88272058 as Chemistry Unit 2 Notes

AS Chemistry Unit 2 Notes Shapes of molecules and Ions 

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



Finding the shape: 1. Draw dot and cross 2. Count the number of electron pairs  – bond pairs and lone pairs 3. Decide the shape adopted by the electron pairs 4. Look at the number of lone pairs and decide the shape adopted by the atom 5. Draw shape, including bond angles

ELECTRON PAIRS

SHAPE

EXAMPLE

2 bond pairs

Linear

BeCl2

3 bond pairs

Trigonal Planar

4 bond pairs

Tetrahedral

BOND ANGLES AND 3D SHAPE

180

o

120

o

BCl3

CH4

o

109.5 5 bond pairs

Trigonal Bipyramidal

PCl5

6 bond pairs

Octahedral

SF6

90



o

Lone pairs repel more than bond pairs because they are attracted to a single nucleus and not shared by two atoms



Lone pairs reduce bond angles between bonding pairs. Each lone pair reduces predicted bond angle between bonding electrons by 2 .5 degrees.

4 election pairs on the central atom  – based on the tetrahedral shape: +

o

4 bond pairs = tetrahedral (e.g. CH 4 and NH4 ) - 109.5

ELECTRON PAIRS

SHAPE

EXAMPLE

3 bond pairs and 1 lone pair

Trigonal Pyramidal

NH3

BOND ANGLES AND 3-D SHAPE

o

107

2 bond pairs and 2 lone pairs

Bent/non-linear

H2O

Organic molecules: 

Tetrahedral around carbon if saturated e.g. C 3H8 or trigonal planar around carbon if there is a C=C bond.



In C2H4 , the double bond reduces the bond angle further (its electron rich)

Multiple Bonds: 

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



E.g. CO2 is linear:

Carbon Structures: 

Carbon has several allotropes  – different molecular structures due to differences in bonding.



Diamond: Each carbon atom forms 4 identical bonds to neighbouring carbon atoms giving a tetrahedral arrangement. Because of the strong covalent bonds, diamond has a VERY high melting temperature, is extremely hard (the hardest known substance) and cannot conduct electricity – no free e-.



Graphite: Carbon atoms in layers. Within a layer, each carbon atom is bonded to 3 other carbons – the 4th outer e- is delocalised and free to move: conducts electricity.

Layers of graphite are weakly bonded to each other  –  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 is large enough, electrons will be transferred  – IONIC BOND. 3+

Small, highly charged cations (e.g. Al ) are highly polarizing , and will pull electrons toward -

then very strongly, especially from a large anion (e.g. I ), resulting in a covalent bond (AlI 3 ) 

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 a permanent dipole.



Polar bonds will deflect a stream of water (because water is polar) e.g. CH 3Cl deflects, CCl 4 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 i ntermolecular forces: 

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

increases, so more London forces.

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 H 2O2 where 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 together in a reaction.



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 MnO 4 ions is obtained from the two

-

-

half equations: -

+

-

2+

MnO4  + 8H + 5e

Mn + 4H2O And

-

2I 

I2 + 2e

-

-

For oxidising agents that contain OXYGEN, e.g. MnO 4 , you will need H+ on the LEFT and H 2O on the RIGHT (oxygen can’t swim)



-

-

-

-

The MnO 4  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 they both have 10e 

-

They are then added together to give: -

+

2MnO4 + 16H  + 10I 

-

2+

2Mn  + 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 with Oxygen, Water and Chlorine: Reactivity INCREASES

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

down group, as the I.E decreases:

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

2MgO(s)

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)

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

Ca (OH) 2 (aq) + H2 (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 

They neutralise dilute acids e.g.: HCl or HNO 3



Form the corresponding salt and water e.g.:

MgO(s) + 2HCl (aq)

MgCl2 (aq) + H2O (l)

CaO(s) + 2HNO3 (aq) Hydroxides are the same:

Ca (NO3)2 (aq) + H2O (l)

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 

2-)

Compounds with doubly charged -ve ions (e.g. SO 4

DECREASESolubility in solubility of   down group. HYDROXIDES

INCREASES down the group Mg (OH) 2 Insoluble Ca (OH) 2 Sr (OH) 2

Solubility of SULFATES DECREASES down the group MgSO4 Most soluble CaSO4 SrSO4 BaSO4

Insoluble

Thermal Stability of group 1 and 2 CARBONATES and NITRATES 

Thermal decomposition: when a substance decomposes when heated



The more 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)

Potassium

Potassium

Nitrate

Nitrite

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) 4LiNO3(s)

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

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

How long it takes until a brown gas - NO 2 – 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.

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.

Halogen

Physical state and colour

Appearance in water

Fluorine

Pale yellow gas

N/A

Appearance in hydrocarbon solvent N/A

Chlorine

Green gas

Bromine

Red-brown liquid

Pale yellow/green solution Red/brown/orange

Pale yellow/green solution Red/brown/orange

Iodine

Grey solid

Brown

Pink/violet

Reactions of halogens: 1. Disproportionation with alkalis – NaOH or KOH COLD alkali to give halide and halate (I) ions: X2 + 2NaOH -

X2(g) + 2OH (aq)

HOT alkali to give halide and halate (V) ions:

NaXO + NaX + H2O -

-

XO (aq) + X (aq) +H2O

O.S:

0

+1

-1

e.g:

I2 + 2NaOH

NaIO + NaI +

3X2 + 6NaOH -

3X2 (g) + 6OH (aq) 0

H2O

3Br2 + 6NaOH

NaXO3 + 5NaX +3H2O -

XO3 +5

(aq) +

-

5X + 3H2O -1

NaBrO3 + 5NaBr + 3H 2O

Sodium

Sodium

Sodium

Sodium

iodate (I)

iodide

bromate (V)

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:



Chlorine is reduced: 3Cl2 + 6e



Overall equation: 3Cl2(g) + 2Fe(s) 



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

3+

2Fe

-

2Fe  + 6e -

-

6Cl

2FeCl3(s)

to +1 and chlorine is reduced to -1) 

S8(s) + 4Cl2(g)

4S2Cl2(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.



2Fe

2+ (aq)

2Fe

3+

(aq) +

-

2e

Reactions of Halides: 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) + H2SO4(l)

KHSO4(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) + H2SO4(l)

KHSO4 (s) + HBr(g)

This reaction gives misty fumes of hydrogen bromide gas, and the HBr is strong enough to  reduce the H2SO4 in a redox reaction. Then this reaction: 2HBr + H 2SO4 (l)

Br2(g) + SO2(g) + 2H2O(l)

O.S of Br:

0

-1

O.S of S:

+6

OXIDATION +4

KI with H2SO4:  

KI(s) + H2SO4(l)

KHSO4(S) + HI(g)

2HI + H2SO4(l)

I2(s) + SO2(g) + 2H2O(l)

REDUCTION



Same first two reactions, but because iodine is a very strong reducing agent, it goes further, and reduces SO2 to H2S :



O.S of I:

6HI(g) + SO2(g)

H2S(g) + 3I2(s) + 2H2O(l)

-1

O.S of S:

0 +4



OXIDATION

-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 make STRONG 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.

Br2(aq) + 2KI(aq)

2 KBr(aq) + I2(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 for halides in solution:

1. Add dilute nitric acid – this removes ions that could interfere with test and ppt. 2. Add silver nitrate solution (AgNO 3(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: 

Chloride Cl- : White ppt which dissolves in dilute NH 3 (aq) and darkens in sunlight



Bromide Br- : Cream ppt, dissolves in concentrated NH3 (aq) and darkens in sunlight



Iodide I : Yellow ppt, insoluble in concentrated NH3 (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 the activation 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 Concentration (solution) Pressure (gas) Surface area (solids)

Temperature

Catalyst

How it affects rate Increasing conc./pressure increases rate

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

Increasing surface area increases rate

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

Increasing temperature increases rate

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

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. Lower temperature Higher temperature

Only molecules in this region can react – molecules have a higher energy than the activation energy

Total number of gas molecules under the curve

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.

Chemical Equilibia 

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



Dynamic equilibrium : When the rates of the forward and reverse re actions 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 NO 2 exists in equilibrium with pale yellow N2O4 : N2O4

2NO2



The forward reaction is endothermic.



If the position of equilibrium shifts to left the mixture pales



If the position of equilibrium shifts to right the mixture darkens

ORGANICS – Alcohols:



General formula: CnH2n+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:

Reactions of alcohols:

1. Combustion of alcohols: C2H5OH(l) + 3O2(g)

2CO2(g) +3H2O(g)

2. Reaction with Sodium:

Observations made: -

2Na(s) + 2CH3CH2OH (l)

+

2CH3CH2O Na + H2 (g)

Sodium fizzes, bubbles form, sodium

sodium ethoxide + hydrogen

Sodium + ethanol

disappears, and white

And the longer the hydrocarbon chain, the less reactive with sodium.

solid product forms

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 K2Cr2O7. 6+ The orange colour is due to the Cr  ions in K 2Cr2O7. 6+ If it oxidises (i.e. the Cr ions become reduced) then the solution t urns green. 6+

Cr (aq) Orange



+

3e

-

3+

Cr (aq) Green

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

alcohols cannot be oxidised, therefore remains orange.

Oxidation of primary alcohols: 

A primary 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.

HEAT



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

No reaction

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. CH 2BrCH2Br = 1,2-dibromoethane

Reactions of Halogenoalkanes: 

Halogenoalkanes contain polar 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.



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

The overall equation with water:

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

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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

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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

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o

o

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

o

2. Halogenoalkanes react with alcoholic ammonia to form amines 

Ammonia NH3 has a lone pair of electrons, and can therefore act as a nucleophile

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Alcoholic ammonia – ammonia dissolved in ethanol.

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Heated under reflux In the second step, and ammonia

MECHANISM

molecule removes hydrogen from the NH 3 +

group to form an ammonium ion (NH 4 ) Step 1

-

This can then react with the Br ion from step 1, to form ammonium bromide: NH4Br Overall reaction: with ethanol and under

Step 2

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

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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)

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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.

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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:

Hydrogen bromide to alkenes:

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Free radical substitution of chlorine in alkanes:

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Initiation, propagation, termination

Predicting the type of mechanism: 

Polar bonds always break heterolytically

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A nucleophile can attack the d atom in a polar bond

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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): -

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Nucleophiles: OH (aq), NH3 (alcoholic), PCl5, NaBr/H2SO4 or PBr5, P/I2

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Electrophiles: H2, X2, HX

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Oxidation [O]: KMnO4 /H , K 2Cr2O7/H

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Other: KOH in hot ethanol, Cl 2 / u.v light, RO –OR/ u.v light (polymerisation)

+

+

+

H = acidified

OZONE 

Ozone molecules – O3

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The ozone layer is at the edge of the stratosphere

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It filters out most of the harmful UV radiation which can damage DNA in cells causing skin cancer and can also cause eye cataracts.

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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*

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* = free radical

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.

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

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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 O 2

The Cl free radical is regenerated

3) The overall reaction is:

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Cl*(g) + O3 (g)

O2 (g) + ClO*(g)

ClO*(g) + O3 (g)

2O2 (g) + Cl*(g)

2O3 (g)

3O2 (g)

and goes on to attack other ozone molecules. This shows that one CFC molecule can destroy thousands of ozone molecules

(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

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Vibrations occur in one of 2 ways, a stretching vibration or a bending vibration

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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 

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Oxygen O2 , and Nitrogen N 2 don’t absorb infrared radiation, but CO2 , H2O, nitric acid (NO) , and methane (CH 4) 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  

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Identification of functional groups: 

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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 2500 - 3300

O- H

Carboxylic acids (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, CH 3OH is shown below:

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The peak at 3230 - 3500 represents an O - H group in alcohols.

Aldehydes and ketones:

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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: 

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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

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  

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+

+

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. The one below is for ethanol. It has a m/z of 46 which is also its Mr.

Fragmentation patterns:  

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 

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: 

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 

+

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: 

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+

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. These peaks have been labelled with a l etter:

 

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 ‘g reener’ 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.

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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.

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The relative greenhouse effect of a gas varies because molecules absorb IR differently.

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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. CO 2 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 CO 2, methane and other gases over relatively short timescales.

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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 CO 2 in its manufacture and burning equals the absorption of CO 2 as the raw material is grown or the fuel formed. Only certain biofuels can be considered carbon neutral

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

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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.

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The fuel petrol is definitely not carbon neutral - releases CO 2 into atmosphere which was trapped in the earth millions of years ago.

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Bioethanol is more or less carbon neutral- produced by fermentation of sugar from crops. It’s thought of as being carbon neutral as the CO 2 released when burnt was removed by the

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

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Hydrogen gas can be carbon neutral .