Organic ChemistryWhat do we mean by “organic” chemistry?What do the following things below have in

common?

Organic-dictionary definitionOxford English dictionary says:“organic” means derived from living matter,

i.e. not produced artificially For example, “organic food” refers to the

food that is produced without the use of chemical fertilisers, pesticides or other artificial chemicals.

Organic- chemical definition Of all the elements in the periodic table, one is

much more versatile (stands out) from the restCarbon can form more compounds than any

other elementChemical definition: Organic chemistry is the

study of compounds containing carbon (other than simple binary compounds and salts) and chiefly or ultimately of biological origin.

Question: Is carbon itself organic or inorganic?

Carbon-revisited hopefully!In group 4 of the periodic tableElectronic configuration: 2,4 (SL) or 1s2

2s2 2p2 (HL)Has a valency (combining power) of 4 with

4 valence (outer shell) electronsAchieves the noble gas configuration of

neon by forming 4 covalent bonds Take methane for example: carbon is

bonded covalently to 4 hydrogen atoms to achieve its octet

Carbon- continuedThe valency of 4 gives rise to a unique property called

catenationCatenation: spontaneous linking of atoms of certain chemical

elements, such as carbon atoms, to form long chains or stable rings by forming covalent bonds with itself

C-C bond enthalpy = 348 kJ mol-1

C-H bond enthalpy = 412 kJ mol-1

Catenation allows billions of organic compounds to be formed Carbon compounds > ∑ all other compounds of all the

elements in the periodic table except hydrogen (since almost all organic compounds also contain hydrogen)

7 million Organic Compounds1.5 million Inorganic Compounds

Organic chemistry- link to natureLife is based on organic compoundsThe 4 major classes of biomolecules are all

organic (contain carbon)Carbohydrates, Proteins, Lipids & Nucleic

AcidsOrganic chemistry is an excellent foundation

for biochemistryBiochemistry is just APPLIED organic

chemistry!Just like physics is applied maths!

HydrocarbonsA compound which consists of carbon &

hydrogen ONLYAre the basis for most organic compoundsIncludes alkanes, alkenes, alkynes (aliphatic

compounds), & arenes (aromatic compounds)

Organic compound families- homologous series

Can be grouped into different families with a common functional group (part of the molecule which gives rise to common reactivity)

Homologous series: a group of organic compounds that follow a regular structural pattern and have the same general molecular formula, differing only by the addition of a methylene, -CH2- group, they have almost identical chemical properties, with physical properties e.g. boiling point increasing gradually as the number of carbon atoms increase

Homologous seriesTake the alkanes for example:The 1st 4 are methane, ethane, propane,

butane

Homologous series- properties1) Successive compounds differ from each

other by a -CH2- unit (methylene group)2) The compounds can all be represented by

a general formula (e.g. alkanes CnH2n+2; if n=3 then the formula is C3H8

3) The compounds have similar chemical properties

4) Successive compounds have physical properties that vary in a regular manner as the number of carbon atoms increases

Homologous series- general trends in physical properties

For the alkanes: As the number of carbon atoms increase

the melting/boiling points gradually increase

This is caused by an increase in the molar mass of the molecule and hence there is greater chance of more temporary dipoles being induced in the case of hydrocarbons (Van der Waals forces)

Graph of boiling points for the first 10 alkanes

Curve is initially quite steep, as for small molecules, the addition of an extra carbon has a proportionally larger effect on the molar mass (eg, from CH4 to C2H6 there is an increase of 97.5%) & hence on the strength of the van der Waals’ forces

As the length of the carbon chain increases, the percentage change in molar masses becomes progressively smaller (there is a 10.9% increase in molar mass from C9H20 to C10H22, hence the curve flattens

The trend is the same for other physical properties, such as density & viscosity for the same reasons

Empirical, molecular, structural (condensed & full) formulaeUsing ethane, C2H6, as an exampleEmpirical formula: simplest whole number ratio of atoms

e.g. CH3

Molecular formula: actual number of atoms of each type present in a molecule e.g. C2H6

Full structural formula: show the relative positioning of all the atoms in a molecule & the bonds between them

Condensed structural formula: omits bonds which can be assumed & groups atoms together e.g CH3CH3

Shapes of Alkanes “Straight-chain” alkanes have a zig-zag orientation in 3-D

Formulae- worked exampleWrite the empirical, molecular, full structural

& condensed structural formula for ethanol, CH3CH2OH

Answer:Empirical: C2H6O

Molecular: C2H6OFull structural:

Further functional groupsAlkenes: CnH2n e.g. ethene C2H4

Alkynes: CnHn e.g. ethyne C2H2

Alcohol: CnH2n+1OH e.g. methanol CH3OHAldehydes: RCHO, e.g. methanal (formaldehyde) HCHOKetones: RCOR’ (R’ can be the same alkyl group as R or

different e.g. propanone (acetone), CH3COCH3

Carboxylic acids: RCOOH, e.g. methanoic (formic) acid, HCOOH

Haloalkanes: RX (X=F,Cl,Br,I) e.g. iodomethane, CH3I

Amines: RNH2 e.g. methylamine, CH3NH2

Esters: RCOOR’ e.g. methyl methanoate HCO2CH3

Arenes: based on the phenyl group (C6H5-) e.g. benzene C6H6

IsomersCompounds with the same molecular formula

but with different arrangement of atoms in the molecule

For the alkanes: from butane (4 carbon atoms), there is more

than 1 structure possibleC4H10 can refer to either butane,

CH3CH2CH2CH3 or methylpropane, CH3CH(CH3)CH3

Structural IsomersDifferent connectivity of atoms E.g.: Butane & methylpropane are structural isomers

of each other

3 types:1)Chain isomerism2)Position isomerism3)Functional group isomerism

Chain isomerismDifferent arrangement of the carbon skeletonFor example butane & methylpropane are

chain isomers of each other Have similar chemical properties but

different physical ones; the branched isomer has a lower melting/boiling point than the straight-chained one

C C C C

H

H

H

H

H

H

H

H

H

H

C C C

H

H

H

CH3

H

H

H

H

butane methylpropane

Position IsomerismHave functional group placed at a different

position along the carbon skeletonE.g. 1-bromopropane & 2-bromopropaneHas similar physical & chemical properties

C C C

Br

H

H

H

H

H

H

H C C C

H

H

H

Br

H

H

H

H

1-bromopropane 2-bromopropane

Functional group IsomerismThe molecules have different functional groups

(hence different chemical properties)E.g. ethanol & methoxymethane all have the

same molecular formula of C2H6O, however ethanol is an alcohol while methoxymethane is an ether

ethanol methoxymethane

Naming organic compounds International Union of Pure & Applied Chemistry (IUPAC)

devised nomenclature method It is irregular for up to 4 carbon atoms, just have the prefixes

meth-, eth-, prop-, but-From 5 carbon atoms & above the naming becomes

systematic, like those of geometrical shapes, e.g. pent- (pentagon), hex- (hexagon), hep- (heptagon), oct- (octagon) etc

The name of any organic compound is usually made up of 3 parts: prefix (substituents), stem (number of carbon atoms in main chain) & suffix (homologous series of main carbon chain)

e.g. methylpropane methyl- CH3 substituent

prop- 3 carbon atoms in main chain -ane- belongs to the alkane homologous series

MonkeysEatPeeledBananas

Counting to Ten in Organic01 = meth Mother02 = eth Enjoys03 = prop Peanut04 = but BUTter05 = pent PENTagon06 = hex HEXagon or HEX nut07 = hept HEPTember (Roman sept is Greek

hept)08 = oct OCTober09 = non NONember (Roman nov is

Greek non)10 = dec DECember

Naming side chainsIf there are alkyl groups (R groups) attached in

isomers, the prefix for the alkyl groups must be used

Prefix Name of alkane

Name of Alkyl group

Structure of alkyl group

Meth- Methane Methyl CH3-

Eth- Ethane Ethyl CH3CH2-

Prop- Propane Propyl CH3CH2CH2-

But- Butane Butyl CH3CH2CH2CH2-

Pent- Pentane Pentyl CH3CH2CH2CH2CH2-

Hex- Hexane Hexyl CH3CH2CH2CH2CH2CH2-

Steps in naming alkanes- step 1 For the molecule CH3CH(CH3)CH(CH3)CH3

Identify the longest continuous carbon chain (This may not be the most obvious, straight one), this gives the stem, given by the 4 blue carbon atoms

Steps in naming alkanes- step 2CH3CH(CH3)CH(CH3)CH3

Identify & name the side-chains/substituent groups as the prefix of the name.

Here there are 2 different methyl groups

Steps in naming alkanes- step 3CH3CH(CH3)CH(CH3)CH3

Where there is more than 1 side-chain of the same type, like here, use the prefixes di-, tri-, tetra- and so on, to indicate this.

If there are several side-chains within a molecule, put them in alphabetical order, seperated by dashes.

There are 2 methyl groups- hence the prefix is dimethyl

Steps in naming alkanes- step 41CH3

2CH(CH3)3CH(CH3)4CH3

Identify the position of the side chains. The carbon chain is numbered from the

end which will give the substituent groups the smallest number.

Here 1 methyl group is attached to carbon number 2; the other to carbon number 3.

The numbers precede the name and each digit is separated by a comma from the next digit

Hence the name of this compound is 2,3-dimethylbutane

Some exercisesName the following compound:

CH H

C C CC C

C

CH

H

H H H

H H

H

H

H

H

H

H

H

H

H

2 methyl groups

2 methyl groups

1

2

3 4 5 6

1234

5

6

Using the green numbering gives the lower substituent number of 3,3-dimethylhexane

Using the orange numbering gives the higher substituent number of 4,4-dimethylhexane

Hence green is correct

AnswerHence the name of the molecule is 3,3-

dimethylhexane

CH H

C C CC C

C

CH

H

H H H

H H

H

H

H

H

H

H

H

H

H

Methylpropane

methylbutane

dimethylpropane

IUPAC common

IUPAC: 2,3,5,4,6-PentahydroxyhexanalCommon: glucose

IUPAC: 3-carboxy-3-hydroxypentanedioic acidCommon: citric acid

Alkene isomersHave the general formula CnH2n

Have the C=C functional group within the chain

Simplest alkene is ethene, C2H4

If molecule is longer than 3 carbon chains, the double bond can be in more than 1 position

ethene propene

Naming alkene isomersSame 1st 4 steps as alkanes except the

name (suffix) ends in -ene instead of -aneStep 5- The position of the double bond

C=C is shown by inserting the numb er of the carbon atom at which C=C starts

E.g. for the isomers of C4H8

CH3CH2CH=CH2 is but-1-ene

CH3CH=CHCH3 is but-2-ene

CH2=C(CH3)2 is 2-methylprop-1-ene

Practice problemWhat is the name of the following alkenes:1)CH3CH=CHCH2CH2CH3

2)CH3CH2CH(CH3)CH=CH2

3)CH2=C(CH3)CH2CH=CH2

Answer:1)hex-2-ene 2)3-methylpent-1-ene3)2-methylpent-1,4-diene

Naming alcohols (ROH)Always end in –olSimplest is methanolLike alkenes, the position of the –OH group must be

specified after ethanole.g. CH3CH2CH2OH is propan-1-ol

CH3CH(OH)CH3 is propan-2-ol

ethanol Propan-2-olPropan-1-ol

Practice problemsName the following alcohols: 1) CH3C(OH)(CH3)CH2CH2CH3

2)

Answers1)2-methylpentan-2-ol2) propan-1,2,3-triol

Naming aldehydes (RCHO)Always end in –al NOT (ol)!Simplest is methanal-CHO group is always at the end, so this carbon

must be carbon 1 so unnecessary to specify location

E.g.

methanal ethanal propanal

Practice problemsName the following aldehydes:1)CH3CH2CH2CHO

Answer:

1)butanal2)butanal

CH2CH3

2) HCOCH2

Naming ketone (RCOR’)Always ends in suffix –oneSimplest is propanone (acetone)The C=O (carbonyl) group can be inserted anywhere

along the hydrocarbon chain except at the end (why?)After butanone, the position of the carbonyl group

must be shownE.g pentan-2-one & pentan-3-one

propanone butanone

Practice problems- ketonesName the following ketones:1) CH3COCH2CH2CH2CH3

2) CH3CH2CH2COCH3

OC CH3

CH2CH2CH3

Answer:1)hexan-2-one2)pentan-2-one3)pentan-2-one

Naming carboxylic acid (RCOOH)End in –oic acidLike aldehydes, COOH is always the

terminal group & hence this carbon is always carbon number 1

Methanoic acidEthanoic (acetic) acid

Propanoic acid

Practice problems- carboxylic acidsName the following carboxylic acids:1)CH3CH2CH2CO2H

2)HOOCCH2CH2CH3

3)CH3CH2CH(CH3)CH2COOH

Answer:1)butanoic acid2)butanoic acid3)3-methylpentanoic acid

Naming Haloalkanes (RX)Have the prefix halo- e.g fluoro-, chloro-, bromo-. Iodo- Involve substituting a halogen atom into an alkane Same numbering as alcohols & ketonesNumbers must be used after 2 carbon (from propane)

atomsE.g. CH3CHClCH3 is 2-chloropropane

CH3CH2CHClCH2CH2Br is 1-bromo-3-chloropentaneUse the prefixes, di-, tri-, tetra- is there are more than 1

of the same type of halogen atom e.g.

chloromethane dichloromethane Trichloromethane (chloroform)

Tetrachloromethane (carbon tetrachloride)

Practice problemsName the following haloalkanes:1) CH3CH2FCHCH2CH3

2) CH3CH2CH2CH2CHBrCH3

Answers:1)3-fluoropentane2)2-bromohexane

Further functional groupsHomologous

series

Functional group

Prefix Suffix Example(s)Structural formulae

Amine R-NH2 Amino- -amine

Methylamine CH3NH2

2-aminobutaneCH3CH(NH2)CH2C

H3

EstersRCOOR’

AlkylAlkanoat

e

Ethyl ethanoate CH3CO2CH2CH3

Propyl ethanoate

CH3CO2CH2CH2CH

3

Aromatic compounds

C6H5- phenyl- -benzeneBenzene C6H6

methylbenzene C6H5CH3

Primary compoundsPrimary carbon atom attached to a

functional group & also to at least 2 H atoms & 1 alkyl (R) group e.g.

C OH

H

R

H

General structure

C OH

H

CH3

H

ethanol

Secondary compoundsSecondary carbon atom attached to a

functional group & just 1 H atoms but 2 alkyl (R) group e.g.

C OH

H

R

R’

C OH

H

CH3

CH3

C OH

H

CH3

CH2CH3

General Propan-2-ol Butan-2-ol

Tertiary compoundsTertiary carbon atom attached to a

functional group & also 3 alkyl (R) groups with no H attached e.g.

C OH

CH2CH3

CH3

CH2CH3

C OH

R’’

R

R’

C OH

CH3

CH3

CH3

General 2-methylpropan-2-ol 3-methylpentan-3-ol

C OH

CH2CH2CH3

CH3

CH2CH3

3-methylhexan-3-ol

Practice questionsAre the following molecules primary,

secondary or tertiary?1)1-chlorobutane2)2-bromobutane3)2-chloro-2-methylbutane

1-chlorobutane

C* attached to 2 H & 1 alkyl groupHence it’s primary

C* CH2CH2CH3

H

H

Cl

2-bromobutane

C* attached to 1 H & 2 alkyl groupsHence it’s secondary

C* CH2CH3

H

CH3

Br

2-bromo-2-methylbutane

C* attached to no H & 3 alkyl groupsHence it’s tertiary

C* CH2CH3

CH3

CH3

Br

VolatilityMeasure of how easily a compound evaporatesDepends on intermolecular forces as kinetic

energy which hold the molecules together must be overcome

Compounds with stronger intermolecular forces will evaporate less readily, hence have higher boiling points

The following factors influence volatility:1)Molecular size2)Shape of molecule (branched/linear)3)Functional group

Volatility 2Compounds with longer carbon chains & hence

greater molar mass have higher boiling points & lower volatility

A molecule with greater molar mass has more electrons present hence more temporary dipoles can be induced leading to more Van der Waals’ forces

The early members of a series e.g. methane to octane are gases & liquids respectively, at room temperatue

The latter members are more likely to be solids e.g. C40-

Volatility 3Also depends on the number of points of

contact between moleculesBranched isomers usually have lower

boiling point/higher volatility than straight-chained linear ones

Less surface area for attraction in branched isomers

Hence methylpropane has lower boiling point than butane

Volatility 4Molecules with functional groups that

contain hydrogen bonds are likely to have the highest boiling point e.g. ethanol (78C)

Molecules with functional groups that contain dipole-dipole attractions are likely to have intermediate boiling points,e.g. ethanal (20.2 C)

Molecules which just have van der Waals’ forces usually have the lowest boiling point e.g propane, (−42.1 °C)

Fair testWhen comparing molecules for boiling

points/volatility in different homologous series, it is important to compare molecules of similar molar mass

This rules out molar mass as being a factor for different boiling points since it becomes the “constant” variable

Hence ethanol (Mr=46) was compared with ethanal (Mr=44) & propane (Mr=44)

Solubility in water Depends on nature of functional group & length of hydrocarbon chain Polar functional groups with hydrogen bonding e.g. ethanol are usually the

most soluble = most hydrophilic Hydrocarbon chains e.g. alkanes are likely to be least soluble = most

hydrophobic Depends on relative length of hydrocarbon chain e.g. very long hydrocarbon chain can overwrite effects of polar functional

group e.g. for the very long chained fatty acids which are solids i.e. butter not oil

AlkanesHydrocarbonsGeneral formula= CnH2n+2

Are saturated hydrocarbonsSaturated- an organic molecule which only

has C-C single bonds & no C=C multiple bonds

Unreactive in general

Low reactivity of alkanesOnly have strong C-C & C-H bondsNeed a large energy input to break these strong

bondsHence stable/unreactive under most conditionsC-C & C-H bonds nonpolar as have similar

electronegativities (EN= 0.4)Hence no electron-rich or electron-deficient sites2 main reactions:1)Combustion2)Halogenation

Combustion of alkanesVery exothermicDue to high strength of C=O in carbon dioxide & O-H

bond in waterBurns in excess oxygen to give carbon dioxide & water CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ mol-1

In a limited supply of oxygen, carbon monoxide & water are produced:

CH4(g) + 1.5O2(g) CO(g) + 2H2O(l)• In an extremely limited supply of oxygen, carbon itself

produced with waterCH4(g) + O2(g) C(s) + 2H2O(l)

Combustion of alkanes- questionsWrite the equation for the reaction of propane in

excess oxygen, limited oxygen & extremely limited oxygen:

Excess oxygen: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Limited oxygen: C3H8(g) + 3.5O2(g) 3CO(g) + 4H2O(l)

Trace amounts of oxygen: C3H8(g) + 2O2(g) 3C(s) + 4H2O(l)

Halogenation of alkanesIf subjected to UV light, methane reacts with chlorine

to form chloromethane & hydrogen chloride

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) In excess chlorine, further reaction takes place to

produce dichloromethane, trichloromethane (chloroform) & tetrachloromethane

Dichloromethane: CH4(g) + 2Cl2(g) CH2Cl2(g) + 2HCl(g)

Trichloromethane: CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

Tetrachloromethane: CH4(g) + 4Cl2(g) CCl4(g) + 4HCl(g)

Halogenation of alkanes-problemsWrite the equation for the reaction between

ethane and 1 mol of bromine:

C2H6(g) + Br2(g) C2H5Br(l) + HBr(g)

General reaction:Alkane + halogen haloalkane + hydrogen

halideRH + X2 RX + HX

R= alkyl group X= halogen atom

Halogenation of alkanes - mechanism

Substitution reaction (chlorine atoms replace hydrogen atoms in the methane molecule)

Substitution reaction- a reaction where 1 atom or group of atoms is replaced by another atom or functional group

Mechanism is free radical substitutionDegree of substitution hard to control, often

mixture of products formed e.g.Ethane reacts with excess bromine to form a

mixture of dibromoethanes: C2H6(g) + 2Br2(g) C2H4Br2(g) + 2HBr(g)

Free radicalsSpecies with unpaired electron e.g. Cl., Br.

Extremely reactive Formed by the homolytic fission of a

molecule: Cl-Cl 2Cl. (produced by action of UV light) Homolytic fission- the breaking of a covalent

bond so that 1 electron from the bond is left on each atom, resulting in the formation of 2 free radicals

Free radical substitution mechanism • Is a substitution reaction as a halogen atom

replaces a hydrogen/alkyl group.• 3 steps:1)Initiation (net increase in radicals)2)Propagation (chain reaction- amount of

radicals stays the same)3)Termination (net decrease in radicals)

Mechanism in detailInitiation: Cl-Cl(g) 2Cl.(g)

Propagation: Cl.(g) + CH4(g) .CH3(g) + HCl(g)

.CH3 (g) + Cl2(g) CH3Cl(g) + Cl.(g)

Termination: Cl.(g) + Cl.(g) Cl2(g) Cl.(g) + .CH3(g) CH3Cl(g) .CH3(g) + .CH3(g) C2H6(g)

Overall reaction: CH4(g) + Cl2(g) CH3Cl(g) + HCl(l)

UV

ExerciseWrite the free radical reaction mechanism for the

reaction of bromine with ethane:Initiation: Br-Br(g) 2Br.(g)

Propagation: Br.(g) + C2H6(g) .C2H5(g) + HBr(g) . C2H5 (g) + Br2(g) C2H5Br(g) + Br.(g)

Termination: Br.(g) + Br.(g) Br2(g) Br.(g) + . C2H5(g) C2H5Br(g) . C2H5(g) + .C2H5(g) C4H10(g)

Overall reaction: : C2H6(g) + Br2(g) C2H5Br(l) + HBr(g)

AlkenesUnsaturated hydrocarbons with C=C double bondUnsaturated molecule- a molecule with 1 or more C=C

double bondsGeneral formula: CnH2n

C=C double bond stronger & shorter than C-C single bond

Double bond consists of a stronger sigma () bond & weaker pi () bond

Quite reactive due to double bondTake part in addition reactionsAddition reaction- a reaction where 2 (or

more)molecules combine together to form a single molecule

Hydrogenation of alkenesAlkene + hydrogen alkaneE.g.C2H4(g) + H2(g) C2H6(g)

H

C C

H

H

H

CC

H

HH

H

+ H2 H

H

ethene ethane

Ni, 180C

Hydrogenation of alkenes-exerciseWrite the reactions for the hydrogenation of

but-1-ene & but-2-ene in the presence of a nickel catalyst & high temperature.

1) With but-1-eneH2C=CHCH2CH3 + H2 CH3CH2CH2CH3

2) With but-2-eneCH3CH=CHCH3 + H2 CH3CH2CH2CH3

NB the original position isomerism is lost in this reaction

Used to convert cooking oils to margarine

unsaturated saturated

Halogenation of alkenesAlkene + Halogen DihaloalkaneAddition reaction

C C

Br

H

H

CC

H

HH

H

+ Br2 H

H

ethene 1,2-dibromoethane

Br

Halogenation of alkenes 2Alkene + Halogen Dihaloalkane

C C

Cl

H

H

CC

CH3

HH

H

+ Cl2 CH3

H

propene 1,2-dichloropropane

Cl

Reaction with hydrogen halidesAlkene + Hydrogen halide haloalkaneAddition reaction

C C

Cl

H

H

CC

H

HH

H

+ HCl H

H

ethene chloroethane

H

Reaction with hydrogen halides 2Reactivity order HI>HBr>HClWeaker strength (longer length) of H-X bond as

descending group 7

C C

Br

CH3

H

CC

CH3

HH

CH3

+ HCl

CH3

H

but-2-ene 2-bromobutane

H

Hydration of alkenesReaction where a water molecule reacts with an

unsaturated compound in an addition reaction Alkene + water alcoholRC=CR + H2O ROH

In Industry:

CC

H

HH

H

+ H2O

ethene ethanol

C C

H

H

H

OH

H

HH3PO4

300C60 atm

C C

H

H

H

CC

H

HH

H

+ H2SO4 OSO3H

H

ethene Ethyl hydrogensulfate

H

Also addition reaction- H+ & HSO4- added across double bond

C C

H

H

H

OSO3H

H

Ethyl hydrogensulfate

H

H2OC C

H

H

H

OH + H2SO4

H

H

ethanol

Note sulphuric acid is reformed- hence it’s a catalyst

Distinguishing between alkanes & alkenesUse halogenation addition reaction to distinguish

between these 2 functional groups

Used as a test for unsaturation

Addition polymerization of alkenesMonomer PolymerAlkene Polyalkene

MonomerEthene

PolymerPoly(e)thene

n can be 1000 or more

Polymerisation

Polymerisation problemDraw the reaction for the polymerisation

of propene:n(Propene) (polypropene)n

CC

CH3

HH

H

n CC

CH3

HH

H

n

propene polypropene

CC

Cl

HH

H

n CC

Cl

HH

H

n

Chloroethene(vinyl chloride)

Polychloroethene(PVC)

CC

C6H5

HH

H

n CC

C6H5

HH

H

n

phenylethene(styrene)

Polyphenylethene(Polystyrene)

Ethene (from cracking)

React with steam React with benzenepolymerise

ethanol

React with chlorine

polymerise

Poly(e)thenePhenylethene

(styrene)Chloroethene(vinyl chloride)

polymerise

Poly(phenylethene)(polystyrene)

Polychoroethene(PVC)

AlcoholsGeneral formula: CnH2n+1OHHave polar hydroxyl (-OH) groupSoluble in water due to hydrogen bonding

from hydroxyl group2 types of main reactions: 1) Combustion 2) Oxidation

Combustion of alcoholsAlcohol + oxygen carbon dioxide + waterROH + O2 CO2 + H2O Methanol: 2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(l) small Hc

Ethanol: 2C2H5OH(l) + 7O2(g) 4CO2(g) + 6H2O(l) medium Hc

Propanol: 2C3H7OH(l) + 9O2(g) 6CO2(g) + 8H2O(l) large HcAlcohol Ratio

Alcohol CO2

Methanol 1 1

Ethanol 1 2

Propanol 1 3

As the alcohol becomes larger the CO2:alcohol ratio is larger, hence more energy is released (due to the strength of C=O bond)

Combustion of alcohols-exercisesWhich of the following alcohols is likely to

release more energy upon complete combustion, hexanol or butanol

Answer:Hexanol as has higher molar mass.

Alcohols vs alkanes for fuelsIn general, alkane hydrocarbons release more

energy upon combustion as they are in a more reduced state (more hydrogen atoms attached)

Hence petrol (mainly octane) releases more energy than ethanol upon combustion

However, alcohols have advantage of being able to be produced from renewable sources e.g. fermentation

Glucose ethanol + carbon dioxideC6H12O6 → 2C2H5OH + 2CO2

Oxidation of alcoholsHydroxyl (-OH) group can be oxidised Oxidation depends on nature of alcohol

(primary, secondary or tertiary)Oxidising agent is potassium dichromate (VI),

K2Cr2O7

Cr2O72- is reduced to Cr3+

Hence alcohol is the reducing agent

Oxidation of alcohols-primary alcoholsOxidised in 2 stages:1)Aldehyde (RCHO)2)Carboxylic Acid (RCOOH)For ethanol:CH3CH2OH + [O] CH3CHO + H2O

H

C C

H

H

H

OH

H

H+/Cr2O72-

[O]C C

H

H

HH

O

+ H2O

Ethanol C2H6O Ethanal C2H4O

NB 2H atoms removed hence oxidation

Oxidation of alcohols-primary alcohols- stage 2CH3CHO + [O] CH3COOH

H+/Cr2O72-

[O]

C C

H

H

HH

O

ethanal

C C

H

H

HOH

O

Ethanoic acid

C2H4O C2H4O2NB: 1 O atom added hence oxidation

Oxidation of alcohols-primary alcohols-controlling the degree of oxidationHow do we control the reaction so that the aldehyde

is formed instead of the carboxylic acid or vice versa?If the aldehyde is the desired product:Heat reaction mixture in excess alcohol with

distillation apparatus (to distill off aldehyde as it’s being formed)

If carboxylic acid is the desired product:Heat reaction mixture under reflux for a longer

period of time with excess oxidizing agent(a vertical condenser to ensure the aldehyde drops back into reaction vessel for further oxidation to the carboxylic acid)

Distillation Reflux

Summary of oxidation of primary alcoholsIn general:

R C

O

H

H+/Cr2O72-

[O]R C OH

H

HH+/Cr2O7

2-

[O]R C

O

OH

Primary alcohol

aldehyde Carboxylic acid

Oxidation of secondary alcoholsOnly 1 product formed (ketone) as there is

only 1 oxidisable hydrogen attached to the secondary carbon

e.g. with propan-2-ol:CH3CH(OH)CH3 + [O] CH3COCH3 + H2O

C OH

H

CH3

CH3

H+/Cr2O72-

[O]C CH3

O

CH3

Propan-2-ol (C3H8O) Propanone (C3H6O)

2 H atoms lost hence oxidation

C OH

H

R

R’

H+/Cr2O72-

[O]C R’

O

R

Secondary alcohol

Ketone

Oxidation of tertiary alcoholsCannot be oxidised as there is no hydrogen

atom attached to the hydroxyl carbon.E.g. for 2-methylpropan-2-ol

NB: tertiary alcohols can only be oxidised under harsh conditions where the carbon skeleton is broken

C OH

CH3

CH3

CH3

H+/Cr2O72-

[O]

No oxidation possible as no H atom on C atom bonded to alcohol group No colour change

Cr2O72- stays orange

Tests to distinguish between aldehydes & ketones All have carbonyl (C=O) group To distinguish aldehydes/ketones from other groups:1) Aldehydes & ketones both form orange precipitate with 2,4-

dinitrophenylhydrazine2) precipitate can be recrystallised & its melting point found (the melting

point of the crystals can be used to identify the particular aldehyde/ketone)

To distinguish aldehydes from ketones:1)Treat the sample with Fehling’s solution (alkaline copper(II) sulfate) or

Tollens’ reagent (silver nitrate in ammonia)2) Only aldehydes react with them; the aldehyde is oxidsed to the

carboxylic acid as an orange-brown precipitate of copper(I) oxide is formed with Fehling’s solution & a silver mirror is formed with Tollens’ reagent as metallic silver is deposited on the side of the test tube

3) Nothing happens with ketones

Fehling’s solution- the test tube on the right has aldehyde as an orange precipitate is formed

Tollen’s reagent- the test tube on the left has aldehyde as a silver mirror is formed

HaloalkanesGeneral formula: CnH2n+1X (X=halogen atom,

F, Cl, Br, I)Are very useful as can be used to synthesize

an array of other organic moleculesUsually oily liquids

Halogenoalkanes - substitution reactions • X replaced by another group• C-X bond reactive due to the polarity

difference (X is more electronegative than C)• Carbon atom attached to halogen has partial

positive charge (electron deficient) & is prone to attack by electron-rich species (nucleophiles)

Nucleophiles Nucleo + phile

A species (molecule/anion) which has a lone pair of electrons which can be donated to an electron-deficient centre in an organic molecule to form a coordinate (dative covalent) bond

Examples include, -OH, H2O, NH3

Nucleus loving

Mechanisms for nucleophilic substitutionThe substitution of an group/group of atoms

with a nucleophile as the attacking species; can occur via an SN1 (subsitution nucleophic unimolecular) or SN2 (subsitution nucleophic bimolecular) mechanism

Mechanism depends on nature of haloalkane (primary, secondary, tertiary)

Primary haloalkanes- mechanisms for nucleophilic substitution Take the following reaction: CH3CH2Br(aq) + OH-(aq) CH3CH2OH(aq) + Br-(aq)

Rate equation found to be rate = k[CH3CH2Br(aq)] [OH-(aq)] Hence 2 species are involved in the rate-determining step, so it’s

bimolecular, hence SN2 In the transition state, the C-Br bond is broken at the same time

as the C-O bond is being formed

Hydoxide ion (nucelophile) & bromoethane

Transition state

Ethanol & leaving group (Br-)

Tertiary haloalkanes- mechanisms for nucleophilic substitutionTake the following reaction:CH3C(CH3)2Br(aq) + OH-(aq) CH3C(CH3)2OH(aq) + Br-(aq)

Rate = k[CH3C(CH3)2Br(aq)]

Hence it’s unimolecular as only 1 species involved in rate determining step

SN1 mechanism

Different mechanism caused by steric bulk by 3 methyl groups- the nucelophile (OH-) cannot approach the electron-deficient carbon atom

2-bromo-2-methylpropane

SN1 mechanism Step 1 (rate-determining step): Br- departs as the leaving group; heterolytic fission

Step 2: nucleophile attacks carbocation intermediate

carbocation

Some definitionsCarbocation- an organic ion with a positive

charge on an electron-deficient carbon atomHeterolytic fission- the breaking of a covalent

bond so that 1 of the atoms/groups takes both of the bonding electrons & becomes negatively charged, leaving the other atom/group positively charged

Compare with homolytic fission (met earlier during the free radical mechanism of alkanes)

Homolytic fission- the breaking of a covalent bond so that 1 electron from the bond is left on each atom, resulting in the formation of 2 free radicals

Other factors which favour SN1With tertiary haloalkanes, the carbocation

intermediate is stabilised by the positive inductive effect of the 3 alkyl groups (alkyl groups are have an electron-donating effect) which help to reduce the positive charge on the positive carbon e.g.:

Tertiary > Secondary > Primary

Nuceleophilic substitution of secondary haloalkanes & relative reactivity Go via a mixture of SN1 & SN2 mechanisms,

depending on the reaction conditions, or some intermediate mechanism

Relative reactivity of different haloalkanes depends on the strength of the C-X bond

Strength: C-F > C-Cl > C-Br > C-IReactivity: C-I > C-Br > C-Cl > C-F

Reaction pathwaysIn organic chemistry, usually a desired

product cannot be made from available starting materials (reactants) in a single step; hence the need for reaction pathways

The production of new organic compounds from raw starting materials is called organic synthesis/synthetic organic chemistry

alkane dihaloalkane

trihaloalkanetetrahaloalkane

haloalkane

alcohol

ketone

alkene

aldehyde

Poly(alkene)

Carboxylic acid

M1- free radical subsitution

M2-nucleophiliic subsitution

1

1

2

23

4 4

4 Key:1- subsitution halogenation 2- addition halogenation3- polymerisation4- oxidation

NB M1 & M2 mechanism required

Reaction pathway puzzlesHow can butanone be synthesised using 2-

bromobutane as one of the starting materials?

Answer:

2-bromobutane butan-2-ol butanone

Reflux with NaOH(aq) Reflux with H+/Cr2O72-

Reaction pathway puzzles 2How can ethanoic acid be synthesised using

ethene as one of the starting materials?

Answer:

ethene ethanol ethanoic acid

Conc. H2SO4/H2O Reflux with H+/Cr2O72-