88272058 as Chemistry Unit 2 Notes

AS Chemistry Unit 2 Notes AS Chemistry Unit 2 Notes Shapes of molecules and Ions

Shapes of molecules and Ions



 Pairs of electrons will repel each other as far as possible (due to electrostatic repulsion)Pairs of electrons will repel each other as far as possible (due to electrostatic repulsion) 

 Finding the shape:Finding the shape:

1.

1. Draw dot and crossDraw dot and cross 2.

2. Count the number of electron pairsCount the number of electron pairs  – – bond pairs and lone pairs bond pairs and lone pairs 3.

3. Decide the shape adopted by the electron Decide the shape adopted by the electron pairspairs 4.

4. Look at the number of Look at the number of lone pairs and decide the shape adopted lone pairs and decide the shape adopted by the atomby the atom 5.

5. Draw shape, including bond anglesDraw shape, including bond angles

ELECTRON

ELECTRON PAIRS PAIRS SHAPE SHAPE EXAMPLEEXAMPLE

BOND ANGLES AND BOND ANGLES AND

3D SHAPE 3D SHAPE 2

2 bond bond pairs pairs Linear Linear BeClBeCl22

180 180oo

3

3 bond bond pairs pairs Trigonal Trigonal Planar Planar BClBCl33

120 120oo 4

4 bond bond pairs pairs Tetrahedral Tetrahedral CHCH44

109.5 109.5oo 5

5 bond bond pairs pairs Trigonal Trigonal Bipyramidal Bipyramidal PClPCl55

6

6 bond bond pairs pairs Octahedral Octahedral SFSF66

90 90oo



 Lone pairs repel more than bond pairs because they are attracted to a single nucleus and notLone pairs repel more than bond pairs because they are attracted to a single nucleus and not

shared by two atoms shared by two atoms



 Lone pairs reduce bond angles between Lone pairs reduce bond angles between bonding pairs. Each lone pair bonding pairs. Each lone pair reduces predictedreduces predicted

bond angle between bonding electrons by 2

bond angle between bonding electrons by 2 .5 degrees..5 degrees. 4 election pairs on the central atom

4 election pairs on the central atom – – based on the tetrahedral shape: based on the tetrahedral shape: 4 bond pairs = tetrahedral (e.g. CH

4 bond pairs = tetrahedral (e.g. CH44and NHand NH44++) - 109.5) - 109.5oo

ELECTRON

ELECTRON PAIRS PAIRS SHAPE SHAPE EXAMPLE EXAMPLE BOND BOND ANGLES ANGLES AND AND 3-D3-D SHAPE

SHAPE

3 bond pairs and 1 lone 3 bond pairs and 1 lone

pair pair

Trigonal

Trigonal Pyramidal Pyramidal NHNH33

107 107oo

2 bond pairs and 2 lone 2 bond pairs and 2 lone

pairs pairs Bent/non-linear H Bent/non-linear H22OO Organic molecules: Organic molecules: 

 Tetrahedral around carbon if saturated e.g. CTetrahedral around carbon if saturated e.g. C₃₃HH₈₈ or trigonal planar around carbon if there is or trigonal planar around carbon if there is

a C=C bond. a C=C bond.



 In CIn C22HH44, the double bond reduces the bond angle further (its electron rich), the double bond reduces the bond angle further (its electron rich)

Multiple Bonds: Multiple Bonds:



 Count as one bond pair of electrons for purpose of determining the shape.Count as one bond pair of electrons for purpose of determining the shape. 

 E.g. COE.g. CO₂₂is linear:is linear:

Carbon Structures:

Carbon Structures:



 Carbon has several allotropesCarbon has several allotropes  – – different molecular structures due to differences in different molecular structures due to differences in

bonding. bonding.



 DiamondDiamond: Each carbon atom forms 4 identical bonds to neighbouring carbon atoms: Each carbon atom forms 4 identical bonds to neighbouring carbon atoms

giving a tetrahedral arrangement. Because of the strong covalent bonds, diamond giving a tetrahedral arrangement. Because of the strong covalent bonds, diamond has a VERY high melting temperature, is extremely hard (the hardest known has a VERY high melting temperature, is extremely hard (the hardest known substance) and cannot conduct electricity

substance) and cannot conduct electricity – – no free e-. no free e-.



 GraphiteGraphite: Carbon atoms in: Carbon atoms in layers.layers. Within a layer, each carbon atom is bonded to 3Within a layer, each carbon atom is bonded to 3

other carbons

other carbons – – the 4 the 4thth outer e- is delocalised and free to move: conducts electricity. outer e- is delocalised and free to move: conducts electricity. Layers of graphite are weakly bonded to each other

Layers of graphite are weakly bonded to each other  – –  London forces. Also has a very high  London forces. Also has a very high melting point.

 Fullerenes: Consist of 32+ carbon atoms. Buckminsterfullerene has 60 carbon

atoms. Ball-shaped molecules. The fourth outer e- is delocalised, so conduct electricity.

 Nanotubes: Fullerenes in the form of tubes. Very small and stiffer than other known

materials. If embedded in polymers they may produce materials with good electrical conductivity and strength.

Intermediate bonding and bond polarity:

 Electronegativity: ability of an atom to attract an electron pair in a covalent bond. Increases

ACROSS the period (Fluorine is the most EN element) and decreases down groups.

 Differences in elecronegativity between two elements will result in electrons being pulled

further to one end, and there will be POLARITY in the bond e.g:

 If the difference islarge enough, electrons will be transferred – IONIC BOND.

 Small, highly charged cations (e.g. Al3+) are highly polarizing, and will pull electrons toward

thenvery strongly, especially from a large anion (e.g. I-), resulting in a covalent bond (AlI3)

 If molecule is SYMMETRICAL, there is no overall polarity. E.g. CCl_4.The dipoles

cancel out.

 Unsymmetrical molecules containing polar bonds will be polar molecules  –  describes as

having apermanent dipole.

 Polar bonds will deflect a stream of water (because water is polar) e.g. CH₃Cl deflects, CCl₄

doesn’t.

Intermolecular Forces:

 3 types of forces BETWEEN molecules: London forces/Van der Waals (weakest), Permanent

dipole-dipole and Hydrogen bonding(strongest)

1. London/VdW: Found in ALL molecules. Caused by an unequal distribution of electrons which makes a temporary dipole. This affects surrounding atoms causing induced dipoles. The net result is a weak attractive force. Everything has London forces, and the MORE electrons, the STRONGER/LARGER the force.

2. Permanent dipole-dipoles: delta-plus of one molecule is attracted to delta-minus of another molecule:

3. Hydrogen bonding: The attraction between a hydrogen

attached to Fluorine, Oxygen or Nitrogen on one molecule and an F, O, N atom on another molecule. e.g.: hydrogen bonding in water:

Trends in physical properties by intermolecular forces:

 Alkane boiling points increase with carbon chain length, because the number of electrons

 Branched chain alkanes have LOWER boiling points than straight chains because they can’t

pack as closely together, whereas straight chains can pack together closely (greater surface area in contact) therefore the IMF forces are greater and they have higher boiling points.

 Alcohols have very high boiling points (lower volatility – harder to evaporate) due to strong

hydrogen bonding.

 HF has a high boiling point due to hydrogen

bonding. The graph dips down to HCl, HBr and HI, which all have dipole- dipole interactions but the number of electrons is increasing, so there are additional London forces which raise the boiling points.

Solubiliy:

 Affected by bonding, and usually a substance will only dissolve if the strength of the new

bonds formed is the same, or greater than the strength of the bonds that are broken.

 Ionic compounds dissolve in polar substances such as water, because the ions a re attracted

to the polar molecules and they surround the ions and pull them away from the ionic lattice. This releases energy known as the hydration enthalpy. This can only happen if the hydration enthalpy is big enough to overcome the lattice enthalpy. (Hydration vs. Lattice)

 Alcohols are soluble in water, because they form hydrogen bonds with it.

 Non-polar molecules wont form hydrogen bonds with water, so don’t dissolve in it.   E.g.

halogenoalkanes like chlorobutane.

 Generally ‘like dissolves like’

Example question: State and explain the solubility of hexane in water

Hexane molecules are held together by London forces. Water molecules are held together by hydrogen bonds. Hexane can’t _{make hydrogen bonds with water, so the two liquids do not mix or} dissolve in each other –immiscible.

Redox

 Oxidation number: the number of electrons that need to be lost or gained to become a

neutral atom.

 Uncombined elements are 0

 F is always -1, group 1 are +1, group 2 are +2, oxygen is -2 (except in peroxides H₂O₂where

its -1), H is +1 (except metal hydrides where its -1)

 Oxidation numbers in a neutral compound add up to zero, and in a polyatomic ion add up to

the charge.

 Ionic half equations are used for redox processes – when oxidation and reduction take place

 If species are reduced, electrons are on the LEFT  If species are oxidised, electrons are on the RIGHT

 Half equations are then added together for the full redox equation

 E.g. : The overall equation for the oxidation of I-ions by MnO₄-ions is obtained from the two

half equations:

MnO4- + 8H++ 5e- Mn2++ 4H2O

And

2I- I2+ 2e

- For oxidising agents that contain OXYGEN, e.g. MnO₄-, you will need H+ on the LEFT and H₂O

on the RIGHT (oxygen can’t swim)

 The MnO₄- half equation has 5e-but the I- equation has 2e-, so to make them both have the

same number of electrons (so they can cancel out when the equations are added together), the MnO4

- equation has to be multiplied by 2, and the I- equation multiplied by 5, so that theyboth have 10e

- They are then added together to give:

2MnO4

-+ 16H+ + 10I- 2Mn2+ + 8H2O + 5I2

 Disproportionation: when one species is both oxidised AND reduced at the same time

e.g.: Cl2+ H2O HCl + HClO

0 -1 +1

The periodic table – Group 2

 Have their highest energy electrons in an s sub-shell, hence they are called s-block elements.

Ionization energy (I.E) trends:

 Going down the group, there is an extra electron shell compared to the element above, and

the atomic radius is increasing

 The outer electrons are increasingly further away from the nucleus; therefore the attractive

force is less.

 The extra inner shells shield the outer electrons from the attraction of the nucleus

 Therefore, the ionization energies DECREASE down the group (gets easier to remove an e-)

Reactions of group 2 elements withOxygen, Water and Chlorine:

1. Burn in Oxygen to formsolid oxides, often burning with a bright flame

e.g.: 2Mg(s)+ O2 (g) 2MgO(s)

Reactivity INCREASES down group, as the I.E decreases: Be doesn’t react Mg with steam Ca steadily Sr fairly quick Ba rapidly

2. React with water to form metal hydroxide and hydrogen:

e.g.: Ca(s)+ 2H2O (l) Ca (OH) 2 (aq)+ H2 (g)

Mg reacts rapidly with steam: Mg(s) + H2O (g) MgO + H2

3. React with chlorine to form solid metal chlorides:

e.g.: Mg(s) + Cl2 (g) MgCl2(s)

Reactions of group 2 OXIDES and HYDROXIDES:

1. Group 2 oxides react with water to form metal hydroxides, which dissolve. They are also alkaline

e.g.: CaO(s)+ H2O (l) Ca (OH) 2 (aq)

2. Group 2 oxides and hydroxides are BASES

 Theyneutralise dilute acids e.g.: HCl or HNO₃  Form thecorresponding salt and water

e.g.: MgO(s)+ 2HCl (aq) MgCl2 (aq) + H2O (l)

CaO(s)+ 2HNO3 (aq) Ca (NO3)2 (aq)+ H2O (l)

Hydroxides are the same: Ca (OH)2 (aq)+ 2HCl(aq) MgCl2 (aq) + H2O (l)

Solubility trends of hydroxides and sulphates:

 Generally compounds of group 2 elements that contain singly charged negative ions (e.g.

OH-) INCREASE in solubility down group

 Compounds with doubly charged -ve ions (e.g. SO₄2-)DECREASE in solubility down group._Solubilityof 

HYDROXIDES

INCREASESdown the group Mg (OH) 2 Insoluble Ca (OH) 2 Sr (OH) 2 Solubility of SULFATES DECREASES

down the group MgSO4 Most soluble

CaSO4

SrSO4

Thermal Stability of group 1 and 2 CARBONATES and NITRATES

 Thermal decomposition: when a substance decomposes when heated

 Themore thermally stable a substance is, the more heat it requires to break it down.  The carbonate and nitrate ions are LARGE and can be made UNSTABLE by a cation. The

greater the polarising power of the cation, the greater the distortion and the LESS stable the anion.

 The further down the group, the larger the cations and less distortion caused therefore the

MORE stable the carbonate/nitrate anion. Thermal stability increases down a group.

 Group 2 compounds are LESS THERMALLY STABLE than group 1 (more distortion by +2

cation) Group 1

 Carbonates: From sodium carbonate down group 1, the carbonates will NOT DECOMPOSE

on heating – thermally stable.

 Nitrates: From sodium nitrate down group 1, the nitrates decompose to form the nitrite

and oxygen

e.g.: KNO3(s) 2KNO2(s)+ O2 (g)

Group 2

 Carbonates: Lithium and group 2 carbonates decompose to form an oxide and carbon

dioxide

e.g.: CaCO3(s) CaO(s) + CO2 (g)

Li2CO3(s) Li2O(s) + CO2 (g)

 Nitrates: Lithium and group 2 nitrates decompose to form the oxide, nitrogen dioxide and

oxygen.

e.g.: Ca(NO3)2 (s) 2CaO(s)+ 4NO2 (g)+ O2(g)

4LiNO3(s) 2Li2O(s)+ 4NO2 (g)+ O2(g)

Testing thermal stability of nitrates and carbonates: 1. Nitrates:

 How long it takes until a brown gas - NO₂ – is produced. It is toxic, so must be done in fume

cupboard 2. Carbonates:

 How long it takes for carbon dioxide to be produced – tested using limewater which turns

cloudy.

Potassium Nitrite Potassium

Flame tests:

1. Mix small amount of compound with few drops of hydrochloric acid 2. Heat a platinum or nichrome wire in hot flame to clean it.

3. Dip the wire into the compound and hold it in hot flame.

Electrons are being excited to higher energy levels by the heat energy. W hen the electrons return to the lower energy levels, they emit energy in the form of visible light.

Flame colours of group 1 and 2 compounds:

Group 1: Lithium – RED Group 2: Magnesium – WHITE

Sodium – YELLOW Calcium – BRICK RED

Potassium – LILAC Strontium – CRIMSON RED

Barium – GREEN The periodic table – group 7, the HALOGENS

 Non-metallic elements, VERY reactive.  Diatomic covalent molecules

 OXIDISING agents (they are reduced themselves), and become less oxidising, or reactive

down the group.

Reactions of halogens:

1. Disproportionation with alkalis – NaOH or KOH

COLD alkali to give halide and halate (I) ions: HOT alkali to give halide and halate (V) ions: X2+ 2NaOH NaXO + NaX + H2O 3X2+ 6NaOH NaXO3+ 5NaX +3H2O

X2(g)+ 2OH-(aq) XO-(aq)+ X-(aq)+H2O 3X2 (g)+ 6OH-(aq) XO3-(aq)+ 5X-+ 3H2O

O.S: 0 +1 -1 0 +5 -1

e.g: I2+ 2NaOH NaIO + NaI + H2O 3Br2+ 6NaOH NaBrO3 + 5NaBr + 3H2O

Halogen Physical state and colour Appearance in water Appearance in hydrocarbon solvent

Fluorine Pale yellow gas N/A N/A

Chlorine Green gas Pale yellow/green

solution

Pale yellow/green solution Bromine Red-brown liquid Red/brown/orange Red/brown/orange

Iodine Grey solid Brown Pink/violet

Sodium iodate (I) Sodium iodide Sodium bromate (V) Sodium bromate

2. Oxidise metals, non-metals and ions

 Metals: e.g. fluorine and chlorine react with iron to form iron (III) halides  Iron is oxidised: 2Fe 2Fe3+ + 6e

- Chlorine is reduced: 3Cl₂+ 6e- 6Cl - Overall equation: 3Cl_2(g)+ 2Fe_(s)  2FeCl_3(s)

 Non metals: e.g. chlorine reacts with sulphur to form sulphur (I) chloride. Sulphur is oxidised

to +1 and chlorine is reduced to -1)

 S_8(s)+ 4Cl_2(g) 4S₂Cl_2(l)

 Ions: e.g. all halogens except iodine (weak oxidising agent) will oxidise iron (II) ions to iron

(III) ions in solution. The solution will change colour from green to orange.

 2Fe2+_(aq) 2Fe3+_(aq)+ 2e

-Reactions ofHalides:

1. Potassium halides with concentrated sulphuric acid:

 React to give a hydrogen halide.

 The trend in strength of the halide ions as reducing agents is: I-> Br- > Cl

-KCl and H2SO4:

 KCl_(s)+ H₂SO_4(l) KHSO_4(s)+ HCl_(g)

But hydrogen chloride is not a strong enough reducing agent to reduce the sulphuric acid, so

reaction stops there. Misty fumes of hydrogen chloride gas will be seen when it comes into contact with moisture in air. This is NOT a redox reaction – O.S of halide and sulphur stay the same (-1 and +6)

KBr and H2SO4:

 KBr_(s)+ H₂SO_4(l) KHSO_{4 (s)}+ HBr_(g)

This reaction gives misty fumes of hydrogen bromide gas, and the HBr is strong enough to reduce the H2SO4in aredox reaction.

Then this reaction: 2HBr + H2SO4 (l) Br2(g)+ SO2(g)+ 2H2O(l)

O.S of Br: -1 0 OXIDATION

O.S of S: +6 +4 REDUCTION

KI with H2SO4:

 KI_(s)+ H₂SO_4(l) KHSO_4(S)+ HI_(g)  2HI + H₂SO_4(l) I_2(s)+ SO_2(g)+ 2H₂O_(l)

 Same first two reactions, but because iodine is a very strong reducing agent, it goes

further, and reduces SO2to H2S :

 6HI_(g)+ SO_2(g) H₂S_(g)+ 3I_2(s)+ 2H₂O_(l)

O.S of I: -1 0 OXIDATION

O.S of S: +4 -2 REDUCTION

 H₂S is a toxic gas, and gives a bad egg smell.

2. Hydrogen Halides with ammonia and water

 Hydrogen halides are colourless gases. They are very soluble, and dissolve in water

to makeSTRONG acids:

HCl(g) H+(aq) + Cl-(aq) (dissociation)

 Hydrogen chloride forms hydrochloric acid; hydrogen bromide forms hydrobromic

acid and so on.

 With ammonia: react to form white fumes of the corresponding ammonium halide:

NH3 (g)+ HCl(g) NH4Cl(s)

3. Displacement by more reactive halogens

 The oxididising strengths of the halogens can be seen in their displacement reaction

with halides.

 E.g. Br_2(aq)+ 2KI_(aq) 2 KBr_(aq)+ I_2(aq)

 The bromine displaces the iodine ions (it oxidises them) giving iodine I_2(aq)and

potassium bromide

 A halogen will displace a halide from solution if the halide is below it in the periodic

table

4. Silver nitrate solution, and silver halides solubility in ammonia and reactions with sunlight:

 To test forhalides in solution:

1. Add dilute nitric acid – this removes ions that could interfere with test and ppt. 2. Add silver nitrate solution (AgNO3(aq))

 A precipitate of the silver halide will form, the reaction is:

Ag+(aq)+ X

-(aq) AgX(s)

The colour of precipitate identifies the halide, and they have different solubility’s in ammonia solution:

 Bromide Br-: Cream ppt, dissolves in concentrated NH_3 (aq) and darkens in sunlight  Iodide I-: Yellow ppt, insoluble in concentrated NH_{3 (aq)}and does NOT darken in

sunlight.

The reaction of silver halides with sunlight (decomposition) is:

2AgBr 2Ag + Br2

Making predictions about fluorine and astatine from trends in group 7:

 Number of electrons increases down group, so London forces will increase. Astatine

will be a solid and have the highest boiling temperature.

 Electronegativity decreases down group, so astatine will have lowest EN value.  Fluorine will be most oxidising

Kinetics

 Reactions only happen when: Particles collide in the correct orientation, and they possess

theactivation energy (minimum amount of kinetic energy particles need to react). This is the collision theory.

 Enthalpy profile diagram:

 Factors affecting the rate of reaction: concentration, temperature, pressure, surface area

and catalysis.

Factor How it affects rate Explanation Concentration (solution) Pressure (gas) Increasing conc./pressure increases rate

The particles becomemore crowded, therefore collide more times which increases the reaction rate.

Surface area (solids)

Increasing surface area increases rate

The smaller the size of reacting particles, the greater the total surface area. Increasing surface area means larger area isexposed for reaction and more collisions.

Temperature

Increasing temperature increases rate

Increasing temperature means the averagespeed of reacting particles increases, therefore more collisions per second.

Catalyst

Speeds up the reaction

Lower the activation energy by providing an alternative route. If activation energy is lower, more particles will have enough energy to react.

Maxwell-Boltzmann distribution:

 Shows distributions of molecular energies in a gas

 When temperature is increased, particles will have more kinetic energy and move faster.

This means that more particles will have energies greater than the activation energy and will react. This changes the shape of the Maxwell Boltzmann distribution curve pushing it to the right, with a peak lower than the original.

Catalysts:

 Increase the rate of a reaction by providing an alternative reaction pathway  with a lower

activation energy. It is chemically unchanged at the end of the reaction.

 Homogenous catalysts: in the same state as the reactants.

 Forms intermediates with the reactants, which the products are then formed from.  The activation energy needed to form the intermediates and the products from the

intermediates is lower than that needed to make the products directly from the reactants.

Only molecules in this region can react –

molecules have a higher energy than the activation energy

Total number of gas molecules under the

curve

Higher temperature Lower temperature

Chemical Equilibia

 Many reactions do not go to completion because the reaction is reversible

 Dynamic equilibrium: When the rates of the forward and reverse reactions are equal. It’s

dynamic because individual molecules react continuously. It is at equilibrium because no net change occurs (overall concentrations remain constant

 Equilibrium can only happen in a CLOSED system.

The effect of conditions on the position of equilibrium:

 Controlled by Le Chatelier’s principle: When a

system at equilibrium is subjected to a change, it will behave in such a way to counteract that change.

 Temperature is a very important way to

control industrial processes, because it is the most effective factor (general rule – increase in 10K doubles the rate of reaction.

 Pressure is very expensive to use in

equilibrium processes.

 The red-brown gas NO₂ exists in equilibrium

with pale yellow N2O4:

N2O4 2NO2

 The forward reaction is endothermic.

 If the position of equilibrium shifts to left the

mixturepales

 If the position of equilibrium shifts to right the mixture darkens

ORGANICS – Alcohols:

 General formula: C_nH_2n+1OH where the functional group is C-OH  Examples: CH₃OH – Methanol – used for fuels and plastics.

CH3CH2OH – Ethanol – fuels, alcoholic drinks

CH3CH2CH2OH – Propan-1-ol

CH3CHOHCH3- Propan-2- ol

 Can be primary secondary or tertiary alcohols:

1. Combustion of alcohols:

C2H5OH(l)+ 3O2(g) 2CO2(g)+3H2O(g)

2. Reaction with Sodium:

2Na(s)+ 2CH3CH2OH (l) 2CH3CH2O-Na++ H2 (g)

Sodium + ethanol sodium ethoxide + hydrogen

And the longer the hydrocarbon chain, the less reactive with sodium. 3. Substitution reactions to form halogenoalkanes:

Alcohols react with PCl5(Phosphorus (v) Chloride), releasing hydrogen chloride gas which

forms misty fumes in air

CH3CH2OH(l) + PCl5 CH3CH2Cl(l) + POCl3 (l)+ HCl (g)

 The OH is swapped for the Cl, and this reaction can be used as a test for an – OH group. The

steamy fumes that are produced turn blue litmus paper red (because HCl dissolves to form a strong acid)

 To make a chloroalkane, just mix a tertiary alcohol (most reactive) and hydrochloric aci d

together. This will give an impure chloroalkane which can be purified.

4. Oxidation of alcohols:

Must be familiar with these functional groups:

 To oxidise alcohols we use acidified potassium dichromate solution.

 This is orange in colour and is a mixture of sulphuric acid, H₂SO₄ and K₂Cr₂O_7.  The orange colour is due to the Cr6+ ions in K₂Cr₂O₇.

 If it oxidises (i.e. the Cr6+ions become reduced) then the solution t urns green.

Cr6+(aq) + 3e- Cr3+(aq)

Orange Green

 The results show that only primary and secondary alcohols can be oxidised, and tertiary

alcohols cannot be oxidised, therefore remains orange.

Observations made: Sodium fizzes, bubbles form, sodium disappears, and white solid product forms

Oxidation of primary alcohols:

 Aprimary alcohol can be oxidised to an aldehyde and then to a carboxylic acid. This is

carried out using an oxidising agent: Mixture of  sulphuric acid, H2SO4 (souce of H+) and

potassium/sodium dichromate, K2Cr2O7

 To stop oxidising at the aldehyde, you must’ allow the product to distil over’  To get the carboxylic acid, you heat under reflux

Primary alcohol to aldehyde

 This is the distillation apparatus.

 The aldehyde has to be distilled off as it forms

as it can be oxidised further

 Distillation evaporates and condenses liquids at

different temperatures. Collect the liquid you want around its boiling point and discard any others

Primary alcohol to carboxylic acid

 When making the carboxylic acid the mixture is refluxed.  Heated strongly with an excess of the acidified potassium or

sodium dichromate, and the alcohol will be completely oxidised passing through the aldehyde stage to form a carboxylic acid.

 Refluxing allows you to heat / boil volatile liquids for a long time. The condenser stops the

volatile liquids evaporating off  , because any vaporised compounds are cooled, condense and drip back down to the reaction mixture

Oxidation of secondary alcohols:

 Secondary alcohols are oxidised to ketones ONLY. Do not undergo further oxidation.  This can be done by refluxing the secondary alcohol with acidified sodium/potassium

dichromate.

Summary:

Primary alcohol Aldehyde Carboxylic acid

Secondary alcohol Ketone

No reaction

Tertiary alcohol

Halogenoalkanes

 Halogenoalkanes have the general formula CnH2n+1X. X is a halogen.  Can also be primary, secondary and tertiary like alcohols.

 When naming, the halogen part is named first (prefix chloro-, bromo-, iodo-) followed by

name of alkane

 E.g. CH₃Cl =Cloromethane

CH3CH2Br = Bromoethane

 If there is more than one halogen di- and tri- are used to indicate the number of halogens

present, e.g. CH2BrCH2Br = 1,2-dibromoethane

Reactions of Halogenoalkanes:

 Halogenoalkanes containpolar bonds because the halogen is

more electronegative than the carbon. This leaves a carbon with a delta + charge, making it open to attack by

 Nucleophiles: attracted to electron deficient atom, d+ and donate a pair of electrons to

form a new covalent bond

 The halogen will be replaced by the nucleophile, which gives a substitution reaction, giving a

new functional group.

1. Halogenoalkanes react with aqueous alkalis to form ALCOHOLS

 Aqueous hydroxide ions need to substitute the halogen. Sodium hydroxide NaOH_(aq) or

potassium hydroxide KOH(aq)can be used.

 The reaction is called hydrolysis and usually carried out under reflux

 Hydrolysis: is a reaction with water or aq hydroxide ions that break a chemical compound

into two compounds

Mechanism:

Water can act as a nucleophile too, but it is a much slower reaction:

First step

Second step

 If water with dissolved silver nitrate is used, this can tell us about the reactivities of

halogenoalkanes

 When water and an alcohol react, and an alcohol is formed, the silver nitrate will react with

the halide ions when they form giving a silver halide precipitate

 The precipitate that forms first indicates which halogenoalkanes hydrolyses first:

Tertiary halogenoalkanes – precipitate forms immediately

Secondary halogenoalkanes – precipitate forms after several seconds Primary halogenoalkanes – precipitate forms after several minutes

 This shows that the reactivity is tertiary 3o > secondary 2o> primary 1o

2. Halogenoalkanes react with alcoholic ammonia to form amines

 Ammonia NH₃has a lone pair of electrons, and can therefore act as a nucleophile  Alcoholic ammonia – ammonia dissolved in ethanol.

 Heated under reflux

3. Alcoholic alkali to form alkenes

When a halogenoalkane reacts with alcoholic alkali, e.g. potassium hydroxide, KOH in hot ethanol, an alkene is made

 This is an elimination reaction  Heated under reflux

Uses of halogenoalkanes:

 Halogenoalkanes are used as fire retardants and refrigerants

MECHANISM

Step 1

Step 2

In the second step, and ammonia molecule removes hydrogen from the NH3

group to form an ammonium ion (NH4+)

This can then react with the Br-ion from step 1, to form ammonium bromide:

NH4Br

Overall reaction: with ethanol and under reflux

 Chlorofluorocarbons (CFCs) used to be used in the past because of their unique properties

(non-toxic, non-flammable, unreactive), but it was found that they deplete the ozone layer in the atmosphere, so are being phased out (see notes later)

 Other halogenoalkanes such as hydrofluorocarbons (HFCs) are now used as safer

alternatives. Mechanisms:

Free radical – species with an unpaired electron Electrophile – species that accepts a pair of electrons Nucleophile – species that donates a pair of electrons Substitution – one species is replaced by another

Addition – joining two or more molecules together to make a larger molecule Elimination – when a small species is eliminated from a larger molecule

Oxidation – loss of electrons. Also is the gain of oxygen/loss of hydrogen Reduction – gain of electrons. Also is the loss of oxygen/gain of hydrogen Hydrolysis – Splitting up using water (usually in form of OH- ions)

Polymerisation – joining together monomers into long carbon-chain polymers. Redox – any reaction where electrons are transferred between two species

Bond breaking – homolytic and heterolytic:

 Homolytic – when the bond breaks evenly, and one electron moves to each atom. This

forms two free radicals as both atoms now have an unpaired electron. The unpaired electron makes free radicals very reactive.

 Heterolytic – when the bond breaks evenly, and both electrons from the shared electron

pair move to one atom. This forms two different species: a positively charged cation – an electrophile, and a negatively charged anion – a nucleophile

When drawing curly arrows – double headed arrow shows movement of electron pair; single headed arrow shows movement of single electron.

Should be able to recall these reaction mechanisms from unit 1: Electrophilic addition and free radical substitution:

 Free radical substitution of chlorine in alkanes:  Initiation, propagation, termination

Predicting the type of mechanism:

 Polar bonds always break heterolytically

 A nucleophile can attack the d+ atom in a polar bond

 An electrophile can attack an electron- rich part of a molecule – e.g. the C=C bond in alkenes

All reagents used in AS chemistry (helpful to learn them):

 Nucleophiles: OH-_(aq), NH_{3 (alcoholic),} PCl_5, NaBr/H₂SO₄or PBr_5,P/I₂  Electrophiles: H_2, X_2, HX

 Oxidation [O]: KMnO₄ /H+, K₂Cr₂O₇/H+

 Other: KOH in hot ethanol, Cl₂/ u.v light, RO –OR/ u.v light (polymerisation)

OZONE

 Ozone molecules – O₃

 The ozone layer is at the edge of the stratosphere

 It filters out most of the harmful UV radiation which can damage DNA in cells causing skin

cancer and can also cause eye cataracts.

 Ozone is formed when UV radiation from the sun hits oxygen molecules. This forms two free

radicals. The free radicals then combine with other oxygen molecules to form ozone molecules

O2+ U.V O* + O*

O2+ O* O3

 The ozone layer is constantly being replaced, and there is a natural balance between

formation of new ozone and breakdown of ozone molecules : O2+ O* O3

 It was discovered that the ozone layer is thinning in places, and a hole in the ozone was

discovered over Antarctica – this means that more harmful UV will reach the earth.

 The decrease in ozone concentrations is due to CFCs – chlorofluorocarbons.

 Because of their un-reactivity, CFCs don’t decay and reach the upper atmosphere and the

ozone layer, where several reactions happen:

1) CFCs are broken down by UV light, forming chlorine free radicals CCl3F2 (g) CCl2F*(g) + Cl*(g)

2) The free radicals are catalysts, and react with ozone to form an intermediate – ClO*, and O2

H+= acidified

Cl*(g)+ O3 (g) O2 (g)+ ClO*(g)

ClO*(g)+ O3 (g) 2O2 (g) + Cl*(g)

3) The overall reaction is: 2O3 (g) 3O2 (g) (Cl is the catalyst)

 Nitrogen oxides are produced from car and aircraft engines and thunderstorms. Like

chlorine radicals, NO* also act as catalysts :

NO* + O3 O2 + NO2*

NO2* + O3 2O2+ NO*

IR

 Some molecules absorb energy from infrared radiation. This makes the bonds vibrate  Vibrations occur in one of 2 ways, a stretching vibration or a bending vibration

 Every bond vibrates at its own unique frequency depending on:

1. Bond strength 2. Bond length

3. Mass of atom at either end of the bond

 Oxygen O₂, and Nitrogen N₂don’t absorb infrared radiation, but CO₂, H₂O, nitric acid (NO) ,

and methane (CH4) do absorb infrared radiation. They absorb IR because they change their

polarity as they vibrate (due to the movement of dipoles in polar bonds)

 Gases that do absorb IR radiation are called greenhouse gases – they stop some of the

radiation emitted by the earth from escaping into space. What the spectrum look like:

 The spectrum gives us 'peaks' which are actually absorbance troughs.

 These troughs are caused by a frequency of IR light being absorbed from a bond vibrating

bond.

 Each 'peak' is characteristic to a specific bond / atoms

The Cl free radical is regenerated and goes on to attack other ozone

molecules. This shows that one CFC molecule can destroy thousands of ozone molecules

Identification of functional groups:

 We have just seen that the peak on an IR spectra are due to specific bonds (and atoms)

vibrating or stretching.

 The frequency at which you find an absorbance peak is therefore unique to bonds and

atoms at each end of the bond.

 This means that functional groups will give specific peaks.

Bond Functional group Wavenumber/frequency

C=O Aldehydes, ketones, carboxylic acids 1640 - 1750

C- H Organic compounds 2850 - 3100

O- H Carboxylic acids

2500 - 3300 (very broad) O- H Alcohols (hydrogen bonded) 3200 - 3550 (broad)

N-H Amines 3200-3500

C-X Halogenoalkanes 500-1000

Alcohols:

The IR spectrum for methanol, CH3OH is shown below:

 The peak at 3230 - 3500

represents an O - H group in alcohols.

Aldehydes and ketones:

 The IR spectrum for propanal,

CH3CHO is shown:

 The peak at 1680 - 1750

represents a C=O group in aldehydes and ketones.

Aldehydes and ketones:

The IR spectrum for propanoic acid, CH3CH2COOH is shown:

 The peak at 2500 - 3300

represents an O - H group in a carboxylic acid.

 The peak at 1680 - 1750

represents a C=O group in a carboxylic acid.

Mass spec

 Ionisation in a mass spectroscope is usually done by electron bombardment.

 Electron bombardment knocks another electron out of the molecule producing a positive

molecular ion -M+. C2H5OH + e - _ C2H5OH +  + 2e

- The molecular ion has the same mass as the Mr of the molecule.

 As we have a mass and a charge we can use a mass spectrometer to determine the Mr (m/z).

Fragmentation:

 Excess energy from the ionisation process causes bonds in the organic molecule to vibrate

and weaken.

 This causes the molecule to split or fragment into smaller pieces.

 Fragmentation gives a positively charged molecular fragment ion and a neutral molecule:

C2H5OH CH3 + CH2OH +

 The fragment ion, CH₂OH+ has a mass and charge so we can use a mass spectrometer to

determine the Mr (m/z) of that fragment.

 Fragment ions can be broken up further to give a range of m/z values.  The m/z values correspond to the Mr's of the molecule and its fragments.

 The Mr of the molecule is always the highest m/z value - i.e. this molecule has not been

fragmented so it must have the highest Mr.

Fragmentation patterns:

 Mass spectroscopy is used to identify and determine the structures of unknown compounds.  Although 2 isomers will have exactly the same M

+

 peak, the fragmentation patterns will be unique to that molecule, like a fingerprint.

 In practice mass spectrometers are linked to a database and the spectra is compared until an

exact match is found:

 These are the mass spectra for pentane and a structural isomer of pentane, 2 methylbutane.  The M

+

 peak is the same for each but the fragmentation patterns are different. Identifying fragment ions:

 When you look at a mass spectrum, other peaks seem to look more important than the M+

peak.

 These fragment peaks give clues to the structure of the compound.  Even simple structures give common peaks that can be identified:

m/z value

Possible identity of the fragment ion 15 CH3+ 29 C2H5 + 43 C3H7+ 57 C4H9 + 17 OH+

 Functional groups are a good place to start, OH = m/z of 17

 Some fragments are more difficult to identify as these will have undergone molecular

rearrangement.

Identification of organic structures:

 A mass spectrum will not only tell you the Mr (from the M+ peak), but it can also tell you

some of the structural detail.

 The mass spectrum above has been produced from hexane.

 The following reactions show how the molecule could fragment to form the fragment ions

57 and 43:

Green Chemistry

Processes in the chemical industry are being reinvented to make them more sustainable or ‘greener’ by:

1. Changing to renewable resources: e.g. plastics made from crude oil can be made from plant products

2. Making more efficient use of energy. E.g. in the pharmaceutical industry microwave

radiation is used to heat the reacting mixture directly rather than using conventional heating systems which heat the reaction vessel which passes on heat to the reaction mixture  – less efficient.

3. Finding alternatives to very hazardous chemicals e.g. some chemicals can harm humans, other living organisms or the environment.

4. Discovering catalysts for reactions with higher atom economies. This is important because a high atom economy means less waste is produced and this makes the best use of resources.

5. Reducing waste and preventing pollution of the environment. E.g. creating recyclable or using products to conserve raw materials and where possible waste should be recycled or biodegradable.

Effects of greenhouse gases:

 Infrared radiation IR from the sun has a short wavelength and most of it passes through the

atmosphere and is absorbed by the earth’s surface.

 The earth heats up, and re-emits longer wavelength IR. Any greenhouse gases in the

atmosphere effectively reflect the longer wavelength IR which warms the atmosphere.

 The relative greenhouse effect of a gas varies because molecules absorb IR differently.  The global warming potential of a gas combines it ability to absorb IR with its lifetime in

the atmosphere. The concentration of a gas in the atmosphere also affects GWP. E.g. CO2

has a low global warming potential, but the concentrations of it are increasing. CFCs have a much higher GWP but the overall concentrations are very low.

Anthropogenic and natural climate change:

 Anthropogenic: results from human activities, e.g. burning fossil fuels and deforestation.

These increase levels of CO2, methane and other gases over relatively short timescales.

 Natural climate change: natural processes such as dissolving of CO₂ in sea water or

formation of carbonates in rocks over hundreds of years. Volcanic eruptions can also cause climate change.

Carbon neutrality and carbon footprint:

 A carbon neutral fuel is one for which the release of CO₂ in its manufacture and burning

equals the absorption of CO2 as the raw material is grown or the fuel formed. Only certain

biofuels can be considered carbon neutral

 A carbon neutral process occurs when there is no overall carbon emission into the

atmosphere.

 A carbon footprint in general is a measure of the amount of carbon dioxide emitted through

the use of fossil fuels. It is often measured in tonnes of carbon dioxide, and can be calculated for an individual, a household, an organisation or over a product lifecycle for manufactured goods.

 The fuelpetrol is definitely not carbon neutral - releases CO₂into atmosphere which was

trapped in the earth millions of years ago.

 Bioethanol is more or less carbon neutral- produced by fermentation of sugar from crops.

It’s thought of as being carbon neutral as the CO2 released when burnt was removed by the

crop as it grew. However, there are still carbon emissions when considering the whole process.

 Hydrogen gas can be carbon neutral .