CONTENTS Structure of alcohols Nomenclature Isomerism Physical properties Chemical properties of alcohols Identification using infra-red spectroscopy Industrial

CONTENTS

• Structure of alcohols

• Nomenclature

• Isomerism

• Physical properties

• Chemical properties of alcohols

• Identification using infra-red spectroscopy

• Industrial preparation and uses of ethanol

• Revision check list

THE CHEMISTRY OF ALCOHOLSTHE CHEMISTRY OF ALCOHOLS

Before you start it would be helpful to…

• Recall the definition of a covalent bond

• Recall the difference types of physical bonding

• Be able to balance simple equations

• Be able to write out structures for simple organic molecules

• Understand the IUPAC nomenclature rules for simple organic compounds

• Recall the chemical properties of alkanes and alkenes

THE CHEMISTRY OF ALCOHOLSTHE CHEMISTRY OF ALCOHOLS

CLASSIFICATION OF ALCOHOLSCLASSIFICATION 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 ALCOHOLSNAMING ALCOHOLS

STRUCTURAL ISOMERISM IN ALCOHOLSSTRUCTURAL 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 ALCOHOLSBOILING 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





Mr bp / °C



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





Mr bp / °C



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

Boiling temperature comparision

Alcohol london forces and Hydrogen bonding

Halagenoalkane C-F bonds stronger than C-C

Alkenes/alkanes only london forces

BoilingTemp.

No. of carbon atoms

SOLVENT PROPERTIES OF ALCOHOLSSOLVENT 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 ALCOHOLSCHEMICAL 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)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

NoteThis is potentially an extremely dangerous preparation because of the close proximity of the very hot concentrated sulphuric acid and the sodium hydroxide solution.


Reagent/catalyst Aluminium Oxide ( Al2O3)

Conditions Heat

Product alkene

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

This is a simple way of making gaseous alkenes like ethene. If ethanol vapour is passed over heated aluminium oxide powder, the ethanol is essentially cracked to give ethene and water vapour.


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 ALCOHOLSOXIDATION 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 ALCOHOLSOXIDATION OF PRIMARY ALCOHOLS

Primary alcohols are easily oxidised to aldehydes

e.g. CH3CH2OH(l) CH3CHO(l) + H2O(l)

ethanol ethanal

It is essential to distil off the aldehyde before it gets oxidised to the (Carboxilic) acid

CH3CHO CH3COOH

ethanal ethanoic acid

Practical details

• 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

K2Cr2O7 + DIL. H2SO4

Heat under distillation

Note : K2Cr2O7 Oxidisong agent


OXIDATION OF SECONDARY ALCOHOLSOXIDATION OF SECONDARY ALCOHOLS

Secondary alcohols are easily oxidised to ketones

CH3CHOHCH3(l) CH3COCH3(l) + H2O(l)

propan-2-ol propanone

Prolonged treatment with oxidising agent with secondary alcohol does not produce carboxylic acid

OXIDATION OF TERTIARY ALCOHOLSOXIDATION OF TERTIARY ALCOHOLS

Tertiary alcohols are resistant to normal oxidation


Heat under reflux

(CH3) 3COH NO REACTION


Heat under reflux/distillation

OXIDATION OF PRIMARY ALCOHOLSOXIDATION OF PRIMARY ALCOHOLS

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

PRIMARY ALCOHOLS OXIDATION TO ALDEHYDES DISTILLATION

SECONDARY ALCOHOLS OXIDATION TO CARBOXYLIC ACIDS REFLUX

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

OXIDATION OF SECONDARY ALCOHOLSOXIDATION OF SECONDARY ALCOHOLS

Reflux condenser prevents volatile compounds like ethanal, ethanol and ethanoic acid vapours from leaving the flask

Note : volatile compoundsChemical compounds that transition to gas at low temperatures


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.


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

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°

OTHER REACTIONS OTHER REACTIONS OF ALCOHOLSOF ALCOHOLS

Reactions involving Oxygen Reactions involving Sodium Reactions involving hydrogen halides

OTHER REACTIONS OF ALCOHOLSOTHER REACTIONS OF ALCOHOLS

OXYGEN

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

SODIUM

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

room temperature

Ignite/spark

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

* Observation : Bubbles will form , Sodium dissolves , White solid formed

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

The general reaction looks like this:

Reactions involving hydrogen halides

ROH + HX RX + H2O

CHLORINATION OF ALCOHOLSCHLORINATION OF ALCOHOLS

When PCl5 is added to dry alcohol, clods of hydrogen cloride fumes are produced

CH3CH2OH + PCl5 CH3CH2Cl + POCl3 + HCl (g)

Hydrogen chloride testHydrogen chloride gas forms a white smoke with ammonia.

BROMINATION OF ALCOHOLSBROMINATION OF ALCOHOLS

C2H5OH + HBr C2H5Br + H2O

Dry conditions

Room temp

NaBr / KBr + 50% CONC. H2SO4

Heat under reflux

3C2H5OH + PBr3 3C2H5Br + H3PO3

Moist red Phosperous + Br2

KBr + H2SO4 ---> KHSO4+ HBr

Heat under reflux

IODINATION OF ALCOHOLSIODINATION OF ALCOHOLS

3C2H5OH + PI3 3C2H5I + H3PO3

Moist red Phosperous + I2

In this case the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H3PO4, and the iodoalkane is distilled off.

The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide which reacts with the alcohol.

Phosphoric(V) acid is used instead of concentrated sulphuric acid because sulphuric acid oxidises iodide ions to iodine and produces hardly any hydrogen iodide.

A similar thing happens to some extent with bromide ions in the preparation of bromoalkanes, but not enough to get in the way of the main reaction. There is no reason why you couldn't use phosphoric(V) acid in the bromide case instead of sulphuric acid if you wanted to.

C2H5OH + HI C2H5I + H2O

NaI / KI + CONC. H3PO4

H3PO4 + KI ----> KH2PO4 + HI

Heat under reflux

SECONDARY ALCOHOLS WITH HALOGENSSECONDARY ALCOHOLS WITH HALOGENS

+ PCl5

Butane-2-ol + PCl5 2-cloro-butane

Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid,

but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance.A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature.

A tertiary halogenoalkane (haloalkane or alkyl halide) is formed

TERTIARY ALCOHOLS WITH HALOGENSTERTIARY ALCOHOLS WITH HALOGENS

Haloalkanes can be prepared from the vigorous reaction between cold alcohols and

phosphorus(III) halides

SUMMARY

QUESTION TIME

Propanoic acid may be prepared by oxidizing propan-1-ol in acidic conditions.

X + H2SO4CH3CH2CH2OH CH3CH2COOH

Procedure1. Pour 10 cm of distilled water into a boiling tube and add 12 g of oxidizing agent X.Shake the mixture and leave X to dissolve.

2. Pour 3 cm of propan-1-ol into a round-bottom flask and add 10 cm of distilled water and a few anti-bumping granules. Set up the apparatus for heating under reflux.

3. Add 4 cm of concentrated sulfuric acid, drop by drop, to the propan-1-ol. While the mixture is still warm, add the solution of oxidizing agent X, drop by drop. Theenergy released from the reaction should cause the mixture to boil without external heating.

4. When all of the solution of X has been added, use a low Bunsen burner flame to keep the mixture boiling for 10 minutes, not allowing any vapour to escape.

5. Distil the mixture in the flask using the apparatus shown below. Collect 5–6 cm of distillate, which is an aqueous solution of propanoic acid.

(a) Suggest, by name or formula, a suitable oxidising agentPotassium dichromate ((VI))/ K2Cr2O7Sodium dichromate ((VI))/ Na2Cr2O7

(b) What colour change does X undergo when it oxidizes propan-1-ol?From

Orange to green

(c) Draw a labelled diagram showing the apparatus for heating under reflux.

(d) Give two reasons why the escape of vapour in step 4 should be prevented.

Reason 1 Yield would be reduced/reactants and or products would be lost complete oxidation could not occurReason 2 Vapour is flammable/toxic/hazardous/harmful/ acidic/irritant

(e) How does the reflux apparatus prevent escape of vapour?Mixture being heated returns to the flaskThe vapour is (cooled and) condensedThe water in the condenser is cold (and flowing)

(f) Some water can be removed from the distillate in step 5 by adding a solid drying agent. The solution of propanoic acid can then be decanted leaving the drying agent behind.(i)Suggest a suitable solid drying agent.(Anhydrous) calcium chloride/ (Anhydrous) magnesium sulfate/(Anhydrous) sodium sulfate Silica gel

(ii) Suggest why removing excess solid drying agent by decanting, rather than filtering through filter paper, improves the yield.Some product is absorbed BY/ INTO filter paper

(g) In a larger scale preparation of propanoic acid, 10.0 g of propan-1-ol was used.(i)Calculate the maximum mass of propanoic acid which could be formed from 10.0 g of propan-1-ol.

Propan-1-ol Molar mass = 60.1Propanoic acid Molar mass =74.1

Mol propanol = (10/60.1) = 0.166/ 0.17 = (mol propanoic acid) Mass propanoic acid = (0.166 x 74.1) = 12.32945 = 12.33/12.3 (g)If 0.17 mol then 12.597/12.6 (g)

8.1066 (g)

(ii) After purification, 6.0 cm of dry propanoic acid was obtained.Calculate the percentage yield in the preparation.The density of propanoic acid is 0.99 g cm.Mass propanoic acid = 6 x 0.99 = 5.94 (g) % yield = (5.94/ 12.33)x100 = 48.17% =

(h) In another experiment, the same reaction mixture (propan-1-ol, X and concentrated sulfuric acid) was heated in the apparatus shown in step 5. Identify the main organicproduct which would be collected and explain why propanoic acid is not produced.Product Propanal/CH3CH2CHO

ExplanationProduct removed as formed/Incomplete oxidation

This question is about the alcohol, propan-1-ol.(a)Give two observations when propan-1-ol reacts with a small piece of sodium.

Observation 1 Sodium dissolves/disappears/gets smaller Observation 2 Bubbles/effervescence/fizzes

(b) A student investigated the rate of reaction of propan-1-ol with sodium.Suggest one suitable measurement which could be made to determine the rate of this reaction.Measure volume of gas in fixed time/measure time to collect avolume of gas/measure time for sodium to dissolve

(c) A small amount of phosphorus(V) chloride (phosphorus pentachloride), PCl5, is added to propan-1-ol in a test tube.(i)Describe the appearance of the fumes at the mouth of the test tube.misty (fumes)white (fumes)

(ii) An open bottle of concentrated ammonia is held near the mouth of the tube.Describe what would be seen at the mouth of the test tube.White smoke

INFRA-RED SPECTROSCOPYINFRA-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 SPECTROSCOPYINFRA-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

INDUSTRIAL PREPARATION OF ALCOHOLSINDUSTRIAL 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 (H3PO4)

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 (H3PO4)

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 ALCOHOLSUSES 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

REVISION CHECKREVISION 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? CAN YOU DO ALL OF THESE? YES YES NONO

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Documents

CONTENTS Structure of alcohols Nomenclature Isomerism Physical properties Chemical properties of alcohols Identification using infra-red spectroscopy Industrial