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Organic Chemistry. Introduction. Important definitions HOMOLOGOUS SERIES – a family of organic compounds which all fit the same general formula, neighbouring members differ from each other by CH 2 , have similar chemical properties and show trends in physical properties.

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introduction
Introduction

Important definitions

HOMOLOGOUS SERIES – a family of organic compounds which all fit the same general formula, neighbouring members differ from each other by CH2, have similar chemical properties and show trends in physical properties.

e.g. the alkanes all fit the general formula CnH2n+2, members of the family include methane (CH4), ethane (C2H6) and propane (C3H8).

slide3

H

H

H

H

H

H

H

C

C

C

C

C

H

H

H

H

H

EMPIRICAL FORMULA – the simplest ratio of moles of atoms of each element in a compound.

MOLECULAR FORMULA – the actual number of moles of atoms of each element in a compound.

STRUCTURAL FORMULA (aka displayed formula or graphical formula) – shows all of the atoms and the bonds between them. e.g. pentane (C5H12).

slide4

A condensed structural formula can also be used which omits the bonds. So for pentane we can use:

CH3CH2CH2CH2CH3

or

CH3(CH2)3CH3

The formulas in the data book are called SKELETAL formulas. These must NOT be used in the exam.

STRUCTURAL ISOMERS – compounds with the same molecular formula but a different arrangement of atoms. e.g. C4H10 represents

slide5

H

H

H

H

H

H

C

C

C

C

H

H

H

H

butane and

slide6

H

H

C

H

H

H

C

C

C

H

H

H

H

H

2-methylpropane

slide7

FUNCTIONAL GROUP – atom or group of atoms which gives an organic compound its characteristic chemical properties.

slide8

C

H

Alkanes

suffix – ane

For example

CH3CH3

ethane

slide9

C

C

Alkenes

suffix – ene

For example

ethene

CH2=CH2

slide10

C

X

Halogenoalkanes

prefix halo -

X = Cl, Br or I

For example

chloroethane

CH3CH2Cl

slide11

C

OH

Alcohols

suffix – ol

or prefix hydroxy-

For example

CH3CH2OH

ethanol

2-hydroxypropanoic acid

CH3CH(OH)COOH

slide12

O

C

H

Aldehydes

suffix – al

For example

ethanal

CH3CHO

slide13

C

O

Ketones

suffix –one

or prefix oxo-

For example

propanone

CH3COCH3

3 – oxobutanoic acid

CH3COCH2COOH

slide14

O

C

OH

Carboxylic acid

suffix – oic acid

For example

ethanoic acid

CH3COOH

slide15

C

NH2

Amines

suffix – amine

or prefix amino -

For example

CH3NH2

methylamine

aminoethanoic acid

H2NCH2COOH

slide16

O

C

OR

Esters

suffix – oate

For example

CH3COOC2H5

ethyl ethanoate

aromatic compounds

H

C

H

C

C

H

Delocalised electrons

C

C

H

H

C

H

Aromatic compounds

Contain the BENZENE ring

formula C6H6

some examples

CH3CHCH2CH3

nitro group

NO2

COOH

1

Br

Cl

1

CH3

2

3

4

Cl

Some examples

(1-methylpropyl) benzene

nitrobenzene

4-chloro-3-methylbenzenecarboxylic acid

1-bromo-2-chlorobenzene

slide19

NH2

Sometimes the benzene ring is not regarded as the key part of the molecule.

In these cases it is referred to as the PHENYL group.

For example

phenylamine

slide20

CH3

O

O

CH3CHCH2C

CH3CH2CH2C

OH

H

O

CH3CH2CCH2CH3

aldehyde

butanal

ketone

pentan-3-one

carboxylic acid

3-methylbutanoic acid

slide21

alkene

alcohol

ketone

ester

carboxylic acid

slide22

carboxylic acid

ester

aldehyde

ether

amine

nitrile

physical properties of organic molecules
Physical Properties of Organic Molecules

Alkanes worksheet

What type of intermolecular force would you expect to find between alkanes, halogenoalkanes, aldehydes, ketones, alcohols and carboxylic acids?

Use this information to deduce the relative boiling points of these homologous series and their solubility in water.

the alkanes
The Alkanes

The alkanes is an homologous series where all members fit the general formula CnH2n+2.

They have trends in physical properties e.g. density and m.p. and b.p. all increase with Mr.

They all undergo similar chemical reactions.

Alkanes are SATURATED HYDROCARBONS. i.e. they contain only single C to C bonds and are made up of C and H atoms only.

slide25

Alkanes are obtained from crude oil by fractional distillation.

They are mainly used as fuels.

The large Mr alkanes do not ignite easily so there is little demand for them as fuels so they are CRACKED to make smaller more useful alkanes and alkenes.

Apart from combustion alkanes undergo few chemical reactions. This is for two main reasons:

slide26

The bonds in alkanes are relatively strong.

  • The bonds have a relatively low polarity as the electronegativity of C and H is similar.
  • As a consequence alkanes can be used as lubricating oils, although they do degrade over time.
reactions of alkanes
Reactions of Alkanes:

Combustion:

Alkanes burn exothermically to produce carbon dioxide and water if there is a plentiful supply of oxygen. This is known as complete combustion.

e.g. CH4 + 2O2 CO2 + 2H2O

Write equations for the complete combustion of butane and octane.

slide28

If there is a limited supply of oxygen incomplete combustion occurs and carbon monoxide or carbon are formed instead of carbon dioxide.

e.g. CH4 + 1½O2 CO + 2H2O

CH4 + O2 C + 2H2O

What problems do the gases released on combustion of alkanes cause?

slide29

Chlorination

Methane does not react with chlorine in the dark but in the presence of ultraviolet light reacts to form hydrogen chloride.

CH4 + Cl2 CH3Cl + HCl

The mechanism for this reaction is known as free radical substitution.

Substitution = replacement of an atom or group of atoms by a different atom or group of atoms.

slide30

Cl

Cl

Free radical = species with an unpaired electron.

Free radicals are formed by homolytic fission of bonds. In homolytic fission one electron from the shared pair goes to each atom.

So

slide31

Cl

+

Cl

unpaired electron

or Cl2 2Cl.

Heterolytic fission of Cl – Cl would result in the formation of Cl+ and Cl-.

There are three steps in the mechanism: initiation, propagation and termination

free radical substitution
Free radical substitution

CH4 + Cl2

CH3Cl + HCl

chlorination of methane

i.e. homolytic breaking of covalent bonds

Overall reaction equation

Conditions

ultra violet light (breaks weakest bond)

excess methane to reduce further

substitution

free radical substitution mechanism
Free radical substitution mechanism

Cl2

Cl + Cl

CH4 + Cl

CH3 + HCl

CH3 + Cl2

CH3Cl + Cl

CH3 + Cl

CH3Cl

CH3 + CH3

CH3CH3

UV Light

initiation step

two propagation steps

termination step

minor termination step

Also get reverse of initiation step occurring as a termination step.

further free radical substitutions
Further free radical substitutions

CH3Cl + Cl2

CH2Cl2 + HCl

CH2Cl2 + Cl2

CHCl3 + HCl

CHCl3 + Cl2

CCl4 + HCl

ultra-violet light

excess chlorine

Overall reaction equations

Conditions

slide35

Write down two propagation steps to explain the formation of dichloromethane.

Methane reacts in exactly the same way with bromine to form hydrogen bromide together with bromomethane, dibromomethane, tribromomethane and tetrabromomethane.

Write down the mechanism for the reaction between chlorine and ethane to form chloroethane. Use the mechanism to explain why small amounts of butane are formed. How could the formation of further substitution products be minimised?

the alkenes
The Alkenes

All fit the general formula CnH2n.

Are unsaturated hydrocarbons as they contain a C = C.

Much more reactive than alkanes.

Industrial importance of alkenes:

  • Making polymers (plastics)
  • Hydrogenation of vegetable oils to make margarine
  • Hydration of ethene to make ethanol.
slide37

When naming alkenes have to include position of double bond, for example:

CH3CH=CHCH3 is but - 2 - ene and

CH3CH2CH=CH2 is but -1- ene

Draw out and name all of the alkenes with the molecular formula C6H12.

Alkenes undergo ADDITION reactions. Two substances combine to form one new substance. Unsaturated molecules are converted to saturated molecules.

reactions of alkenes

H

H

H

H

C = C

H – C – C – H

+ H2

H

H

H

H

Reactions of Alkenes
  • Addition of hydrogen (hydrogenation) Alkenes react with hydrogen in the presence of a nickel catalyst at 180 °C to form an alkane. e.g.

C2H4 + H2 C2H6

ethene

ethane

slide39

Most oils are esters of propane-1,2,3-triol (aka glycerol) with 3 long chain carboxylic acids (aka fatty acids). The esters are sometimes called triglycerides. Hydrogenation of these oils produces margarine.

  • The common fatty acids include
  • octadeca-9-enoic (oleic) acid – unsaturated acid found in most fats and olive oil
  • octadeca-9,12-dienoic (linoleic) acid – unsaturated acid found in many vegetable oils such as soyabean and corn oil
slide40

CH2OOC(CH2)7CH=CH(CH2)7CH3

CHOOC(CH2)7CH=CH(CH2)7CH3

CH2OOC(CH2)7CH=CH(CH2)7CH3

Above is the triglyceride formed between propan-1,2,3-triol and oleic acid. Hydrogenation using a nickel catalyst and slight pressure removes some of the C=C. This enables the chains to pack together more closely which increases the van der Waals forces and hence m.p. so the oils are solidified forming margarine.

slide41

H

H

H

H

C = C

+ Br2

H – C – C – H

H

H

Br

Br

  • Addition of halogens (halogenation)
  • Halogens react with alkenes at room temperature and pressure in a non-polar solvent to form a dihalogenoalkane.
  • e.g.

C2H4 + Br2 C2H4Br2

ethene

1,2-dibromoethane

slide42

Bromine water is used as a test for unsaturation.

In the presence of an alkene, bromine water turns from red brown to colourless. Alkanes do not react with bromine water.

slide43

C = C

+ Br2(aq)

– C – C –

Br

Br

Bromine Water Test For Alkenes

colorless

amber

colorless

slide44

H

H

C = C

H

H

3. Reaction with hydrogen halides (hydrohalogenation)

Alkenes react with hydrogen halides (HCl, HBr etc.) to form halogenoalkanes. The reaction occurs at room temperature and pressure. e.g.

H

H

+ HBr

H – C – C – H

H

Br

C2H4 + HBr  CH3CH2Br

ethene

bromoethane

slide45

H

H

H

H

C = C

+ H2SO4

H – C – C – H

H

H

OSO3H

H

  • Hydration (reaction with water)
  • This can be done in two ways:
  • a) Addition of concentrated sulphuric acid to form an alkyl hydrogensulphate. Water is then added to hydrolyse the product and produce an alcohol. The sulphuric acid is regenerated.
slide46

H

H

H

H

H – C – C – H

H – C – C – H

+ H2O

H

OH

H

OSO3H

+ H2SO4

C2H4 + H2SO4 CH3CH2OSO3H

CH3CH2OSO3H + H2O C2H5OH + H2SO4

ethanol

slide47

H

H

H

H

C = C

+ H2O

H – C – C – H

H

H

H

OH

b) Alkenes can also undergo direct hydration to form an alcohol. Ethene can be converted to ethanol by reaction with steam in the presence of a phosphoric(V) acid (H3PO4) catalyst at a pressure of 60 – 70 atm and a temperature of 300 °C.

What advantages and disadvantages does this method have over production of ethanol by fermentation?

slide48

5. Addition Polymerisation

The formation of polymers involves alkenes reacting with themselves to form a long chain molecule called a polymer. The individual molecules used to make the polymer are called monomers. Ethene is polymerised to form poly(ethene)

nCH2 = CH2 CH2 CH2

n

CH2 CH2

is the repeating unit

n = about 100 to 10 000

slide49

CH2=CHCH3

- CH2 – CH -

CH3

- CH2 – CH -

CH2=CHC6H5

C6H5

slide50

poly(chloroethene)

polyvinylchloride

PVC

CH2=CHCl

- CH2 – CH -

Cl

CF2=CF2

- CF2 – CF2 -

Poly

(tetrafluoroethene)

PTFE

Non-stick coating (Teflon)

alcohols
Alcohols
  • Alcohols are the homologous series with the general formula CnH2n+1OH.
  • They all contain the functional group, OH, which is called the hydroxyl group.
  • Alcohols can be classified as primary, secondary or tertiary, depending on the carbon skeleton to which the hydroxyl group is attached.
slide52

RCH2OH

1 alkyl group on C next to OH so primary alcohol, 1°

R2CHOH

2 alkyl groups on C next to OH so secondary alcohol, 2°

R3COH

3 alkyl groups on C next to OH so tertiary alcohol, 3°

Draw out the structure, name and classify all the alcohols with the formula C4H9OH.

slide53

H

H

C

OH

H

H

H

H

H

H

H

H

C

C

C

H

H

C

C

C

C

H

H

H

H

H

H

OH

Butan-1-ol

primary

Butan-2-ol

secondary

slide54

H

H

C

OH

CH3

OH

H

H

H

H

H

C

H

C

C

C

C

CH3

H

H

H

H

2-methylpropan-1-ol primary

2-methylpropan-2-ol tertiary

reactions of alcohols
Reactions of Alcohols
  • Combustion
  • In countries such as Brazil, ethanol is mixed with petrol and used to power cars. Ethanol is less efficient as a fuel than petrol as it is already partially oxidised but does make the country less reliant on supply of petrol. As it can be produced by fermentation of sugar beet, many consider ethanol a carbon neutral fuel.
slide56

O

H

C

H

H

H

H

H

C

C

C

OH

H

H

H

  • Oxidation of Alcohols
  • Primary alcohols are oxidised first to aldehydes, such as ethanal. A suitable oxidising agent is acidified potassium dichromate(VI)

Cr2O7/H+

+

H2O

ethanal

ethanol

slide57

O

O

H

H

C

C

H

H

H

OH

C

C

H

H

An aldehyde still has one hydrogen atom attached to the carbonyl carbon, so it can be oxidised one step further to a carboxylic acid.

Cr2O7/H+

ethanal

ethanoic acid

slide58

In practice, a primary alcohol such as ethanol is dripped into a warm solution of acidified potassium dichromate(VI).

The aldehyde, ethanal, is formed and immediately distils off, thereby preventing further oxidation to ethanoic acid, because the boiling point of ethanal (23 °C) is much lower than that of either the original alcohol ethanol (78 °C) or of ethanoic acid (118 °C). Both the alcohol and the acid have higher boiling points because of hydrogen bonding.

If oxidation of ethanol to ethanoic acid is required, the reagents must be heated together under reflux to prevent escape of the aldehyde before it can be oxidised further.

slide59

H

H

H

H

H

H

H

C

H

H

C

C

C

C

C

O

OH

H

H

H

H

Secondary alcohols are oxidised to ketones. These have no hydrogen atoms attached to the carbonyl carbon and so cannot easily be oxidised further.

Cr2O7/H+

propan-2-ol

propanone

slide60

Distinguishing between 1°, 2° and 3° alcohols

When orange acidified potassium dichromate(VI) acts as an oxidising agent, it is reduced to green chromium(III) ions.

Primary and secondary alcohols both turn acidified dichromate(VI) solution from orange to green when they are oxidised, and this colour change can be used to distinguish them from tertiary alcohols.

Tertiary alcohols are not oxidised by acidified dichromate(VI) ions, so they have no effect on its colour, which remains orange.

halogenoalkanes
Halogenoalkanes

Named by using the name of the alkane from which they are derived with the prefix chloro-, bromo- or iodo-.

For example:

CH3CH2Br is bromoethane

(CH3)2CHCH2Cl is 1-chloro-2-methylpropane

slide62

Remember the position of the halogen atom must be indicated using the appropriate number so

CH3CH2CH2Cl is 1-chloropropane and

CH3CHClCH3 is 2-chloropropane

Halogenoalkanes can be classified in the same way as alcohols.

Draw out, name and classify all the isomers with the formula C5H11Br.

slide63

Key feature of halogenoalkanes is

C X

where X = Cl, Br or I

What is notable about this bond compared with say, C – C and C – H?

The halogen atom is more electronegative than C so the bond is polarised:

+

-

C X

slide64

ORDER OF BOND POLARITIES:

+

-

+

-

+

-

C Cl

>

C Br

>

C I

So is order of reactivity:

chloroalkane > bromoalkanes > iodoalkanes?

Is there another factor that ought to be considered before reaching a conclusion?

BOND ENERGIES

slide65

Bond energies:

346

290

234

This suggests that the order of reactivity is:

iodoalkane > bromoalkanes > chloroalkanes

slide66

There’s only one way to find out which is best!

Do an experiment, not fight!

But what do halogenoalkanes react with?

The + carbon atom is susceptible to attack by NUCLEOPHILES.

A nucleophile is a species with a lone pair of electrons.

e.g. OH-, NH3, CN-.

When attack by a nucleophile occurs, the carbon – halogen bond breaks releasing a halide ion.

A suitable nucleophile for experimentation is OH- from an aqueous solution of an alkali such as sodium hydroxide.

slide67

CH3CH2X + OH-  CH3CH2OH + X-

X = Cl, Br or I.

OH has replaced the X so overall we have NUCLEOPHILIC SUBSTITUTION

How can we follow the rate of this reaction so that we can determine the order of reactivity?

Halide ions  coloured precipitates when silver nitrate is added.

So we can measure how long it takes for a precipitate to form.

See EXPERIMENT SHEET.

mechanisms for nucleophilic substitution
Mechanisms for nucleophilic substitution

SN1 = unimolecular nucleophilic substitution (only one species in the slow step of the mechanism, rate determining step)

SN2 = bimolecular nucleophilic substitution (two species in the slow step of the mechanism, rate determining step)

Use of curly arrows:

Curly arrows are used in reaction mechanisms to show the movement of electron pairs.

slide69

X

TAILS come from

eithera bond pairof electrons

ora lone pairof electrons

HEADS point

eithernext to an atom

orat an atom

to form a bond pair of electrons

to forma lone pairof electrons

X

slide70

H

H

H

+

+

C

C

CH3

CH3

Br

C

CH3

H

H

H

-

Br

H

-

OH

C

OH

CH3

H

Heterolytic fission of C – Br bond

slow

+

Intermediate carbocation

fast

SN1

slide71

-

H

H

HO

Br

C

C

CH3

Br

H

CH3

H

-

H

Br

CH3

C

-

HO

OH

H

Transition state

+

SN2

slide72

Which is best? SN1 or SN2?

For primary halogenoalkanes – SN2

For tertiary halogenoalkanes – SN1

3° halogenoalkanes cannot undergo the SN2 mechanism as 5 bulky groups would not fit around the C in the transition state - steric hindrance.

1° halogenoalknes are less likely to undergo SN1 as this would involve the formation of a primary carbocation as an intermediate. Alkyl groups push electron density to the C atom they are attached to (positive inductive effect) which stabilises the positive charge. More alkyl groups mean a more stable carbocation.

slide73

2° halogenoalkanes react via a mixture of SN1 and SN2. The mechanism predominating depends upon the nature of the alkyl groups and the nature of the solvent.