51Example Predict the feasibility of the following reactions under standard conditions.a. MnO2 + 4 H+ + 2 Cl- → Mn2+ + 2 H2O + Cl2b. Zn + Sn4+ → Sn2+ + Zn2+Solutiona. The two half-cells involved are:Reduction : MnO2 (s) + 4 H+(aq) + 2e →Mn2+(aq) + 2 H2O (l) Eored= + 1.23 VOxidation : 2 Cl-(aq) → Cl2(g) + 2e Eooxi= - 1.36 VOverall : MnO2 + 4 H+ + 2 Cl- → Mn2+ + 2 H2O + Cl2E°cell = E°oxi + E°red = (+ 1.23) + (- 1.36) = - 0.13 VThe reaction is not feasible (non-spontaneous reaction)b. The two half-cells involved are:Oxidation : Zn(s)→ Zn2+(aq) + 2e Eooxi = + 0.76 VReduction : Sn4+ (aq) + 2e → Sn2+ (aq) Eored = + 0.15 VOverall : Zn(s) + Sn4+(aq) → Sn2+(aq) + Zn2+(aq) Eo = (+0.76) +(+0.15) = + 0.91 VThe reaction is feasible (spontaneous reaction)2.2.5 NERNST EQUATION AND EQUILIBRIUM CONSTANT (K)Nernst equation is used to calculate cell potential (Ecell) at non-standard condition and at any chosen concentration (concentration ≠ 1M).Ecell = E°cell -0.0592log[product]xn [reactant]yn = no. of electrons involved
52Example Given thatFe3+ + e Fe2+ E° = + 0.771 VDetermine the half –cell potential when the concentration of Fe2+ is five times that of Fe3+SolutionEcell = E°cell -0.0592log[product]xn [reactant]yGiven that [Fe2+] = 5[Fe3+]Ecell = 0.771 -0.0592Log 5 1 = 0.771- 0.041 = 0.730 VEffect of Concentration on Half Cell / Cell PotentialConcentration significantly affects half-cell and cell potential (voltage) via the Nernst equation, which shows potential decreases as product concentrations rise (or reactant concentrations fall), moving towards equilibrium where cell potential becomes zero and the reaction quotient (Q) equals the equilibrium constant (?). Increasing reactant concentration or decreasing product concentration makes the cell potential more positive (more spontaneous), while the equilibrium constant (?) itself only changes with temperature, not concentration, but its value dictates the standardpotential (?∘) and spontaneity.How it works:• Decreasing reactants/Increasing products: (Q) increases, (ln (Q)) increases, so Ecell decreases (becomes less positive/more negative).• Increasing reactants/Decreasing products: (Q) decreases, (ln (Q)) decreases, so Ecell increases (becomes more positive).
53+ −2.3 ELECTROLYSIS2.3.1 Electrolytic cellElectrolysis is the decomposition of an electrolyte by an electrical current.▪ Electrochemical cell converts chemical energy to electrical energy. ▪ Electrolytic cell uses electricity to produce chemical reactions.The diagram below (figure 2.7) shows the arrangement of a simple electrolysis cell. Figure Electrolytic cell• The electrodes are dipped/immersed into the same electrolyte.• Molten salt and aqueous ionic solution are usually used as electrolytes. • The electrode connected to the positive terminal of the battery is called anode. The electrode connected to the negative terminal of the battery is called cathode.• Oxidation takes place at the anode, while reduction takes place at the cathode.• Negative ions (anions) are attracted to the anode where they undergo oxidation by giving out electrons.• Positive ions (cations) are attracted to the cathode where they receive electrons and undergo reduction.• Electron flows from anode to cathode.cationanionanodecathodeelectrolyte A
542.3.2 Predicting the product of electrolysisFactors affecting the products of electrolysis:▪ The standard electrode potential (SRP) values of the species concerned▪ Concentration of the ions2.3.2.1 Electrode potential valuesThe reaction taking place at the anode of an electrolysis cell is oxidation. Substances which lose electrons easily (i.e. strong reducing agents) will be oxidized first.Hence, the more negative the standard electrode potential, the easier it is to be discharged at the anode.The reaction taking place at the cathode of an electrolysis cell is reduction Substances which accept electrons readily (i.e. strong oxidizing agents) will be reduced first.Hence, the more positive the standard electrode potential, the easier it is to be discharged at the cathode.Example Predict the electrolysis reaction when CuSO4 (aq) is electrolyzed using platinum electrodes SolutionThe platinum electrodes are inert, so we need to consider only the species present in the solution.The ions present in the solution are:CuSO4 (aq) → Cu2+ (aq) + SO42−(aq)H2O (l) H+ (aq) + OH−(aq)At Cathode: The standard electrode potential for the two half-cell are:Cu2+ (aq) + 2e Cu (s) E = + 0.34 V2 H+ (aq) + 2e H2(g) E = 0.00 VCu2+ is a stronger oxidizing agent than H+. Hence, copper (II) ions will be reduced in preference to H+.The reaction at the cathode is: Cu2+ (aq) + 2e → Cu (s)
55At Anode, the two half-cell are:2 H2O (g) + O2 (g) + 4e 4 OH−(aq) E = + 0.40 VS2O82−(aq) + 2 e 2 SO42−(aq) E = + 2.01 V OH−ion is a stronger reducing agent than SO42−(its E value is more negative). Hence, OH−ion will be oxidized to O2:4 OH−(aq) 2 H2O (g) + O2 (g) + 4e Generally, in electrolysis using inert electrodes;• Species with the most positive E value will be discharged first at the cathode.Species with the most negative E value will be discharged first at the anode.2.3.2.2 Concentrationa) Electrolysis of molten saltMolten salt consists of pure molten ionic compound. There is no presence of water in the salt. Hence, only ions will undergo reduction and oxidation. No competition between water and ions of molten salt.Electrolysis of molten salt requires high temperature. The example is the electrolysis of molten NaCl (l). Dissociation of molten NaCl will give Na+(l) and Cl-(l) ion. Half-cell equation:Anode : 2Cl-(l) → Cl2(g) + 2eCathode : 2Na+(l) + 2e- → 2Na (s)Overall : 2Cl-(l) + 2Na+(l) → Cl2 (g) + 2Na(s)Electrolysis of molten NaCl gives sodium metal deposited at cathode and chlorine gas evolved at anode. Electrolysis of molten NaCl is industrially important. The industrial cell is called ‘Downs Cell’.
56Downs Cellb) Electrolysis of aqueous solutionElectrolysis of aqueous solution is more complex in the presence of water as the water itself can be oxidized and reduced instead of the ions of the aqueous solution only. Competition exists between water with anion and cation. • Possible cathode half-reactions (reduction)a) Reduction of H2Ob) Reduction of cations in the solution• Possible anode half-reactions (oxidation)a) Oxidation of H2Ob) Oxidation of active metal electrodesc) Oxidation of anions in the solutionIf the water is selected to react at anode and cathode, the half cell of the water will be:If oxidized (anode) : 2H2O(l) → O2(g) + 4H+(aq) + 4e- E° = + 1.23VIf reduced (cathode) : 2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = - 0.83VElectrolysis in diluted solutionConsider the electrolysis of aqueous NaCl (aq). At each half-cell:Cathode : 2 competing reactions (Na+ and H2O are attracted to cathode) 2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = - 0.83V 2Na+(aq) + 2e- → 2Na(s) E° = - 2.71VBetween the two half-cell at cathode, ✓ Eofor water molecules is more positive.✓ H2O easier to be reduced.
57So, cathode : 2H2O (l) + 2e- → H2 (g) + 2OH- (aq) E° = - 0.83VHydrogen gas is released.Anode : 2 competing reactions (2Cl- and H2O are attracted to anode) O2 (g) + 4H+ (aq) + 4e- → 2H2O (l) E° = + 1.23VCl2 (g) + 2e- → 2Cl-(aq) E° = + 1.36VBetween the two half-cell at cathode, ✓ Water will be selected for oxidation because of its lower Eo✓ H2O easier to oxidizeSo, anode : 2H2O (l) → O2(g) + 4H+(aq) + 4e- E°oxi = - 1.23VOxygen gas is released.Electrolysis in concentrated solutionConsider the electrolysis of aqueous NaCl(aq). At each half-cell:Cathode : 2 competing reactions (Na+ and H2O are attracted to cathode) 2H2O (l) + 2e- → H2(g) + 2OH-(aq) E° = - 0.83V 2Na+(aq) + 2e- → 2Na(s) E° = - 2.71VBetween the two half-cell at cathode, ✓ E for water molecules is more positive.✓ H2O easier to be reduced even though the concentration of Na+ is higher.This is because the difference between the E values of the two ions is too large. Furthermore, if sodium (a very reactive metal) is formed it will immediately reacts with water to form the Na+ ions again.Sodium metal is unstable in water, 2 Na(s) + 2 H2O (l) → 2 NaOH (aq) + H2(g)So, cathode : 2H2O (l) + 2e- → H2 (g) + 2OH- (aq) E° = - 0.83VHydrogen gas is released.Anode : 2 competing reactions (2Cl- and H2O are attracted to anode) O2 (g) + 4H+ (aq) + 4e- → 2H2O (l) E° = + 1.23VCl2 (g) + 2e- → 2Cl-(aq) E° = + 1.36VBetween the two half-cell at cathode, ✓ In concentrated solution, chloride ions will be oxidized because of its high concentration.So, anode : 2Cl-(aq) → Cl2(g) + 2e- E°oxi = - 1.36VChlorine gas is released.
582.4 Quantitative Electrolysis and Faraday’s LawFirst Faraday’s LawState that the mass (amount) of substance produce during electrolysis is directly proportional to the quantity of electricity (electric charges; measured in coulombs, C) that flows through the electrolytem Q , Q=It , where:m = mass of substance , Q = amount of charges that flow through electrolyte.I = current (measured in ampere, A)t = times (s)Faraday’s constant (F) is the total charge in one mole of electron.1 F = 96500 C = 1 mole electronExample Using this half equation of Mg2+ and Al3+ to prove the second Faraday’s law.SolutionMg2+ + 2e → Mg so that ½Mg2+ + e → ½ Mg Al3+ + 3e → Al so that 1/3 Al3+ + e → 1/3 AlBoth are the same quantities of electricity. The amount of Al less than Mg. Otherwise charge of Al is +3 but for Mg is +2.Example What weight in grams of Cu will be plated out from a solution of Cu2+ on passing 0.15 Faradays?SolutionCu2+ + 2e → CuTwo faradays plate out one mole, soWeight = [0.15 faraday] x [½ moles / Faraday] x [ 63 g/mole] = 4.73 g CuExample We can use electrolysis to determine the gold content of a sample.The sample is dissolved, and all the gold is converted to Au3+ (aq), which is then reduced back to Au(s) on a weighed electrode. The reduction half-reaction is Au3+(aq) + 3e → Au (s).What mass of gold will be deposited at the cathode in 1.00 hour by a current of 1.50A?[Ar (Au) = 197 , 1F= 96500 C ]
59SolutionAmount of charge = time x current, Q = It = 1 x 60 x 60 x 1.5 = 5400 C1 mol electron = 96500 C, so for 5400 C the mole of electron is:n of electron = 5400 C / 96500 C = 0.05595 mol of electronUse a half equation to relate moles of electrons to moles of the products,Au3+ (aq) + 3e → Au (s)3 moles electron = 1 mol AuSo, the number of moles of Au = 0.05595 /3 = 0.01865 mol AuTherefore, the mass of Au in gram is = 197 g/mol x 0.01865 mol= 3.68 g Au
60CHAPTER 3 REACTION KINETIC
61SUBTOPICS EXPLANATORY NOTES3.1 Rate of reaction and stoichiometrya. Define reaction rate.b. Write differential rate equationc. Determine the reaction rate based on rate differential equation3.2 Rate Law a. Define rate law, order of reaction and half lifeb. Write the rate law with respect to the order of reaction.c. Determine the order of reaction involving reactant using:i. Initial rate methodii. Half life based on graph of concentration against time.3.3 relationship between reactant, concentration, and time half-life of a reaction.a. Deduce the order of a reaction by the initial rate method; deduce zero- , first- and second- order reactions; deduce the order of reaction from concentration-time graphs.b. Use integrated form of rate equation to determine zero- , first- and second- order reactions involving a single reactant.c. Calculate t½ for a first order reaction.3.4 Collision theory a. Explain qualitatively, in terms of collision theory, the effect of concentration, and temperature on the rate of reaction.3.1 Rate of reaction and stoichiometryRate of reaction is referred to as the change of the concentration of reactant or product with time. Consider the reactionA B The equation tells us that during the course of reaction, reactant molecules (A) are consumed while product molecules (B) are formed. As a result concentration of the reactant decreases while concentration of the product increases.
62The change in Concentration with Time for the Reaction A → BInstantaneous rate is determined from a graph of time versus concentration by drawing a line tangent to the curve at the point that corresponds to a particular time.Br2(aq) + HCOOH(aq) 2Br-(aq) + 2H+(aq) + CO2(g)The instantaneous rate of the above reaction at t = 100s, 200s and 300s are given by the slope of the tangents at these times in Figure 2.2.Graph time versus concentrationAverage rate = [Br2]final - [Br2]initialtfinal – tinitial
63Reaction Rate and StoichiometryFor stoichiometrically simple reaction such as, A → B▪ The minus sign indicates concentration of A decreases.▪ The plus sign indicates concentration of B increases.Unit: mol L−1 s−1For more complex reactions, 2A → B2 of moles of A disappear for each mole of B that forms. The rate of disappearance of A is twice as fast as the rate of appearance of B.Rate =- 1 d [A]=+ d [B]2 dt dtIn general, for the reaction aA + bB → cC + dDWhere a, b, c and d are stoichiometric coefficient.Rate =- d [A]=+ d [B]dt dtRate =- 1 d[A]=- 1 d[B]=+ 1 d[C]=+1 d[D]a dt b dt c dt d dt
643.2 Determination of Rate LawAn expression relating the rate of a reaction to the rate constant and the concentrations of the reactants. For a general reaction of the type,aA + bB → cC + dDthe rate law takes the form; rate = k [A]x [B]ywhere ; [A] = concentration of A in molar[B] = concentration of B in molark = rate constantx = order with respect to Ay = order with respect to Bx + y = reaction orderNote: the value of x and y is NOT taken from a and b.Definition of reaction order▪ The sum of the powers to which all reactant concentrations appearing in the rate law are raised.▪ The value of x and y must be determined experimentally.▪ The value of x and y CANNOT be deduced from stoichiometric coefficientExperimental Determination of Rate LawFor reaction: A → Brate = k[A]xi) For a certain reaction x = 0, the rate law for the reaction is, rate = k [A]0 rate = k • Rate of reaction is always constant.• Therefore this reaction is zero order with respect to A.
65ii) For a certain reaction where x = 1, the rate law for the reaction is, rate = k [A]1 Assume that intially [A] = 1.0 M rate1 = k (1.0 M) If we double the concentration of A from 1.0 M to 2.0 M rate2 = k (2.0 M) = 2k (1.0 M)Hence, rate2 = 2 rate1• Thus, doubling the concentration of A will double the rate of reaction.• Therefore, this reaction is first order with respect to A. iii) For a certain reaction x = 2, the rate law for the reaction is, rate = k [A]2 Assume that intially [A] = 1.0 M rate1 = k (1.0 M)2 = k (1.0 M) 2 If we double the concentration of A from 1.0 M to 2.0 M rate2 = k(2.0 M)2 = 4k(1.0 M)Hence, rate2 = 4 rate1• Thus doubling the concentration, the rate will increase by a factor of 4.• Therefore this reaction is second order with respect to A.
66The initial rate method2 I − + S2O82− → I2 + 2 SO42−The kinetic data for the above reaction (in the presence of Fe2+ catalyst) is shown below.Experiment [I−]/moldm-3[S2O82−]/moldm-3Initial rate/moldm-3s-11 0.05 0.05 8.20 x 10-42 0.05 0.10 1.64 x 10-33 0.10 0.10 1.64 x 10-4a) Determine the order of reaction with respect to I − and S2O82−.b) Calculate the rate constant.c) Calculate the initial rate when the concentration of I− and S2O82− are 0.12 mol dm-3 and 0.15 mol dm-3respectively.Solutiona. Let the rate equation be:rate = k [I−]x [S2O82−]yComparing experiment 2 and 1;1.64 x 10-3=(0.05)x(0.1)y8.20 x 10-4(0.05)x(0.05)y 2 = (2)yy = 1
67Comparing experiment 2 and 3;1.64 x 10-3=(0.05)x(0.1)y8.20 x 10-4(0.10)x(0.10)y( ½)x = 1 x = 0The reaction is first order with respect to S2O82− and zero order with respect to I−.b. Considering experiment 1:8.20 x 10-4 = k (0.05)0(0.05)1k =8.2 x 10-4= 1.64 x 10-2 s-10.05c. Rate = (1.64 x 10-2)(0.12)(0.15)0= 1.97 x 10-3 mol dm-3 s-1.3.3 Relationship between reactant, concentration, and half-life of a reaction3.3.1 Zero Order ReactionA zero order reaction is a reaction independent of the concentration of reactant.A→ productThe rate law is given by- d dt[A]= k [A]o - [A] = kt Unit of k for zero order reaction Ms-1
68Half life of a zero order reactionHalf life (t½) is defined as the length of time required for the concentration of a reactant to decrease to half of its initial value.Substituting t = t½ , and [A] =2[A] into the zero order reaction, gives [A]o - [A] = kt [A]o2[A]0= kt½ Solving for t½ gives t½ = 2k[A]0Characteristic graph for zero order reactionRate[A]Zero orderGraph of rate versus [A][A]0 −A]tGraph of [A] versus t
693.3.2 First Order ReactionsA first order reaction is a reaction which rate depends on the concentration of reactant raised to the first power. From the rate law,rate = k[A]To obtain the units of k k = s-1By using calculus - ln[A] + ln [A]oor ln [A][A]0= kt Or when expressed in common logarithm log [A][A]0 = 2.303ktHalf-life of a first order reactiont½ = kln 2Characteristic graph for first order reactionRate[A]Graph of rate versus [A][A]tGraph of [A] versus tt1 t2
70t1 = t2 = 1st order reactionExampleFor the first order decomposition of H2O2 (aq) given the k = 3.66 x 10-3 s-1 and [H2O2]o = 0.882 M, determinea) the time at which [H2O2] = 0.600 Mb) the [H2O2] after 225 s.Solutiona) ln [H O ][H O ]2 22 2 0 = kt ln 0.6000.882= 3.66 x 10–3 s-1 x t ln 1.47 = 3.66 x 10–3 s-1 x tln [A]tGraph of ln[A] versus tln([A]o/[A]) = ktln[A]o– ln[A] = ktln[A] = −kt + ln[A]o y = mx + cln[A]o / [A]tGraph of ln versus tln([A]o/[A]) = kt y = mx
71 t = 33.66x10ln1.47− = 0.385 3.66 x 10-3 = 105.263sb) ln[H O ][H O ]2 22 2 0= kt ln[H O ]0.8822 2 = 3.66 x 10–3 s-1 x 225 s ln [H O ]0.8822 2 = 0.824 [H O ]0.8822 2= e0.824 [H O ]0.8822 2 = 2.280 [H2O2] = 2.2800.882=0.387 M Example The decomposition of ethane C2H6 to methyl radicals is a first order reaction with a rate constant of 5.36 x 10-4 s-1 at 700oC. C2H6 (g) → 2CH3 (g)Calculate the half-life of the reaction in minutes.Solution t½ = kln 2 = 45.36x100.693− = 1.29 x 103 s = 21.5 min
723.3.3 Second Order ReactionsA second order reaction is a reaction which rate depends on the concentration of one reactant raised to the second power or on the concentration of two different reactants each raised to the first power.The simpler type involves only one type of reactant molecule. A→ productwhere; rate = −ddt[A]= k[A]2 From the rate law rate = k[A]2As before, we can determine the units of k by writing k = M-1s-1Using calculus we obtain the following expression [A]1= o[A]1 + kt Half life of a second order reaction t½ = o[A]1k
73b) Characteristic graph for second order reactionExample Iodine atoms combine to form molecular iodine in the gas phase I(g) + I(g) → I2(g)This reaction follows second order and has a high rate constant 7.0 x 109 M-1s-1i) If the initial concentration of iodine was 0.086 M, calculate the concentration after 2 min.ii) Calculate the half life of the reaction if the initial concentration of iodine is 0.06 M and 0.42 M respectively.Solutioni)[A]1= o[A]1 + kt [A]1 = [0.086]1 + (7.0 x 109 M-1s-1)(2 x 60) = 11.628 + (8.4 x 1011) = 8.4 x 1011 [A] = 1.190 x 10-12 MRate[A]Second orderGraph of Rate versus [A]tGraph of versus t
74ii) [I2] = 0.06 Mt½ = 0[A]1k = 7.0x10 M s (0.060M)19 −1 −1 = 2.4 x 10-10s [I2] = 0.42 M t½ = o[A]1k = 7.0x10 M s (0.042M)19 −1 −1 = 3.4 x 10-10 s3.4 Activation energy and enzymes as catalysts The collision theory is a theory developed from the kinetic theory to account for the effects of concentration and temperature on reaction rates. The collision theory is based on three ideas:a. Molecules frequently collide with one another. Chemical reactions occur as a result of collision between reacting molecules. The rate of a reaction is directly proportional to the number of collisions per second. b. There is some minimum collision energy below which no reaction occurs. In order to react, the colliding molecules must have total potential energy equal to or greater than the activation energy (Ea), which is the minimum amount of energy required to initiate a chemical reaction. The species temporarily form by the reactant molecule as the result of the collision before they form the product is called the activated complex. Collisions are known as effective collisions.rate number of collisions
75Reaction profile and activation energy for (a) exothermic (b) endothermic reactionc. We must also consider the “orientation factor”, that is how the reacting molecules are oriented relative to each other. Molecules must collide in the right orientation in order that the collisions will result in a reaction. This is sometimes called the steric factor.K + CH3I KI + CH3This reaction is most favorable only when the K atom collides with the I atom in CH3I head-on. Otherwise, few or no products are formed.3. 5 Factors affecting the rate of chemical reactionThe rate of a chemical reaction depends on • The concentration of reactant• Temperature • Size of the reactant particles (surface area)• Pressure (if reactants are in gaseous state)• CatalystThe reaction rate depends on the frequency of effective collisions. The frequency of effective collisions depends on the total collision frequency and the fraction of reactant molecules having sufficient energy to react to collision.
76Frequency of effective collision = total collision frequency x fraction of molecules with sufficient energy.Any change in conditions which increases either the total collision frequency or the fraction of molecules with enough energy will increase the reaction rate.a) ConcentrationIf concentration is doubled the number of collisions would also double, because in any given volume there would be twice as many molecules that could collide. Consequently, the rate would increase by a factor of 2.b) TemperatureThe average kinetic energy of molecules is proportional to the temperature. The higher the temperature the higher the rate of reactionKinetic energy distributions for a reaction mixture at two different temperatures.The above figure shows Maxwell-Boltzmann speed distribution for a gas at temperature T1 and temperature T2. Note that the curve flattens at the higher temperature. The shaded areas represent the number of molecules traveling at a speed equal to or greater than a certain speed u1. The higher the temperature, the greater the number of molecules moving at high speed.As temperature increases, the average kinetic energy of the particles increases,resulting in more collisions per unit time. More importantly, when the temperature is increased, there is an increase in the total number of particles with energy
77equal to or greater than the activation energy, Ea. These combined effects caused the rate of reaction to increase.c) Size of particles (reactant)The smaller the size of reacting particles, the greater is the total surface areaexposed for reaction and consequently the faster the reaction. In the case of heterogeneous systems, in which the reactants are in different phases, the area of contact between the reacting substances will influence the reaction rate considerably. d) PressureChange in pressure only affects gaseous reactions. When pressure is applied, volume decreases. The molecules are brought closer together. The number of molecules per unit volume increases resulting in more collisions per second. The number of effective collision increases so the rate of reaction also increases.e) CatalystA catalyst is substance that increases the rate of a chemical reaction without itselfbeing consumed. A catalyst provides an alternative pathway for the reaction to occur. This alternative pathway has a lower activation energy compared with the one without the catalyst. A catalyst increases the rate by lowering the activation energy for the reaction.Comparison of the activation energy of the catalyzed and uncatalyzed reaction
78CHAPTER 4 ALCOHOL
79SUBTOPICS EXPLANATORY NOTES4.1 Classification of alcoholsnaming, physical propertiesa) Draw, classify, and name the hydroxyl compounds according to IUPAC nomenclature.b) Explain boiling point of alcohol 4.2 Reaction to form halogenoalkanes,reaction with sodium, oxidation, dehydration, Esterificationa) Explain the chemical properties with reference to reaction(i) substitution using HCl, PX3, PX5 (X=Cl,Br) and SOCl2(ii) with Na(iii) esterification(iv) dehydration with concentrated H2SO4/H3PO4(v) oxidation with KMnO4, CrO3 / H+, PCC and Cr2O72-/ H+4.3 Iodoform test, Lucas’ testa) Identify methyl alcoholb) Identify 1o, 2o and 3o alcohols using Lucas reagent, i.e HCl / ZnCl24.0 HYDROXY COMPOUND4.1 INTRODUCTION▪ Hydroxy compounds contain the hydroxyl group, -OH, which as the functional group, determines the characteristic properties of this family.▪ 2 types of hydroxy compounds –i) aliphatic alcoholii) phenol (aromatic alcohol) i) Aliphatic alcohol✓ Cn H2n+1 OH✓ -OH is attached to alkyl group, R✓ IUPAC nomenclature e (alkane) → ol (alcohol)✓ Example:CH3CH2OHEthanolCH3CH2CH2OHpropanolCH2OHPhenylmethanolH3C CH2OH(1,4-methylphenyl)methanol
80ii) Phenol✓ -OH group is attached directly to the aromatic ring✓ The –OH group in phenols is known as the ‘phenolic’ group and not hydroxyl group.✓ IUPAC nomenclature: phenols.✓ Example:OHPhenolOHCH32-methylphenol CLASSIFICATION OF ALCOHOLS▪ primary alcohol (1°) – OH group attached to 1° carbon (eg: ethanol)▪ secondary alcohol (2°) – OH group attached to 2° carbon (eg: 2-butanol)▪ tertiary alcohol (3°) – OH group attached to 3° carbon (2-methyl-2-propanol)NOMENCLATURE OF ALCOHOLIUPAC system is widely used through the following rules:The parent alcohol is determined by choosing the longest carbon chain with OH group attached to it, and the lowest number is given. The prefix and suffix also use if alkyl groups are existed more than one times When there are two or more –OH groups present, the name ends with diol, triol and so on.OH OH2,4-pentandiolOHOHOH1,2,2-propanetriolOH2-butanolOH4,4- dimethyl-2-pentanolOHCyclopentanolOH3-pentanol
81BOILING POINT OF ALCOHOL▪ As molecular weight increases, van der Waals forces increase, boiling point will increase too.▪ For isomeric alcohol (same molecular weight) :- increased in boiling point with increasing carbon number. decreasde in boiling point with branching. boiling point in descending order: 1° alcohol > 2° alcohol > 3° alcohol 4.2 CHEMICAL PROPERTIES OF ALCOHOLSAs Weak acid (Cleavage of RO─H bond)A. Reaction with Sodium• Alcohol reacts with Sodium to form sodium alkoxide and hydrogen gas.• This reaction shows the acidic properties of alcohol.• General equation:RO-H + Na → RO-Na+ + ½ H2• Example:CH3CH2OH + Na → CH3CH2O-Na+ + ½ H2 (sodium alkoxide)B. Reaction to form an ester (esterification)Reaction with carboxylic acids.• General equation: RCOOH + R'OH RCOOR' + H2Ocarboxylicacidalcohol ester water• CH3COOH + CH3CH2OH → CH3COOCH2CH3 + H2O etanoic acid ethanol ∆• During this reaction, the O-H bond in ethanol is broken.• The reaction is slow, so concentrated sulphuric acid is used as a catalyst and also as a dehydrating agent since the removal of water will push the equilibrium to the right.Cleavage of R-OH BondA. Dehydration of alcohols to yield alkenes• Example:C CR''HR'RR''' OH + H2SO4 (conc.) C CR'''R'' R'R+ H2O∆
82HH• Refer to the Saytzeff Rule to predict major alkene product.**Saytzeff Rule = states that the major product is the more stable alkene. The more stable alkene is the one with the greatest number of alkyl groups attached to the C-C• Examples:B. Reaction with Hydrogen Halides• When alcohol react with a hydrogen halide, a substitution takes place producing alkyl halide and water.• General equation:R - OH + H X → R - X + H2O• Example:CH3CH2-OH + HCl → CH3CH2-Cl + H2OC. Reaction with Phosphorus Halides (PX3/PX5) and Thionyl Chloride (SOCl2)• All these reactions result in cleavage at the C-O bond of alcohol to form alkyl halides.• Example:C(CH3)3-OH + PBr3 → C(CH3)3-Br + H3PO3CH3CH2-OH+ PCl5 → CH3CH2-ClCH3CH2CH2-OH + SOCl2 → CH3CH2CH2-ClReaction that may also involve breaking the C-C or C-H bonds.A. Oxidation reaction of alcohol• Oxidation of methanol (simplest alcohol) using KMnO4 and K2Cr2O7 can produce methanoic acid (carboxylic acid), CO2 and water.• Oxidation methanol using PCC can produce methanal (an aldehyde).• The reagant commonly used are KMnO4, K2Cr2O7 (as a strong agents) give carboxylic acid and PCC as a weak agent give an aldehyde.• Simplest alcohol (methanol) will produce methanal and carboxylic acid, primary alcohol (carboxylic acid), secondary alcohol (ketone) and tertiary alcohol (no reaction) C H3CCCH3C H3OHH2S O4 H2C C H2CCH3+minor product major productH3C C CCH3C H3
83Class of alcohol Reaction with ProductPrimary i) K2Cr2O7 / H2Cr2O7 ii) PCC i) Carboxylic acidii) AldehydeSecondaryi) K2Cr2O7 / H2Cr2O7ii) KMnO4iii) PCCi) ketoneii) ketoneiii) ketoneTertiary i) K2Cr2O7 / H2Cr2O7ii) PCC / KMnO4i) No reactionii) No reaction• Example:CH3CH2OHKMnO4/ H+CH3COOH1 colourlessoCH3CHCH3OH KMnO4 / H+CH3CCH3O2 colourlessoCH3CCH3OHCH3KMnO4/ H+no reaction(purple)3o• In the reaction, the purple colour of potassium manganate (VII) is decolourised, while the orange colour of potassium dichromat (VI) turns green.• Tertiary alcohols (and phenols) are resistant to oxidation because it involves the breaking of strong carbon-carbon bonds.Note: any alkyl group that is attached to the benzene ring will be oxidized to carboxylic acid (-COOH).4.3 IDENTIFICATION TEST OF ALCOHOLLucas Test to classify alcohols.▪ Lucas Reagent – a mixture of concentrated HCl and zinc chloride (HCl/ZnCl2)▪ primary alcohol – does not turn cloudy ▪ secondary alcohol - the solution turn cloudy within 5 minutes▪ tertiary alcohol - the solution turn cloudy immediately.
84OHl▪ Example:CH2OHOHOHCH3HCl / ZnCl2HCl / ZnCl2HCl / ZnCl2does not turns cloudyturns cloudy in 5 minutesturns cloudy immedietlyIodoform to classify alcohols.• Iodoform reagent: I2 in NaOH(aq) @ Na2CO3, heat • NaIO (Sodium iodate(I))• Formation of yellow precipitate, CHI3 , indicate that organic compound consist of methyl carbonyl or methyl alcohol:- ll l H• Example: CH OHH3CR+ 3I2base (OH-)C O-OR+alcoholCHI3yellow precipitate(iodoform)-C-CH3 @ -C-CH3 O-C-CH3 @ -C-CH3O OHH
85 CHAPTER 5 PHENOL
86SUBTOPICS EXPLANATORY NOTES5.1 Naming and physical propertiesa) Name the compounds according to IUPAC nomenclature.b) Explain the boiling point of phenol.c) Compare the acidity of phenol, alcohol and water5.2 Chemical ReactionExplain the chemical properties of phenol with reference to reaction with:a) sodiumb) sodium hydroxidec) nitration (include orto, para director)d) halogenation i) Br2 in inert solvent (include orto, para director)ii) Br2 in wateriii) Br2 in FeCl35.1 INTRODUCTION▪ Formula: C6H5OH ▪ Differ from alcohols in having the –OH group attached directly to an aromatic ring.5.1.1 NOMENCLATURE▪ Generally named as derivatives of the simplest member of the family.Example:HO CH34-methylhydroxybenzene(4-methylphenol)HO OH1,3-benzenediolBrNH2HO3-amino-5-bromohydroxybenzene(3-amino-5-bromophenol)O2NClHO4-chloro-2-nitrohydroxybenzene(4-chloro-2-nitrophenol)OH OH Kekulé structure
875.1.2 PHYSICAL PROPERTIES OF PHENOL• High boiling points because of intermolecular hydrogen bonding.• Acidity (phenol is stronger acid than water and alcohol because it can form phenoxide ion which is resonance stabilized).5.2 CHEMICAL REACTION 5.2.1 Reaction with SodiumExample:OH+ NaO-Na++ 1/2 H25.2.2 Reaction with NaOH Example:OH+ NaOHO-Na++ H2ONote: aliphatic alcohol does not react with NaOH because aliphatic alcohol is a weakeracid compared to phenol (phenoxide ion is more stable compared to alkoxide ion). O H O H O HHydrogen bondingOH-H+O O O Ophenoxide ion
88Chemical test between phenol and aliphatic alcoholOH+ NaOHO-Na++ H2OsolubleOH+ NaOHinsoluble in water(2 layers form)5.2.3 Nitration• With dilute HNO3, the reaction will form a mixture of ortho and para products.OHdilute HNO3OHNO2+OHNO2o-nitrophenol@2-nitrophenolp-nitrophenol@4-nitrophenolNote: The – OH group is ortho-para director• But with the treatment of concentrated HNO3, 2,4,6-trinitrophenol is formed.OHconcentratedOHNO22,4,6-trinitrophenolO2NNO2HNO3
895.2.4 HalogenationA) Reaction with Bromine in inert solventTreatment of phenols with bromine in non-polar solvent results in replacement of hydrogen at ortho and para position (2 products). OHBr2/ CH2Cl2OHBr+OHBro-bromophenol(o-bromohydroxybenzene)p-bromophenol(p-bromohydroxybenzene)B) Reaction with Bromine in water• Treatment of phenols with aqueous solutions of bromine results 2,4,6-tribromophenol. This reaction can be used as one of the identification test of phenol as 2,4,6-tribromophenol is observed as white precipitate.OHBr2/ H2OOHBr BrBr2,4,6-tribromophenol (white precipitate)C) Reaction with FeCl3 (aq)• Phenol turns the yellow FeCl3 solution to purple complex. Ordinary alcohols donot react. This test is best used to distinguish phenols from alcohols.OH OHFeCl3(aq)FeCl3purple complex(yellow)
90Exercises1. For the following scheme,W(I)HydrationCH2CCH3OHH(II)CH2CCH3BrHa. Identify W in the reaction sequence given above. b. Name the alcohol produced in step I and classify it. c. Suggest a reagent that can be used for the conversion in step II.2. Write the structural formula of the compound formed in each of the following reactions:a.CH2OHCH2OH + Na25CAb.OH180oCBconc.H2SO4c.CH2(OH)CH(OH)CH2OH + HCld.CH3CH(CH3)CH2OH DK2Cr2O7H+e.CH2(OH)CH2CH2(OH) + HCOOH Ef.(CH3)3COH + PCl5 F3. A few organic compounds are represented by the alphabets, P to T.H3C CHOHP: CH3OHQ.R:CH2=CHCH3S:H3C C OHCH3HT: OHWhich of the above compounds;a. decolorize bromine at room temperature? b. Release hydrogen gas when reacted with sodium? c. Forms a yellow precipitate when heated with alkaline iodine? d. Forms a purple coloration when a drop of FeCl3 is added? e. Forms an acidic solution when it is strongly oxidized by acidified potassium manganate (VII)
91CHAPTER 6 SYNTHETIC POLYMERS
92SUBTOPICS EXPLANATORY NOTES6.1 Introduction to polymerb. Explain the terms: monomer, polymer, homopolymer, copolymer, straight chain polymer and crosslink polymer.c. State some commons examples of synthetic polymers.6.2 Addition polymerization a. Explain the preparation of synthetic polymers through:addition polymerization to produce polyalkenes such as polyethene, polyvinyl chloride and polystyrene. 6.3 Condensation polymerization a. Explain the preparation of synthetic polymers through:Condensation polymerization to produce polyamides such as nylon 6, nylon 6, 6 and Kevlar and polyester such as Dacron and terylene.6.1 INTRODUCTION▪ All living organisms are primarily made up of polymeric materials. Fingernails, hair and most parts of our body are polymers. Polymer is a macro molecule that is made up of small repeating units called monomers.▪ The process that combines monomers to form polymers is called polymerization. ▪ Polymers that are built from the same type of monomers are called homopolymers. For example: polyethene and polypropene.▪ Polymers that are built by different monomers are called copolymers.Table 6.1 shows the monomers used for making some common polymers and the structure of the repeating units.Monomer PolymerRepeating unit(monomer)CH2 = CH2ethene-CH2-CH2-CH2-CH2-Poly(ethane)Polythene @ PECH2 CH2nCF2 = CF2tetrafluoroethene-CF2-CF2-CF2-CF2-Poly(tetrafluoroethene)Teflon @ PTFECF2 CF2nCH3CH=CH2propeneC C C CHCH3HHHCH3HHC CHCH3HH n
93Poly(propene)Vectra @ PPCH2=CHC≡NC C C CHHHCNHHHCNPoly(propenenitrile)OrlonC CHHHCN n6.2 SYNHTETIC POLYMERS• Synthetic polymers are polymers that are prepared in industries from monomers that have gone through polymerization process.▪ Synthetic polymerSynthetics polymers can be classified base on structures of the polymersThere are three types of structures in synthetic polymers:a. linear polymersb. cross-linked polymer
94a. Linear polymers• consist of monomers that are linked in straight and long chain. Linear or straight-chain polymer can be folded back upon themselves in a random fashion. Linear polymer is recycleable because it is soft and can be reformed when heated. The figure below shows the example of linear polymer. • An example of a linear polymer is the homopolymer, poly(ethane)— CH2- CH2- CH2 – CH2—• The general formula of the polymer of a homopolymer can be given as -( A-A-A-A-A-A-A-A)-n• Monomers link together in a straight chain • The repeating units in a linear copolymer can be arranged alternately, randomly or in blocks, for example:Note: Where A and B are the monomers or repeating units of the copolymer.• Example of linear copolymer:C CO OOH2CH2COPolyestherb. Cross-linked polymers • contain branches that connect linear polymer chain, as shown in the figures below. Cross-linked polymer is harder (rigid) and more elastic polymer compared to linear polymer. • This polymer cannot be remelted or remolded again.
95 A A A A A A A A A AA A A A A A A A A AX X Long polymer chain cross-linked by branched• Example of cross-linked polymer is vulcanized rubber and bakelite plastic.Some examples of synthetic polymers are Nylons in textiles and fabrics, Teflon in non-stick pans, Bakelite for electrical switches, polyvinyl chloride (PVC) in pipes, etc. The common PET bottles are made of a synthetic polymer, polyethylene terephthalate. The plastic kits and covers are mostly made of synthetic polymers like polythene.6.3 Type of polymerization▪ Two types of chemical reactions that are used to form polymer:A. addition polymerizationB. condensation polymerization.A) Addition polymerization• Addition polymerization involved the addition reaction of unsaturated monomers without elimination of any small molecule. Therefore, addition polymer always involves the polymerization of monomers which have double bond within the monomers. • The example of this reaction is as below to produce polyethane, polyvinyl chloride and polystyrene.i. Reaction to produce polyethaneH2C CH2peroxideCH2CH2H2C CH2 H2C CH2 H2C CH2nCH2CH2n
96ii. Reaction to produce polyvinyl chlorideHC CH2peroxideHC CH2ClnCliii. Reaction to produce polystyreneHC CH2peroxideHC CH2nExamples of addition polymers:Polymer Monomer EquationLow density polyethene (LDPE) 0.92 gcm-3EtheneCH2=CH2C CHHHHnO2, 200°C1200 atmC CHHHH nPhysical characteristics:Comprises of branched chains. Hence, the packing is not close. Low density, melting point 105°C.Uses:To make plastic film, squeeze bottles, electric wire insulation.Polymer Monomer EquationHigh density polyethene (HDPE)0.95 gcm-3EtheneCH2=CH2C CHHHHn60°C 1 atmTiCl4, R3AlC CHHHH nPhysical characteristics:Comprises of linear chains. Hence, the packing is close. Higher density, melting point 135°C.Uses:To make toys, bottles, milk jugs, plastic pails.Polymer Monomer EquationPolyisobutylene 2-methyl propeneCH2=CCH3 I CH3C CHHCH3CH3nO2, 200°C1200 atmC CHHCH3CH3 n Physical characteristics:Synthetic rubber. Air tight and gas impermeable. Flexible.Uses:Inner tubes, sealants, stoppers for medicine bottles, chewing gum.
97Polymer Monomer EquationPolypropene(Polypropylene)PropeneCH2=CH I CH3C CHHHCH3n C CHHHCH3 n60°C 1 atmTiCl4, R3AlPhysical characteristics:Tough plastic material. Resists moisture, oils and solvents. Melting point 121°C.Uses:To make indoor and outdoor carpets, packaging material, cling film, toys.B) Condensation polymerization• Condensation polymers are formed when two different monomers combine together with elimination of a small molecule such as water or methanol. • The monomers of condensation polymers (or also known as step-growth polymers) must have at least two reactive functional groups on both ends, which can react with each other or with other functional groups.• There are two types of condensation polymers:i) Polymers formed from a single monomerwhich has two different functional groups. For example, proteins are formed from amino acids which have the amino, -NH2 and the carboxyl, -COOH groups. Nylon 6 is also formed from only one monomer, 6-aminohexanoic acid.H2N (CH2)5 COOH6-aminohexanoic acidN (CH2)5 CONH H(CH2)5 COnn@N (CH2)5 CH On+ n H2O+ n H2ONylon 6
98ii) Polymers formed from two different monomerseach having two identical functional groups. For example, polyesters can be formed from diols (dihydric alcohols, with two –OH groups) and dicarboxylic acids (with two –COOH groups).HO C C OHHHHHC (CH2)4 COOHO+ HOethane-1,2-diol hexanedioc acidC (CH2)4 CO OO C C OHHHHn nn+ n H2OpolyesterExamples of condensation polymers are polyesters and polyamides.Polymer MonomerPolyamide a. Diamine + dicarboxylic acid, HOOC-(CH2)n –COOHb. Diamine + alkane dioyl dichloride, ClOC-(CH2)n -COClPolyester a. Diol + dicarboxylic acid , HOOC-(CH2)n-COOHb. Diol + alkane dioyl dichloride , ClOC – (CH2)n-COCl① Polyamides ▪ Example of polyamide polymers are Nylon 6,6 and Kevlar. i. Formation of Nylon 6
99ii. Formation of Nylon 6,6 N (CH2) n 6 N + C (CH2)4 COOHOHOhexanedioc acidnHHHH1,6-hexanediamineN (CH2)6 N + n H2OnHCH O(CH2)4 CONylon 6,6 iii. Formation of Kevlar N N + COCOn H n Cl ClHHHC CO ONHNHn+ n HClKevlar1,4-diaminobenzene terephthalic acid② Polyester▪ Another type of condensation polymer is polyester. As the name implies, the repeating functional groups in this polymer chain are ester. ▪ The most familiar polyester is polyethylene terephthalate known as Dacron and Terylene. ▪ The polymer is formed by the reaction of ethylene glycol with methyl ester of terephtalic acid. In the process, a molecule of methanol is split out for each new ester group formed. The equation below show the formation of Dacron and Terylene.
100i. Formation of Dacron HO C C OHHHHH+ethane-1,2-diolCOO C C OHHHHnnn CH3OHC CO OH3CO OCH3CO+Dacrondimethyl terephthalatemethanol ii. Formation of Terylene.HO C C OHHHHH+ethane-1,2-diolCOO C C OHHHHnnn H2OC CO OHO OHCO+Terylenebenzene-1,4-dicarboxylic acid