polymer revolution organic functional group. learning outcomes (d) recognise and write formulae for...

Polymer Revolution

Organic Functional Group

Learning Outcomes

• (d) recognise and write formulae for alkenes and use systematic nomenclature to name and interpret the names of alkenes;

(e) recognise members of the following homologous series: aldehydes, ketones, carboxylic acids;

(f) recall the difference between primary, secondary and tertiary alcohols from their structures and identify examples of them;

Answer

• Aldehyde

Starter Activity

CLASSIFICATION OF ALCOHOLS

Aliphatic • general formula CnH2n+1OH - provided there are no rings

• the OH replaces an H in a basic hydrocarbon skeleton




Aromatic • in aromatic alcohols (or phenols) the OH is attached directly to the ring • an OH on a side chain of a ring behaves as a typical aliphatic alcohol

The first two compounds are classified as aromatic alcohols (phenols) because the OH group is attached directly to the ring.




Aromatic • in aromatic alcohols (or phenols) the OH is attached directly to the ring • an OH on a side chain of a ring behaves as a typical aliphatic alcohol

The first two compounds are classified as aromatic alcohols (phenols) because the OH group is attached directly to the ring.

Structuraldifferences • alcohols are classified according to the environment of the OH group

• chemical behaviour, eg oxidation, often depends on the structural type

PRIMARY 1° SECONDARY 2° TERTIARY 3°

Alcohols are named according to standard IUPAC rules

• select the longest chain of C atoms containing the O-H group;

• remove the e and add ol after the basic name

• number the chain starting from the end nearer the O-H group

• the number is placed after the an and before the ol ... e.g butan-2-ol

• as in alkanes, prefix with alkyl substituents

• side chain positions are based on the number allocated to the O-H group

e.g. CH3 - CH(CH3) - CH2 - CH2 - CH(OH) - CH3 is called 5-methylhexan-2-ol

NAMING ALCOHOLS

STRUCTURAL ISOMERISM IN ALCOHOLS

Different structures are possible due to...

A Different positions for the OH group and

B Branching of the carbon chain

butan-1-ol butan-2-ol

2-methylpropan-1-ol2-methylpropan-2-ol

BOILING POINTS OF ALCOHOLS

Increases with molecular size due to increased van der Waals’ forces.

Alcohols have higher boiling points thansimilar molecular mass alkanes

This is due to the added presence ofinter-molecular hydrogen bonding.More energy is required to separate the molecules.

Mr bp / °C

propane C3H8 44 -42 just van der Waals’ forces

ethanol C2H5OH 46 +78 van der Waals’ forces + hydrogen bonding

BOILING POINTS OF ALCOHOLS

Increases with molecular size due to increased van der Waals’ forces.

Alcohols have higher boiling points thansimilar molecular mass alkanes

This is due to the added presence ofinter-molecular hydrogen bonding.More energy is required to separate the molecules.

Mr bp / °C

propane C3H8 44 -42 just van der Waals’ forces

ethanol C2H5OH 46 +78 van der Waals’ forces + hydrogen bonding

Boiling point is higher for “straight” chain isomers.

bp / °Cbutan-1-ol CH3CH2CH2CH2OH 118

butan-2-ol CH3CH2CH(OH)CH3 100

2-methylpropan-2-ol (CH3)3COH 83

Greater branching = lower inter-molecular forces

SOLVENT PROPERTIES OF ALCOHOLS

SolubilityLow molecular mass alcohols are miscible with water

Due to hydrogen bonding between the two molecules

Heavier alcohols are less miscible

Solventproperties Alcohols are themselves very good solvents

They dissolve a large number of organic molecules

Show the relevant lone pair(s) when drawing hydrogen bonding

CHEMICAL PROPERTIES OF ALCOHOLS

The OXYGEN ATOM HAS TWO LONE PAIRS; this makes alcohols...

BASES Lewis bases are lone pair donors Bronsted-Lowry bases are proton acceptors

The alcohol uses one of its lone pairs to form a co-ordinate bond

NUCLEOPHILES Alcohols can use the lone pair to attack electron deficient centres

ELIMINATION OF WATER (DEHYDRATION)

Reagent/catalyst conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions reflux at 180°C

Product alkene

Equation e.g. C2H5OH(l) ——> CH2 = CH2(g) + H2O(l)

Mechanism

Step 1 protonation of the alcohol using a lone pair on oxygenStep 2 loss of a water molecule to generate a carbocationStep 3 loss of a proton (H+) to give the alkene

AlternativeMethod Pass vapour over a heated alumina (aluminium oxide) catalyst

ELIMINATION OF WATER (DEHYDRATION)

MECHANISM

Step 1 protonation of the alcohol using a lone pair on oxygenStep 2 loss of a water molecule to generate a carbocationStep 3 loss of a proton (H+) to give the alkene

Note 1 There must be an H on a carbon atom adjacent the carbon with the OH

Note 2 Alcohols with the OH in the middle of a chaincan have two ways of losing water.

In Step 3 of the mechanism, a proton can be lostfrom either side of the carbocation. This gives amixture of alkenes from unsymmetrical alcohols...

OXIDATION OF ALCOHOLS

All alcohols can be oxidised depending on the conditions

Oxidation is used to differentiate between primary, secondary and tertiary alcoholsThe usual reagent is acidified potassium dichromate(VI)

Primary Easily oxidised to aldehydes and then to carboxylic acids.

Secondary Easily oxidised to ketones

Tertiary Not oxidised under normal conditions.They do break down with very vigorous oxidation

PRIMARY 1° SECONDARY 2° TERTIARY 3°

OXIDATION OF PRIMARY ALCOHOLS

Primary alcohols are easily oxidised to aldehydes

e.g. CH3CH2OH(l) + [O] ——> CH3CHO(l) + H2O(l)

ethanol ethanal

it is essential to distil off the aldehyde before it gets oxidised to the acid

CH3CHO(l) + [O] ——> CH3COOH(l)

ethanal ethanoic acid


Primary alcohols are easily oxidised to aldehydes


ethanol ethanal

it is essential to distil off the aldehyde before it gets oxidised to the acid

CH3CHO(l) + [O] ——> CH3COOH(l)

ethanal ethanoic acid

Practical details

• the alcohol is dripped into a warm solution of acidified K2Cr2O7

• aldehydes have low boiling points - no hydrogen bonding - they distil off immediately• if it didn’t distil off it would be oxidised to the equivalent carboxylic acid• to oxidise an alcohol straight to the acid, reflux the mixture

compound formula intermolecular bonding boiling point

ETHANOL C2H5OH HYDROGEN BONDING 78°C

ETHANAL CH3CHO DIPOLE-DIPOLE 23°C

ETHANOIC ACID CH3COOH HYDROGEN BONDING 118°C


Controlling the products


then CH3CHO(l) + [O] ——> CH3COOH(l)

Aldehyde has a lower boiling point so distils off before being oxidised further

OXIDATION TO ALDEHYDESDISTILLATION

OXIDATION TO CARBOXYLIC ACIDSREFLUX

Aldehyde condenses back into the mixture and gets oxidised to the acid

OXIDATION OF SECONDARY ALCOHOLS

Secondary alcohols are easily oxidised to ketones

e.g. CH3CHOHCH3(l) + [O] ——> CH3COCH3(l) + H2O(l)

propan-2-ol propanone

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol.

OXIDATION OF SECONDARY ALCOHOLS

Secondary alcohols are easily oxidised to ketones

e.g. CH3CHOHCH3(l) + [O] ——> CH3COCH3(l) + H2O(l)

propan-2-ol propanone

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol.

OXIDATION OF TERTIARY ALCOHOLS

Tertiary alcohols are resistant to normal oxidation


Why 1° and 2° alcohols are easily oxidised and 3° alcohols are not

For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.




H H

R C O + [O] R C O + H2O

H H

1°




H H

R C O + [O] R C O + H2O

H H

H H

R C O + [O] R C O + H2O

R R

1°

2°




H H

R C O + [O] R C O + H2O

H H

H H

R C O + [O] R C O + H2O

R R

This is possible in 1° and 2° alcohols but not in 3° alcohols.

1°

2°




H H

R C O + [O] R C O + H2O

H H

H H

R C O + [O] R C O + H2O

R R

R H

R C O + [O]

R

This is possible in 1° and 2° alcohols but not in 3° alcohols.

1°

2°

3°

Answer:

ESTERIFICATION OF ALCOHOLS

Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions reflux

Product ester

Equation e.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

ethanol ethanoic acid ethyl ethanoate

Notes Concentrated H2SO4 is a dehydrating agent - it removes water

causing the equilibrium to move to the right and increases the yield

ESTERIFICATION OF ALCOHOLS

Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions reflux

Product ester

Equation e.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

ethanol ethanoic acid ethyl ethanoate

Notes Concentrated H2SO4 is a dehydrating agent - it removes water

causing the equilibrium to move to the right and increases the yield

Uses of esters Esters are fairly unreactive but that doesn’t make them uselessUsed as flavourings

Naming esters Named from the alcohol and carboxylic acid which made them...

CH3OH + CH3COOH CH3COOCH3 + H2O

from ethanoic acid CH3COOCH3 from methanol

METHYL ETHANOATE

OTHER REACTIONS OF ALCOHOLS

OXYGEN Alcohols make useful fuels

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

Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources

OTHER REACTIONS OF ALCOHOLS

OXYGEN Alcohols make useful fuels

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

Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources

SODIUMConditions room temperature

Product sodium alkoxide and hydrogen

Equation 2CH3CH2OH(l) + 2Na(s) ——> 2CH3CH2O¯ Na + + H2(g) sodium ethoxide

Notes alcohols are organic chemistry’s equivalent of waterwater reacts with sodium to produce hydrogen and so do alcoholsthe reaction is slower with alcohols than with water.

Alkoxides are white, ionic crystalline solids e.g. CH3CH2O¯ Na+

BROMINATION OF ALCOHOLS

Reagent(s) conc. hydrobromic acid HBr(aq) orsodium (or potassium) bromide and concentrated sulphuric acid

Conditions reflux

Product haloalkane

Equation C2H5OH(l) + conc. HBr(aq) ———> C2H5Br(l) + H2O(l)

Mechanism The mechanism starts off similar to that involving dehydration(protonation of the alcohol and loss of water) but the carbocation(carbonium ion) is attacked by a nucleophilic bromide ion in step 3

Step 1 protonation of the alcohol using a lone pair on oxygen

Step 2 loss of a water molecule to generate a carbocation (carbonium ion)

Step 3 a bromide ion behaves as a nucleophile and attacks the carbocation

INFRA-RED SPECTROSCOPY

Chemical bonds vibrate at different frequencies. When infra red (IR) radiation is passed through a liquid sample of an organic molecule, some frequencies are absorbed. These correspond to the frequencies of the vibrating bonds.

Most spectra are very complex due to the large number of bonds present and each molecule produces a unique spectrum. However the presence of certain absorptions can be used to identify functional groups.

BOND COMPOUND ABSORBANCE RANGE

O-H alcohols broad 3200 cm-1 to 3600 cm-1

O-H carboxylic acids medium to broad 2500 cm-1 to 3500 cm-1

C=O ketones, aldehydes strong and sharp 1600 cm-1 to 1750 cm-1

esters and acids

INFRA-RED SPECTROSCOPY

IDENTIFYING ALCOHOLS USING INFRA RED SPECTROSCOPY

Differentiation Compound O-H C=O

ALCOHOL YES NO

ALDEHYDE / KETONE NO YES

CARBOXYLIC ACID YES YES

ESTER NO YES

ALCOHOL ALDEHYDE CARBOXYLIC ACID PROPAN-1-OL PROPANAL PROPANOIC ACID O-H absorption C=O absorption O-H + C=O absorption

Answer:

INDUSTRIAL PREPARATION OF ALCOHOLS

FERMENTATION

Reagent(s) GLUCOSE - produced by the hydrolysis of starch

Conditions yeastwarm, but no higher than 37°C

Equation C6H12O6 ——> 2 C2H5OH + 2 CO2


FERMENTATION

Reagent(s) GLUCOSE - produced by the hydrolysis of starch

Conditions yeastwarm, but no higher than 37°C

Equation C6H12O6 ——> 2 C2H5OH + 2 CO2

Advantages LOW ENERGY PROCESSUSES RENEWABLE RESOURCES - PLANT MATERIALSIMPLE EQUIPMENT

Disadvantages SLOWPRODUCES IMPURE ETHANOLBATCH PROCESS


HYDRATION OF ETHENE

Reagent(s) ETHENE - from cracking of fractions from distilled crude oil

Conditions catalyst - phosphoric acid high temperature and pressure

Equation C2H4 + H2O ——> C2H5OH


HYDRATION OF ETHENE

Reagent(s) ETHENE - from cracking of fractions from distilled crude oil

Conditions catalyst - phosphoric acid high temperature and pressure

Equation C2H4 + H2O ——> C2H5OH

Advantages FASTPURE ETHANOL PRODUCEDCONTINUOUS PROCESS

Disadvantages HIGH ENERGY PROCESSEXPENSIVE PLANT REQUIREDUSES NON-RENEWABLE FOSSIL FUELS TO MAKE ETHENE

Uses of ethanol ALCOHOLIC DRINKSSOLVENT - industrial alcohol / methylated spiritsFUEL - petrol substitute in countries with limited oil reserves

USES OF ALCOHOLS

ETHANOL

DRINKSSOLVENT industrial alcohol / methylated spirits (methanol is added)FUEL used as a petrol substitute in countries with limited oil reserves

METHANOL

PETROL ADDITIVE improves combustion properties of unleaded petrolSOLVENTRAW MATERIAL used as a feedstock for important industrial processesFUEL

Health warning Methanol is highly toxic

LABORATORY PREPARATION OF ALCOHOLS

from haloalkanes - reflux with aqueous sodium or potassium hydroxide

from aldehydes - reduction with sodium tetrahydridoborate(III) - NaBH4

from alkenes - acid catalysed hydration using concentrated sulphuric acid

Details of the reactions may be found in other sections.

REVISION CHECK

What should you be able to do?

Recall and explain the physical properties of alcohols

Recall the different structural types of alcohols

Recall the Lewis base properties of alcohols

Recall and explain the chemical reactions of alcohols

Write balanced equations representing any reactions in the section

Understand how oxidation is affected by structure

Recall how conditions and apparatus influence the products of oxidation

Explain how infrared spectroscopy can be used to differentiate between functional groups

CAN YOU DO ALL OF THESE? YES NO

polymer revolution organic functional group. learning outcomes (d) recognise and write formulae for...

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