CHAPTER 9THERMOCHEMISTRY

CALORIMETER- Heat absorb/release concept- Bomb calorimeter/coffee-cup

EXOTHERMIC/ENDOTHERMIC- Comparison

- Enthalpy diagram

BORN-HABER CYCLE- Lattice energy

- enthalpy solution

HESS LAW- Diagram method- Algebraic method

STOICHIOMETRY

DEFINITION ENTHALPY

Standard enthalpy of combustion

(i) Heat released when 1 mole of substances completely combusted in O2 gas at standard states (25oC, 1 atm)1) Heat release/absorb2) 1 mole …..3) Method4) At standard state (25oC, 1 atm)

CALORIMETER

Heat energy release

Heat energy absorb

Question 1A mass 50.0 g mass of a metal was heated to 100oC and the plunged into 100 g of water at 24oC. The temperature of the resulting mixture became 28oC.(A) How many joules did the water absorb?(B) How many joules did the metal lose?(C) What is the specific heat capacity of the metal?

Question 1 (A)

Question 1 (b)

Question 1 (C)

Question 2

When 2.28 g octane, C8H18 was burnt in a bomb calorimeter, the temperature of 1000 cm3 water increased by 24.2oC. What is the enthalpy of combustion of octane? [Specific heat capacity of solution = 4.20 Jg-1oC-1: density of solution = 1.0 g cm-3]. (Mr octane=114)

[Ans: - 5082 kJ mol-1]

STEP 1 : CALCULATE HEAT RELEASE, q

q heat = - (q cal + q water) = - (C T + mc T) (unit C & c – Joule, J) = - ( 1000 g x 4.20 x 24.2) = -101640 J

STEP 2: EQUATION AND MOL SUBSTANCE

C8H18(g) + 25/2 O2 (g) 8CO2(g) + 9H2O(l)

Mol substance = mass/ molar mass = 2.28 / 114 = 0.02 mol

STEP 3: ENTHALPY COMBUSTION, H

0.02 mol C8H18 needed -101640 kJ

1 mol C8H18 needed -101640 x 1 mol 0.02 mol

= - 5082000 JHeat of combustion = - 5082 kJ/mol

Question 3100 cm3 of 0.5 mol dm-3 KOH and 100 cm3 of CH3COOH of the same concentration were mixed at 20oC in a polystyrene beaker. The final temperature of the experiment was 23.25oC. Calculate the standard enthalpy of neutralization. [Specific heat capacity of solution = 4.20 Jg-1K-1: heat capacity of the polystyrene beaker can be ignored].

STEP 1 : CALCULATE HEAT RELEASE, q

q heat = - (q cal + q soln) = - (C T + mc T) (unit C & c – Joule, J) = - ( 200 g x 4.20 x (23.5-20oC)) = - 2940 J

STEP 2: EQUATION AND MOL SUBSTANCEMol = MV 0.1 X 0.5 = 0.05 molCH3COOH(g) + KOH(g) CH3COOK(g) + H2O(l)

0.05 mol 0.05 mol 0.05 mol

STEP 3: ENTHALPY COMBUSTION, H

0.05 mol H20 needed -2940 J

1 mol C8H18 needed -2940 x 1 mol 0.05 mol

= - 58800 JHeat of combustion = - 58.8 kJ/mol

STOICHIOMETRY

Question 4 2Al2O3 (s) 4Al (s) + 3O2 (g) Hrxn = 3351 kJ

Based on the Hrxn value of the reaction above, determine:(a) the enthalpy of formation Al2O3 [Ans: -1675.5 kJ mol-1]

(b) the heat release when 81.0 g of aluminium is combusted in pure oxygen [Ans: - 2513.25 kJ]

(c) how is the mass of aluminium must be combusted in pure oxygen to produce heat – 26808 kJ.

[Ans: 864 g]

HESS’S LAWALGEBRAIC METHOD

Use the following reactions and their respective enthalpy changes to calculate the enthalpy change for the oxidation of ethyne.

C2H2 (g) + 5/2 O2 (g) 2CO2 (g) + H2O (l)

C (s) + O2 (g) CO2 (g) : Hof = -394 kJmol-1

2C (s) + H2 (g) C2H2 (g) : Hof = +227 kJmol-1

H2 (g) + 1/2 O2 (g) H2O (l) : Hof = -286 kJmol-1

Target Equation

Reference Equation

❶ ❷ ❸

QUESTION 5

Calculate the standard enthalpy of combustion of methanol, CH3OH if the standard enthalpies of formations. Given:Hof CH3OH(l) : -238.5 kJ/mol,

Hof CO2 (g) : -393.5 kJ/mol Hof H2O (l) : -285.6 kJ/mol

• Target: CH3OH(l) + 3/2 O2 (g) CO2(g) + 2H2O (l)Enthalpy formationC (s) + 2H2(g) + ½ O2(g) CH3OH(l) [-238.5]C (s) + O2(g) CO2(g) [-393.5] H2(g) + ½ O2(g) H2O(l) [-285.6] Enthalpy= (238.5) + (-393.5) + (2 x -285.6) Ans: -726.2 kJ/mol

BORN HABER CYCLE

H1 = - 411Na (s) + ½ Cl2 (g) NaCl (s)

H2 = +108 H4 = +121

Na (g) Cl (g)

H3 = +496 H5 = -349

Na+ (g) + Cl-(g)

H6 = ?

H formation = sum all enthapies + Lattice energy

Question 6

(a) Calculate enthalpy solution MgF2 if, enthalpy of hydration Mg 2+ ion = –1903 kJ/mol and enthalpy of hydration F- ion = - 461 kJ/mol. Lattice energy of MgF2

is -3121 kJ/mol

(b) What would you expect regarding the lattice energy? Explain your answer. (i) CaCl2 and CaF2

(ii) MgO and NaCl

Question 6(a)MgF2(g) Mg2+ (aq) + 2F- (aq)

H soln

Mg2+ (g) + 2F- (g)

- Lattice energy

+3121 kJ -1903 kJ-461 x 2 kJ

Hydration Mg 2+

Hydration F -

Hsoln = +296 kJ ………

Hsoln = (-L.E) + Hhyd

Question 6 (b) (i) & (ii)

(i) Lattice energy CaF2 > CaCl2 …………

Because ionic size of F- is smaller than Cl-

So, the attraction of Ca2+ and F- is stronger than Ca2+ and Cl-. …….

(ii) Lattice energy MgO > NaCl ………… Because ionic charge of Mg2+ and O2- is bigger than Na+

and Cl-

So, the attraction of Mg2+ and O2- is stronger than Na+ and Cl-. …….

CHAPTER 10ELECTROCHEMISTRY

GALVANIC CELL- Galvanic cell diagram

- SHE diagram- calculate standard electrode potential

- Nernst equation

ELECTROLYTIC CELL- Selective charge(molten, dilute and

concentrated)- calculate Faraday’s

constant

AGENTS- oxidizing/reducing

agents

GALVANIC CELL

Question 1

A galvanic cell consists of a Mg electrode in 1.0 M Mg(NO3)2, a Sn electrode in 1.0 M Sn(NO3)2 and a KCl salt bridge. Draw and label the cell diagram

1.0 M Sn(NO3)2 1.0 M Mg(NO3)2

1.0 M Sn2+1.0 M Mg2+

@

Mg (s)electrode

Sn (s) electrode

e->

Anode (-) Cathode (+)

Sn 2+ (aq)

K+Cl-

Question 2

Draw and label a schematic figure of a standard hydrogen electrode to measure standard reduction potential of a silver electrode

E ocell H2/H+ = 0.00 vEo cell Ag+ /Ag = + 0.80 Reduction - cathode

@ Ag+ (aq)

QUESTION 3

The cell notation of an electrochemical cell is given as below:

Zn (s)/Zn2+(aq, 1 M) // H+(aq, x M) /H2 (g, 1 atm)/Pt (s)

Eo Zn 2+/Zn = 0.76 V , Eo

H+/H2 = 0.00 V ,

(I) Write an equation at anode and cathode(II) If cell voltage for this cell is 0.67 V, determine the concentration

of H+ (hydrochloric acid)(III) What is the pH of the acid?(IV) Suggest two method to increase cell voltage, E cell

QUESTION 3(i) and (ii)

Anode: Zn (s) Zn2+ (aq) + 2e-

Cathode: 2H+ (aq) + 2e- H2 (g)

Overall: 2H+ (aq) + Zn (s) H2 (g) + Zn2+ (aq)

E0cell = E0

cathode – E0 anode

=(0.00 V)- ( -0.76 v) Ans = 0.76 V

E cell = Eo cell – 0.0592 log [Zn2+ ] . P2H2

2 [H+ ]2

0.67 = ( 0.76 V) – 0.0592 log [1.0 M] . (1.0 atm) 2 [H+ ]2

= 0.03 M

QUESTION 3(iii)

pH = - log [H+] = - log (0.03 M) = 1.52

QUESTION 3(v)E cell = Eo cell – 0.0592 log [Zn2+ ] . P2H2

2 [H+ ]2

if the Eo cell (+ value) Example: E cell = (+0.76 V) – 0.0592 log [Zn2+ ] . P2H2

2 [H+ ]2

So, (i) lowering temperature (ii) Increasing concentration [H+ ]

(iii) decrease concentration [Zn2+ ]

QUESTION 4

Calculate equilibrium constant, K for this reaction

2H+ (aq) + Zn (s) H2 (g) + Zn2+ (aq)

Eo Zn 2+/Zn = 0.76 V , Eo

H+/H2 = 0.00 V ,

Anode: Zn (s) Zn2+ (aq) + 2e-

Cathode: 2H+ (aq) + 2e- H2 (g)

E0cell = E0

cathode – E0 anode

=(0.00 V)- ( -0.76 v) Ans = 0.76 V

E cell = Eo cell – 0.0592 log K

2 E cell = 00 = (0.76 V) -0.0592 log K

2K = 4.7389 x 1025

OXIDIZING AND REDUCING AGENT

QUESTION 5 The standard reduction electrode potential for

several half-reactions at 298 K are shown below:

Electrode reaction Eo (V) Ag+ + e- Ag + 0.80 Co3+ + e- Co2+ + 1.92 Zn2+ + e- Zn - 0.76(a) Can Zn oxides Ag in the reaction?(b) Write cell notation between Ag and Co in

above standard reduction electrode.

Question 5 (a)

Arrange the substances according strength ofreducing agent(arrange based on SRP table)

Co3+ + e- Co2+ + 1.92Ag+ + e- Ag + 0.80Zn2+ + e- Zn - 0.76

Ans: No, because Zn more negative Eocell than Ag .

Reduction

oxidation

Question 5 (b)

Co3+ + e- Co2+ + 1.92Ag+ + e- Ag + 0.80

Reduction (cathode)

Oxidation (anode)

Anode: Ag(s) Ag + (aq) + e- Cathode: Co3+ (aq) + e- Co2+ (aq) Overall: Co3+ (aq) + Ag (s) Co2+ (aq) + Ag+ (aq)

Ans:

Ag(s) / Ag + (aq,1 M) // Co3+ (aq, 1M) ,Co2+ (aq, 1M) /Pt (s)

Cu2+(aq) + 2e- Cu(s) E = 0.34 V o

2H+(aq) + 2e- H2(g) E = 0.00 V o

Zn2+(aq) + 2e- Zn(s) E = – 0.76 V o

Strength of oxidizing agent:

Cu2+ > H+ > Zn2+

Strength of reducing agent:

Zn > H2 > Cu

QUESTION 6

OXIDIZING AGENT ? REDUCING AGENT ?

ELECTROLYTIC CELL

QUESTION 7

A concentrated aqueous solution of sodium chloride is electrolysed for an electrolytic cell.

(i) Write the possible oxidation/reduction at anode and cathod.

(ii) Predict the products of the electrolysis of an aqueous solution of sodium chloride in a cell using platinum electrodes. Give reasons.

QUESTION 7 (i)(i) NaCl (aq) -----concentrated

Na+ , Cl- ,H2O

Anode(Oxidation): 2Cl-(aq) Cl2(g) + e-

2H2O(aq) O2(g) + 4H+ + 4e-

Cathode(Reduction): Na+(aq) + e- Na(s)

2H2O(aq) + 2e- H2(g) + 2OH-

Possible Equation

QUESTION 7 (ii)

Anode (oxidation) 2Cl-(aq) Cl2(g) + e-

• Because Cl- can be oxidized due to the high concentration.• Product – Cl2(g)

Cathode (reduction) 2H2O(aq) + 2e- H2(g) + 2OH-

• Because Na is active metal. Metal cannot be reduced because E0

cell > negative.

• product- H2(g)

QUESTION 8

If 3.00 A of current is used for 20 minutes during electrolysis concentrated ZnCl2, calculate the quantity of substances that are formed at both electrodes at room temperature.

[Mr Zn = 65 g/mol]

[Ans: 1.22 g and 0.448 dm3]

Step 1: ANodeCalculate: Quantity charge, Q Q = It

= (3 A)(20 x 60s)= 3600 C

Step 2: Equation (depend on question)Anode (oxidation)

2Cl-(aq) + 2e- Cl2(g)

Step 3: Stochiometry1 mol Cl2 produced

(2 x 96500) = 193000C 1 mol Cl2 deposited 193000C (equation)

………… mol Cl2 deposited 3600 C

2 F

(Q=It)

193000 C ------------- 1 mol3600 C -------------,

= 0.0187 mol

Step 4: 1 mol---------24 dm3 (room temp)0.0187 mol…………. 0.0187 x 24/ 1mol = 0.448 dm3

3600 C x 1 mol 193000C Volume STP/

room temperature

Step 1: CathodeCalculate: Quantity charge, Q Q = It

= (3 A)(20 x 60s)= 3600 C

Step 2: Equation (depend on question)

Cathode: Zn2+ + 2e Zn (s)

Step 3: Stochiometry1 mol Cr produced (2 x 96500) = 193000C 1 mol Cr deposited 193000C (equation) ………… mol Cr deposited 3600 C

2 F

(Q=It)

289000 C ------------- 1 mol18000 C -------------,

= 0.0187 mol

Step 4:Mass = mol x mr = 0.0187 mol x 65 g/mol

= 1.22 g

3600 C x 1 mol 193000C

CHAPTER 11RATE OF REACTION

FACTOR AFFECT RATE OF REACTION- Surface areas

- concentration/pressure- catalyst

- temperature

RATE OF REACTION- Differential rate

- Order of reaction (table)- rate constants

ACTIVATION ENERGY- Graph

- Formula

CONCENTRATION & HALF-LIFE- Graph

- Formula

MAXWELL-BOLTZMANN DISTRIBUTION

RATE OF REACTION

Hydrogen sulfide burns in oxygen to form sulfurdioxide and water.

2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g)

(i) Write the differential rate equation(ii) If sulfur dioxide (SO2) is being formed at a

rate of 0.30 molL–1s–1, what are the rates ofdisappearance of hydrogen sulfide (H2S) and

oxygen (O2)? (iii)If rate=k[H2S], what is the order of reaction

with respect to O2.

QUESTION 1

2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g)

rate = d[SO2]dt

12

= – d[H2S]dt

12

= – d[O2]dt

13

= 0.30 molL–1s–1

=d[SO2]

dt

So, rate of disappearance of H2S = 0.30 molL–1s–1

d[H2S]dt–

2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g)

rate = d[SO2]dt

12

= – d[H2S]dt

12

= – d[O2]dt

13

= d[SO2]dt

32

= (0.30 molL–1s)32

= 0.45 molL–1s–1

So, rate of disappearance of O2 = 0.45 molL–1s–1

d[O2]dt–

• If rate=k[H2S], what is the order of reaction with respect to O2.

zero order for O2

Question 2

Consider the decomposition of N2O5(g)

2N2O5(g) 4NO2(g) + O2(g)

Make a graph that shows concentrations of N2O5(g), NO2(g), and O2(g) as a function of time, all on the same set of axes and roughly to scale.

Time

Conc

entr

ation

[NO2]

[N2O5]

2N2O5(g) 4NO2(g) + O2(g)

[O2]

Experiment [F2] [ClO2]Initial Rate

(M/s)

1 0.10 0.010 1.2 x 10-3

2 0.10 0.040 4.8 x 10-3

3 0.20 0.010 2.4 x 10-3

Determine the rate law and calculate the rate constant for the following reaction from the following data:

F2 (g) + 2ClO2 (g) 2FClO2 (g)

QUESTION 3

Rate = k[F2]x[ClO2]

y Find order in F2 with [ClO2] constant:

Rate 3 k[F2]x[ClO2]y

Rate 1 k[F2]x[ClO2]y=

2.4 x 10-3 (0.2)x

1.2 x 10-3 (0.1)x=

(2.0)1 = (2.0)x x = 1 first order in F2

Experiment [F2] [ClO2]Initial Rate

(M/s)

1 0.10 0.010 1.2 x 10-3

2 0.10 0.040 4.8 x 10-3

3 0.20 0.010 2.4 x 10-3

Rate = k[F2]x[ClO2]

y Find order in ClO2 with [F2] constant:

Rate 2 k[F2]x[ClO2]y

Rate 1 k[F2]x[ClO2]y=

4.8 x 10-3 (0.04)y

1.2 x 10-3 (0.01)y=

(4.0)1 = (4.0)y y = 1 first order in ClO2

Experiment [F2] [ClO2]Initial Rate

(M/s)

1 0.10 0.010 1.2 x 10-3

2 0.10 0.040 4.8 x 10-3

3 0.20 0.010 2.4 x 10-3

Rate = k[F2]x[ClO2]y

x = 1, y = 1

rate = k[F2][ClO2] (rate law)

Use any experiment data (example Exp. 1) to find k

1.2 x 10-3 M s–1 = k(0.1 M)(0.01 M)

k = 1.2 M–1s–1 (rate constant)

rate = k[F2][ClO2]

overall order = x + y = 1 + 1 = 2 (second order)

CONCENTRATION AND HALF-LIFE

QUESTION 4

The reaction 2NO (g) N2 (g) + O2 (g) is a second order reaction. If it has a rate constant of 3.6 x 10-2 M-1s-1 at room temperature, calculate

(i) The concentration of NO after 15 minutes if the initial concentration is 1.20 M

(ii) The half-life of this reaction

* Let try question about percentages

If question involves:concentration and half-life

Order Rate Law

Concentration-Time Equation (Integrated

Rate Law) Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]21

[A]=

1[A]o

+ kt

[A] = [A]o – kt

t½ln2k

=

t½ = [A]o

2k

t½ = 1k[A]o

[A]o

[A]ln = kt

k = 3.6 x 10-2 M-1s-1

The reaction is second–order overall.

=1

1.2 M + (3.6 x 10-2 M-1s-1 ) x 900

[A] = 0.030 M

1[A]

=1

[A]o

+ kt[A]o = 1.2 M

[A] = ? M t = 15 minute x 60 = 900 s

(a)

1[A]

k = 3.6 x 10-2 M-1s-1

The reaction is second–order overall.

(b)

t½ = 1k[A]o

[A]o = 1.2 M

=1

3.6 x 10-2 M-1s-1 x 1.2 M

= 23.1 s

QUESTION 5The decomposition of hydrogen peroxide, H2O2, is a second order reaction.

H2O2 (aq) H2O (l) + ½ O2 (g)

The following table lists the changes in the concentration of hydrogen peroxide with time.

t (time) 0 1 5 10 15 20

[H2O2] (M) 0.300 0.215 0.110 0.068 0.045 0.039

(i) Write the rate law for the decomposition of hydrogen peroxide.(ii) Plot a graph using the appropriate axes to determine the rate constant for this reaction.

(ii) The reaction is second–order overall.

1/ [H2O2]

time (min)

slope = k = 6.18 x 10–4 min–1

k = 6.18 x 10–4 min–1

(i) Rate = k [H2O2 ]2

t½ = 1k[A]o

[A]o = 0.3 M

=1

6.18 x 10–4 min–1 x 0.3 M

= 23.1 min

Plot ln[reactant] vs. time:linear graph first order

Plot 1/[reactant] vs. time:linear graph second order

Plot [reactant] vs. time:linear graph zero order

ACTIVATION ENERGY

The decomposition of HI has rate constants k = 0.079 Lmol–1 s–1 at 508oC k = 0.24 Lmol–1 s–1 at 540oC. What is the activation energy of this reaction inkJmol–1?

QUESTION 6

SIL, 3 ed, p.671RAY, 7 ed, p.217BRA: 3 ed., p.602*PET: 8 ed., p.300

k1 = 0.079 Lmol–1 s–1 T1 = 508oC = (508 + 273)K = 781 K

k2 = 0.24 Lmol–1 s–1 T2 = 540oC = (540 + 273)K = 813 K

Ea = ?

= –Ea

R1T2

1T1

–k2 k1

ln

= – Ea

8.314 Jmol–1K–1 813 K1 – 781 K

1

Ea = 183311 Jmol–1

= 183.3 kJmol–1

= –Ea

R1T2

1T1

–k2 k1

ln

0.24 Lmol–1 s–1

0.079 Lmol–1 s–1ln

Consider the decomposition of NO2 into NO and O2:

2NO2(g) 2NO(g) + O2(g)

The following data were collected for the reaction.Rate constant, k

(Lmol–1 s–1 )Temperature

(oC)

7.8 400

10 410

14 420

18 430

24 440

Determine graphically the Ea for the reaction.

QUESTION 7

k(Lmol–1 s–1 )

ln kT

(oC)T

(K)1/T

(K–1)

7.8 2.05 400 673 1.49 x 10–3

10 2.30 410 683 1.46 x 10–3

14 2.64 420 693 1.44 x 10–3

18 2.89 430 703 1.42 x 10–3

24 3.18 440 713 1.40 x 10–3

Plot graph: ln k versus 1/T

*MAKE SURE GRAPH MUST BE: ln K vs 1/T

1/T (K–1)

ln k

ln A

Ea 1ln k = ln A –

R T

slope = Ea–R

Slope = – Ea/R = – 1.44 x 104 K

Ea = ( 8.314 J/K•mol )( 1.44 x 104 K)

= 1.19 x 105 J/mol

= 1.2 x 102 kJ/mol

MAXWELL-BOLTZMANN DISTRIBUTION

MAXWELL−BOLTMANN’S DISTRIBUTION

CHAPTER 12INTRO TO ORGANIC CHEMISTRY

FUNCTIONAL GROUP

STRUCTURE- Expanded structure

- Condensed structure- Skeletal structure

CARBON CHIRALITYENANTIOMER DIASTERIOMER

RACEMIX MIXTURE

NUCLEOPHILE/ELECTROPHILE- Homolytic/heterolylitic cleavage

- Nucleophilic/Electrophilic- Type of reactions

ISOMERISM- chain isomers

- positional isomers- functional group

isomers

FUNCTIONAL GROUP

QUESTION 1Identify (circle) the functional groups in the following molecule:

CH3O

NH

NH

O

CH3O

NH

NH

O

AROMATIC RING

AMINO

CARBON-CARBON DOUBLE BOND

AMIDE

ALKOXY

ISOMERISM

How many structural isomers does pentane, C5H12, have?. Draw all the isomers.Compare their boiling point [8 Marks]

QUESTION 2

pentane

CH3—CH2—CH2—CH2—CH3

CH3—CH—CH2—CH3

CH3

2–methylbutane

CH3—C—CH3

CH3

CH3

2,2–dimethylpropane

Molecular formula = C5H12

2,2–dimethylpropane , 2–methylbutane , pentane

Increase boiling point

Write the structural formula for functional groupisomers with molecular formula:

(a) C3H6

(b) C3H8O

QUESTION 3

(a) C3H6

(b) C3H8O

CH2 CHCH═ 3

propene cyclopropane

CH3CH2CH2–OH

1–propanol

CH3–O–CH2CH3

ethyl methyl ether

ELECTROPHILE AND NUCLEOPHILE

Identify the species I and II as electrophile ornucleophile in the reactions:

SPECIES I SPECIES II

(a) CN– CH3Br+ CH3CN + Br–

(b) HC≡CH NH2–+ HC≡C– + NH3

(c) C2H5Br AlBr3+ [C2H5]+ [AlBr4]–

Cl+ Cl2

FeCl3 + HCl(D)

QUESTION 4

(a) CN– CH3Br+ CH3CN + Br–

(b) HC≡CH NH2–+ HC≡C– + NH3

nucleophile electrophile

nucleophileelectrophile

(C) C2H5Br AlBr3+ [C2H5]+ [AlBr4]–

nucleophile electrophile

Cl+ Cl2

FeCl3 + HCl(D)

nucleophile electrophile

QUESTION 5

Show the cleavage and name the type of reaction

CH3CH2—O—CH3 (a) CH3CH2—Br + –OCH3

+ Br–

(b)+ HCl

Cl

(c)

+ (CH3)3C–ClC(CH3)3

+ HClAlCl3

CH3CH2—O—CH3 (a) CH3CH2—Br + –OCH3

+ Br–

(b)+ H-Cl

Cl

(c)

+ (CH3)3C–ClC(CH3)3

+ HClAlCl3

nucleophile electrophile

nucleophile electrophile

nucleophileelectrophile

nucleophilic substitution

electrophilic addition

electrophilic substitution

QUESTION 6

Show the homolytic cleavage and heterolytic cleavage.

x is halogen group

C XCH3

CH3

CH3

(i) Homolytic cleavage

C XCH3

CH3

CH3

(ii) Homolytic cleavage

C XCH3

CH3

CH3

CARBON CHIRALITYENANTIOMER

DIASTERIOMERRACEMIX MIXTURE

Draw enantiomer of the following molecules

(a) CH3CHOHCOOH lactic acid(b) CH3CH(NH2)COOH alanine (amino acid)(c) CH3CHOHCHOHCH2CH3

Optical active?

QUESTION 7

(a) CH3CHOHCOOH COOH

CH3—C—H

OH

C CH3 H

COOH

OH

CCH3

COOH

H

OH

Enantiomers:

(b) CH3CH(NH2)COOH

COOH

CH3—C—H

NH2

Enantiomers:

C CH3 H

COOH

NH2

C H

COOH

CH3

NH2

(c) CH3CHOHCHOHCH2CH3

H

CH3—C—C—CH2CH3

OH

H

OH

QUESTION 7Consider the molecule

(I) Identify the chiral carbon or stereogenic centre

with an asterisk (*)(II) Draw a pair of enantiomers and

diastereomers. Give your answers

CH3CHOHCHOHCH2CH3

CH3CHOHCHOHCH2CH3

H

CH3—C—C—CH2CH3

OH

H

OH

Enantiomers:

Diastereomers:

(a) ClCH CHCl═

C CCl

H

Cl

H

cis isomer trans isomer

C CH

Cl

Cl

H

QUESTION 8

Draw the geometrical isomers

(b) Draw the structure of cis–trans isomer of1,2–cyclobutanediol.

1,2–cyclobutanediol

OHHO

cis–1,2–cyclobutanediol

OHOH

HH

trans–1,2–cyclobutanediol

OHH

HOH

CHAPTER 13HYDROCARBON

PHYSICAL PROPERTIES ALKANE

- React with oxygen (limited/excess)

- solubility in water- boiling point

NOMENCLATURE

ALKANE

ALKENE

CHEMICAL REACTION ALKANE

- *Halogenation (Mechanism –Free

Radical)

NOMENCLATURE

CHEMICAL REACTION ALKENE

-Hydrogenation-*Hydrohalogenation- *Hydration- Halogenation (H2O/inert solvent)- Oxidative cleavage (Ozonolysis/KMnO4)- *Add H2SO4

PREPARATION ALKENE- *Dehydration

- Dehydrohalogenation

ALKANE

CH3—CH2—CH3

Example: propane

PHYSICAL PROPERTIES ALKANEALKANE

Insoluble in watera) non-polar molecules

b) can’t form hydrogen bond

combustion

CH3CH2CH3 + 5O2 3CO2 + 4H2O + heat

CH3CH2CH3 + 7/2 O2 3CO + 4H2O + heat

Excess oxygen

limited oxygen

Boiling point

List the following hydrocarbons in order of decreasing Boiling points:

3,3–dimethylpentane 2–methylheptaneheptane

Boiling point

3,3–dimethylpentane

MOR: 6 ed., p.123*

2–methylheptane

heptane

C–C–C–C–C

C

C

7C

C–C–C–C–C–C–C

C

8C

C–C–C–C–C–C–C 7C

2–methylheptaneHeptane ,3,3–dimethylpentane ,

Increasing boiling point

2–methylheptane highest boiling point - more carbon so highest molecular weight- Van der Waal’s strongest- Highest boiling point

Heptane higher boiling point than 3,3–dimethylpentane - Bigger surface area- So, Van der Waal’s strongest- Higher boiling point

CHEMICAL REACTION OF ALKANE

HALOGENATION(MONOSUBTITUTION)

*MECHANISM

Draw all structural isomers formed by monochlorination of 2–methylbutane?

QUESTION 1

CH3–CH–CH2–CH3

CH3

hn

CH3–CH–CH2–CH2Cl

CH3

+ Cl2

1–chloro–3–methylbutane

+ CH3–CH–CHCl–CH3

CH3

2–chloro–3–methylbutane

+ CH3–CH–CH2–CH3

CH2Cl1–chloro–2–methylbutane

2–methylbutane

+

2–chloro–2–methylbutane

CH3–C–CH2–CH3

CH3

Cl

(a) Draw the products

CH3CH2CH3light

Cl2

propaneCH3CH2CH2–Cl

1–chloropropane

+

CH3CHCH3

Cl2–chloropropane

Two types H atoms (1o and 2o)

Two monochloro products formed

QUESTION 2

(b) C5H12

CH3–C–CH3

CH3

CH3

2,2–dimethylpropane

light

Cl2 CH3–C–CH2Cl

CH3

CH3

1–chloro–2,2–dimethylpropane

Overall reaction:

CH4 + Cl2 CH3Cl + HCllight

Steps:

• Initiation

• Propagation

• Termination

The most important step is propagation!

MECHANISM OF HALOGENATION