UNIT 4

Entropy

- Entropy is the measure of order or disorder

Increase in entropy (S > 0); thus disorder increases

Decrease in entropy (S < 0); thus disorder decreases

- S system = S products - S reactants - S surroundings = - H/ T

- S total = S system + S surroundings or S system - H/ T (Standard Unit of Energy is Joules) Reaction S system S surroundings Feasibility Example

Exothermic ++Always feasibility C (g) + O2 (g) --> CO2 (g)

Endothermic --Not feasible CO2 (g) --> C (g) + O2 (g)

If -H/ T > S sys; the reaction is feasible More likely at low temp H2O (g) --> H20 (l)

If -H/ T < S sys; the reaction is feasible

More likely at high temp H2O (l) --> H20 (g)

When looking at the changes in entropy of the system and the surroundings you have to consider both the enthalpy change in both and the changes in physical states in the system

Enthalpy of Hydration

- Enthalpy change that takes place when one mole of a gaseous ion dissolves in sufficient water to give an infinitely dilute solution

M + (g) + aq --> M + (aq) H hyd

X- (g) + aq --> X + (aq) H hyd

Rates

- Rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time

rate = C / T (mol dm-3 s-1)

Techniques to measure the rate of a reaction (4-8 GF)

- Measuring the volume of a gas produced

- Measuring the change in mass of a reaction mixture

- Colorimetry

- Titrimetric analysis

Conductrimetric analysis

Measuring the volume of a gas produced accurately

Rate equation & Order of Reaction aA + bB --> cC + dD Rate [A] x [B] y Rate equation = k [A] x [B] y k - rate constant x - order wrt A y - order wrt B x + y - overall order of reaction * The power to which the concentration of a given reactant is raised in the rate equation is defined as the order with respect to that reactant * The sum of the powers of the concentration terms in the rate equation is called the overall order * The proportionally constant in the rate equation is called the rate constant * If a reactant does not appear in the rate equation; the reaction zero order with respect to that reactant This is because the reaction has more than one step and that particular reactant enters the mechanism after the rate determining step * The rate equation cannot be deduced merely from the stoichiometric equation because the rate equation depends on the mechanism of the reaction; therefore the rate equation should be found experimentally and not by simply looking at the equation * Even if a reactant affects the rate of a reaction if will not appear in the rate equation it its in large excess so that its concentration is relatively unchanged during the reaction (pseudo zero order) * Overall order of reaction: The sum of the powers to which the reactant concentrations are raised to in the experimentally determined rate equation Deduction of the order of a reaction 1 Initial rate method - Initial rate is the rate at the instant that the chemicals are mixture - When concentration of a reactant is changed if the rate does not change the order with respect to that reactant is zero Reactant Concentration Rate Order with respect to that reactant

DoubledRate is doubled; increases by a factor of 2 1

Tripled Rate is tripled; increases by a factor of 3 1

DoubledRate is quadrupled; increase by a factor of 4 (2^2) 2

Tripled Rate is increased by a factor of 9 (3^2) 2

Concentration - Time Graphs - The shape of a concentration - time graph depends on the order of the reaction The rate of the reaction is equal to the gradient of a concentration-time graph If the concentration-time graph is a curve as shown in the above graphs the order is either 1 or 2 The half life is calculate to find the exact order Half - life is the time taken to halve the concentration a reactant If two consecutive half-lives are the same; the reaction is first order wrt to that reactant and if it doubles reaction is second order wrt to that reactant Rate - Concentration Graphs Activation Energy - The minimum kinetic energy that should be present in colliding molecules for a collision to be successful - A reaction with a smaller activation energy occurs faster than a reaction with a higher activation energy Reaction profile diagrams for typical exothermic and endothermic reactions are given below: Exothermic Endothermic * The rate constant of a reaction depends on 3 factors: 1) Temperature 2) The activation energy of the reaction 3) Orientation factor (constant for a particular reaction) (Since activation energy and the orientation factor remains constant (standard) for a particular reaction; only temperature is seen to affect the rate constant) Ea - activation energy R - gas constant T - absolute temperature A - Arrhenius constant - The activation energy can be calculated from the gradient of the graph Mechanism and the rate determining step - Rate equation obtained from experimental data can be used to support a proposed mechanism for a particular reaction Example: Hydrolysis of tertiary halogenoalkanes #[(CH3)3CI][OH]-Rate/ Mol dm-3 s-1

10.10.11.11 X 10^-5

20.20.12.22 X 10^-5

30.30.13.33 X 10^-5

40.10.21.11 X 10^-5

50.10.31.11 X 10^-5

- When the concentration of [(CH3)3CI] is doubled ( #1 to #2) the rate increases by a factor of two

Therefore the order wrt to [(CH3)3CI] is 1

- When the concentration of [OH]- is doubled; the rate does not increase

Therefore the order wrt to [OH]- is 0

Overall order = 1; Rate = k [(CH3)3CI]

* If the total volume of a mixture is constant; volume concentration

- This means that the hydroxide ions enter the mechanism after the rate determining step; so there should be at least 2 steps in the mechanism

- The experimentally determined rate equation is consistent with the proposed method:

SN1

Step 1:

Step 2:

- The reactants that take part in the rate determining step and the steps prior to that appear in the rate equation

- If the first step is the rate determining step; the rate equation for the overall reaction is the same as that for the rate determine step

Hydrolysis of Primary Halogenoalkanes

[CH3Br] (moldm-3)[OH]- (moldm-3)Rate (moldm-3s-1)

0.10.11.5 x 10^-5

0.20.13.0 x 10^-5

0.30.14.5 x 10^-5

0.10.23.0 x 10^-5

0.10.34.5 x 10^-5

- When the concentration of [CH3Br] is doubled; the rate increases by a factor of 2

Therefore the order wrt to [CH3Br] is 1

- When the concentration of [OH]- is doubled; the rate increases by a factor of 2

Therefore the order wrt to [OH]- is 1

Overall order: 2 Rate = [CH3Br] [OH]-

- The kinetics of this reaction suggests that the rate determining step involves both halogenoalkanes and the hydroxide ions

- This rate equation is consistent with the proposed mechanism

SN2

The derivation of the rate equation is more complex when the second or a subsequent step is rate determining step

Three molecules cannot collide together & C enters the mechanism after the rate equation

A + 2B + C --> D + E

Rate = k [A] [B]^2

A + B --> Intermediate 1

Intermediate 1 + B --> Intermediate 2 (rds)

Intermediate 2 + C --> D + E

According to the rate equation 1 molecule of A and 2 molecules of B should take art in the rate determining reaction step or before that

Powers of concentration terms in a single step are the same as the stoichiometry in that step

Example: A + 2B + 3C --> Products Rate = [A] [B]2 [C]3

A + C --> Intermediate 1 (AC)

Intermediate 1 + B --> Intermediate 2 (ACB)

Intermediate 2 + C --> Intermediate 3 (AC2B)

Intermediate 3 + B --> Intermediate 4 (AC2B2)

Intermediate 4 + C --> Products (AC3B2)

Reaction of Iodine with Propanone

CH3COCH3 + I2 --> CH3COCH2I + H+ + I-

Experiments show that the reaction is first order wrt propanone and first order wrt to H+; but zero order wrt to Iodine

This shows that iodine is not involved in the rate determining step or any step before that

Step 1:

Step 2:

Step 3:

Altering the order by the method of excess reagents

A + 2B --> C + D

Rate = k [A]^p [B]^q

* If the concentration of B is made 10 times that of A;the change in concentration of B during the reaction is negligible

Therefore when excess of B is used that concentration becomes virtually constant; so:

Rate = k [A]^p

Catalysts A catalysts is a substance which speeds up a chemical reaction by providing an alternate pathway with a lower activation energy and remains chemically unchanged

- Catalysts can be divided into two groups:

Homogenous catalysts

* Heterogenous catalysts Homogenous Catalysts

- Catalysts & the reactants are in the same phase:

Example: Chlorine free radicals act as homogenous catalysts in the gas phase

Cl + O3 --> ClO. + O2

ClO. + O3 --> Cl. + 2O2

H2SO4 (aq)

CH3COOH (aq) + C2H5OH (aq) --> CH3COOC2H5 (aq) + H20 (l)

Esterification is given in the above equation and sulfuric acid is used as the catalyst

Heterogenous Catalysts - A heterogenous catalyst is in a different phase than the reactants

N2 + 3H2 --> 2NH3

(Refer Unit 5 Catalysts)Chemical Equilibria

In a dynamic equilibrium; the rate of the forward reaction and the rate of the reverse reaction are the same

Therefore there is no further change in concentrations of reactants or products

Equation Law

aA + bB cC + dD

Kc = [C]c [D] d [A]a [B] b Kc - equilibrium constant - The equilibrium constants depends only on temperature

- The units of equilibrium constant changes depending on the equation

- The equilibrium constant of the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction

Kc = 1/ Kc

Worked Example:

0.256g of HI was heated at 764K in a sealed flask of volume 100cm3 (0.01 dm3) ; when equilibrium was reached the flask was cooled quickly

It was found that the mixture contained 0.00028 mol of I2

Calculate the value of Kc

2HI (g) I2 (g) + H2 (g)

HI

I2

H2 Start (Moles) 0.002

0

0

(0.256/127.9)

Change (Moles) 0.00144

0.00028

0.00028

(0.002 - 0.0056)

Equilibrium

0.0144

0.0028

0.0028 Concentration

(Moles / Volume)

Kc = 0.0028 X 0.0028

0.01442

= 0.038Concentration of Water in Equilibrium Expressions

- When water is a reactant or product; but not a solvent; it is included in the Kc expression

- When water is the solvent; even if it is a reactant or product; the concentration of water is not included in the Kc expression

- When used as a solvent; the concentration of water does not change because it is in large excess so the concentration term remains constant and its value is not incorporated into Kc Worked Example:

1 mole of ethanol and 2 moles of ethanoic acid were mixed in a sealed flask and left to reach equilibrium

The equilibrium mixture contains 1.15 moles of acid. Calculate Kc

CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l)Start (Moles) 2

1

0

0

Change (Moles) 1.15

0.15

0.85

0.85

Equilibrium

1.15/ V

0.15 / V

0.85 / V 0.85 / VConcentration

(Moles / Volume)

Kc = 0.7225/ V2

0.1725/ V2

= 4. 19

The equilibrium constant in terms of Partial Pressure (Kp) - The pressure of a gas is due to the collisions of its molecules with the wall of the container its molecules with the wall of the container

* The partial pressure of a gas in a mixture of gases is the pressure that that gas would exert if only that gas occupied the whole volume of the contained at a particular temperature

- The partial pressure of a gas A; P(A) is equal to the product of the mole fractions of A and the total pressure

P(A) = n (A) x total pressure

total number of moles

- However every gas molecule in the mixture contributes to the total pressure

So the total pressure; P is equal to the sum of all the partial pressures of the gases in a mixture

- Since only gases have partial pressures; solids and liquids are not included in the Kp equation

Example: aA (s) + bB (g) cC (g) + dD (g)

Kp = P(C)c x P(D)d

P(B)b

Worked Example

Ammonium hydrogen sulfide decomposes when heated according to the following equation:

NH4HS (s) NH3 (g) + H2S (g)

Kp = 0.142 atm2 at 50C a) State the expression for Kp

Kp = P(NH3) P(H2S)

b) Calculate the Partial Pressure of both gases at 50C and hence the Total Pressure

Since the molar ratios of both gases are the same; the Partial Pressures of both gases are the same; hence:

Kp = P(NH3) P(H2S)

Kp = X2

0.142 = X2

X = 0.377 atm (Partial Pressure of each gas)

Therefore Total Pressure; P = 0.377 atm x 2 = 0.754 atmEquilibrium Constant & Entropy Change

- As we already have learnt; the more positive the value of Stotal; the more and more the reaction goes to the right hand side

- Stotal decides the magnitude of the equilibrium constant (K)

It is related by the following equation

Stotal = RlnK

R - gas constant Worked Example:

Use the data below to calculate the value of the equilibrium constant at a temperature of 25C

AgCl (s) Ag+ (aq) + Cl- (aq)

Hsol = + 66 KJmol-1

Stotal = Ssystem + Ssurroundings

Ssurroundings = -H/ T = -(66 x 1000)/ 298 = -221.48 Jmol-1K-1

Ssystem = Sproducts - Sreactants

= (73 + 57) - 96

= 34 Jmol-1K-1

Stotal = 34 Jmol-1K-1 + (-221.48 Jmol-1K-1)

= - 187.48 Jmol-1K-1

Stotal = RlnK

- 187.48 = 8.31lnK

e (- 187.48/ 8.31) = K

K = 1.6 X 102 mol2dm-6 Distribution of a substance between two immiscible layers

I2(water) I2 (hexane)

K = [I2] hexane

[I2] water - K is known as Partition Constant or Distribution Constant

- Equilibrium constant for different reaction are calculated based on data from experiments

ICT can be used to present and analyze data; in order to calculate K

Reaction Quotient (Q)xA + yB --> mC + nD

Kc = [C]m [D]n [A]x [B]y- The right hand side of the above equation is called the concentration term or the reaction quotient; Q

- Q = K; the system is in equilibrium

No further change in concentrations of reactants or products

- Q < K; system is not in equilibrium

The concentration of reactants is greater than that of products

Reactants reach to produce move products in order to increase the numerator of Q

Equilibrium shifts to the right

- Q > K; system is not in equilibrium

The concentration of products is greater than that of reactants

Products will react to give more reactants; increasing the denominator of Q and reducing the value of Q

Equilibrium shifts to the left

Applications of Rates & Equilibrium

Many important industrial reactions are reversible

Chemists and chemical engineers are devising new methods by applying the theoretical ideas and models that explain how fast reactions go (rates) and in which direction and how far (equilibrium)

Factors affecting a system at equilibrium

1 Temperature:

For an exothermic reaction; the equilibrium constant decreases with increasing temperature

If the reaction is endothermic; the equilibrium constant increases with increasing temperature

Therefore temperature changes also affects equilibrium position

- For exothermic reactions:

An increase in temperature causes Kc to become smaller; Q will be larger than the new value of Kc and the system is not at equilibrium

The system reacts to restore the equilibrium; it is done by making more reactant molecules until Q = K; therefore the position of equilibrium shifts to the left

- For endothermic reactions:

An increase in temperature causes Kc to become bigger; therefore Q is small than the new value of Kc and the system is not at equilibrium

The system reacts to restore the equilibrium

It is done by making more product molecules until Q = K; therefore the position of equilibrium shifts to the right

Example: [Cu(H2O)6]2+ + 4Cl- CuCl42- + 6H2O (blue)

(yellow)

At room temperature - the solution contains a mixture of both ions Cooling - produces move blue colour Heating - produces more yellow colour The forward reaction is endothermic When the temperature is increases the endothermic reaction is favoured (more yellow) * However the rate of a reaction increases with increasing temperatureThis is true for both exothermic and endothermic reactions 2 Pressure: - Pressure has a marked effect on gaseous equilibria - However the pressure of a gaseous system or the volume of the container has no effect on the value of the equilibrium constant; K - The effect of pressure on the equilibrium position depends on the number of moles of gaseous molecules on each side of the equation Increase in pressure: This is usually done by decreasing the volume of the container in which the reactants are mixed Example: N2 (g) + 3H2 (g) 2NH3 (g) Kp = P (NH3)2 P(N2) x P(H2)3

- When the pressure is increases; the partial pressure of reactants increases by a greater factor

- Therefore Q becomes smaller than Kp and the system is not in equilibrium anymore

- To restore equilibrium; the position of equilibrium shifts to the right so increase the partial pressure of the productsDecrease in pressure:

This is usually done by increasing the volume of the container in which the reactants are mixed- When the pressure decreases; the partial pressure of the reactants decrease by a greater factor

- Kp remains constant and Q becomes greater than Kp; so the system is not at equilibrium

- The system reacts to restore the equilibrium by moving the position of equilibrium to the left to reduce the partial pressure of the products

* When an increase pressure causes the numerator and denominator of the reaction quotient term rise by the same factor. The value of Q does not change and Kp = Q

Thus the system is still in equilibrium and the position of equilibrium does not change

3 Concentration:

2SO2 (g) + O2 (g) 2SO3 (g)

Kc = [SO3]2 [SO2]2 [O2]2

- If the concentration of reactants is increased; the denominator of Q increases; causing Q to decrease

Kc remains the same so Kc > Q; so the system is not in equilibrium

- To resort the equilibrium; the position of equilibrium shifts to the right

4 Catalysts:

- A catalyst has no effect on the equilibrium constant or the reaction quotient

Therefore the position of equilibrium does not change

- A catalyst speeds up both the forward and reverse reactions equally; therefore equilibrium is reached faster * The application of rates and equilibria is important in industrial and pharmaceutical processes; because it enables us to:

1) Maximise the percentage conversion of reactant to product (increases the equilibrium yield)

2) Make this amount of product as quickly as possible (increases the rate of reaction)

3) Lower costs

4) Have a high atom economy

- Industrial processes are not in equilibrium since the products are removed from the reaction mixture (eg: Haber Process)

Chemical industries usually use a continuous flow method

Acid - Base Equilibria - Acids were first defined as substances with a sour taste

- Bases were first defined as soapy substances

- However taste and field tests are not recommended; because many acid and bases can be harmful

- Later on Arrhenius proposed a theory

The Arrhenius Theory - Acids are substances that produce H+ ions when dissolved in water

- Bases are substances that product OH- ions when dissolved in water

- The Arrhenius Theory was limited to aqueous solutions; therefore Bronsted & Lowrie extended the Theory to include non-aqueous solvents

The Bronsted - Lowrie Theory

- An Acid is a substance that gives a proton (H+ ion) to a base

- A base is a substance that accepts a proton (H+ ion) from an acid

Acid - Base Conjugate Pairs

- An Acid is a proton donor

- A Base is a proton acceptor

- Acid - Base equilibria involves a transfer of protons

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Forward Reaction - acid: HA

base: H2O Reverse Reaction - acid: H30+

base: A-

A- is the base derive from the acid HA by the loss of a proton; therefore A- is known as the conjugate base of HA

H3O- is the acid derived from H20 after accepting a proton; therefore H3O+ is known as the conjugate base of H2O

Base (1) Conjugate Acid (1) HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Acid (2) Conjugate Base (2) Examples:

1) Give the formulae of the conjugate bases of:

Remember: a conjugate base is formed by the loss of a proton

Acid Conjugate Base

HCl Cl-

H2O H3O-HSO4- SO42-

2) Give the formulae of the conjugate acids of:

Remember: a conjugate acid is formed by the gain of a proton

Base Conjugate Acid

H2O H2OOH- H2O SO42- HSO4- HSO4- H2SO4-

- Some substances can act as acids and bases; these are known as amphoteric substances (example: water)

H2O + HCl --> H3O+ + Cl- water acts as a base

H2O + NH3 --> NH4+ + OH- water acts as an acid

Strong & Weak Acids - A strong acid is an acid which is completely ionised in an aqueous solution; forming hydrated hydrogen ions (H3O+)

HCl + H2O --> H3O+ + Cl-

- A weak acid is an acid that is only slightly (DO NOT SAY PARTIALLY) ionised in aqueous solutions and forms a small proportion of H3O+ ions CH3COOH + H2O CH3COO- + H3O+

Strong & Weak Bases

- A strong base is a base which is completely ionised in a aqueous solution; forming OH- ions

NaOH + H2O --> H3O+ + Cl-

- A weak base is a base that is only slight (DO NOT SAY PARTIALLY) ionised in aqueous solutions and forms a small proportion of OH- ions

NH3 + H2O OH- + NH4+

Acid & Base Dissociation Constants - A weak acid is in equilibrium with its conjugate base

HA + H2O H3O+ + A-

Kc = [A-] [H3O+] [HA] [H2O]

- Since water is the solvent; the concentration of water remains constant

So Kc x [H2O] = [A-] [H3O+]

[HA]

Ka = [A-] [H3O+] (acid dissociation constant)

[HA]

- A weak base is in equilibrium with its conjugate acid

NH3 + H2O NH4+ + OH-

Kc = [NH4+] [OH-]

[NH3] [H2O]

- Since water is the solvent; the concentration of water remains constant

So Kc x [H2O] = [NH4+] [OH-]

[NH3]

Kb = [NH4+] [OH-]

(base dissociation constant)

[NH3]

Ka and Kb depend on the temperature and the nature of the reaction (Stotal = RlnK)

The pH Scale - The pH scale is used to show the degree of acidity of a substance

It runs from -1 to +14

pH = - log [H3O+]

- The lower the pH (the higher the H+ / H3O+ concentration) so the higher the acidity

Example:

[H3O+] moldm-3

pH

0.001

3y

0.01

2

0.1

1

1

0

- pH has no units

- The H+ / H3O+ concentration can be found using the pH value

Since; log [H3O+] = pH

[H3O+] = 10 - pH

Auto ionization of Water

- Water can act as an amphoteric substance; therefore one water molecule can protonate another

Base (2) Conjugate Acid (2)

H2O + H2O --> H3O+ + OH-

Acid (1) Conjugate Base (1)

Kc = [H3O+] [OH-] [H2O]2

- Only a small amount of water is ionised at any time, so its concentration remains virtually unchanged (constant)

So the ionic product of water (Kw):

Kw = [H3O+] [OH-]

- The value of Kw at 25C is 1.0 x 10-14 mol2dm-6- Pure water has the same concentration of H3O+ and OH-

Worked Example (pH of pure water at 25C):

(at 25C) [H3O+] [OH-] = 1.0 x 10-14

(for pure water) [H3O+] = [OH-]

[H3O+] 2 = 1.0 x 10-14

[H3O+] = 1.0 x 10-7

pH = - log 1.0 x 10-7 = 7

- Since Kw = [H3O+] [OH-]

logKw = {-log[H3O+]} + {-log[OH-]}

Worked Example (pKw of pure water at 25C):

[H3O+]/ [OH-] = 1.0 x 10-7

pH/ pOH = - log 1.0 x 10-7 = 7

pKw = 7 + 7 = 14

- A neutral solution is one in which the concentration of H3O+ and OH- ions are equal

- In all aqueous solutions at 25C (not necessarily neutral solutions):

[H3O+] [OH-] = 1.0 x 10-14

- Acidic Solutions: [H3O+] > [OH-]

[H3O+] > 1.0 x 10-7

pH < 7

- Alkaline Solutions:[H3O+] < [OH-]

[H3O+] < 1.0 x 10-7

pH > 7

Worked Example:

At a higher temperature; Kw was found to be 1 X 10 -12 mol2dm-6

a) Is this ionization of water exothermic or endothermic

Endothermic; Kw increases with increasing temperature (as the ionization of water is endothermic, so an increase in temperature shifts equilibrium to the right).

b) At this temperature; calculate the pH of a neutral solution (in a neutral solution the concentration of H3O+ and OH- ions are equal) So:

[H3O+] [OH-] = 1 x 10-12 [H3O+] = 1 X 10 -12

[H3O+] = 1 X 10 -6

pH = - log [1 X 10 -6] = 6

pH of a Weak Acid

- A weak acid is in equilibrium with its conjugate base as given below

HA + H2O H3O+ + A-

Ka = [H3O+] [A-]

[HA]

- [HA] in the above equation is the equilibrium concentration of HA

However; since the ionization of a weak acid is negligible; the initial concentration of HA can be used in these calculations

Therefore we assume:

[HA]initial = [HA]equilibrium Worked Example:

Calculate the pH of a 0.135 moldm-3 solution of ethanoic acid

Ka = 1.8 X 10 -5 moldm-3 at 25C

CH3COOH + H2O H3O+ + CH3COO-

Ka = [ H3O+ ] [ CH3COO-] [CH3COOH]CH3COOHH3O+CH3COO-

Initial Concentration 0.13500

Equilibrium Concentration 0.135 - X X X

1.8 X 10 -5 = X2

0.135 (since we assume [CH3COOH] initial = [CH3COOH] equilibrium) X = 1.56 X 10-3

pH = -log 1.56 X 10-3pH = 2.81

* Assumptions made in this calculation:

1) Ethanoic acid (weak acid) is only slightly ionised; therefore the initial concentration and the equilibrium concentration are the same

2) The auto ionization of water is negligible (which could affect the H3O+ concentration)

pKa Calculations - pKa = -logKa

- Ka = 10 -pKa

Worked Example:

The pKa of methanoic acid is 3.75

Find Ka Ka = 10 -pKa

Ka = 10 -3.75 = 1.78 X 10 -4

- Strong Acids Ka > 1

pKa < 0

- Weak Acids Ka < 1

pKa > 0

Diprotic Acids

- A diprotic acid produce two H+ ions per molecule; example: H2SO4

- The first dissociation of the acid is complete

H2SO4 --> H+ + HSO4-

0.1M 0.11M 0.1M

- HSO4- is a weak acid; so the second dissociation is a slight ionization

HSO4- --> H + +SO4-2

Ka = [ H+ ] [ SO4 -2] [HSO4-]

Ka = [ H+ ] [ SO4 -2] (H+ concentration = HSO4- concentration)

[0.1]

Given that Ka = 1 X 10-2

[ H+ ] = 0.0316 moldm-3 * The calculated value obtained for the concentration of H+ in the equilibrium mixture is higher than the actual value of H+ ions at equilibrium resulting from the ionization of HSO4-

This is because in reality; the H+ ions produced by the first (complete) ionization suppresses the formation of H+ from the second ionization (as the higher concentration of H+ ions would cause equilibrium to shift to the left)

Therefore the overall calculated H+ i ion concentration is less than the calculated value

Salts of Weak Bases

Example: NH4Cl

- Ammonium chloride solution is completely ionised as given below

NH4Cl (aq) --> NH4+ (aq) + Cl- (aq)

- Ammonium ions can act as acid (conjugate acid os NH4Cl)

NH4+ (aq) +H2O (aq) NH3 (aq) + H3O- (aq)

Ka = [NH3] [H3O+]

[NH4+]

- When calculated; the pH of NH4Cl is less than 7; so a solution of NH4Cl is acidic

The weaker the base; the stronger its conjugate acid and the more acidic a solution of its salt becomes

Salt of a Weak Acid

Example: CH3COO-Na+

- CH3COO-Na+ dissociates completely in water

CH3COONa --> CH3COO- + Na+

- CH3COO- ions can act as a base (conjugate base)

CH3COO- + H2O CH3COOH + OH-

Kb = [CH3COOH] [OH-] [CHCOO-]Worked Example:

Calculate the pH of 0.1 M of Sodium Ethanoate solution (at 25C)

Ka of CH3COOH = 1.75 x 10-5

Since Ka x Kb =Kw(1.75 x 10-5) x Kb =Kw (1.75 x 10-5) x Kb = 1 x 10-14 Kb = 5.7 x 10-10

Kb = [CH3COOH] [OH-] [CHCOO-]

5.7 x 10-10 = [CH3COOH] [OH-] (We assume that [CH3COOH] = [OH-]) 0.1

[OH-] = 7.56 x 10-6

[OH-] x [ H3O+ ] = 1 x 10-14

[H3O+ ] = 1 x 10-14

7.56 x 10-6[H3O+ ] = 1.32 x 10-9

pH = -log 1.32 x 10-9

pH = 8.88

- When calculated; the pH of CH3COO-Na+ is more than 7; so a solution of CH3COO-Na+ is alkaline

The weaker the acid; the stronger its conjugate base and the more alkaline a solution of its salt

(Refer George Facer page 103) - Salts of strong acids and bases dissolved in water without any reaction taking place

Therefore these salts are neutral

Example: NaCl --> Na+ + Cl-

Na+ and Cl- do not react with water as their electronic configurations make them relatively stable and are therefore inert

Buffer Solution - A solution which resists pH changes due to the addition of small amounts of an acid or a base is known as a Buffer Solution

There are two main types of buffer solutions:

1) A mixture of a weak acid with its salt

Example: CH3COOH & CH3COONa

2) A mixture of a weak base with its salt

Example: NH3 and NH4Cl

Ethanoic Acid & Sodium Ethanoate Mixture

- Sodium ethanoate is fully ionised in water

CH3COONa (aq) --> CH3COO- (aq) + Na+ (aq)

- Ethanoic acid is slightly ionised in water and this ionization is suppressed due to the presence of a high concentration of ethanoate ions

CH3COOH + H2O --> CH3COO- + H3O+

- Therefore the mixture acts as a large reservoir of CH3COOH and CH3COO- ions

-When a small of acid is added;

CH3COO- + H3O+ --> CH3COOH + H2O - When a small amount of base is added;

CH3COOH + OH- --> CH3COO- + H2O - The ratio of acid concentration to salt concentration does not change

[CH3COOH] : [CH3COO-]

- Therefore the pH of the solution remains virtually constant

* For the buffer to perform well; the ethanoic acid and ethanoate ions must be present in equal concentrations

Such a mixture can easily be prepared by half neutralizing ethanoic acid using a strong base (NaOH)

- The half neutralization point is when an acid or base is Worked Example:

50CM3 of 1M CH3COOH is neutralised by 25CM3 of 1M NaOH

CH3COOH + NaOH --> CH3COO-Na+ + H2O Initial

0.05 0.025 -

-

Concentrations Final 0.025 - 0.025 0.025

Concentrations

* The half neutralisation technique is very useful in determining the dissociation constant

CH3COOH + H2O H3O+ + CH3COO-

Ka = [H3O+] [CH3COO-]

[CH3COOH]

At the half neutralisation point:

[CH3COO-] = [CH3COOH]

So:

Ka = [H3O+] (as the concentrations of CH3COO- and CH3COOH cancel out)

pKa = pH

- When a weak acid is half neutralised by a strong base pH = pKa

- Ionisation of the weak acid is suppressed by the CH3COO- formed by the complete ionisation of the salt

Therefore CH3COO- ion concentration is equal to the salt concentration (this is an assumption)

- Since the ionisation of the weak acid is suppressed

[CH3COOH] initial = [CH3COOH] at equilibrium

- Since Ka = [Salt] [H3O+]

[Acid]pKa = pH - log [Salt]

[Acid]pH = pKa + log [Salt]

[Acid]- The above equation can be used for buffer solutions

Worked Example:

- 1/3 of the number of mole of a weak acid HA was neutralised by adding aqueous NaOH

The pH of the buffer was 4.2

Calculate Ka

HA + NaOH --> NaA + H2O (Since the initial moles are unknown we take it as X)

Initial

X 1/3 x - - Moles Final Moles (X - 1/3X) - 1/3X 1/3X (concentration in terms of X) [HA] = 2 [A-]

HA + H2O A- + H3O+

Ka = [A-] [H3O+]

[HA]

Ka = [A-] [H3O+]

2[A-]

Ka = [H3O+]

2

4.2 = -log [H3O+]

10-4.2 = [H3O+]

[H3O+] = 6.3 X 10-5

Ka = 6.3 X 10-5/ 2

= 3.15 X 10-5Ammonia - Ammonium Chloride Buffer - NH4Cl is fully ionised as given below

NH4Cl (aq) --> NH4+ (aq) + Cl- (aq)

- The weak base ammonia is only partially ionised and this ionisation is suppressed due to the high concentration of NH4 + in the medium

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

- Therefore the buffer solution acts as a large reservoir of NH4+ ions and NH3

- When a small amount of acid is added:

H3O+ + H2O --> NH4+ + H2O - When a small amount of base is added:

OH- + NH4+ --> NH3 + H2O - As a result the concentration ratio of NH3: NH4+ stays the same

So the pH becomes virtually constant

The Rule of Two - The rule of two gives a approximate pH of weak acids; weak bases and their salts

- pH of a 0.1M Strong Acid is 1 and 0.1M Strong Base is 13

out

Solution (example) pH

Strong AcId (HCl) 1

Strong Base (NaOH) 13

Salt of a strong base and strong acid (NaCl) 7

Weak Acid (CH3COOH) 1 + 2 = 3

Weak Base (NH3 (aq)) 13 - 2 = 11

Salt of a weak acid and a strong base (CH3COONa)7 + 2 = 9

Salt of a strong acid and weak base (NH4Cl) 7 - 2 = 5

Salt of weak acid and a weak base 7

Tip: you add or subtract 2 from the standard pH values depending on the strength of various substances; for example if there is a strong acid and a weak base your final pH value must be less than 7 as the acid is stronger so you subtract 2 from the standard pH value of 7

(The standard values are the first three values listed in the table)

pH changes during Acid - Base Titrations

- The end point of a titration is when the appropriate colour change takes place

- The equivalence point is when the two reactants are mixed in the exact proportions as indicated by the chemical equation

This is also called the neutral point

Strong Acid - Strong Base Titrations - 25cm3 of 0.1M HCl is placed in a conical flask and 0.1M NaOH is filled into a burette

- The equivalence point is reached when 25cm3 of NaOH is added

- The salt present at the equivalence point is NaCl

It is not hydrolyzed

therefore the solution is neutral at the equivalence point so the pH is 7

Weak Acid - Strong Base Titrations Strong Acid - Weak Base Titrations Weak Acid - Weak Base Titrations No vertical section as the pH does not change rapidly at the equivalence point

Therefore the end point cannot be detected

Summary

Conical FlaskBurette Initial pHVertical RangepH value at equivalence point Final pH

Strong Acid Strong Base 13 - 11 7Just below 13

Weak Acid Strong Base 37 - 11 9Just below 13

Strong Acid Weak Base 13 - 75Just below 11

Weak Acid Weak Base 3 - 7Just below 11

Remember: we measure the pH of the conical flask so the initial pH depends only on the strength of the acid and the final pH depends mostly on the strength of the base

For the vertical range the values range from +/- 4 (2x2) from 7 towards a more acidic pH for strong acid (for the lower value) and a more alkaline pH for a strong base (for the higher value) A weak acids or weak base does not have an effect on the range so that particular value (either the lower value or higher value) will stay at 7 (the neutral value)

If both a weak acid and a weak base is mixed there will be no vertical range

The pH value at the equivalence point is usually the mean of the lowest value and the highest value of the range (other than for Weak Acids and Weak Bases where the pH is 7)

Acid - Base Indicators - Indicators are use to detect the end point of titrations

me

* Acid-base indicators are either weak acids or weak bases

- Therefore they are partially ionised in water

HIn + H2O H3O+ +In-

^ Colour 1 ^ Colour two

In - weak acid indicator

- If an acid is added to a solution which contains the indicator; the position of equilibrium moves to the left producing more HIn

Therefore colour 1 produced

- If an alkali is added to the mixture; added OH- reacts with H3O+. As a result the position of equilibrium shifts to the right producing more of colour 2

* Therefore the colour of the indicator changes depending on the pH

Ka =Kind = [H3O+] [In-]

[HIn]

pH = pKind + log [In-]

[HIn]

* For the eye to detect one colour; the concentrations of the substances should be as given below:

For Colour 1: [HIn] = 10 [In-]

pH = pKind + log [In-]

10[In-]

pH = pKind + log 10-1

pH = pKind - 1 For Colour 2: [In-] = 10[HIn]

pH = pKind + log 10[Hln]

[Hln]

pH = pKind + log 10 pH = pKind + 1 - Therefore for an indicator to give a sharp colour change at the end point; the range of pKind +/- 1 must lie within the vertical section of the pH curve

- Since weak acids- weak base titrations do not have a vertical section on its pH curve; an indicator cannot be used to detect the end point

- Due to the sensitivity of the human eye; the pH range of an indicator is usually +/- 2 pH units * The ideal indicator of a given acid-base titration:

1) Has a pKaindicator value exactly equal to the pH of the half equivalence point 2) Kind = [H3O+] [In-] = Ka

[HIn]

In the middle point of the colour change of the indicator

[HIn] = [In-]

Therefore pH = pKa

* The best indicator has a pKa value equal to the pH value of the equivalence point

pKa = pHequivalence 3) A good indicator must show a rapid colour change at the end point

Worked Example: For each of the titrations listed below; state which of the indicators (A; B or C) could be used to find the equivalence point

IndicatorpKind

A3.7

B7.1

C9.6

a) H2SO4 & KOH: Indicator B b) H2SO4 & NH3: Indicator A c) HCl & NaOH: Indicator B Isomerism

- Isomers are different compounds that have the same molecular formula

Structural Isomers

- Compounds with the same molecular formula but different structural formulae

There are three types of Structural Isomers:

1) C-Chain Isomers:

Example:

2) Positional Isomers:

Example:

3) Functional Group Isomers:

Example:

Stereo Isomers

- Compounds with the same structural formula but different spacial arrangements

There are two types of Stereo Isomers:

1) Geometric Isomers (Cis-trans & E-Z)

2) Optical Isomers

Geometric IsomerismGeometric isomerism arises due to:

1) Restricted free rotation around a double bond

2) Each double bond carbon is attached to two different groups

Optical Isomerism

- If two objects are mirror images of each other and they are not super imposable; they are said to be chiral

- Chiral molecules are asymmetrical

- Most chiral compounds have a carbon atom attached to 4 different atoms or groups

- A chiral centre in a molecule leads to two optical isomers known as enantiomers

- An enantiomer is an isomer than it non-superimposable on its mirror image

- The two forms of lactic acid have similar chemical and physical properties; except for their effect on plane polarised light

However the biochemical reactions of optical isomers can be different

Plane - Polarised Light

- Unpolarised light has oscillations or vibrations in all directions/ in all planes

- When unpolarised light is passed through a polaroid/ polarizer the waves will have oscillations in one place

This is known as Plane Polarised Light

- When plan polarised light is passed through a single enantiomer; it will rotate the plane of plane polarised light because it is optically active

- One enantiomer rotates the plane of the plane polarised light clockwise and it is the (+)-isomer and the other one will rotate it anti-clockwise so it known as the (-)-isomer

- The two enantiomers are name D & L as well

- A solution obtaining equimolar amounts of the enantiomers is called a racemic mixture

- It does not rotate the plane of the plane polarised light

- The extent of the rotations (angle of rotation) can be measured using a polarimeter (refer page 126 George Facer)

- The angle of rotation depend on:

1) The nature of the enantiomer

2) The concentration of the enantiomer

- If a molecule has 2 different chiral centers there are four optical isomers:

The four optically active isomers are:

A+ B+

A+ B-

A- B+

A- B-

Optical isomerism and reaction mechanism

- The optical activity of reactivity of product and reactant of organic reactions can help chemists to determine the mechanisms of reactions

Example: Nucleophilic substitution of halogenoalkanes can take place either through SN1 or SN2 reactionsSN2

- Optically active alcohol with an inverted stereo structure is obtained

SN1

- In this mechanism a carbo-cation is formed as the intermediate and it is a planar structure

If the nucleophile attacks from the top; a enantiomer is formed and if its attacks from the bottom the other enantiomer is formed

- The two enantiomers are formed in equal amounts so it will be a racemic mixture with no optical rotations

- Optical activist of the halogenoalkanes and the products of the reaction can be used as evidence for the mechanismCarbonyl Compounds

- This group is known as the carbonyl group

- There are two important groups of carbonyl compounds

1) Aldehydes

2) Ketones Examples of Aldehydes:

Examples of Ketones:

Physical Properties of Carbonyl Compounds

- Methanol is a gas and other carbonyl compounds are liquids at room temperature

- Carbonyl compounds are polar; therefore dipole-dipole and london forces are formed between molecules

Therefore carbonyl compounds have boiling and melting temperatures than alkanes and alkenes with the same number of electrons

- Intermolecular hydrogen bonds are not formed in carbonyl compounds as + hydrogen is not present

Therefore boiling and melting temperatures lower than the alcohols with the same number of electrons

Solubility of Carbonyl Compounds

- Lower members of carbonyl compounds are soluble in water; however the solubility decreases with the increasing carbon chain length (the bulkiness of the R-group negates the hydrogen bonds between the - O and + H)

- These compounds dissolve in water by making hydrogen bounds

- Propanone is an excellent solvent for organic substances as well

- Lower members of aldehydes have pungent smells

- Ketones have much sweeter smells

- Complex aldehydes and ketones are used in perfumes

Reactions of Carbonyl Compounds

- The polar nature of C=O makes the carbon atom susceptible for nucleophilic attack

- A nucleophile is a species with a lone pair of electons that can make a convalent bond with + atoms

Reaction 1: Addition reaction with HCN

- Aldehydes and ketones undergo addition reactions with HCN to produce hydroxynitriles (cyanohydrines)

- In this reaction HCN molecule is added across the C=O groups

- HCN is a weak acid and provides very few CN- ions

HCN H+ + CN-

- Therefore a base is added to produce more nucleophile ions

HCN + OH- CN- + H2O - This happens in the presence of KCN in cold alkaline solution

- If the pH is very low; there wont be enough CN- ions for the first step

- If the pH is very high; there wont be enough HCN for the second step - The mechanism for a ketone is the sam; but the reaction takes place at a slower rate

Example: Reaction with propanone

Stereochemistry of the Reaction

- Aldehydes and ketones are planar around the carbonyl there CN- can attack from the top or bottom

- The product of the above addition reaction is chiral

Therefore we get a racemic mixture of the two optical isomers

Reduction of Aldehydes and Ketones (gives an alcohol) - Aldehydes and ketones can be reduced using lithium tetrahydride aluminate (lithium aluminum hydride; LiAlH4) dissolved in ether (sodium tetrahydride borate; NaBH4)

Aldehydes: reduced to primary alcohols

Ketones: reduces to secondary alcohols

- This reduction takes place in two steps:

1) H- is added to + carbon atom

(dry conditions are needed for this step; therefore LiAlH4 is dissolved in ether)

2) Addition of an aqueous solution of an acid to protonate the O- formed in the first step

- LiAlH4 reacts specifically with polar - bonds therefore it reduced the C=O group in aldehydes; ketones; acids; acid derivatives and CN groups (cyanides) but it doesn't reduce the -bond in the C = C double bond

- However the H2/ Pt can reduce both C =O and C = C

Bradys Reagent - Bradys reagent is 2, 4 - dinitrophenylhydrazine ( 2, 4 - DNP)

- Aldehydes and Ketones react with Bradys reagent to form yellow or orange crystalline solids

These derivatives are known as 2, 4 - dinitrophenylhydrazones

- The above reaction is a condensation reaction

It can be used as a test for carbonyl compounds

- When Bradys reagent is added a yellow precipitate is given by simple aldehydes and ketones and an orange precipitate is given by aromatic aldehydes and ketones

- The melting temperature of the solid can be determined after the purification and recrystalisation

- The carbonyl compound can be identified by comparing the experimental melting point with the melting point in the data booklet

Procedure with Brady's Reagent

Step 1: React the carbonyl compound with a solution of 2, 4 - dinitrophenylhydrazine

Step 2: Filter off the solid

Step 3: Re-crystallize the solid using a minimum volume of hot ethanol

Step 4: Dry the purified produce and determine the melting point

Step 5: Refer to the data booklet and identify the solid

Oxidation - Only aldehydes are oxidised

- Oxidation of aldehydes can be carried out using different methods

Procedure for the Oxidation of Aldehydes

Method 1: Warm the solution with acidified potassium dichromate solution (K2Cr2O7)

(Acidified K2Cr2O7 turns from orange to green)

Method 2: Add potassium manganate (VII)

For an acidic solution of KMnO4:

MnO4- (purple) is reduced to Mn2+ (colourless) For an alkaline or neutral solution of KMnO4:

Brown precipitate of MnO2 formed

* In an alkaline solution the final colour can be green as well

Method 3: Warm with Fehlings solution (Test for aldehydes) The deep blue solution gives a red-brown precipitate

Fehlings solution is a mixture of copper sulfate and sodium potassium tartarate in alkali

Method 4: Warm with Tollens Solution

- Tollens solution is made by adding a few drops of NaOH to silver nitrate and dissolving the precipitate in dilute ammonia

- Ag(NH3)2+ is present in the solution

- When the aldehyde is warmed with the above solution; a silver mirror is formed inside the test tuber

- Ketones do not undergo oxidation so the above methods can be used to distinguish between aldehydes and ketones

Iodoform Reactions

- Ethanal and methylketones undergo this iodoform reactions

- A mixture of I2 and NaOH is warmed with the compound

- Compounds with give a yellow precipitate of tri-iodomethane (iodoform) - The following functional groups give positive results with iodoform tests:

1) Aldehydes: ethanal only

2) Ketones: methyl ketones

3) Alcohols: ethanol and methyl secondary alcohols

Iodoform reactions can be used to reduce the carbon number of an organic compounds by one

Carboxylic Acid and Carboxylic Acid Derivatives

- Carboxylic acids are compounds with a - COOH group

General Formula: CnH2n+1CO2H

Number of Carbon Atoms Name Structure Formula

1Methanoic Acid

2Ethanoic Acid

3Propanoic Acid

- The physical and chemical properties of carboxylic acids are largely governed by two factors:

1) Polarity of the carboxyl group

2) Length of the carbon chain

Physical Properties

- Members in this homologous series are liquids up to the 9th member

Thereafter they are solids

- Carboxylic acids have strong smells example: Vinegar (ethanoic acid)

Rancid Butter (butanoic acid)

Carboxylic acids with a few carbons are soluble in water due to the formation of hydrogen bonds with water molecules

- The solubility of carboxylic acid decreases as the hydrocarbon length increases

Carboxylic Group: Hydrophilic

Hydrocarbon Group: Hydrophobic

(Refer page 146 Table 7.2 GF)

- Benzoic acid is a solid at room temperature; sparingly soluble inc old water and soluble in hot water

- The unusual pattern in melting point is due to the dimerisation of the acid

Methanoic acid and ethanoic acid are found as dimers due to the hydrogen bonding

This does not happen in acids with more carbon atoms

- The boiling point changes as we expect; because the hydrogen bonds break when the acid melts

Therefore the boiling temperature increases with increasing carbon atoms

Worked Example:

Explain why Methylpropanoic acid has a lower melting point than butanoic acid

- Methylpropanoic acid has fewer point of contact (for the formation of hydrogen bonds) than butanoic acid

Further; there are stronger dispersion forces in butanoic acid

Therefore butanoic acid has a higher melting point

Preparation of Carboxylic Acids

Method 1: Oxidation of Primary Alcohols or Aldehydes

Method 2: Hydrolysis of nitriles

- Nitriles are compounds with - CN group

- When a nitrile is heated under reflux with a dilute acid a carboxylic acid is formed

CH3CN + H+ + 2H2O -------------> CH3COOH + NH4+

- If the nitrile is heated under reflux with an aqueous alkali; a salt of the carboxylic acid is formed

Then the carboxylic acid can be obtained by adding a strong dilute acid

CH3CN + OH- + H2O ------------> CH3COO- + NH3

CH3COO- + H+ ------------> CH3COOH

Method 3: Iodoform

Reactions of Carboxylic Acids 1) Reaction with water:

CH3COOH + H2O ------------> CH3COO- + H3O+

pH 3 2) Reduction - Lithium tetrahydriodoaluminate dissolved in dry ether is used as the reducing agent

- H- is a powerful nucleophile

It attacks the carbonyl group to produce a primary alcohol after a series of reactions

- An aqueous solution of an acid must be added to get the final product?

3) Neutralisation - Carboxylic acids react with bases to form salt and water

CH3COOH + NaOH ------------> CH3COO-Na+ + H2O (Basic Salt)

4) Reaction with metal carbonates:

2CH3COOH + Na2CO3 ------------> 2CH3COO-Na+ + H2O + CO25) Reaction with metals:

Carboxylic acids form hydrogen gas:

2CH3COOH + Zn ------------> (CH3COO)2Zn + H2

6) Esterification: - Carboxylic acids react with alcohols in the presence of concentrated H2SO4 catalysts in a reversible reaction to form an ester

Conc H2SO4Carboxylic acid + alcohol ester + water Heat

- Esters are substances with pleasant fruity smells

- When an acid chloride is used instead of carboxylic acid; the reaction is irreversible

- Advantages of this reaction:

1. The reaction is not reversible

2. HCl gas is formed and leaves the reaction mixture

7) Halogenoalkanes Carboxylic Acids react with anhydrous PCl5 to form acyl chlorides

- Steam fumes of HCl form

Carboxylic Acid Derivatives

1) Hydrolysis of Esters:

Esters can be hydrolyzed under acidic or alkaline conditions

- When the ester is refluxed with dilute HCl or dilute H2SO4 it give a carboxylic acid and an alcohol

- The yield of this reaction is low; because it is reversible

- When an ester is boiled under reflux with aqueous NaOH; a salt of the carboxylic acid and the alcohol forms

- The organic acid can be obtained by adding a dilute strong acid like HCl or H2SO4 to the salt of the carboxylic acid

- This reaction gives a higher yield as it is not reversible

Natural Esters - Fats and oils are collectively called lipids; they are esters of glycerol and long chained carboxylic acids called fatty acids

Therefore natural esters are referred to as triglycerides

- When fats and oils are boiled with concentrated NaOH solution; the ester link is hydrolyzed producing sodium salts of fatty acid and glycerol

- Sodium salts of long chained carboxylic acids can be precipitated from the aqueous solution by treating it with saturated NaCl

This is known as Salting Out

- The sodium or potassium salts of long chained carboxylic acid are the main chemical substances in salts

The alkaline hydrolysis of esters is referred to as trans-esterification

Trans-esterification (refer GF pg 153 - 155) - When an ester reacts with another carboxylic acid

This acid part of the ester is replaced by the new acid

CH3COOC2H5 + HCOOH HCOOC2H5 + CH3COOH

- This type of reactions are used in the manufacture of low fat margarine

- In this reaction the incoming acid is saturated (no double bonds present)

- The advantage of this method is that it doesntt form trans-fats - Just like the above reaction the alcohol part can be replaced by an alcohol part

CH3COOC2H5 + CH3OH CH3COOCH3 + C2H5OH

- This type of reaction is used in the manufacture of biodiesel where the incoming alcohol is methanol or ethanol and the product alcohol is glycerol

- The natural oil ester is mixed with methanol and a catalyst

Trans-esterification takes place forming the methyl ester

Polyesters

- Polyesters are condensation polymers

- A polyester can be produced by reaction a dicarboxylic acid with a diol

- The most common polyester is terylene

Acyl Chlorides (Acid Chlorides) Functional group:

- Most acid chlorides are volatile liquids at room temperature

- They dissolve in several organic solvents

- They react with water; therefore soluble in water as well

- Methanoyl chlorides cannot be isolated as it decomposes at room temperature into CO & HCl

- Acid chlorides can be prepared using the reaction of carboxylic acid with PCl5 (in dry conditions)

CH3COOH + PCl5 --------> CH3COCl + POCl3 + HCl

- The carbon atom in the - COCl group is +; therefore it is susceptible to nucleophilic attack

- C-Cl bon is weaker than C-O- bond as Cl is a larger atom than oxygen atom

- Therefore C-Cl bond breaks more easily; furthermore Cl- is a good leaving group

- Therefore the molecule is very susceptible to nucleophilic attack in substitution reactions and it takes place via addition and elimination Reactions of Acyl Chlorides

1) Reaction with water:

CH3COCl + H2O --> CH3COOH + HCl

- This reaction occurs at RTP

- Steamy fumes of HCl are formed

* Acyl chlorides must be kept away from water as they react with water

2) Esterification CH3COCl + CH3OH --> CH3COOCH3 + HCl

(methyl ethanoate)

- Acyl chlorides react with alcohol to form esters and misty fumes of HCl

- The reaction takes place at room temperature without any catalysts

- It gives a higher yield; because the reaction is non-reversible

- Good evidence for the higher reactivity of acyl chlorides than carboxylic acids

3) Reaction with ammonia:

- When an acid chloride reacts with concentrated ammonia at RTP; an amide and hydrogen chloride are formed

- HCl reacts with ammonia to produce a white smoke of ammonium chloride

4) Reaction with Lithium Aluminum Hydride:

CH3COCl + 4[H] ------------> CH3CH2OH + HCl

Spectroscopy

- Different regions of the electromagnetic spectrum is used in spectroscopy

- Electromagnetic radiation consists of an oscillating electric and magnetic field of a wide range of frequencies

- When electromagnetic radiation is passed through a substance; some of the radiation is absorbed (Refer GF 160 - 168 Pg) Mass Spectroscopy

Worked Example:

a) Explain how a sample is ionised in mass spectroscopy

The gaseous sample is bombarded with high energy electrons

An electron gets knocked out from the outermost shell by the high energy electron

X (g) + e- -------> X+ (g) + 2e-

b) How are positive ions accelerated and deflected

Through the electric field and magnetic field respectively

Microwaves

Microwaves are important in heating as it can cause the rotation of polar molecules such as water

Microwaves cannot cause the rotation of non-polar molecules such as CCl4 and O2

The rotation of polar molecules increases as they become aligned with the oscillating electric filed produced by microwave. As a result the molecules collide and rotation energy is converted into kinetic energy and the way heats up

If water is frozen, molecules do not absorb microwaves, because the molecules cannot rotate at fixed positions

Defrosting in a microwave oven works because the surface of frozen food is covered with a thin layer of liquid water. When this thin layer heats up nearby ice melts and this happens until the whole food is defrosted

Microwaves are used in pharmaceutical industries to heat up various chemicals

The advantage of microwave heating is that theres less heat loss and less chance of damaging fragile molecules

The following diagram explain we can use microwaves in a chemistry lab

The microwaves can selectively heat up the aqueous layer, but not the organic layer. Therefore the solubility of a substance in the aqueous layer increases. So microwaves are useful in analytical sciences

Nuclear Magnetic Resonance (NMR) Spectroscopy

NMR Spectroscopy uses the radio frequency region in the electromagnetic spectrum

Proton is another name given to a hydrogen nucleus as there are no neutrons

The hydrogen nucleus can spin clockwise and anticlockwise and this spinning nucleus can produce a magnetic field

Both the spin states are degenerate (have the same energy)

When an external magnetic field is applied the degeneracy is lost. That is the 2 spin states will have two different energies (split in energy)

The spin aligned state (the spin state which is aligned with the magnetic field) has a lower energy level than the spin opposed state

E matches with the size of the energy in the radio frequency region. When the exact frequency is applied a resonance takes place. This means that nuclei (protons) will jump from a lower spin state to a higher spin state.

The magnitude of E depends on the magnitude of the applied magnetic field. The stronger the magnetic field, the greater the split

E also depends on the extent of shielding of the nucleus by the electron cloud around it. If the nucleus is more shielded it will experience the effect of the magnetic field to a lesser extent

As an example consider a proportional molecule:

There are three different hydrogen nuclei, they are in three different chemical environments.

Type 3 is most shielded or least deshielded

Type 1 is least shielded or most deshielded

In an NMR Spectrometer a final radio frequency is maintained and the strength of the magnetic field is increased from a lower value to a higher one. The 3 types of protons will resonate at 3 different strengths of the magnetic field. It will produce 3 peaks in the NMR Spectrum in the 1:2:3 intensity ratio.

Under a given magnetic field:

Shielded protons split low Deshielded protons split high

We use a standard known as TMS (Tetra methyl silane). Protons in TMS produce one peak in the NMR Spectrum. It is used to be the origin on the O value in the scale that shows the chemical shift of the NMR Spectrum. Chemical shift () is measured in ppm.

The extent to which a certain peak differs from the TMS peak is called a chemical shift

TMS is chosen because:

- only a trace amount is needed - It is a volatile liquid that can be evaporated off easily - Molecules of TMS do not interact with molecules of the sample

Protons in most organic compounds show nuclear resonance at chemical shift value that are positive with respect to TMS protons

The sample being investigated is dissolved in a solvents such as D2O or CCl4 as these solvents do not show resonance due to the absence of hydrogen atoms

Low Resolution NMR Spectrum of CH3CH2CHO

A low resolution NMR give the number of different types of hydrogen atoms in the molecules (number of peaks) and the ratio of the different types of the hydrogen atoms (ratio of the peak heights).

Worked Example:

Give the number of peaks and the approximate ratio of their heights in the low resolution NMR Spectrum of the following molecules

1) Propanal: 3 peaks in the ration 1:2:3

2) Propane: 2 peaks in the ration 1:3

Worked Example:

Draw the low resolution NMR of:

High Resolution NMR Spectroscopy

In a high resolution NMR Spectrum, a peak produced by the hydrogen atoms in a given chemical environment is split into a set of sub-peaks by the neighboring hydrogen atoms.

If the number of neighboring hydrogen atoms in n. the peak is split into (n + 1) sub-peaks, this is known as the (n + 1) rule.

High resolution NMR Spectroscopy is more informative in structure determination.

The ratio of the heights of the sub-peaks are given below

nAfter SplittingRatio

0Singletn/a

1Doublet1:1

2Triplet1:2:1

3Quartet1:3:3:1

4Quintuplet (multiplet)1:4:6:4:1

Hydrogen atom in OH, CHO and COOH are not split and they do not cause any splitting

Splitting due to neighboring hydrogen atoms is called coupling

Magnetic Resonance Imaging (MRI)

The same principle used in NMR is used

Useful to detect the condition of soft tissues such as brain and heart tissue without causing any damage. Therefore known as a non-invasive technique

Therefore the patient does not suffer any ill effect.

The chemical shift of the hydrogen atoms in soft tissues are different to those in hard tissues. This is used in MRI scanners.

Chromatography

It is a separation technique.

Chromatography of all types have the same basic principle. In every chromatography technique the sample is dissolved in a suitable solvent and washed through stationary phase by a mobile phase known as the eluent.

There are different types of chromatography methods:

Paper Chromatography

Thing layer chromatography

High performance (pressure) liquid chromatography (HPLC)

Gas- Liquid Chromatography ( GLC or GC)

Stationary phase is a solid or a liquid, whereas the mobile phase is a liquid or a gas

High Performance (Pressure) Liquid Chromatography (HPLC)

A column packed with a solid of uniform particle size is used as the stationary phase.

The sample dissolved in a suitable solvent is added to the top of the column

Then the liquid used as the mobile phase is forced through the column under high pressure.

The time taken for a particular component in the sample to pass through the column is known as the retention time.

The components in the same have different retention times depending on the different strengths of interactions of the component with the stationary phase and the mobile phase.

Example: A component which is more soluble in the mobile phase and less absorbed onto the stationary phase can pass through the column faster and will have a lower retention time . A high pressure is used in order to increase the speed at which the eluent passes through the column. This will reduce the diffusion of component molecules spreading out of the bands.

If the column is connected to an IR Spectrometer the components can be identified.

Usually a polar stationary phase and non-polar eluent was used in early chromatography was used in early chromatography techniques.

If the two phases are reversed it is known as Reserve Phase High Performance Liquid Chromatography (RPHPLC)

RPHPLC is used in pharmaceutical industry to separate and identify the purity of drugs.

Retention time is a characteristic for a particular component. It can be used to separate different components.

Gas Liquid Chromatography

The sample has to evaporate and then it is injected into the top of chromatography column.

The injected sample is forced through the column by the movement of an inert gas (gaseous mobile phase).

The column is an inert solid and the liquid stationary phase is adsorbed onto the surface of the column.

The mobile phase is an inert gas such as hydrogen, argon, oxygen, nitrogen or air. This is known as a carrier gas. Depending on the strength of the interaction between the sample and the liquid stationary phase, different substances pass through the column at different rates.

The detector of the instrument can identify the thermal conductivity changes of the carrier gas after it is mixed with the sample.

GLC is also coupled with IR Spectroscopy in order to identify the components in the sample mixture.

This method can be used to separate mixtures of volatile liquids.

Remember NaOH(aq) means

NaOH + H2O

Remember H3O+ can be written as H+

pKw = pH + pOH

pKw = pH + pOH

Proving that Ka x Kb = Kw

HA + H2O--------> H3O+ + A-

Ka= [H3O+] [A-] / [HA] -----------------(1)

A- + H2O---------> HA + OH-

Kb= [HA] [OH-] / [A-] -----------------(2)

(1) * (2)

Ka x Kb = [[H3O+] [OH-]

So:

Ka x Kb =Kw