2 x 10 -7 mol -1 dm 3 s -1 A t = 2.6 x 10 9 mol -1 dm 3 s -1 E p = 34 kJ E t = 10 kJ A i = 5.6 x 10 14 s -1 E i = 126 kJmol -1 R = 8.314 Jmol-1K-1 Ans: 2.63

20. (Odian p275 3 rd Edition)

21. Degree of polymerisation - How many monomer is converted to polymer? Average degree of polymerisation, D p = M p /M o Ie. ratio of monomer that form polymer with the amount of monomer present initially.(how many monomer becomes polymer) But D p = a ν where ν is kinetic chain length ie.average number of monomer molecules consumes (polymerised) per each radical. (ie if there are a lot of initiator then polymer chain reduced then degree of polymerisation reduced.This assupmtion holds only at steady state condition) a is constant relating to termination factor. Higher kinetic chain length means higher degree of polymerisation and vice versa (1)

22. M1M1 MaMa M1MaM1Ma M1MaM1Ma KpKp M2M2 KpKp MbMb M2MbM2Mb M2M2 McMc KpKp M3McM3Mc

23. ie. number of times a monomer propagate per initiation step. Propagation increase chain length but increase number of initiator will reduce chain length. = -d[M]/dt d[R ]/dt = k p [M] [M i ] 2k i [I] Σ ν = rate of propagation rate of initiation [Σ M i ] = k i [I] 1/2 k t but from steady state assumption,

24. Substituting for Σ[ M i ], v = k p [M] 2(k i k t ) 1/2 [I] 1/2 so, D p = a k p [M] 2(k i k t ) 1/2 [I] 1/2 Substitute in (1) M p = M o D p = ie. M p is proportional to concentration of monomer but inversely proportional to initiator concentration -Hence methods of controlling molecular weight. Increase monomer concentration increase rate of polymerisation and average polymer molecular weight Increase initiator concentration increase rate of polymerisation but reduce average polymer molecular weight.

25. 1.Temperature Increase in temperature, will increase rate of initiator productions. Under steady state condition, this will increase the termination step (why?) ie. -d[Mi ] = 2k t (Σ[M i ]) 2 dt Since Σ[Mi ] 2 = k i [I] k t and [Mi ]  [I] 1/2 Polymer chain molecular weight M p = M o a k p [M] 2(k i k t ) 1/2 [I] 1/2 ie. M p  Factors affecting degree of polymerisation 1.Temperature 2.Chain transfer 3.Regulators/modifier 4.Retarders 5.Inhibitor

26. Relation between degree of polymerisation with temperature Substitute k i =A i exp(-E i /RT) k p =A p exp(-E p /RT) k t =A t exp(-E t /RT) Normally positive Conclusion: Temperature increase will decrease in degree of polymerisation.

27. Q2. Calculate change in degree of polymerisation when temperature of a radical polymerisation is increased from 50 o C to 80 o C given that A p = 2.2 x10 7 mol -1 dm 3 s -1 A t = 2.6 x 10 9 mol -1 dm 3 s -1 A i = 5.6 x 10 14 s -1 E p = 34 kJmol -1 E t = 10 kJmol -1 E i = 126 kJmol -1 (Hint: Use Arrhenius equation and substitute into average degree of polymerisation equation) Ans:0.34

28. Chain transfer -Growing chain terminated prematurely. Rate of polymerisation not change but degree of polymerisation reduced. -can occur to monomer, initiator, solvents, transfer agent Eg. 1 - monomer Back-biting in PE, polyvinyl acetate result in branching. Less crystalline with lower Tm and difficulty on processing transfer constant = C M = k tr /k p k tr

29. Eg 2 – solvent Polystyrene + CCl 4 Eg.3 - Initiator M + RO – OR M - OR + RO R = alkyl, acyl M + H - OOR M – H + ROO k tr transfer constant = C S = k tr /k p transfer constant = C I = k tr /k p

30. Structure and reactivity Transfer constant, C = (k tr/ k p ) for chain transfer agent Transfer agentstyreneVinyl acetate benzene0.0231.2 cyclohexane0.0317.0 t-butyl benzene0.0689.9 Ethyl benzene0.6755.2 Toluene0.125 t-butyl chloride0.04 t-butyl bromide0.06 t-butyl iodide1.85 CCl 4 110 CBr 4 22,000 t-butyl mercaptans210,000480,000 t-butyl peroxide0.00076 – 0.00092 t-butyl hydroperoxide0.035 Cumyl hydroperoxide0.063

31. 1/DP [transfer agent] CBr 4 CCl 4 Conclusions i. D p inversely proportional to [transfer agent] ii.D p is dependent on type of transfer agent Effect of transfer agent on degree of polymerisation -Transfer agent affect reduced degree of polymerisation - rate of polymerisation depends on k p, k tr and k i

32. Factors: C – H bond – due to resonance benzene very stable due to aromatic property. But for toluene, radical that is form at methyl group can be delocalize with the adjacent aromatic group. C – H versus C – Hal bond strength C-Halogen stronger than C-H bond. C – Cl > C – Br > C – I bond strength Size of Cl < Br < I. So strength decrease in reverse order.

33. Chain regulators/modifiers -Use to vary degree of polymerisation using different chain transfer agent or different concentration of a specific transfer agent. - Use high k tr for low MW but low k tr for high MW Eg. 1 Polymerisation of low MW ethylene in CHCl 3 (k tr = 5.6) to give,a low fluorinated end-chain polyethylene lubricant Eg. 2 Polymerisation of ethylene in benzene (k tr = 0.023) to give a high MW PE Eg. 3 n-dodecyl mercaptan is used industrially to produce SBR rubber. It gives a lower viscosity hence better processing.

34. Retarders and inhibitor (Odian p259 3 rd Ed) Retarder – rate and degree of polymerisation decreased eg. nitrobenzene Inhibitor – rate and degree of polymerisation stop. eg. benzoquinone normal retarder inhibitor Induction period % conversion Mechanism: Retarder slow polymerisation rate by producing inactive initiator. Inhibitor eat away all active radicals and converting them complete inactive. The inhibitor itself become inactive thus resume polymerisation process hence induction period. During storage/ transport, inhibitor is added to styrene. Before use, this inhibitor has to be clean with NaOH, water and drying using CaCl 2

35. Inhibitor(with styrene)Inhibitor constant( k z /k p ) Nitrobenzene (+ styrene)0.326 Nitrobenzene (+ vinyl acetate) 11.2 p-benzoquinone (+ styrene)518 p-benzoquinone (+ MMA)5.7 p-benzoquinone (+ nitrile)0.91 Oxygen( + styrene)14,600 Fcators affecting retarders/inhibitors p-benzoquinone Terminate by coupling or disproportionation With electron rich monomer eg. styrene act as inhibotor but with electron poor monomer eg. methyl methtacrylate and acrylonitrile act as retarder.

36. nitrobenzene Vinyl acetate: inhibitor due to electron rich Styrene: retarder due to electron rich MMA: no effect

37. Unreactive radical centre Molecular oxygen M + O 2 M OO Diphenyl picryl hydrazyl very stable

38. Different between step and chain polymerisation % conversion Molecular weighht Molecular weight - In chain polymerisation, propagation occurs by successive addition of monomer onto the active radical/ionic center. The speed of monomer addition is very high. Degree of polymerisation DP is high and formation of high molecular weight polymer is very rapid and at low % conversion. Once the stationary state is achieved, i.e. when the number of free radicals being formed is equal to the number disappearing, the value of DP remains constant. - In step polymerisation, each monomer react with another monomer to give a dimer, then this dimer react with another dimer to give a tetramer and so on. High molecular weight only occur after sometimes from the start of the reaction and at high % conversion.

39. Chain growth polymerisation Step growth polymerisation 25% conversion 50% conversion100% conversion 50% conversion 25% conversion

40. Step-growth polymerisation (polycondensation) Definition: Reaction between organic base (eg. alcohol or amine) and organic acids (eg. carboxylic acid, acid chloride) with the elimination of small organic molecules( eg. H 2 O) (RJ Young p9 3 rd Ed) Example - condensation H+H+ Acetic acid ethanol Ethyl acetate water polycondensation If bifunctional, linear chain, if polyfunctional cross-link.

41. Kinetics -Kinetics same as small molecular units. Example: Rate of reaction = rate of disappearance for RCOOH = R’OH = H + = X Case I: At stoichiometric amount of acid and base without catalyst, rate of reaction = -d[RCOOH] = -d[ROH] dt dt = k[RCOOH] [ROH][H + ] = k[X] 3 since the H + (catalyst) derived from RCOOH - autocatalysis This is 3 rd order reaction, hence upon integrating and rearranging, rate = 1

42. P = degree of conversion ie the proportion of number of molecules reduces as the reaction take place. X = X o (1 - P) Substitute in the rate of reaction equation 1 Plot of against t gives linear slope of gradient 2kX o 2 t 1 2kX o 2 (X = number of molecules remain)

43. Case II: With the presence of catalyst H + where k’ = k[H + ] which is constant At constant [H + ] and equal molar of monomers,[X] This is second order reaction. Upon integration and rearranging, Substitute the expression for X into above equation A plot of against t gives gradient X o k’ intercepting at +1 t +1 X o k’

44. Q1. The following data is obtained for polyesterification of Time/h[HO-R-OH] 03.10 0.51.30 1.00.83 1.50.61 2.00.48 2.50.40 3.00.34 Using equimolar of reactants, the concentration of etheylene glycol is monitored using spectroscopic method as shown. Determine the rate constant and overall order of reaction. Is this an autocatalysis mechanism? Determine degree of conversion after 1 h and 5 h. Ans:2.43 x 10-7 m3/molsec

45. Ans 2 nd order: 3 rd order: t 1/(1-P)1/(1-p) 2 t Not autocatalysis k= 2.43x10 -7 M -1 s -1 Extend of reaction after 1 h, p = 0.731 after 5 h, p = 0.931

46. 2. The following data are given for the polycondensation of an acid, HO – (CH 2 ) 14 – COOH where the concentration of the acid was measured at different times by titration: t/(hours):0 0.5 1.0 1.5 Acid/[M]: 3.10 1.30 0.83 0.61 Determine the rate constant and order of reaction. Calculate the degree of conversion after 1 hour

47. MW Carothers equation – relates molecular weight to the degree of conversion of a step growth polymerisation Case 1: stoichiometric ratio 1:1 Degree of conversion, If N = 0, P = 1 % conversion = P. 100 % conversion Number average degree of polymerisation, (N o = initial,N = group left)

48. Example: P: 0.50 0.90 0.99 0.999 0.9999 D p : 2 10 100 1000 100000 Combine the two equation and rearranging, Carothers Equation M o = mean molecular weight of repeat unit of the polymer (Above equation used for 100% conversion)

49. Control MW by changing the mole ratio of the reacting monomers Case II: Non-stoichiometric ratio If N A = number of mole of monomer A N B = number of mole of monomer B If r = 1, D p similar to case I If almost all A reacted, P 1 where r ≤ 1 Q. Calculate D p at P=0.99 with r = 0.98 and 0.99

50. 1. What is the Mn of the polymer below at extent of reaction 35%, 90% and 98%

51. Mean molecular weight, Mo = 192/2 = 96 At P = 0.90, Mn = 960 P = 0.98, Mn = 4800 * Only above 95% degree of conversion the MW gets very large

52. 1. Vinyl acetate at a concentration of 4 M in benzene is polymerized at 60°C using benzoyl peroxide (0.05 M) as an initiator having decomposition rate constant (ki) of 1.12*10 - 6 s-1 and initiation efficiency of 0.75 and a termination by disproportionation only. Calculate the average degree of polymerization. The rate constants at 60°C, kp = 1.0x103 l mol-1 s-1; kt = 3.2x107 l mol-1 s-1.

53. Homework Ans: r = 0.9949

54. Polyethylene glycol(MW=6000) is to be reacted with dicyclohexylmethane diisocyanate to produce polyurethane of Mn = 15500. Calculate mole ratio required for reacting the two. If use 3g of the isocyanate how much required for glycol?

55. MW of diisocyanate = 262 Mo= 6000 + 262 = 6262/ 2 = 3131 Mn = Dp Mo ; Dp = Mn / Mo = 15500/3131 = 5 For complete conversion, r = 2/3

56. No. of mole of diisocyanate = 3/262 = 1.145 x 10 -2 Since Mn dependent mostly on glycol, therefore No. of mole of glycol = 1.145 x 10-2 x 2/3 = 7.6 x 10 -3 MW of glycol required = 6000 x 7.6 x 10 -3 = 45.8g

57. Polydispersity -Distribution of molecular weight with respect to number of the polymer chain. -During synthesis, the chain length and weight of all polymer chain are not similar. -Physical and mechanical properties dependent on molecular weight. The average molecular weight,M n, shows the highest number of polymer chain. But this could be just 5% of all the total number of chain and 95% have molecular weight more or less than Mn. So, the properties of polymer is also dependent on the polydispersity Molecular weight Tensile strength MnMn solubility Molecular weight amount

58. Molecular weight determination 1.Gel permeation chromatogarphy(GPC) (size exlusion chromatography) 2.Solution viscosity 1.GPC -Coulumn packed with highly cross-link polystyrene -Polymer solution flow through column. -Small size polymer penetrate into the pore of column and their flow retarded. -Big size polymer cannot enter pores hence flow out of column at faster rate. -Obtained M n, M w and polydispersity from data. Polymer eluted Retention volume

59. 2. Solution viscosity -Ubbelohde viscometer - obtain: i.Relative viscosity: η r = ii.Specific viscosity: η sp = iii. Reduced viscosity: η red = iv. Inherent viscosity: η inh = v. Intrinsic viscosity: = ln = η int Q. Can molecular weight determine from viscosity measurement ? C 0

60. YES η int c Reduce viscosity Inherent viscosity η int = K. M a Mark-Houwink Equation Values for K and a obtained from referenced data which is temperature and solvent dependent eg. polystyrene at 25 o C using toluene solvent, K = 7.5 x 10 -3 ml/g and a = 0.75. If not available then has to calibrate ownself. 0

61. The following viscosity data were obtained for solutions of poly(methyl methacrylate) in chloroform at 298 K. Conc(g ml -l x10 -2 ) 0.0000.03540.05150.06490.100 Flow time(s) 170.1178.1182.0185.2194.3 [For poly(methyl methacrylate) in chloroform at 298 K, the Mark-Houwink parameters, a = 0.80 and K = 0.48 x 10 -2 ]

62. g/100ml0.000.03540.05150.06490.100 t(sec)170.1178.1182.0185.2194.3 η sp 00.0470.0710.0890.142 η red 01.331.361.371.42 η int η int = 1.18 η int = K. M a 1.18 = 0.48 x 10 -2 M 0.80 M = 977

63. Copolymerization -Formation of new polymer chains through chemical process having more than one type of monomer units. -eg. A + B A – B - B – A – B – A – A - -Differ from blending/composite copolymerisation blending A B A B A - Covalent forces between different monomer units – chemical interaction. -Can tailor-made monomer to suit the desired final properties. -Secondary froces between chains eg. hydrogen bonds, polar interaction and van der Waals ie physical mixing -Problem of mixing and need compatibilizer or coupling agent

64. Why copolymerisation Improve physical/mechanical properties better than the parent homopolymers. i. ekonol (polyester) T m > 400 o C High level of crystallinity T m = 260 o C Low level of crystallinity Rigid Flexible tough + flexible High impact polystyrene ii. Polystyrene-polybutadiene

65. iii. PLLA- PCL copolymer Tensile strength/MPa Elongation at break/% Tm/ o C 703.3169 30337.259 PLLA-PCL copolymer 40288.4144 polycaprolactone polylactide (N.M.Luan, MSc Thesis, May 2008, USM)

66. Type of copolymer; 1. Random copolymers 5. Stereo block copolymer A – B – B – B – A – A – B - 2. Alternating copolymers A – B – A – B – A – B – A - 3. Block copolymers A – A – A – A – A – B – B – B – B – B - 6. terpolymer 4. Graft copolymers A – C – A – B – B - B - C - C A – A – A – A – A – A – A – A – B B B

67. CH 2 = CHCNCF 2 = CF 2 Draw copolymer structure

68. + What are the monomers? x, y?

69. Aims of this section To define reactivity ratio To determine when does monomer copolymerise or homopolymerise To study factors which affect the composition of each monomer in a copolymer chain – compositional drift To study factors which affect the compositional drift

70. Kinetics of chain copolymerisation (radicals, cationic & anionic) Consider monomer A giving active centre A* monomer B giving active centre B* 4 types of reaction occurs: A* + A AA* self propagation A* + B A*B cross propagation B* + B BB* self propagation B* + A B*A cross propagation k11k11 k 12 k 11 k 22 k 21

71. Rearranging and substitution, If reactivity ratio r 1 = k 11 /k 12 ; r 2 = k 22 /k 21, then This is known as copolymer equation. Hence at any one time, the ratio of monomer in the chain is given by the above equation. It does not necessarily describe the structure of the copolymer produced at the end of the reaction since [A] and [B] can vary during the course of the reaction.

72. What is reactivity ratio? -Ratio of activities of each species with its own monomer compared to its reacvtivity with other type of monomers ie: if r > 1homopolymerise(k 11 > k 12 ) if r < 1copolymerise (k 11 < k 12 ) if r = 1homo or copolymerise (k 11 = k 12 ) r = 10.5 r = 0.1 Polystyrene prefer to homopolymerise rather than copolymerise with methyl methacrylate Eg. in cationic polymerisation:

73. M1M1 M2M2 r1r1 r2r2 Styrene Methyl methacrylate 0.5 StyreneAcrylonitrile0.40.05 StyreneVinyl acetate550.01 Styrene1,3-butadiene0.81.4 StyreneVinyl chloride170.02 Methyl methacrylate Vinyl acetate200.015 Methyl methacrylate Vinyl chloride12~0 Methyl methacrylate Acrylonitrile1.20.15 Maleic anhydride Vinyl acetate00.019 (RJ. Young, pg. 69) – Radical mechanism

74. Copolymerisation of styrene (r 1 ) and methyl methacrylate (r 2 ): i.Radical polymerisation r 1 = 0.52; r 2 = 0.46 ii.Anionic polymerisation r 1 = 0.12; r 2 = 6.4 iii.Cationic polymerisation r 1 = 10; r 2 = 0.1 i.r 1 =r 2 =1ideal situation. No preference to homo or copolymerise hence random coplymerisation ii.r 1 = r 2 = 0No preference to homopolymerise so alternating copolymerisation iii.r 1 >>1, r 2 <<1 A monomer prefer to homopolymerise than copolymerise so give block copolymer Conclusion: r 1 = r 2 = 1 random (both almost equal to 1 r 1 r 2 = 0 alternating(both almost equal to zero) Most copolymers are between these two extremes.

75. Why is the copolymerization of styrene with methyl methacrylate preferably performed under radical condition compared to ionic condition?

76. Q-e scheme 2 factors affect reactivity values, r: i. resonance ii. electronic ie. polar k AB = Q A Q B exp (-e A e B ) Where Q A measure resonance contribution of M A radical Q B measure resonance contribution of monomer B e A measure electronic properties of M A radical e B measure electronic properties of monomer B The values obtain is through semi empirical hence they are just an approximate and a guide only… but a good guide!

77. Qe acrylonitrile0.481.23 Butadiene1.70-0.50 Maleic anhydride0.863.69 Styrene1.00-0.800 Methyl methacrylate 0.740.40 Vinyl acetate0.026-0.88 Reference monomer is styrene: Q = 1, e = -0.8

78. Resonance effect In radical copolymerisation, for monomer CH 2 = CHR -A reactive monomer is when substituent group R can stabilised the radical intermediate through resonance. Becomes stable intermediate once coploymerised hence less active radical centre. Eg.styrene Q = 1.00 Stable radical intermediate - unreactive The radical intermediate is stable due to the resonance effect. The propagation rate constant, k is low kpkp

79. An unreactive monomer is when substituent group R cannot stabilised the radical intermediate through resonance. Becomes unstable intermediate once copolymerised hence more active radical centre. Eg. vinyl acetate (Q = 0.026) unstable radical intermediate - reactive The radical intermediate is unstable since no resonance effect. The propagation rate constant, k is high kpkp Styrene and vinyl acetate block

80. Polar effect -If the difference in polarity of 2 monomers is large they will copolymerised efficiently but if the differences is small they will homopolymerised. ie. i. For large polarity differences, eg styrene(r=0.097) with maleic anhydride (r=0.001) Styrene (non-polar) maleic anhydride(polar) (Q=1.00,e = - 0.8) (Q=0.86,e = 3.69) alternating copolymer Draw the copolymer structure?

81. For polymers with identical polarity, r 1 = Q 1 /Q 2 r 2 = Q 2 /Q 1 eg. butadiene(Q 1 = 1.7, e = -0.50) and vinyl acetate (Q 2 = 0.026, e = -0.88) block copolymer Conclusion i. Q and e values almost similar, ideal copolymer ii. Large e value differences of opposite sign, alternating copolymer iii. If Q values very different, block copolymer

82. Q. Base on the Q-e scheme above, predict type of copolymer and draw the polymer formed between: i. styrene + butadiene ii. Maleic anhydride + butadiene

83. Azeotrope: composition of monomer feed (f) = composition of resulting polymer (F) f = F Can also be written as: Where: The dependence of copolymer composition on the proportion of the polymer in the mixture ? f F

84. 0 0.5 1.0 F1F1 iii i ii Copolymerisation of styrene (r 1 ) and methyl methacrylate (r 2 ): Radical polymerisation r 1 = 0.52; r 2 = 0.46 (Azeotrope at f 1 =0.52) Anionic polymerisation r 1 = 0.12; r 2 = 6.4 (No azeotrope) Cationic polymerisation r 1 = 10; r 2 = 0.1 (No azeotrope) f1f1 Conclusion: Copolymer ratio depends on: i.Reactivity ratio ii.Feeding ratio

85. Construct a graph of copolymer composition F 1 against monomer feed composition f 1 when r 1 r 2 = 1 at r 1 = 0.1, 1, 5. Is there any azeotropic feed ratio? F1F1 f1f1 01 1 5 1 0.1

86. 1 0.5 0 Polymer composition, F 0 Conversion(P) 1 r A = 0, r B = 0, f 1 = 0.5 alternating 1 0.5 0 Polymer composition, F 0 Conversion(P) 1 r A = 1, r B = 1, f 1 = 0.5 random monomer A monomer B Χ (black) = 2 alternating = 1 random = 0 block X 2 1 0 X 2 1 0 Compositional drift – change of composition of copolymer at any instantaneous monomer feed ratio. Overall composition equal to initial monomer feed but the local composition along the chain depends on the instantaneous mole ratio and reactivity ratio for each monomers. See: (Macromolecules 1997,30,6727)

87. 1 0.5 0 Polymer composition, F 0 Conversion(P) 1 r A = 5, r B = 0.5, f 1 = 0.5 random 1 0.5 0 Polymer composition, F Conversion(P) r A = 9.55(acrylonitrile), r B = 0.512(methyl methacrylate), f 1 = 0.5 random X 2 1 0 X 2 1 0 monomer A monomer B Χ (black) = 2 alternating = 1 random = 0 block

88. 1 0.5 0 Polymer composition, F Conversion (P) r A = 0.037(styrene), r B = 0.012(maleic anhydride), f 1 = 0.5 alternating X 2 1 0 1 0.5 0 Polymer composition, F Conversion (P) r A = 0.037(styrene), r B = 0.012(maleic anhydride), f 1 = 0.75 alternating – random - block X 2 1 0 monomer A monomer B Χ (black) = 2 alternating = 1 random = 0 block

89. 1 0.5 0 Polymer composition, F Conversion(P) r A = 0.037(styrene), r B = 0.012(maleic anhydride), f 1 = 0.1 alternating - block X 2 1 0 monomer A monomer B Χ (black) = 2 alternating = 1 random = 0 block Therefore preparation for copolymer as expected has to be at low conversion so as to avoid compositional drift

90. Discuss the factors which control rate and degree of chain growth polymerisation. Hand in by 3 Sept 2009 (Your answer should include critical discussion on various factors supported with data taken from references and journals about 2-3 pages together with references. No cut & paste and not from lecture notes).

91. Control in stereoregularity Produce linear Polyethylene Isotactic polypropylene Branched PE Linear PE Isotacticsyndiotactic atactic

92. Coordination polymerisation (ref: Macrogalleria) -Control stereoregularity/tacticity of polymerisation product hence more crystalline eg. Ziegler-Natta catalyst Catalyst system: halides of transition metal and co-catalyst organo-metallic compound in group I – III Eg. TiCl 4 + Al(C 2 H 5 ) 2 Cl Monomers:ethylene, propylene Mechanism: (I) Formation of catalyst active site. -1 empty orbital on Ti filled up by methylene group of catalyst leaving one more empty orbital - Aluminium of catalyst centre loosely bonded to Cl and its methylene group. octahedral

93. (II) Coordination with vinyl monomer eg propylene isotactic Cis-opening of double bond

94. Rearrange equatorial to axial Propagation repeat with all the methyl groups on the growing polymer on the same side of the chain ie isotactic polypropylene. The incoming propylene molecule can only react if it's pointed in the right direction where there is a stereoregularity control ie the methyl group of PP pointed away from active site of the catalyst.

95. Syndiotactic : catalyst system VCl 4 /Al(C 2 H 5 ) 2 Cl. The growing polymer chain stays in equatorial position until another propylene molecule comes.

96. Propylene monomer can only add so that the methyl group is on one side of the chain. The growing polymer chain switches positions with each propylene monomer added, the methyl groups end up on alternating sides of the chain, giving a syndiotactic polymer. Click here

97. Metallocene - high molecular weight eg PE > 7 million - control tacticities A metallocene is a positively charged metal ion sandwiched between two negatively charged cyclopentadienyl anions. Cyclopentadiene cyclopentadienyl anion

98. Why tacticity? Eg. dicyclopentadienylzirconium dichloride. Cocatalyst methylaluminoxane (MAO) – render it soluble and zirconium become +ve charge 1. Clamp-shaped due to the present of 2 chlorine ligands 2. Rigid and chiral centre due to methylene bridge which prevent the ring system swivel 3. Bulky indenyl ligands, pointed in opposite directions Guide the incoming monomers, so that they can only react when pointed in the right direction to give isotactic polymers. The incoming monomer having its methyl/bulky group away from catalytic site. Chiral centre

99. Polymerisation steric hindrance 1 st monomer

100. 2 nd monomer

101. Non-chiralchiral chiral - Atactic - isotactic - syndiotactic

102. 1 In Ziegler-Natta catalysis, a.The central atom form an octahedral geometry during the propagation step. b. Complexation occurs between metal atom and carbonyl group of the monomer. c.Aluminium can be used as the central metal atom at the catalytic site. d.The catalyst produced a highly stereo-specific and branched polymer. 2. Why Ziegler-Natta catalysis only works with  -olefin monomer for example ethylene, propylene, butadiene and styrene. a. During the transition state the  -olefin will be able to stabilized the radical complex form in the intermediate. b. Delocalisation of π electron in  -olefin intermediate structure through resonance will be extensive hence stabilized the metal complex structure. c.  -olefin structure pose a less steric hindrance for entrance into the reactive site of the metal catalyst compared to non-  -olefin monomer. d. Rearrangement from axial to equatorial position at the metal centre is thermodynamically favorable for  -olefin monomer compared to non-  -olefin monomer.

103. 3. The following features contribute to the stereoregular polymer synthesis using metallocene catalyst except: a.Structural rigidity at the catalytic site b.Chiral juxtaposition of reactive group at the catalytic site c.The metal catalyst is sandwich between two parallel aromatic indenyl ligand d.Electron deficient metal catalyst form a complexation with the unsaturated bond of the monomer.

104. Polymerisation techniques: 1.Bulk2. Solution 3. Suspension4. Emulsion 1. Bulk polymerisation Monomer +Polymer Initiator stirring No solvent. Monomer, initiator and polymeric product are mutually soluble hence homogeneous. Advantages: High purity used as such since no other material is present simple set-up

105. Example polymethyl methacrylate(PMMA) and polystyrene N 2 in N 2 out Monomers + initiator stirrer

106. Pre-steady state rate Steady state Gel effect Glassy state (high Tg) Conversion % A Gel effect in bulk polymerisation

107. Disadvantages:Gel effect 1.Pre-steady state: rate of radical production is more than its destruction/termination. So rate of reaction increases with time. 2.Steady state: rate of radical production is equal to its destruction/termination 3.As more chains grow, the reaction medium become viscous. This limit chain mobility and reduces chain termination hence lowered the rate of reaction – gel effect (Region A) 4.High viscosity will increase temperature of reaction medium and reduce heat dissipation to outside and this increase the production of initiator.Rate increase tremendously duringn gel state. 5.Propagation increase and the reaction medium becomes more viscous and more heat produce thus more initiator formed – autoacceleration. 6.Might cause explosion. If high Tg polymer, then the polymeric product in reaction medium turns glassy and gel form. 7.Avoid (i) use low conversion (ii) rigorous stirring (iii) polymeric product can precipitate out.

108. Other disadvantages polymer/monomer solubility – difficulty in separation and purification eg. monomer get trapped in the polymer network Unreacted monomer High MW dispersity. So reaction stop at lower conversion for low polydipersity Conclusion: - Degree of polymerisation: as temperature increased during gel period, radicals production also increased hence lower degree of polymerisation. -Rate of polymerisation: as temperature increase the rate also increased but might deviate from 1 st order monomer dependence. - A high rate of reaction affect a low degree of polymerisation with high polydispersity

109. 40% Variation of first order reaction rate for polymerisation of methylmethacrylate in benzene at 323 o K at different monomer concentrations as the concentration of monomers increase, the rate increase tremendously

110. High molecular weight polymer will not be produces until the very last stage of polymerisation. The viscosity of the reaction medium is rather low hence no gel effect. Why is the bulk polymerisation suitable for step polymerisation?

111. 2. Solution polymerisation Monomer +Polymer Initiator + + stirringSolvent solvent Homogeneous – all constituents dissolve in solvent throughout reaction process eg styrene in toluene Example: Polyvinyl acetate, polystyrene, polyacrylonitrile and ester of acrylic acid Advantages : Control of heat transfer by rigorous stirring and reflux. No autoaccelaration and gel effect. Hence suitable for high temperature reactions.

112. H 2 O in H 2 O out condenser Solvent condense Monomers + initiator + solvent Magnetic stirrer Reflux The reaction mixture heated or stirred vigorously The solvent evaporate and condense in the cold condenser Heat is dissipated to the condenser and the solvent return back into the reactor The cycle continues until reaction complete

113. Advantages – role of solvent Solubilises monomers, catalyst/initiator and polymer product so that homogeneous eg. polystyrene use non-polar aromatic solvent: ethylbenzene or toluene. The product is soluble in the solvent hence propagation continues to give a higher molecular weight. Use high boiling point solvent so that: - enough thermal energy for reaction to occur especially condensation polymerisation - high solvent temperature prevent pre-mature precipitation eg. polyketone use N-methylpyrolidone(NMP) (bp 260 oC) For ionic polymerisation eg. anionic, polar solvent induce ion-pair separation hence increase the reaction rate and degree of polymerisation K+K+ K + + Ion pair Rate in THF 550x than benzene solvent Free ion

114. Disadvantages: (i)Higher dilution result in lower molecular weight product(eg. radical) (ii) Reduce rate of polymerisation(eg. radical). (iii) Chain transfer(radical) eg. (iv) Solvent removal – filtration (solid-liquid) - centrifuge (solid-liquid) - distillation (liquid-liquid) - extraction eg. soxhlet (solid-solid)

115. distillation Rotary evaporator Purification techniques

116. centrifuge Soxhlet extraction

117. 3. Suspension polymerisation Monomer Initiator (monomer soluble) Solvent (usually water) Stabilizer Polymer Water stabilizer stirring A heterogeneous system where monomer droplets of size 10 µm – 10 mm are dispersed in aqueous phase.The monomer is insoluble in water hence act as small bulk polymerisation. Kinetic behaviour similar to bulk polymerisation. Continous aqueuos phase Monomer droplets

118. i.speed of agitator used, ii.the volume fraction of the monomer phase, iii.type and concentration of stabilizer used. If polymer product insoluble in water,precipitation occur in droplets which result in irregular shape/size particles. Size distribution of droplet depends on balance between droplet break-up and droplet coalescence ie: If polymer product soluble in monomer, gel is form than become hard spheres in bead form hence bead polymerisation Monomer droplets gelbeads Monomer droplets precipitate Irregular particles

119. Stabilizer – prevent coalescing of monomer droplets and adhesion of the forming beads. 1 % weight of water content. 2 types: i. water soluble eg. polyvinyl alcohol, hydroxy-propyl cellulose, sodium poly(styrene sulfonate) ii. Water insoluble eg.hydroxyapatite, talc, barium sulfate Initiator – soluble in monomer droplets but insoluble in water. Eg. dibenzoyl peroxide. Water:monomer ratio range from 1:1 to 4:1 Used directly as coating, paints, adhesives which does not require pure products. Example: styrene, acrylic acid, vinyl chloride, Microscopic pictures of PS beads

120. Advantages: Solvent (H 2 O) cheap able to control heat transfer from reactor. able to control viscosity since no gel effect Narrow molecular weight distribution due to small droplet of monomer suspension acting as mini-reactor Products formed in beads hence isolation easy through filtration Disadvantages: Low yield per reactor volume compared to bulk polymerisation Impure product due to adsorption of stabiliser on polymer surface Not used for water sensitive system eg. Ziegler-Natta catalyst, ionic (chain transfer) and condensation polymerisation (hydrolysis)

121. Experimental Styrene and 1, 4-divinylbenzene (the latter as 50-60% solution in ethyl benzene) are destabilized and distilled. A three-necked flask, fitted with stirrer, thermometer, reflux condenser and nitrogen inlet, is evacuated and filled with nitrogen three times. 250 mg of poly (vinyl alcohol) are placed in the flask and dissolved in 150 mL of de-aerated water at 50°C. A freshly prepared solution of 0.25g (1.03 mmol) of dibenzoyl peroxide in 25 mL (0.22 mmol) of styrene and 2 mL (7 mmol) of 1, 4-divinylbenzene is added with constant stirring so as to produce an emulsion of fine droplets of monomer in water. This is heated to 90°C on a water bath while maintaining a constant rate of stirring and passing a gentle stream of nitrogen through the reaction vessel. After about 1 h (about 5% conversion) the cross-linking becomes noticeable (gelation). Stirring is continued for another 7 h at 90° C, the reaction mixture then being allowed to cool to room temperature while stirring. The supernatant liquid is decanted from the beads, which are washed several times with methanol and finally stirred for another 2 h with 200 ml of methanol. The polymer is filtered off and dried overnight in vacuum at 50° C. Yield: practically quantitative. Questions 1.Write the overall reaction scheme for this suspension polymerisation 2.Why need to destabilised and distilled styrene and 1,4-divinylbenzene. 3.What is the function of polyvinylalcohol 4.Why temperature controlled initially at 50 oC than 90 oC. 5.Why passed through N2 gas 6.Why gelation occurs 7.Why use methanol to wash the beads 8.Sketch the experimental set up for this polymerisation.

122. Answer 1

123. Emulsion polymerisation Consist of: i.Water as solvent ii.Initiator – water soluble eg persulphate iii. Emulsifier – forming micelle eg. Anionic: CH 3 (CH 2 ) 6 COONa cationic:CH 3 (CH 2 ) 9 NH 2 HCl Non-ionic:C 12 H 2 O 9 (C 16 H 31 O 2 ) 2 peroxide Sodium lauryl sulfate micelle iv. Monomer – insoluble/sparingly soluble in water hence form emulsion eg. vinyl chloride, styrene

124. CMC – critical micelle concentration ie. attractive forces between hydrophobic and hydrophilic segment exceed solvation energy with water [surfactant] < CMC[surfactant] ≥ CMC Suspension - solution consist of monomer and polymer particles monomer MMA – 2 X Butadiene – 5 X Styrene – 40 x Vinyl acetate – 0.5 X

125. Initial stage -formation of large monomer droplet -surfactant, initiator(H 2 O soluble) and free monomers scattered freely in the emulsion Stage 1 -initiation start and monomer begins to grow -surfactant attached to growing polymer to from micelle.

126. Stage 2 -micelle size swollen as polymer chain grow -more monomer from monomer ‘resevoir’ diffuse into micelle. -’on-off’ mechanism of polymerisation. Stage 3 -more micelle structures form to give latex. - Polymerisation complete with no more monomer particles.

128. Stage I. 2-15% of total polymer conversion. Process facilitated by capturing radicals. Number of particles diffuse into micelles increases, therefore rate increases. Particle number stabilizes at 0.1% of original micelle concentration (10 13 -10 15 particles per mL). As particle grows, absorbs more surfactant. Surfactant conc. falls below CMC. Inactive micelles dissolve into solution. Stage I is finished when all surfactant is adsorbed to particles

129. Stage II Particle number remains constant. Slight increase in rate due to Gel effect. Stage II finished when the droplets disappear Stage III. Particle number remains constant. [M] decreases, therefore rate decreases

130. I-I 2I I + M M On-Off mechanism Initial stage – production of monomer radical in aqueous phase which diffuse into micelle particle 1 st step – entrance of second monomer radical into micelle particle and terminate polymerisation 2 nd step – entrance of 3 rd monomer radical and initiate polymerisation

131. 3 rd step – entrance of 4 th monomer radical and terminate polymerisation Due to size of micelle, effectively 1 radical diffuse at a time. The 1st radical initiate propagation. Then the 2nd radicals enter and cause termination. Polymerisation start again when 3rd radical enter and terminate at 4th radical. So the mechanism is on-off system.On average each particle undergoes half of polymerisation rate.

132. Kinetics Rate of free radical polymerisation: -d[M]/[dt] = kp[M]Σ[M.] where M. is the growing chain radicals. where N = number of particles per unit volume. Substitute into above, On average ½ of particles contains radicals but ½ don’t. So ½ of particle propagate but ½ terminate at any one time. So Σ[M.] = N/2

133. Conversion vs. time for emulsion polymerization of styrene with different concentrations of potassium laurate at 60◦C. Moles of emulsifier: 1 =.0035; 2 =.007; 3 =.014 Effect of emulsion concentration

134. Effect of initiator concentration Conversion vs. time for emulsion polymerization of vinyl chloride at 50◦C with different concentrations of initiator: 1 =.0012%; 2 =.0057%; 3 =.023%.

135. Comparison between bulk and emulsion polymerisation Bulk: Increase [M] result in rate increase but decrease the degree of polymerisation of polymer since production of monomer radical [M ] will increase. Emulsion: rate Rate = Increase of [M] increase rate but the increase of radical monomer [M ] does not affect the degree of polymerisation due to this on-off mechanism but only dependent on number of micelles. important

136. Advantages: Low viscosity Heat removal Disadvantages Surfactant might be water sensitive. Chain length proportional to number of micelles. However due to on-off mechanism, the initiator entering each micelles is constant.Thus increase of surfactant increase rate and molar mass of polymer. Normal homogeneous polymerisation increase in rate will decrease the molecular weight of polymer. In bulk polymerisation, degree of polymerisation not always equal to kinetic chain length due to termination by coupling but in emulsion degree of polymerisation is egual to kinetic chain length due to termination by a long radical polymer chain with a small monomer radical.

137. Quiz ( answer can be more than one) 1.Where are the initiators throughout the polymerisation process a.In aqueous phase b. in micellesc. in monomer droplets 2. Where are the surfactant throughout the polymerisation process a.In aqueous phase b. in micellesc. in monomer droplets 3. Where are the monomer throughout the polymerisation process a.In aqueous phase b. on polymer particlesc. in monomer droplets 4. Where is monomer radicals produced a.In aqueous phase b. in micellesc. in monomer droplets 5. What is inside micelles during polymerisation a.Initiatorb. monomer c. monomer radicald. polymer 6. Why monomer radicals goes into micelles a.Micelles is small size than monomer droplet b.To initiate polymerisation c.Radical monomer is hydrophobic d.Water is hydrophilic

138. 7. Which statement is true a. Initiator is soluble in water so it won’t enter the micelle b. Initiator is soluble in water so it can readily perform polymerisation in aqueous phase 8. Compare the number of micelles particle at stage I, II and III a.I>II>IIIb. I III 9. Why at stage II the rate is almost constant a.No. more monomer radicalb. no. of monomer radical is constant c. no. of micelle is constant 10. Compare the size of micelle at stage I, II and III. a. I>II>IIIb. I III

140. Epoxy resin Use: - Coatings eg. adhesives, industrial flooring, chemical resistant coatings and tank linings, electrical potting, laminates - Structurals eg chip packaging, composite matrices (About 50% of Boeing 787 composites including fuselage and wings, missile casing, tanks), - encapsulant (eg.LED) Properties - good thermal stability(Tg ~ 120 o C cured) - dimensional stability(4.1 x 10 -4 K -1 ) - excellent chemical -corrosion resistance -high tensile strength(83 MPa) and modulus(3.6 GPa) - ease of handling and processability Automotive composite LED encapsulant Electronic packaging For practical aspect of synthesis Ref 1: Polymer Synthesis:Theory and Practice by D. Braun et al QD281 P6P783 Ref 2: Preparative method of polymer chemistry 3 rd Ed by WR Sorenson QD 381 S713 )

141. Synthesis i. Dehydrohalogenation of halohydrin - 2HCl Bisphenol A epichlorohydrin

142. Reaction Condition - catalyst: NaOH in aqeous solvent -Reaction temperature ~50 o C but not >90 o C -Vigorous stirring – 5 h -When reaction complete, let the resin settled, siphoned off solvent, filter then vacuum dry in oven at 80 o C overnight Hot plate Reacting mixture Magnetic bar H 2 O in H 2 O out Water bath vacuum Vacuum filtration residue filtrate Reflux Liebeig condenser

143. Mechanism? Role of catalyst NaOH

144. Molecular weight Ratio of bisphenol to epichlorohydrin – control MW Low molecular weight, ratio of epichlorohydrin:bisphenol higher High molecular weight, ratio of epichlorohydrin:bisphenol lower n ≤ 1 liquid n > 1 solid What is the MW of epoxy when n=0, 2, 9? 1.11 1.15 1.22 1.57 Melting point

145. Epoxy Equivalent weight Defines as grams of resin containing one equivalent of epoxy groups (unit g/eq). ~0.5 g of epoxy resin is refluxed with an excess (50 ml)of pyridine hydrochloride solution (16 ml pure conc. HCl are made up to 1L with pure pyridine) for 20 mins and after cooling is back-titrated with 0.1 N NaOH using phenolphthalein as indicator. 1 mol of epoxy group react with 1 mol of HCl Epoxy equivalent weight is use to determine the ratio of curing agent requires to cure the epoxy resin. Why not based on molecular weight of epoxy resin?

146. Example 1 gram epoxy resin contains 4 x 10 -3 mole epoxy group. Calculate the epoxy equivalent weight? Ans 250 g/eq

147. ii. Epoxidation of cycloaliphatic alkenes with peracid. Percarboxylic acid: Reaction condition: Non-polar solvent eg. benzene, dichloromethane at temperature < 5 o C In situ production: hydrogen peroxide with acetic acid using conc. sulfuric acid as catalyst H 2 O 2 + CH 3 COOH CH 3 COOOH + H 2 O H+H+ (Ref: Advances in Polymer science vol 144 p 50-111, 1999; Also library QD 262 H842, page 292)

148. Reactivity of glycidyl epoxy differ from cycloaliphatic epoxy group hence affect properties after curing Monomers for epoxy resins Bisphenol - F

149. Types of Curing Agents: Primary, secondary amines (tertiary amines ?) eg. ethylene diamine(EDA), triethylene tetraamine(TETA) Anhydride eg methyl hexahydrophthalic anhydride(MHHPA), succinic anhydride MHHPA succinic anhydride Amides Phenols eg novolac Mercaptans cynates ester (ref:Journal of Applied Polymer Science, Vol. 100, 2293–2302 (2006)

150. Curing of epoxy resins The low molecular weight epoxy resin is liquid but high molecular weight is fuseable solid having low thermal property (mostly amorphous, Tg < 120 o C). Uncure epoxy resin soluble in most solvent but cured resin is insoluble. Hence cross-linking process to give a better thermo-mechanical end product. i. through epoxy ring (Ref: PolymerNetbase:Polymeric Material encyclopaedia in library database)

151. Draw the cross-link network for the cured epoxy resin as shown below:

153. Effect of cross-linking agent on curing rate of DGEBA 1. Basicity/nucleophilicity 2. Steric hindrance

154. 1. Determine the monomer used for the following:

155. Epoxy/diamine ratio TgΔ H/Jg -1 100:15137-400 100:16.3146-414 100:18141-396 100:22120-350 DGEBA with 1,2-diamine cyclohexane

156. Gelation: liquid-to-rubber transition, Gelation is a well- characterized,distinct and non-reversible event which occurs at a certain degree of conversion. Gel time is when maximum conversion rate is achived during curing. Vitrification: liquid/rubber-to-glass transition is a gradual, thermo-reversible process. No molecular motion occur. Conversion,  = 0 10% 90% on-set max cure

157. Gel time Plot graph of rate of conversion d  /dt against time for curve 1…

158. Polyimide - Commercial name Kapton, Vespel PVC PI Density 1390 kg/m1430 kg/m Young Modulus 300 MPa3200 MPa Tensile strength 50 – 80 MPa75-90 MPa Elongation break 20 – 40 %4.8% Tg 82 oC>400 oC Tm 100 – 260 oCNone CTE 8 x 10 -5 /K5.5 x 10 -5 /K Water absorption 0.40.32 Price 1.00 є/kg3.5 є/kg

159. 45 exo Tg temp % decomposition Differential scanning calorimetry(DSC) Thermogravimetric analysis(TGA) -Application: High temperature electrical insulation, auto- and aeronautic structural components, printed circuit board as dielectric material Solar cell substrate

160. Room temperature, DMAc 300 o C – H 2 O Polyamic acid Polyimide synthesis i. Polyaddition ii. Polycondensation dianhydride dianiline

161. Reaction condition -Mole ratio dianhydride : dianiline = 1 : 1 - Solvent system: polar aprotic eg. dimethyl formamide(DMF), dimethyl acetamide (DMAc) and n-methyl pyrollidone (NMP) - Inert atmosphere eg. nitrogen gas -2 steps process: i. polyaddition at < 30 o C. Reaction almost complete after 30 mins. Obtain precipitate after washing with water (low MW) or methanol (high MW) ii. Polycondensation ie. cyclisation of ring with elimination of H 2 O. Performed thermally >300 o C or chemically using acetic anhydride as catalyst

162. The pre-polymer polyamic acid is soluble to some organic solvents eg. DMF and NMP hence can be: i. Solution casting ii. Spin coating iii. Electro-spinning During solution phase addition of fillers and other additives is possible. But once it is cured, it is very intractable to processing having Tg > 230 o C Thermal curing is performed in stages eg: 100 o C120 o C250 o C300 o C This is to allows for: fully crosslink, minimize residual stresses that are conducive to crack formation get rid of any trapped air/volatile impurities. 0.5h 2.0h 1.0h

163. Type of monomer DianhydrideDianiline A stable polyimide hence high molecular weight result from a strong base dianiline eg. contains ether and sulfide group. A low molecular weight result from poor base dianiline eg. contains carbonyl and sulfone group

164. Polyimide design 1.Low glass transition X Rotational barrier(Kcal/mol) Tg/ o C CH 2 0.32 285 O 0.41 290 S 0.71 303 CO 0.76 310 C(CF 3 ) 2 1.08 317 SO 2 2.33 335 Factor affecting the Tg is the chain rigidity. Rigid less mobile high Tg Flexiblemore mobilelow Tg

165. 2. Low coefficient of thermal expansion (CTE) α = 10.0 ppm/ o C, rod-like, rigid and crystallizable α = 51.0 ppm/ o C, zig-zag, non-crystallizable (US Patent 4,690,999)

166. ε = 4.28 ε = 3.78 ε = 3.05 3. Low dielectric constant

167. 2. Suggest a cross-link structure for chitosan through imide linkage 1. Suggest the monomer used for the following polyimide.

168. Polysiloxane (silicones polymer) -Silicone in same group as carbon -high temperature stability -Si – O bond very strong but flexible so although low Tg (eg -127 o C) but decomposed at > 300 o C -resistance to moisture (hydrophobic) and other adverse conditions. -high permeability to moisture and vapor, no moisture stays entrained in the material. -Resistence to ozone, UV,ozoneUV Application 1.Aerospace eg. making tiles at space shuttle, grease and sealants 2. Opticals and electronics eg. LED encapsulation, dielectric materials 3. Medicine eg. contact lens, body-implants eg. heart valves, prosthetics parts -Change in molecular architecture results in changes in modulus, glass transition temperature, coefficient of thermal expansion (CTE), eg.(i) low n liquid, high n solid (ii) High phenyl content increase refractive index( ie from 1.38 – 1.53) - C-C-C bond length of 1.54 Å, Si-O-Si bond length of 1.63 Å. C-C-C bond angle of 112˚, Si-O-Si bond angle of 130˚

169. Silicones active centre ++ -- Nucleophilic attack Si Nuc + O - Example: If 3 chlorine, then will give 3-dimensional network - Produce linear and cyclic structure of n = 3 - 9 hydrolysis silanol

170. 1.Cationic polymerisation- catalyst: Bronsted acids eg. H 2 SO 4,CF 3 SO 3 H,CH 3 SO 3 H Lewis acid eg. AlCl3, BF3 i.Initiation – protonation of monomer Linear polysiloxane obtained through ring opening cationic and anionic polymerisation of cyclic oligomers. Oxonium ion Cationic initiator

171. ii. Propagation iii. termination – end capping Hexamethyldisiloxane

172. 2. Anionic polymerisation Catalyst: alkali metal hydroxides eg. KOH, alkoxides, (CH 3 ) 3 SiOK and other bases.Higher MW than cationic polymerisation Initiation propagation

173. Low MW – waxy eg silicone oils High MW – elastomeric. Requires x-linking X-linking 1.Peroxides – radical mechanism 2.Condensation Eg. silanol + 3 H 2 O Cold curing

174. Eg. hydrosilylation

175. Predict the product of the following: X X + Platinum catalyst

176. Q. Predict the polysiloxane product from polycondensation of the following monomers using methane sulfonic acid as catalyst:

177. Answer where m and n depends on mole ratio of the two monomers

178. Assignment (Choose either one): 1.Review the curing technology of epoxy resin 2.Review the synthesis of low dielectric constant of polyimide Hand in by 20 Oct 2009

179. Problem based learning(PBL) Based on the US Patent 5444106: 1.What is the objective of this patent? 2.What are the problems staement faced in these type of materilas? 3.What are the additives added into the formulation? 4.What are the MW of the polysiloxane resin? 5.What are the phenyl content of the polysiloxane resin? 6.Describe the fume silica used as proposed in this patent. 7.What are the curing agent used in this patent? 8.Describe the use of UV absorber in this patent. 9.Describe the use of inhibitor in this patent. 10.Describe the processing of this material into a lens.

180. Polyurethanes Applications: 1.Fibres/elastomer (Tg below room temperature) 2.Adhesives and coatings 3.foams Soft flexible Rigid crystalline elastomer foam Ref:Iranian Journal of Polymer Science and Technology vol1 no 2 August 1992)

181. Polyurethane Synthesis: addition polymerisation between polyol and diisocyanate Polyol diisocyanate Oxygen attack carbon Nitrogen attack hydrogen

182. Modulus,є’ Dynamic mechanical analysis -1000100200 Temp/oC Melting of soft block Melting of rigid block

183. Polyurethane - monomers Aliphatic diisocyanate Aromatic diisocyanate Polyols

184. Disadvantage of PU Poor resistance to hydrolysis Discolouration

185. To produce fibres, adhesives and foam structure, polyurethanes need to be cross-link. Cross-linking agent: water, polyol and diamines Water

186. Diamines

187. Foam formation When the polymeization reaction begins, the mixture is a liquid, and any carbon dioxide that is produced just bubbles away. But as the reaction progresses, and molecular weight increases, the mixture will become more and more viscous, until it becomes a solid. When this happens, the carbon dioxide bubbles will be trapped in the viscous liquid. When the polymer solidifies, they stay there, trapped. These bubbles make the polymer a foam.

188. Fenol-Formaldehyde resin – high molecular thermoset products formed by condensation of phenol and carbonyls compounds eg. formaldehyde. Initially soluble and fuseable but later become cross-linked(cured) by subsequent reaction. -high hardness, good electrical and mechanical properties and chemical stability. ResolNovolac catalystBase eg. NaOH, RNH 2 Acid eg oxalic, sulfuric acids X-link agentSelf x-linkingRequires addition of x- link agent eg. formaldehyde, hexamethylenetetramine MW300 - 700600 - 1500 ratioFormaldehyde > phenol Eg. 1.5:1 Formaldehyde < phenol Eg. 0.8:1

189. Resol synthesis (EBP 216): Mix fenol (10 g) with formalin (12 ml) in a test-tube then add NaOH ( 1 mL). Heat the reaction mixture in oil bath at ~100 o C with stirring for 1 hour. Saparate from aqueous layer. Heat treatment Stage A stage B stage C Linear unX-link partially x-link fully x-link Soluble in EtOH soluble in EtOH insoluble in EtOH Liquid fuseable solid Methylene linkage Ether linkage