All Members of a Homologeous Series Have Similar Chemical Properties

TOPIC 10:

Organic Chemistry

10.1 Homologous Series (SL/HL)

This is the name given to a group of compounds that can be described by a general formula.

NameAlkanesAlkenesAlcohols

GeneralCnH2n+2CnH2nCnH2n+2O

Formula

ExamplesCH3CH3CH2CH2CH3CH2OH

CH3CH2CH3 CH2CHCH3 CH3CH2CH2OH

CH3CH2CH2CH3 CH2CHCH2CH3 CH3CH2CH2CH2OH

All members of a homologeous series have similar chemical properties.

The next member in a homologeous series differs by a –CH2- group(called a repeating unit).

This causes a gradual change in physical properties as the chain length increases. (see below)

10.1Physical Properties in Homologous Series

Boiling Point increases as chain length increases.

This is due to an increase in Van De Waal forces between the molecules as the molar mass (and surface area) increases.

10.1 Representing Organic Structures

Organic structures can be represented in a variety of ways:

Empirical Formula: Simplest ratio of atoms of each elements in a compound.

E.g. Pentene: Empirical Formula = CH2

Molecular Formula: Actual numbers of atoms of each element present in the compound.

E.g. Pentene: Molecular Formula = C5H10

Structural Formula: How the atoms are arranged in the molecule.

Condensed structural formula shows the position of the atoms but omits the bonds.

E.g. Propene:CH2=CHCH2CH2CH3 or CH2CH(CH2)2CH3

Full structural formula/graphic formula/displayed formula = Shows every atom and bond.

E.g. Propene: HH H HH

C=C - C - C - C - H

H H H H

Show the condensed structural formula and the full structural of the following molecule:

Hexane: Molecular formula = C6H14

TOK: Why is it necessary to use different types of formula to represent organic molecules?

10.1 Functional Groups (SL/HL)

The functional group is the reactive part of an organic molecule that largely dictates its chemical properties.

Homologous SeriesFormula of Functional GroupExample

AlkaneR-H

AlkeneC=C

Benzene ring

HalogenoalkaneR-X

(Halide)

AlcoholR-OH

Aldehydes

Ketones

Carboxylic acids

AminesR-NH2

Amides

Amino Acids

Esters

HL ONLY:Nitriles

10.1 Naming Organic Compounds (SL/HL)

Find the longest chain

1 = meth2 = eth3 = prop4 = but5 = pent6 = hex

7 = hept8 = oct

Name any branches

CH3 = methylC2H5 = ethylC3H7 = propyl

Number the positions of the branches (keeping the numbers as low as possible)

E.g.CH3CH2CH(CH3)CH32-methyl butane

When there are more than one side chains, these are indicated as di = 2 tri = 3 and tetra = 4

E.g.CH3C(CH3)2CH2CH32,2-dimethylbutane

CCl3Htrichloromethane

Name the Functional Group.

These are usually indicated by a suffix

Functional group Suffix

Alkane -ane

Alkene-ene

Alcohol-anol

Aldehyde-anal

Ketone-anone

Carboxylic acid-anoic acid

BenzeneBenzene

HL ONLY: Esters - anoate

Amines - amine

Amide - amide

Nitriles - nitrile

A few are indicated by a prefix

Functional groupPrefix

HalogenoalkaneChloro-, bromo-, iodo-.

Benzene Phenyl/benzene

TOK: Why are organic compounds named in this way?

What advantages are there internationally?

This system of naming has meant that many older names of substances (chloroform, toluene, formaldehyde) have become obsolete. What are the advantages and disadvantages of this process?

E.g.CH3OHmethanol

CH3CH2Clchloroethane

CH3CH2CH3CHObutanal

CH3COCH3propanone

CH3CH2CO2HPropanoic acid

CH2CHCH3

CH3CH2CH2OH

HCOOH

CH3CH2CH2I

CH3COCH2CH2CH3

CH3CHO

HL ONLY:CH3CH2CH2NH2

CH3(CH2)3CN

CH3COOCH3

HCOOCH2CH2CH3

CH2CONH2

Extension:C6H5CH3

CH3CH2COONHCH3

C6H5CH2COOH

C6H5OH

10.1 Structural Isomers (SL/HL)

Two organic molecules can have the same molecular formula, but have different structural formulas depending on how the atoms are arranged.

10.1 Structural Isomers of Alkanes (SL/HL)

This is called chain isomerism and is a type of structural isomerism.

Draw and name the chain isomers of C4H10

Draw and name the chain isomers of C5H12 and C6H14.

Note:

Straight chain isomers tend to have higher melting and boiling points than branched chains.

This is due to the reduction in surface area of the branched chains and the reduced Van De Waal forces that this causes.

10.1 Structural Isomers in other Functional Groups (SL/HL)

If the functional group can be in more than one position then it must be numbered and named accordingly.

CH3CH2CH2OH propan-1-ol CH3CH(OH)CH3 propan-2-ol

This is a different type of isomerism called position isomerism.

CH2=CHCH2CH3 but-1-eneCH3CH=CHCH3 but-2-ene

The final type of structural isomerism is called functional group isomerism.

This occurs when a compound has the same molecular formula, but has a different functional group.

E.g.aldehyde/ketone:

C3H7Oaldehyde:CH3CH2CHO (propanal)

Ketone:CH3COCH3 (propanone)

carboxylic acids/esters:

C3H6O2

alcohols/ethers*:

C3H8O

10.1 Primary, Secondary and Tertiary Structures.

This indicates the number of carbon atoms attached to the functional group in the organic molecule.

Primary, secondary and tertiary structures often react very differently (see oxidation of alcohols and nucleophilic substitutions of halogenoalkanes)

PrimarySecondaryTertiary

Draw the condensed structural formula of the isomers, name the molecules and then decide if they are primary secondary or tertiary structures.

C4H10O

C4H10Br

10.1 Physical Properties of Functional Groups (SL/HL)

Boiling Points and Volatility

The longer the carbon chain in an organic molecule, the higher the boiling point. This is due to an increase in molar mass and consequent increase in Van De Waals forces and occurs in each homologous series.

The nature of the functional group plays a large effect in the type of intermolecular force and therefore the boiling point of each series of compounds.

Hydrocarbons have only weaker Van De Waals forces. They tend to have low boiling points.

Aldehydes, ketones, halogenoalkanes and esters have stronger dipole-dipole forces and therefore have higher boiling points.

Amines, amides and more particularly alcohols and carboxylic acids have very high boiling points due to the formation of hydrogen bonds between the molecules.

Volatility (ease of vaporisation) follows the same trends. Compounds that are hydrogen bonded tend to be less volatile, and substances tend to be less volatile as carbon chain increases.

Solubility in water

Organic compounds that can form hydrogen bonds with water tend to be soluble in water (carboxylic acids, alcohols, amines, amides.)

The solubility decreases as the length of the non-polar carbon chain increases.

The most soluble compounds therefore tend to be short chain alcohols and carboxylic acids.

Carboxylic acids are more soluble than alcohols since they can form stronger H-bonds due to the additional electronegativity of C=O group.

10.2 Reactions of Alkanes (SL/HL)

Alkanes are relatively unreactive and inert.

This is due to the strengthand lack of polarity of the C-C and C-H bonds.

The main type of reaction that they will undergo is combustion.

The products of complete combustion when there is a plentiful supply of oxygen are carbon dioxide and water.

E.g.C3H8+5 O23 CO2+4 H2O

Even though the energy required to break the bonds within the alkane are very large, the C=O and O-H bonds formed are even stronger resulting in a lot of energy release and a very exothermic reaction.

The CO2 released can cause environmental problems. It will absorb IR radiation reflected from the Earth and so contribute to global warming.

Incomplete combustion will occur in a limited supply of oxygen. The products may be carbon monoxide or carbon powder.

E.g.C3H8+3.5 O23 CO+4H2O

C3H8+2 O23 C+4H2O

All these combustion reactions are redox reactions. In all cases the oxygen is reduced to O2- in H2O by gaining electrons.

In complete combustion the carbon is oxidised to C+4 in CO2

In incomplete combustion the carbon is only oxidised to C+2 in CO, or C0 in elemental carbon.

The products of incomplete combustion can be serious pollutants. Carbon monoxide is a poisonous gas that is absorbed by the blood instead of oxygen leading to internal suffocation. Carbon powder in the air, if breathed in, will settle on the lining of the lungs causing serious breathing difficulties.

10.2 Free Radical Substitution Reactions of Alkanes (SL/HL)

Alkanes will react with halogens (Cl2 or Br2) in the presence of ultra violet light.

The mechanism for reaction involves free radical substitution,

Initiation: The bond between the halogen atoms is quite weak. The UV light has sufficient energy to cause it to break so that each of the atoms has an unpaired non-bonded electron.

Cl – ClCl*+Cl*

This is called homolytic fission.

The highly reactive species created are called free radicals.

Since this step in the reaction mechanism involves the generation of reactive free radicals it is called the initiation step.

Propagation: These free radicals will react with inert alkane molecules to produce a halogenoalkane and hydrogen halide.

CH4+Cl*CH3*+HCl

CH3* + Cl2CH3Cl + Cl*

Since these steps in the reaction mechanism involve both using and generating free radicals, they are called propagation steps.

Termination: The possibility of two free radicals being used up can stop the reaction.

CH3*+Cl*CH3Cl

This step in the mechanism is called the termination step.

Show and label the steps in the mechanism for the free radical substitution of ethane with bromine.

10.3 Reactions of Alkenes (SL/HL)

The C=C bond in alkenes is said to be unsaturated.

The C=C bond is rich in electrons and tends to be attacked by positive species which are electron deficient (called electrophiles)

The C=C bond is not twice as strong as the C-C and so the conversion of a double into a single bond is relatively energetically favourable with a reasonably low activation energy.

The most common type of reactions of the alkenes involves the (electrophillic) addition of a species to the double bond. (The bond is then said to be saturated).

Common Addition Reactions

With Hydrogen (Hydrogenation)

E.g.

This reaction is industrially important: Hydrogen is added to unsaturated vegetable fats during the manufacture of margarine. This removes some of the C=C bonds and results in an increase in melting point, allowing the margarine to become a solid at room temperature.

With Bromine (Bromination)

E.g.

When a solution of bromine is added to an alkene, the orange colour disappears. This is used as a test for unsaturation (molecules containing C=C bonds).

With Water (Hydration)

E.g.

This reaction is industrially important in the production of ethanol from ethane (obtained by cracking crude oil) as an important alternative to the formation of ethanol through fermentation.

With Hydrogen Halides

E.g.

Note: In examples 3 and 4 it is important to note the symmetrical nature of the alkene. In an unsymmetrical alkene, there are two possible products. IB (SL/HL) does not require this level of understanding.

10.3 Reactions with other Alkenes (Polymerisation) (SL/HL)

An alkene (monomer) will undergo addition reactions with itself to form a long chain (polymer) under heat and pressure.

This process is called polymerisation.

Man-made polymers are called plastics.

E.g. ethene forms polyethane.

E.g. chloroethene forms polychloroethane. (PVC)

The reactions have huge industrial importance due to the desirable and diverse properties of plastics.

Show an equation that illustrates the conversion of propene to polypropene by polymerisation:

10.4Alcohols (SL/HL)

Alcohols can be classed according to their structure as primary, secondary or tertiary.

This class of structure plays an important role in the reactions of the alcohols.

What is the structural difference between a primary, secondary and tertiary alcohol?

Alcohols can be used as fuels and will combust readily in a supply of oxygen.

E.g. CH3OH + 1 ½ O2CO2 + 2H2O

Give an equation that shows the complete combustion of ethanol.

10.4 Oxidation ofPrimary Alcohols (SL/HL)

Primary alcohols can be easily oxidised using acidifiedpotassium dichromate solution [O](a strong oxidising agent see Topic 9).

This causes the chromium to change from +6 state (orange) to +3 (green)

Give the ionic half equation showing the action of this oxidising agent.

The initial product of this oxidation reaction is an aldehyde.

E.g. Ethanol.

Where [O] is used as a symbol for the oxidising agent.

To obtain the aldehyde in good yield it must be distilled from the reaction mixture and away from the oxidising agent as soon as it is produced. (aldehydes have a much lower boiling point than the unreacted alcohol)

Aldehydes can be further oxidised to carboxylic acid if they are kept in the reaction mixture with the oxidising agent. This process is called heating under reflux.

E.g. Ethanal.

Where [O] is used as a symbol for the oxidising agent.

Show all the steps, reactants and reaction conditions involved in the complete oxidation of methanol.

10.4Oxidation of Secondary and Tertiary Alcohols (SL/HL)

Secondaryalcohols:These are oxidised to ketones.

They cannot be further oxidised as there are no more H atoms that can be replaced.

E.g. propan-2-ol

Where [O] is used as a symbol for the oxidising agent.

Tertiary alcohols:Cannot be oxidised by a dichromate solution.

Show all the steps, reactants and reaction conditions involved in the oxidation of butan-2-ol.

10.5 Reactions of the Halogenoalkanes (SL/HL)

Halogenoalkanes can be classed according to their structure as primary, secondary or tertiary.

This class of structure plays an important role in the reactions of the halogenoalkanes.

Halogenoalkanes contain polar C-X bonds that make them easy to convert to other products.

Halogenoalkanes will undergo nucleophilic substitution(SN) reactions.

This involves the attack of a negatively charged species (called a nucleophile) on the slightly positive carbon atom bonded to the halogen.

This causes the polar C-X bond to break so that the halogen is substituted by the nucleophile.

E.g. 1. Reaction with NaOH

There are 2 mechanisms for nucleophilic substitution reactions, called SN1 and SN2. This stands for substitution nucleophilicfirst or second order.

SN1 Mechanism

Tertiary halogenoalkanes react in a SN1 mechanism.

The rate of this reaction is dictated only by one factor, the concentration of the halogenoalkane.

Rate = k [halogenoalkane]

Only the halogenoalkane is involved in the rate expression and therefore involved in the rate determining step (see topic 6 kinetics mechanisms)

This is therefore a first order reaction that is unimolecular process with a molecularity of one.

First stage (rate determining = slow)

The first stage of the mechanism involves the breaking of the C-X bond to give an intermediatecarbocation.
This is bond breaking is called heterolytic fission.
This is a slow process and so is the rate-determining step.

Second stage (Fast)

The second stage involves the nucleophile bonding with the carbocation intermediate. This happens very quickly. It therefore does not appear in the rate expression.

SN2 Mechanism

Primary halogenoalkanes react in this way.

The rate of reaction is dictated by two factors. By both the concentration of the halogenoalkane and the concentration of the nucleophile.

Rate = k[Halogenoalkane] [Nucleophile]

Both the halogenoalkane and the nucleophile are involved in the rate expression and therefore both are involved in the rate determining step (see topic 6 kinetics mechanisms)

This is a second order reaction, a bimolecular process with a molecularity of two.

This mechanism involves the simultaneous attack of the nucleophile and loss of the halogen.

An activated complex is formed during the process.

E.g.

Secondary halogenoalkanes react by a combination of both SN1 and SN2 mechanisms.

10.6 Reaction Pathways

Deduce the two stage reaction pathways for the following organic conversions, given the starting materials and products:

Include the reagents and conditions required.

Starting MaterialProduct

PropenePropanone

MethaneMethanol

BromoethaneEthanal

Extension:

EtheneTetrachloroethane

20.2 Nucleophilic Substitution Reactions of Halogenoalkanes (HL ONLY)

Nucleophiles are chemical species that are able to donate an outer electron pair to an electron deficient species.

In halogenoalkanes the electron deficient species is the carbon atom in the polar C-X bond.

Substitution occurs when the C-X bonds breaks (heterogeneously) and a new bond forms with the nucleophle.

Factors Affecting the Rate of Nucleophilic Substitution

Identity of the halogen.

Bond breaking is an endothermic process.

The stronger the C-X bond, the more energy required to break this bond, the less energetically favourable the reaction.

C-I bonds are the weakest and break the easiest, then C-Br bonds and C-Cl bonds are the strongest

Iodoalkanes react the fastest then bromoalkanes and then chloroalkanes.

Type of Nucleophile.

Bond forming is an exothermic process.

The stronger the bond formed between the nucleophile and the halogenoalkane, the more energetically favourable the reaction.

Hydroxide ions are therefore better nucleophiles than water because no bond breaking is required in the process.

Hydroxide (OH-) ions react faster than water (H2O).

The type of halogenoalkane.

The reaction involving intermediate carbocations (SN 1) reacts faster than the formation of the activated complex (SN2). This is because the formation of an activated complex requires a larger activation energy (collision energy).

Tertiary halogenoalkanes react faster then secondary and primaryreact the slowest.

20.2 Other Nucleophilic Reactions of Halogenoalkanes (HL ONLY)

Nucleophiles have non bonded electron pairs that they are able to donate. The easier they can do this, the more effective the nucleophile.

NH3, CN-, OH- are ‘good’ nucleophiles. Cl- and H2O are not as effective nuclephiles.

The nuclephile attacks the slightly positive carbon atom bonded to the halogen.

This causes the polar C-X bond to break so that the halogen is substituted by the nucleophile.

E.g. 1.Reaction with NH3 to give amines.

E.g. 2.Reaction with KCN to give nitriles.

Note: The nitriles produced in these reactions can be reduced using hydrogen (in the presence of a Ni catalyst) to produce an amine.

E.g.

Substitution of a halogenoalkene with CN- and then reduction of the product with H2 (Ni) is an especially useful reaction pathway, as it increases the length of the carbon chain by one.

These examples of nucleophilic substitution can also be described in terms of SN1 and SN2 mechanisms.

What type of mechanism would be used in the reaction between iodoethane CH3CH2I and potassium cyanide KCN?

Give the rate expression for this reaction.

Draw the mechanism expected in the nucleophilic substitution reaction between bromoethane CH3CH2Br and ammonia NH3. Label the rate determining step.

20.3 Elimination Reactions of Halogenoalkanes (HL ONLY)

Refluxing bromoalkanes with a conc solution of hydroxide(OH-) ions (dissolved in ethanol).

Results in the elimination of water to give an alkene.

E.g. 2-bromopropane and NaOH(in ethanol).

The OH- ion acts as a base. Removing a hydrogen ion to give water and resulting in the release of the bromide ion from the molecule.

Mechanism:

Show the mechanism for the reaction of 1-bromoethane reacting with conc KOH dissolved in ethanol.

20.4 Condensation Reaction (HL ONLY)

a) Carboxylic Acids and Alcohols

Alcohols react with carboxylic acids to producesweet smelling compounds called esters.

During the reaction, the alcohol and acid condense together and eliminate water. This is therefore referred to as a condensation reaction.

E.g. Ethanoic acid and methanol.