unit 2: application of core principles of chemistry · intermolecular & intramolecular hydrogen...

Dr. Catherine Tan

1 +6018 – 375 2833 www.chemistryboutique.com

Unit 2: Application of Core Principles of Chemistry (Edexcel International Advanced Level Chemistry)

Shapes of Molecules and Ions • The electron pairs around the central atom of a molecule dictate the shape of the

molecule.

• Steps in determining the shape of molecule:

1. Decide which atom is the center of a molecule.

2. Determine the no. of electron pairs around the central atom.

i. Look up valence electrons in central atom

ii. Add one electron for each atom joined to the central atom.

iii. Add electron if particle is negatively charged or subtract electron if the

particle is positively charged.

3. Determine the no. of bond pairs and lone pairs.

Valance Shell Electron Pair Repulsion (VSEPR) stated that:

– The electron pairs (either bond pair or lone pair) repel each other and move as far

apart as possible.

lone-pair lone-pair repulsion > lone-pair bond-pair repulsion> bond-pair bond-pair repulsion

• The repulsion between electron pair increased by an increase in electronegativity of the

central atom.

Dr. Catherine Tan


Molecule Electron dot diagram No. of bonding

pair

No. of lone pair Shape Geometry

BeCl2 Linear

H2O Bent

Dr. Catherine Tan


BF3 Trigonal planar

NH3 Pyramidal

Dr. Catherine Tan


CH4 Tetrahedral

PF5 Trigonal bipyramid

Dr. Catherine Tan


SF6 Octahedral

Dr. Catherine Tan


No. of

bonding pair

No. of

lone pair

Shape of molecule Example

2 0 Linear BeCl2

3 0 Trigonal planar BF3

4 0 Tetrahedral CH4

3 1 Trigonal pyramidal NH3

2 2 Bent H2O

5 0 Trigonal bipyramid PF5

6 0 Octahedral SF6

Dr. Catherine Tan


Bond Energy and Bond Length

• Bond energy is defined as the standard enthalpy change for breaking the bond in 1 mol

gaseous molecules.

• Bond length is the distance between the nuclei of two bonded atoms. The shorter the bond

length, the higher it is the bond energy.

Bond Polarity

• Covalent bonds may have some ionic character which resulting in a polar covalent bond.

• A polar covalent bond is formed between two atoms of different electronegativities.

• For example, Hydrogen fluoride consists of a polar covalent bond. This is due to the high

electronegativity of fluorine atom which tends to exert a stronger attraction on the

bonding electrons as compared to hydrogen atom.

• This unequal sharing of electrons is known as polarisation and the covalent bond is said

to be polarised.

• Therefore, a molecule is said to be polar (or has dipole moment) when its bonds are

polarised and it is not symmetrical.

• Ionic bond also contain some covalent character due to polarisation (cation attracts the

negative charge of the anion.

Dr. Catherine Tan


Intermolecular Forces

• High degree of covalency in ionic bond exist when,

– The cation is small

– The anion is large

– The charge on both ions is large

• There are two types of intermolecular forces:

– Van der Waals’ forces

– Hydrogen bonding

– Van der Waals’ forces are forces of attraction can be divided into two,

– Dipole-dipole forces

– Temporary dipole-induced dipole forces

• Dipole-dipole forces exist between polar molecules.

• The positive end of the dipole of one molecule will attract the negative end of the dipole

of another molecule.

• As for non-polar molecules such as oxygen and nitrogen, it is suggested that there are

force of attraction between molecules since they can be liquefied and solidified.

• Temporary dipole-induced dipole attraction is due to the temporary fluctuations in the

electron density of a molecule.

Factor affecting the strength of Van der Waals’ forces

• No. of electrons - The greater the no. of electrons in the molecules, the stronger is the van

der Waals’ forces of attraction.

• Shape of molecule - For instance, strength of van der Waals’ is reduced when there is a

branching because smaller surface area of contact for van der Waals’ forces.

• Hydrogen bond can exists between molecules that contain a hydrogen atom covalently

bonded to a small and highly electronegative atom, for instance fluorine, oxygen and

nitrogen. This is also known as intermolecular hydrogen bonding

Dr. Catherine Tan


• The highly electronegative atom will attract electron density towards itself and causes a

dipole moment.

• Hence, the hydrogen atom of one molecule will possess a partial positive charge and will

be attracted to the fluorine, oxygen or nitrogen atom of another molecule that carries a

partial negative charge.

Intermolecular & Intramolecular Hydrogen Bonds

• In 2-nitrophenol, there is possibility of an intramolecular hydrogen bond forming

between hydrogen atom of hydroxyl group (-OH) and oxygen atom of nitro group (-NO2)

Dr. Catherine Tan


• However, in 4-nitrophenol, the functional groups are on opposite sides of benzene ring.

Hydroxyl group is too far away from nitro group to form intramolecular hydrogen bond.

So, intermolecular hydrogen bonds are formed between molecules

• Intermolecular bonding in 4-nitrophenol is far stronger than in 2-nitrophenol, so boiling

point 4-nitrophenol is higher than 2-nitrophenol

Physical Properties of Hydrogen Bonding

• Hydrogen bonding is responsible for the high boiling point of water and low density of

ice.

• Water might exist as gas at room temperature without existence of hydrogen bonding.

• In the presence of hydrogen bonding, ice has an open structure which account for the

lower density of ice as compared to water.

Dr. Catherine Tan


Giant Covalent Molecules

Diamond

• Carbon has an electronic arrangement of 1s2 2s2 2p2. In diamond, each carbon shares

electrons with four other carbon atoms - forming four single bonds.

The physical properties of diamond

• has a very high melting point (almost 4000°C). Very strong carbon-carbon covalent

bonds have to be broken throughout the structure before melting occurs.

• is very hard. This is again due to the need to break very strong covalent bonds operating

in 3-dimensions.

• doesn't conduct electricity. All the electrons are held tightly between the atoms, and aren't

free to move.

• is insoluble in water and organic solvents. There are no possible attractions which could

occur between solvent molecules and carbon atoms which could outweigh the attractions

between the covalently bound carbon atoms.

Graphite

• Graphite has a layer structure.

Dr. Catherine Tan


• Each carbon atom in graphite uses three of its electrons to form simple bonds to its three

close neighbours. That leaves a fourth electron in the bonding level. These "spare"

electrons in each carbon atom become delocalised over the whole of the sheet of atoms

in one layer. They are no longer associated directly with any particular atom or pair of

atoms, but are free to wander throughout the whole sheet.

The physical properties of graphite

• It has a high melting point, similar to that of diamond. In order to melt graphite, it isn't

enough to loosen one sheet from another. You have to break the covalent bonding

throughout the whole structure.

• It has a soft, slippery feel, and is used in pencils and as a dry lubricant for things like

locks. You can think of graphite rather like a pack of cards - each card is strong, but the

cards will slide over each other, or even fall off the pack altogether. When you use a

pencil, sheets are rubbed off and stick to the paper.

• It has a lower density than diamond. This is because of the relatively large amount of

space that is "wasted" between the sheets.

• Insoluble in water and organic solvents - for the same reason that diamond is insoluble.

Attractions between solvent molecules and carbon atoms will never be strong enough to

overcome the strong covalent bonds in graphite.

• Conducts electricity. The delocalised electrons are free to move throughout the sheets. If

a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet

and be replaced with new ones at the other end.

Silicon Dioxide

• Silicon dioxide is also known as silicon (IV) oxide.

• Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all

you need to do is to modify the silicon structure by including some oxygen atoms.

The physical properties of silicon dioxide

• It has a high melting point - varying depending on what the particular structure is

(remember that the structure given is only one of three possible structures), but around

Dr. Catherine Tan


1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the

structure before melting occurs.

• It is hard. This is due to the need to break the very strong covalent bonds.

• It doesn't conduct electricity. There aren't any delocalised electrons. All the electrons are

held tightly between the

Fullerene

• a spherical molecule and contain 60 carbon atoms

• it consists of hexagonal rings of carbon atoms and alternating pentagonal and heptagonal

carbon rings to allow curvature of the surface producing molecules that have a complete

hollow shape

• They are classed as simple covalent molecules

• Fullerene molecules can be used for drug delivery into the body

Graphene

• A new substance that is a one layer of graphite

• Three covalent bonds per atom and the 4th outer electron per atom is delocalised

• Have very high tensile strength because of the strong structure of many strong covalent

bonds

• Can conduct electricity due to the presence of free moving electrons

Dr. Catherine Tan


Carbon Nanotubes

• Have very high tensile strength because of the strong structure of many strong covalent

bond

• Can conduct electricity well along the tube due to the presence of free and delocalised

electrons

• Can be used as a vehicle to deliver drugs to cells

Metallic Bonding

• It is defined as the electrostatic attraction between the positively charged metal ions and

the ‘cloud’ of delocalised electrons

The physical properties of metal

• Metals can be deformed since the electron ‘cloud’ prevent the repulsion among the

cations.

• Since it electrons are free to move throughout the metal piece, hence metal is a good

electrical and thermal conductors.

• Owing to these typical properties, metal such as aluminum, copper can be used for

packaging, drawn into wires, and their alloys are strong enough for usage in aeroplane,

car, train and building.

Dr. Catherine Tan


Redox Redox Process

o Redox reaction – a reaction in which the reduction of one reactant and the oxidation of

another reactant occurs simultaneously.

o Oxidising agent – a reactant that causes oxidation and itself reduced in the reaction.

o Reducing agent – a reactant that causes reduction and itself oxidised in the reaction.

Redox in terms of Electron Transfer

o Oxidation is defined as loss of electrons whereas reduction is defined as gain of electrons.

o e.g. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

o Zn(s) is a reducing agent and itself oxidised to Zn2+(aq).

o Cu2+(aq) is an oxidixing agent and itself reduced to Cu(s).

o Disproportionation is a redox reaction in which both oxidation and reduction of the same

element occurs.

o e.g. Cu2O(s) + H2SO4(aq) Cu(s) + CuSO4(aq)

+1 0 +2

Rules for working out Oxidation Numbers

o All free atoms in elements have an oxidation number of zero.

E.g. Mg Fe O2 Cl2

o For simple ions, the oxidation number is the same as the charge on the ion.

E.g. Na+ Al3+ Mg2+

o For covalent compound, the covalent bond is changed into ‘ionic bond’ by assuming that the

bonded electrons are on the more electronegative atom. The oxidation numbers are the

charge on the ‘ions’.

E.g. H2O SO3 HCl

o In a neutral molecule, the sum of the oxidation numbers of all the atoms equals zero.

o e.g. Al2O3 2(3+) + 3(-2) = 0

o In an ion, the sum of the oxidation numbers of all the atoms in the formula unit equals the

charge on the ion.

o e.g. MnO4- 1(+7) + 4(-2) = -1

Dr. Catherine Tan


Construct Redox Equations

o Redox equation can be constructed as follow:

• Copy out the two half equations from data booklet.

▪ Write the half-equation with more positive Eө on top.

▪ Write the half-equation with more negative Eө underneath and reversed

so that all the reactants are on the left.

• Multiply the half-equations with appropriate factors so that the number of electrons

involved in each is the same.

• Add two equations together, canceling out the ions or molecules that appear on both

sides of the equations

Dr. Catherine Tan


The Periodic Table – Groups 2 and 7 Group 2

o Atomic radius

- Atomic radium increases down the group

- Atoms have more shells of electrons making the atom bigger

o Melting point

- Melting point decreases

- Metallic bonding weakens as the atomic size increases

- Distance between the positive ions and delocalized increases

- Weaker electrostatic forces between the positive ions and delocalized electrons

o First ionization energy

- First IE decreases

- The outermost electrons are held more weakly as they are successively further from the

nucleus in additional shells

- The outer shell electrons become more shielded from the attraction of the nucleus

o Reactivity

- Increases down the group

- As the atomic radii increase, there is more shielding

- Nuclear attraction decreases

- Easier to remove electron

- Cations form more easily

- Must be stored under oil to protect it from reaction with water and oxygen

- Mg and Be form a protective oxide layer to avoid further corrosion

Dr. Catherine Tan


o Reaction with oxygen

Group 2 element Equation Property

Be 2Be(s) + O2(g)2BeO(s)

Amphoteric (ionic compound

with covalent character)

Mg 2Mg(s) + O2(g)2MgO(s)

Basic

Ca 2Ca(s) + O2(g)2CaO(s)

Basic

Sr 2Sr(s) + O2(g)2SrO(s)

Basic

Ba 2Ba(s) + O2(g)2BaO(s)

Basic

Dr. Catherine Tan


o Reaction with water

Group 2 element Property Equation

Be Has no reaction with either

cold water or steam

-

Mg Has no reaction with cold

water but it reacts with

steam

MgO dissolves slightly in

water to give weakly

alkaline solution

Mg(s) + H2O(g)MgO(s) + H2(g)

MgO(s) + H2O(l)Mg(OH)2(aq)

Ca React vigorously with cold

water

Ca(s) + 2H2O(l)Ca(OH)2(s) +

H2(g)

Sr React vigorously with cold

water

Sr(s) + 2H2O(l)Sr(OH)2 + H2(g)

Ba React vigorously with cold

water

Ba(s) + 2H2O(l)Ba(OH)2 + H2(g)

- Solubility of group II hydroxides increases down the group

o Solubility of sulphates

- Less soluble down the group

- BaSO4 is the least soluble

Dr. Catherine Tan


o Thermal decomposition of group 2 carbonate

- MgCO3(s)MgO(s) + CO2(g) (at 540oC)

CaCO3(s)CaO(s) + CO2(g) (at 900oC)

SrCO3(s)SrO(s) + CO2(g) (at 1290oC)

BaCO3(s)BaO(s) + CO2(g) (at 1360oC)

- Thermal stability increases down the group

- Cationincrease in size

- Charge density decreases

- Polarizing power of the cationdecreases

- Anion electron cloud is less distorted

o Thermal decomposition of group 2 nitrate

- Mg(NO3)2(s)MgO(s) + 2NO2(g) + ½ O2(g)

Ca(NO3)2(s)CaO(s) + 2NO2(g) + ½ O2(g)

Sr(NO3)2(s)SrO(s) + 2NO2(g) + ½ O2(g)

Ba(NO3)2(s)BaO(s) + 2NO2(g) + ½ O2(g)

- Thermal stability increases down the group

- Cationincrease in size

- Charge density decreases

- Polarizing power of the cationdecreases

- Anion electron cloud is less distorted

Group 7

Element Colour and state

at room

temperature

Boiling point/ ͦC Solubility

In water In organic

solvent

Fluorine Pale yellow gas -188 - -

Chlorine Yellow-green gas -35 Moderately

soluble

Yellow solution

Bromine Dark red liquid 58 Slightly soluble Orange-brown

solution

Iodine Black solid 183 Insoluble Violet/purple

solution

Dr. Catherine Tan


o Redox behavior

- Oxidizing power of halogens decreases down the group

F2 Cl2 Br2 I2

- Reducing power of halide ions increases down the group

F-<Cl-< Br-< I-

- Feasible halogen displacement redox reactions

F2 + 2Cl- Cl2 + 2F-

F2 + 2Br- Br2 + 2F-

F2 + 2I- l2 + 2F-

Cl2 + 2Br- 2Cl- + Br2

Cl2 + 2I- l2 + 2Cl-

Br2 + 2I- 2Br- + l2

o Volatility

- Decreases down the group

- Halogen has simple molecular structure consisting of diatomic molecule

- Held together by weak VDW forces

- Down the group, the molecular size increases

- VDW increases

- Boiling point of halogen increases and halogen becomes less volatile

o Reaction of halogen with hydrogen

- H2 + F22HF

H2 + Cl22HCl

H2 + Br22HBr

H2 + I22HI

- Thermal stability of hydrides decreases down the group

- Due to decrease in strength of the H-X bond

- As the size of halogen increases

- Size of halogen increases increase in bond length decrease in bond strength

- HF>HCl>HBr> HI (thermal stability)

Dr. Catherine Tan


- The acid strength of hydrides increases down the group

- Due to decrease in the strength of H-X bond

- HI>HBr>HCl> HF (acid strength)

- Oxidation of hydrides (HX H2 and X2)

HF Electrolysis

HCl KMnO4 (strong oxidizing agent)

HBr H2O2 (weaker oxidizing agent)

HI Readily oxidized at atm O2

o Reaction of halide ions with silver followed by ammonia

- The function of ammonia

✓ Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq)

✓ NH3 acts as a ligand with Ag+ ion

✓ A ligand is ion or molecule that donates its lone pair electron through

coordinate bond/dative bond to metal ion

Dr. Catherine Tan


o Reaction of halide ions with concentrated sulphuric acid

Ion Observation Equation

F-

White fumes of HF

NaF + H2SO4 NaHSO4 + HF

Cl-

White fumes of HCl

NaCl + H2SO4 NaHSO4 + HCl

Br-

White fumes of HBr + red brown

bromine vapor

NaBr + H2SO4 NaHSO4 + HBr

2HBr + H2SO4 Br2 + SO2 + 2H2O

I-

White fumes of HI, black solid

iodine + purple iodine vapor +

bad egg smells hydrogen

sulphide

NaI + H2SO4 NaHSO4 + HI

2HI + H2SO4 I2 + SO2 + 2H2O

8HI + H2SO4 4I2 + H2S + 4H2O

o Reaction of chlorine with NaOH

- With cold NaOH, Cl2 undergoes disproportionation.

0 -1 +1

Cl2 + 2NaOH(aq)NaCl(aq) + NaClO(aq) + H2O(l)

- With hot NaOH, Cl2 undergoes disproportionation.

0 -1 +5

2Cl2 + 6NaOH(aq)5NaCl(aq) + NaClO3(aq) + 3H2O(l)

Dr. Catherine Tan


Kinetics Activation Energy

o Reaction involves collision between more than two species

o Orientation of collision refers to the molecules hit each other in the right way to decide

whether a particular collision will result in a reaction

o Energy of collision is refer to activation energy

o Even if the species are orientated properly, a reaction will not happen unless the particles

collide with a certain minimum energy called activation energy of the reaction

o Activation energy is the minimum energy required before a reaction can occur

Maxwell-Boltzmann Distribution

• In any system, the particles present will have a very wide range of energies

• For gases, this can be shown on a graph called Maxwell-Boltzmann distribution which is

a plot of the number of particles possess each particular energy to react when they collide

• The graph is asymmetry/not normal distribution

• The area under the graph represents the total number of molecules

• At any given temperature, most molecules have an average energy, a small number of

molecules have very high or very low energies

Dr. Catherine Tan


Effect of Temperature On Reaction Rate

• The effect of temperature on reaction rates is illustrated in the Maxwell-Boltzmann

distribution

• An increase in temperature increases the average kinetic energy of molecules and a

greater molecules has higher kinetic energy

• At higher temperature, the number of molecules with energy same or greater than

activation energy increased, so reaction rate increases because the number of frequency

of effective collisions increase

Dr. Catherine Tan


Effect of Catalyst on Reaction Rate

• Adding a catalyst will speed up rate of reaction

• A catalyst provides an alternative route for the reaction to happen with a lower activation

energy

Dr. Catherine Tan


• Properties of a catalyst are as follow:

– Only a small quantity is required to catalyse a reaction

– It does not undergo any chemical changes at the end of reaction, but it can

undergo physical changes

– It is specific in its action and does not catalyse all reactions

– It alters the reaction mechanism that will lower the activation energy

– It does not change the reaction enthalpy or equilibrium constant, but it can change

the order of reaction

Dr. Catherine Tan


Chemical Equilibria Reversible Reactions

o Reversible reaction is one which proceeds in both directions.

o The reactants are not completely converted to products.

o Equilibrium is reached where both reactants and products are present.

o e.g. H2(g) + I2(g) 2HI(g)

o Equilibrium can only occur in a closed system.

o e.g. When few cm3 of water are placed in a flask which is corked, liquid is changing to the

gas and vice versa continuously and the total amount remain the same, hence, equilibrium is

reached.

H2O(l) H2O(g)

Irreversible Reactions

o If the activation energy of the reverse reaction (Eb) is exceptionally high, then the reaction

will be unfavourable known as irreversible reaction.

H2O(l) H2O(g) (open system)

Dr. Catherine Tan


Dynamic Equilibrium

o A reversible reaction can reach dynamic equilibrium when the rate of the forward reaction

becomes equal to the rate of backward reaction.

o e.g. 3H2(g) + N2(g) 2NH3(g)

o As soon as NH3 molecules are formed, they decomposed back to N2 and H2. At this stage, the

total amount of each substance remains the same since the production rate of NH3 equals to

the decomposing rate of H2 and N2.

Le Chatelier’s Principle

o Le chatelier’s Principle states that if a change is made to a system in equilibrium, the system

reacts in such a way as to tend to oppose the change and a new equilibrium is formed.

o e.g. If something is added to the system at equilibrium, then the system will behave so as to

remove it and vice versa.

o It summarises the effect of external factors such as temperature, concentration and pressure

on a system at equilibrium.

Changes in Concentration

• e.g. Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq)

• When extra SCN-(aq) is added, by Le chatelier’s Principle, the position of equilibrium

shifts to the right so as to remove the extra SCN-(aq). Hence, more [Fe(SCN)]2+(aq) is

produced.

• However, when some SCN-(aq) is removed, the position of equilibrium will shifts to the

left so as to replace some SCN-(aq). Therefore, less [Fe(SCN)]2+(aq) is produced.

Changes in Pressure

• Pressure changes can only affect the reaction equilibrium when gases are involved.

• e.g. 2SO2(g) + O2(g) 2SO3(g)

• When pressure increased, the position of equilibrium shifts to the right so as to reduce the

pressure by reducing the number of molecules of gas. The new equilibrium contains more

SO3(g).

• When pressure decreased, the position of equilibrium shifts to the left so as to increase

the pressure by increasing the number of molecules of gas. The new equilibrium contains

more SO2(g) and O2(g).

Changes in Temperature

• e.g. 2SO2(g) + O2(g) 2SO3(g) ∆H= -94.9kJ/mol

• The reaction above is an exothermic reaction, hence, when temperature increased, by Le

chatelier’s Principle, the position of equilibrium will move to a direction that can reduce

the temperature, that is, shifts to the left. So more SO2(g) and O2(g) produced.

Dr. Catherine Tan


• Whereas when temperature decreased, the position of equilibrium shifts to the right in

order to increase the temperature, hence, more SO3(g) is produced.

Equilibrium Constant, Kc & Kp

o The equilibrium constant is a measure of the extent to which the reactants are converted into

products before equilibrium is reached.

o The equilibrium constant is only affected by temperature.

o Changes in concentration, pressure or the presence of catalyst will not affect the equilibrium

constant.

Dr. Catherine Tan


Dr. Catherine Tan


Q: Given the equilibrium constant for following equation is 4.0,

CH3CH2OH + CH3COOH CH3COOCH2CH3(l) + H2O(l)

Calculate the mass of ethanoic acid that must be mixed with 2.0 mol of ethanol to produce 1.5

mol of ethyl ethanoate at equilibrium.

Dr. Catherine Tan


Haber Process

o It is a process in manufacturing ammonia, NH3.

3H2(g) + N2(g) 2NH3(g) ∆H = -92kJ/mol

o Typical conditions are as follow:

– A pressure of 200 atm.

• Le chatelier’s Principle predicts that high pressure gives a higher yield of

ammonia.

• But when pressure is too high, it would results in higher cost and maintanance

of equipment. Hence, a pressure of 200 atm is used.

– A moderate temperature of 500oC.

• Le chatelier’s Principle predicts that low temperature gives a higher yield of

ammonia.

• However, if temperature is too low, the rate of reaction is too slow. Hence, a

moderate temperature of 500oC is used.

– Iron catalyst (finely divided) is also used to increase the rate of reaction so that the

equilibrium is reached faster.

– Small amount of promoters such as K2O or Al2O3 also needed to enhance the

efficiency of the catalyst.

– N2 and H2 are used in a proportion of 1:3. N2 H2 is obtained from fractional

distillation of liquid air and natural gas, respectively.

Contact Process

o This is the main chemical reaction occurring in the Contact process. It is a reversible,

exothermic reaction accompanied by a decrease in volume. Applying Le Chatelier's Principle,

to get a good yield of sulphur trioxide,

o The following points are to be borne in mind: A low temperature is necessary. An optimum

temperature of

o 410-450oC is found suitable. High pressure would favour the forward reaction. It is not

possible to build acid resistant towers which can withstand high pressures. A pressure of 760

- 1520 mm of Hg is used.

o Excess of oxygen is necessary as it would favour the forward reaction. A suitable catalyst is

required.

o Platinum is a more efficient catalyst than vanadium pentoxide. But it is more expensive. It

also gets poisoned by impurities like arsenic (III) oxide, Vanadium pentoxide though less

efficient is cheaper and not affected by impurities. K2O may be used as a promoter to

enhance the activity of the catalyst.

Dr. Catherine Tan


o 410-450oC is found suitable. High pressure would favour the forward reaction. It is not

possible to build acid resistant towers which can withstand high pressures. A pressure of 760

- 1520 mm of Hg is used.

o Excess of oxygen is necessary as it would favour the forward reaction. A suitable catalyst is

required. Platinum is a more efficient catalyst than vanadium pentoxide. But it is more

expensive. It also gets poisoned by impurities like arsenic (III) oxide, Vanadium pentoxide

though less efficient is cheaper and not affected by impurities. K2O may be used as a

promoter to enhance the activity of the catalyst.

Bronsted-Lowry Theory of Acids and Bases

o It states that an acid is a species which donates proton (H+) (proton donor) to a base; a base

is a species that accepts proton (H+) (proton acceptor) from an acid.

e.g. HA + H2O H3O+ + A-

acid base conjugate conjugate

acid base

Strong and Weak Acids

• The relative strengths of acids depend on the extent of dissociation of the acid solution.

• The strongest acid is the one with greatest degree of dissociation.

• Therefore, strong acids are strong electrolytes while weak acids are weak electrolytes.

Dr. Catherine Tan


Organic Chemistry

Alcohols

o Contain –OH group

o Can form H-bond with other alcohol molecules and with water

o Has high MP and BP

o Soluble in water

o Can be used as a solvent

Manufacture of Alcohol

i. Fermentation of sugars

C6H12O6 2C2H5OH + 2CO2

This process has a number of advantages:

o it is a low-technology process, which means it can be used anywhere

o it does not use much energy

o it uses sugar cane as a raw material, which is a renewable resource

There are, however, a few disadvantages associated with this process:

o it is a batch process, which means that once the reaction has finished the vessel needs to

emptied before the reaction can be started again

o it is a relatively slow process

o it produces fairly impure ethanol

ii. Hydration of ethene

o Temperature: 300 oC

Pressure: 60 atm

Catalyst: H3PO4

C2H4 + H2O C2H5OH

This process has a number of advantages:

- it is a relatively fast process

- it is a continuous flow process, which means that ethene can be entered into the vessel

continuously and the reaction never has to be stopped

Dr. Catherine Tan


- it produces pure ethanol

There are also a number of disadvantages associated with this process:

- it requires fairly high technology

- it uses a lot of energy

- the ethene comes from crude oil, which is a non-renewable resource

Primary, Secondary and Tertiary Alcohols

Primary

C C C

H

H

H H

H

HOH

H

The carbon attached to the OH is attached to 0 or 1 other carbon atom

Secondary

H C

H

HC

H

C

H

H

H

OH

The carbon attached to the OH is attached to 2 other carbon atoms.

Tertiary

H C

H

HC

H

C

H

H

OH

H H

C

H

The carbon atom attached to the OH is attached to 3 other carbon

atoms.

Dr. Catherine Tan


Reaction of Alcohols

o Alcohol molecules are saturated and polar, containing a +ve carbon. Thus alcohols

tend to undergo nucleophilic substitution reactions.

o The OH can combine with an adjacent H atom to form a stable H2O molecule. Thus

alcohols can also undergo elimination reactions.

o Alcohols can lose hydrogen and undergo a variety of oxidation reactions.

i. Nucleophilic Substitution

o Alcohol Halogenoalkane

a. Condition: HBr (mixture of NaBr/KBr and conc. H2SO4), heat and distil

HI (mixture of NaI/KI and conc. H3PO4)

o Rate of reaction: tertiary alcohol > secondary alcohol > primary alcohol

o Nucleophilic substitution of tertiary alcohol reacts with conc. HCl at room

temperature

b. Condition: PCl3/PCl5

3 CH3-CH2-CH2-OH + PCl3 3 CH3-CH2-CH2-Cl + H3PO3

CH3-CH2-CH2-OH + PCl5CH3-CH2-CH2-Cl + POCl3 + HCl

o Can be used a test for –OH groups as white fumes of HCl produced

ii. Reaction between Sodium and Ethanol

CH3-CH2-CH2-OH + Na CH3-CH2-CH2-O-Na+ + H2

Dr. Catherine Tan


iii. Elimination

o Condition: Al2O3, heat/ conc. H3PO4

CH3-CH2-CH2-OH CH3-CH2=CH2+ H2O

Dr. Catherine Tan


iv. Oxidation

a. Partial oxidation of primary and secondary alcohol

Condition: acidified K2Cr2O7, heat under distillation

• If a primary alcohol is mixed with an oxidising agent, two hydrogen atoms

can be removed and an aldehyde will be formed:

E.g. CH3CH2OH + [O] CH3CHO + H2O

H H

HH

CC OHH + [O] + H2OC

O

H

H

H

CH

Ethanol Ethanal

• If a secondary alcohol is mixed with an oxidising agent, two hydrogen atoms

can be removed and a ketone will be formed:

Eg CH3CH(OH)CH3 + [O] CH3COCH3 + H2O

C

O

CH3

CH3CH

OH

CH3CH3

+ [O] + H2O

Propan-2-ol Propanone

• Tertiary alcohols are not readily oxidised since they do not have available

H atoms to give up.

b. Complete oxidation of primary alcohol

Condition: acidified K2Cr2O7, heat under reflux

Dr. Catherine Tan


Summary of oxidation reactions of alcohols and carbonyls

primaryalcohols

secondaryalcohols

tertiaryalcohols

aldehydes ketones

carboxylicacids

X

[O]

[O]

[O]

v. Esterification

Condition: Conc. H2SO4 (catalyst)

vi. Iodoform

Condition: Sodium hydroxide and iodine

Positive result: pale yellow precipitate (triiodomethane)

• Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.

• If "R" is a hydrocarbon group, then you have a secondary alcohol. Lots of secondary

alcohols give this reaction, but those that do all have a methyl group attached to the

carbon with the -OH group.

Dr. Catherine Tan


• No tertiary alcohols can contain this group because no tertiary alcohols can have a

hydrogen atom attached to the carbon with the -OH group. No tertiary alcohols give the

triiodomethane (iodoform) reaction.

Halogenoalkane

o Contain a halogen atom e.g. F, Cl, Br, I attached to an alkyl group

o Also called as alkyl halides

o Different kinds of halogenoalkanes

Primary The carbon which carries the halogen atom

is only attached to one other alkyl group

Secondary The carbon which carries the halogen atom

is attached to two other alkyl group

Tertiary The carbon which carries the halogen atom

is attached to three other alkyl group

Dr. Catherine Tan


Physical Properties

o Boiling points below room temperature

o Gases at room temperature

o BP: CH3X < CH3CH2X < CH3CH2CH2X VDW (permanent dipole) gets stronger as

the molecule gets bigger or has more electrons

o BP: CH3F < CH3Cl < CH3Br < CH3I increase in size from F to I (increase in number

of electrons), stronger VDW (permanent dipole)

o BP: tertiary halogenoalkane < secondary < primary attractions are stronger if the

molecules can lie closely together, more VDW forces can be formed

Dr. Catherine Tan


Chemical Reactivity

o Bond strength falls as from C-F to C-I

o Bond strength as bong length

o Size of halogen atom: F < Cl < Br < I

o C-F bond length is the shortest, strongest bond strength

o Rate of reaction: C-F < C-Cl < C-Br < C-I

o Chemical Reactions:

- Nucleophilic Substitution

▪ Nucleophile: a species which is strongly attracted to a region of positive

charge e.g. negative ions or have a strongly - ▪ Nucleophilic substitution can be divided into SN2 & SN1 mechanism

Dr. Catherine Tan


SN2 SN1

Primary & Secondary Secondary & Tertiary

S: substitution N: nucleophilic 2: initial stage of

the reaction involves 2 species

S: substitution N: nucleophilic 1: initial

stage of the reaction involves 1 species

Primary halogenoalkane did not use the SN1

mechanism because primary carbocation is

unstable than the tertiary

Tertiary halogenoalkane can form stable

tertiary carbocation due to the presence of

electron-donating alkyl groups

1. Hydrolysis

Reagent: Aqueous NaOH/KOH

Condition: Heat under reflux

Product: Alcohol

Dr. Catherine Tan


2. Elimination

Reagent: NaOH/KOH in ethanol


Product: Alkene

3. Reagent: NaCN/KCN in ethanol


Product: Nitrile

Dr. Catherine Tan


4. Reagent: ammonia in ethanol

Condition: sealed tube

Product: amine

Dr. Catherine Tan


Mass Spectra and IR

Dr. Catherine Tan


Dr. Catherine Tan


Green Chemistry

Dr. Catherine Tan


unit 2: application of core principles of chemistry · intermolecular & intramolecular hydrogen...

Documents