Chemistry Form 6 Sem 3 05

5.0 Alcohol
~ Organic compound with at least one hydroxyl group (–OH) which act as functioning group.
Alcohol has the general formula of CnH2n+1OH or sometimes CnH2n+2O.
The naming of alcohol end with ~ol. 5.1 Nomenclature of alcohol (naming & classifying alcohol)
The way of naming alcohol is similar to the way of naming alkene [1] Find the longest carbon chain with –OH in it, and name accordingly [2] Identify the alkyl group that attached towards the “parent” chain and name the alkyl [3] Give the prefic of di- ; tri- or tetra based on how many similar alkyl attached toward it [4] Give the numbering of alkyl based on the carbon number based on alcohol

3-methylpentan-2-ol

3,4-dimethylheptan-4-ol

5-ethyl2,5-dimethylheptan-2-ol

CH3C(CH3)(OH)CH2CH(CH2CH3)2

4-methylpentan-1-ol

4-ethyl-2-methylhexan-2-ol

5-ethyl-2methylheptan-3-ol

CH3CH2CH(CH3)CH2OH

C(CH3)3CH(OH)CH2CH3

C(CH3)3CH2C(OH)(CH3)2

2,2-dimethylpentan-3-ol

2,4,4-trimethylpentan-2-ol

3,3-dimethylbutan-2-ol

3-ethyl-2,4-dimethylpentan-2-ol

2-methylbutan-1-ol 4-methylpentan-2-ol

H

H

H

CH2CH3 OH H

C

C

C

H

CH3 H

C

C

CH3 H

H

Isomerism in alcohol.
Alcohol may exhibit structural isomerism and in some case, optical isomerism. For example, butanol, C4H9OH, may have 5 different isomers

H

H

H

CH3 H

C

C

H

OH H

C

H

H

H

OH H

C

C

C

C

H

H

H

H

H

H

H

CH3 H

C

C

C

H

H

H

OH


Practice : write out all the possible isomers for pentanol, C5H11OH H

H

H

H

H

H

H

C

C

C

C

C

H

H

H

H

H

H

H

OH H

H

C

C

C

C

C

H

H

H

H

H

OH

H

H

H

H

CH3 H

C

C

C

C

H

H

H

H

OH

H

H

CH3 H

H

C

C

C

C

H

H

H

H

OH

5.2 Physical properties of alcohol (A) Boiling point of alcohol
Similar to other organic compounds, the boiling point of alcohol increased with number of carbon Alcohol

CH3OH

C2H5OH

C3H7OH

C4H9OH

C5H11OH

Boiling point oC

46

78

90

115

135

Boiling point trend

C6H13OH C7H15OH C8H17OH

152

169

190

BOILING POINT INCREASE DOWN HOMOLOGUS SERIES

increase with
Similar to other organic compounds, the boiling point …………….

the number of carbon as the weak Van Der Waals forces increase with molecular mass of the compound. …………………………
Though, the hydrogen bonding are weaker when goes down to homologous series as the polarity of molecules decrease ……………… as the number of carbon increase.
Similar too, to other organic compound, alcohol with more branch has lower boiling point than a straight-chain molecule

Molecules

Boiling point (0C)

Butan-1-ol

Butan-2-ol

2-methylpropan-2-ol

117

99

82


Straight chain molecules have higher boiling point compare to larger surface branched chain because straight chain molecule has a …………

area than a branched chain molecule. The more branches attached to smaller the surface area ; …………… weaker the forces of the parent chain, the ……………. attraction between molecules ; lower ………… the boiling point.


The number of the hydroxyl group in an organic compound is also one

of the major factor which contribute to its boiling point Molecules Boiling point (0C)

Butan-1-ol

Butan-1,2-diol

Butan-1,2,3-triol

117

208

274


The boiling point of the alcohol increase with the number of –OH. This hydrogen bond formed between –OH of is a result caused by more …………………..

the molecules. So the more the –OH ; stronger the hydrogen bond ; higher the boiling point.


Compare to alkane and haloalkane, alcohol has a higher boiling point

Compound

Ethanol (C2H5OH)

Propane (C3H8)

Fluoroethane (C2H5F)

Relative molecular mass

46

44

48

Boiling point (oC)

78

– 4.2

7


Alcohol has the highest boiling point compare to other organic hydrogen bond between the compound because it forms strong ……..………….

molecules. Fluoroethane has a higher boiling point than propane as polar fluoroethane is a ……………………. molecules and so, the weak Van Der Waals’ forces are stronger than propane since propane is a ………………………… Non-polar …………………..…. molecule.

B)Solubility of alcohol in water
Hydrogen bonding occur between alcohol molecules because of the presence of hydroxy ……………. group. This bring 2 important consequences toward hydrogen where
It cause the boiling points of alcohol higher than those in alkanes and haloalkanes
It cause lower alcohol (methanol and ethanol) to be completely miscible with water.


Solubility decrease with the increase of number of carbon in alcohol.

Butan-1-ol and pentan-1-ol are slightly miscible with water and the rest become more and more insoluble.
This is due to the non-polar properties of alkyl which attached to the – OH, directly influence the efficiency of hydrogen bond, causing the poplar bonding to be more obvious than hydrogen bonding. (dipoledipole interaction between R- and R- are more obvious)
Ethanol is a good solvent for both polar and non polar solute because it alkyl group and a polar group (……………) hydroxyl in it. As contain non polar (………….) a result, ethanol is used as solvent in many industries

(C) Acidity of alcohol
Alcohols are generally a weak acid. Table below shows the pKa value of some alcohols and water Compounds

Methanol (CH3OH)

Ethanol (C2H5OH)

Propan-1-ol (C3H7OH)

pKa

15.5

16.0

18.0

Phenol p-methylphenol (C6H5OH) CH3-C6H5OH 10.0

11.0

Water 14.0


As shown in the table above, alkyl-alcohol is weaker than phenyl-

alcohol. This is a result of the different effect of the group that attached to hydroxyl group –OH. donating
Alkyl is an electron …………..…….. group whereas phenyl is an electron withdrawing group. …………..………

Alcohol

Explanation Ethanol dissociate in water according to the equation

CH3CH2–OH + H2O CH3CH2–O- + H3O+ δ+ δ− Alkyl group, which is an electron donating group, donate CH3CH2–O–H electron to O and caused the electron density of O in R–OH Ethanol increase. As a result, O is more readily to accept proton, which makes the equilibrium favours to left. Phenol dissociate in water according to the equation

Phenol

C6H5–OH + H2O C6H5–O- + H3O+ The phenyl group is an electron-withdrawing group, which withdrawn the electron density from partially negative charge, δ−, from O making O less readily to accept proton. As a result, O is more readily to donate proton which makes equilibrium favour more to right.

5.3 Chemical properties of alcohols 5.3.1 Preparation of alcohol in industries.
Alcohol can be prepared by a few methods in industries / laboratory 1. Fermentation 2. Hydration of alkene (see Chapter 2) 3. Hydrolysis of haloalkane (see Chapter 4) 4. Grignard reagent (see Chapter 4) Name of reaction

Reagent used and condition

Equation

Fermentation of glucose

Zymase enzyme

C6H12O6
2 CH3CH2OH + 2 CO2 Glucose ethanol carbon dioxide

Name of reaction

Reagent used and condition

Equation

Steam (H2O) --------Phosphoric acid, (H3PO4 ) Hydration of At 300oC ; 60 alkene atm @ Concentrated H-2SO4 at 800C. Hydrolysis of haloalkane

NaOH (aq) under reflux

Reaction of Grignard reagent

Aldehyde / ketone with H2SO4

1-chloropropane

sodium Hydroxide

propan-1-ol

H 3O +

CH3CH2MgBr + CH3CH=O → CH3CH2CH(OH)CH3

5.4 Chemical reaction of alcohol
Aliphatic alcohol undergoes 2 types of reaction which involve R–O–H where : ⇒ Fission of O – H

⇒ Fission of C – O • Dehydration of alcohol • Reaction with hydrogen halide • Reaction with phosphorous halide (PX5) or thionyl chloride, SOCl2

• Formation of alkoxide • Formation of ester • Oxidation of alcohol

Name of reaction

Reagent used and condition

Formation of alkoxide

Sodium (Na)

Esterification

Carboxylic acid (R–COOH) catalysed by conc. sulphuric acid (H2SO4)

Equation 2 CH3CH2O–H + 2 Na
2 CH3CH2O–Na+ + H2 (g) Ethanol sodium sodium ethoxide hydrogen

Ethanol

propanoic acid

ethyl propanoate

water

Name of reaction

Reagent used and condition

Oxidation of alcohol

Acidified KMnO4 or acidified K2Cr2O7 + heat

Equation

propan-1-ol

propan-2-ol

Dehydration (removal of water) from alcohol

Excess conc. H2SO4 at 1800C or Alumina (Al2O3) at 350oC

Halogenation of alcohol

Phosphorous pentachloride (PCl5)

propanal

propanoic acid

propanone

CH3CH2CH2OH + PCl5
CH3CH2CH2Cl + POCl3 + HCl

(1) Reaction with sodium metal
When sodium is added to alcohol, a white solid (sodium alkoxide) formed and effervescences occur and hydrogen is released. Example : 2 CH3CH2CH2OH + 2 Na
2 CH3CH2CH2O-Na+ + H2 Propan-1-ol sodium propoxide
Sodium alkoxide formed dissolve readily in water to form back alcohol + sodium hydroxide
The reaction is slower than when sodium reacts with water.
Reactivity decrease with the class increase 30 alcohol < 20 alcohol < 10 alcohol (less reactive)
Sodium hydroxide (NaOH) cannot react with aliphatic alcohol. a) CH3CH(OH)CH3 + Na
CH3CH(O–Na+)CH3 + ½ H2 b) C(CH3)2(OH)CH2CH3 + K
C(CH3)2(O–K+)CH2CH3 + ½ H2 c) CH3C(CH3)(OH)CH2CH3 + Na
CH3C(CH3)(O–Na+)CH2CH3 + ½ H2

(2) Esterification : Formation of ester
When excess alcohol (R–OH) react with carboxylic acid (R”COOH) and catalysed by a few drops of concentrated sulphuric acid and heat under reflux.
R–OH + R”COOH
R”COO–R + H2O Alcohol

Carboxylic acid

CH3CH2OH Ethanol

CH3COOH Ethanoic acid

CH3CH2CH2OH Propan-1-ol

CH3CH2CH2COOH Butanoic acid

CH3CH2CH2CH2OH Butan-1-ol

CH3CH2OH Ethanol

Ester

CH3COOCH2CH3 Ethyl ethanoate

Water

H 2O

CH3CH2CH2COOCH2CH2CH3 propyl butanoate

H 2O

CH3CH2COOH Propanoic acid

CH3CH2COOCH2CH2CH2CH3 butyl propanoate

H 2O

CH3CH2COOH Propanoic acid

CH3CH2COOCH2CH3 ethyl propanoate

H 2O


Esterification can also be achieved by replacing carboxylic acid with

alkanoyl chloride Example :

ethanol ethanoyl chloride ethyl ethanoate hydrogen chloride
Note that in the reaction above, no acidic medium is required. Compare to carboxylic acid, alkanoyl chloride is more reactive than carboxylic acid. Also, the reaction produces a white fume of hydrogen chloride as side product.

(3) Oxidation of alcohol
Using strong oxidising agent such as acidified potassium dichromate [K2Cr2O7 / H+], alcohol can be oxidise to form carbonyl compound and even to carboxylic acid.
Using different categories of alcohol, different type of carbonyl compounds are formed. Class

Example

10 alcohol (methanol )

CH3OH

10 alcohol

CH3CH2CH2OH propan-1-ol

20 alcohol

CH3CH(OH)CH3 propan-2-ol

30 alcohol

CH3C(CH-3)(OH)CH3 2-methylpropan-2-ol

Reaction

No reaction


Note the following changes occur in the oxidation of alcohol
Oxidation of primary (10) alcohol will yield an aldehyde while

oxidation of secondary (20) alcohol will yield a ketone.
Aldehyde formed from 10 alcohol can be further oxidised to form carboxylic acid. For the case of methanal, further oxidation of methanal will yield carbon dioxide and water.
Tertiary (30) alcohol is not oxidised when react with strong oxidising agent as it does not have H attached to the C–OH.
The differences in behaviour of alcohols toward oxidising agents may be used to distinguish between 10 alcohol, 20 alcohol and 30 alcohol. So, this is consider a basic test to distinguish between the class of alcohol used.

Alcohol

Product

CH3CH2CH2COOH

CH3CH2COCH3

No reaction

Alcohol

Product

CH3CH2COCH2CH3

C(CH3)3COOH

CH(CH3)2COCH3


In industries, oxidation of alcohol is carried under catalytic

dehydrogenation, where hydrogen is removed from the alcohol, forming aldehyde, ketone and even an alkene. Class

Example

10 alcohol

CH3CH2CH2OH propan-1-ol

20 alcohol

CH3CH(OH)CH3 propan-2-ol

30 alcohol

CH3C(CH-3)(OH)CH3 2-methylpropan-2-ol

Reaction


Note that, unlike oxidation using acidified potassium manganate (VII),

here, the side product is hydrogen gas. Furthermore, aldehyde and ketone formed are not further oxidised.

(4) Dehydration of alcohol
Dehydration of alcohol is an elimination reaction where water is removed from organic compound.
Dehydration of alcohol can be carried out under these conditions :
Heating mixture of excess concentrated acid such as H2SO4 at

1800C
Passing alcohol vapour over aluminium oxide (Al2O3) as catalyst at 3000C.


Dehydrating 1o alcohol will yield only 1 product whereas dehydrating 2o alcohol will yield 2 products.

Class

Example

10 alcohol

CH3CH2CH2OH propan-1-ol

20 alcohol

CH3CH2CH(OH)CH3 butan-2-ol

Result

Propan-1-ol

propene

CH3 O4 . H 2S c n o c

CH3

30

alcohol

CH3CH2C(CH3)(OH)CH3 CH3CHC 2-methylbutan-2-ol

CH2

H OH H

oC 180

conc .H S 2 O 4 1

80 oC

CH3CH2C

CH2

2-methylbut-1-ene CH3 CH3CH

CCH3

2-methylbut-2-ene

+

H2O


The major/minor products of the alkene formed followed Saytzeff’s

Rule where alkene containing the greater alkyl is predominant. (H atom from a lesser C–H is preferably to be eliminated)
However, if excess alcohol react with concentrated H2SO4, ether is given off.
2 CH3CH2CH2OH
CH3CH2CH2–O–CH2CH2CH3 + H2O Propan-1-ol dipropyl ether
Same result is given off by using aluminium oxide (Al2O3). The ease of dehydration increase in order from 10 alcohol < 20 alcohol < 30 alcohol. Example : Write out the possible products for dehydration of these alcohols 1. CH3CH2CH(OH)CH2CH3
CH3CH2CH=CHCH3 2. CH(CH3)2CH(OH)CH3
CH(CH3)2CH=CH2 3. C(CH3)3OH
C(CH3)2=CH2

+ C(CH3)2=CHCH3

(5) Halogenation of alcohol – formation of haloalkane
As introduced in the earlier chapter, haloalkane can be prepared by adding conc. hydrochloric acid (HCl) with the aid of zinc chloride, ZnCl2 to alkene. This mixture is called as Lucas reagent.
Halogenation can also be carried out using halogen rich compound, such as phosphorous (V) pentachloride (PCl5) or thionyl chloride (SOCl2). CH3CH2CH2OH + HCl (conc)
CH3CH2CH2Cl + H2O propan-1-ol CH3CH2CH(OH)CH3 + PCl5
CH3CH2CH(Cl)CH3 + HCl + POCl3 butan-2-ol CH3C(CH3)(OH)CH3 + SOCl2
C(Cl)(CH3)3 + HCl + SO2 2-methylpropan-2-ol


To prepare a bromoalkane, reagent used is concentrated

hydrobromic acid, HBr, catalysed by concentrated sulphuric acid CH3CH2CH2OH + HBr (conc)
CH3CH2CH2Br + H2O Propan-1-ol

5.4 Phenol and Aromatic alcohol 5.4.1 Manufacturing of phenol
There are 3 methods of making phenol. # The cumene process # The hydrolysis of chlorobenzene / diazonium salt # Alkali fusion with sodium benzenesulphonate (1) Synthesising phenol – cumene reaction
Step 1 : Formation of cumene using benzene ring and propene.

Benzene

propene


Step 2 : Oxidation of cumene.

cumene


Step 3 : Decomposition by sulphuric acid : Migration of phenyl group

cumene hydroperoxide

phenol

propanone

(2) Hydrolysis of chlorobenzene : Dow Process
Phenol has been process using Dow process widely in chemical industries. It generally involve 2 steps. Step 1 : Hydrolysis of chlorobenzene by NaOH to form phenoxide salt.

chlorobenzene

sodium phenoxide

Step 2 : Distillation of phenoxide salt mixed with hydrochloric acid.

Sodium phenoxide

phenol

(3) Hydrolysis of diazonium salt
In laboratory, phenol is prepared by hydrolysis of diazonium salt. Effervescence occur and a colourless gas is given out, along with the white fume of hydrogen chloride

benzenediazonium chloride

water

phenol


The formation of azo will be discussed extensively when we are in

amine later part

5.5 Chemical reaction of phenol
The –OH act as ring activating groups when attached to benzene. As a result, it activates the rings and cause benzene to be more reactive. Consequently, phenol is more reactive toward electrophilic substitution than benzene. (1)

Halogenation of phenol
When bromine water is added to phenol at room temperature, brown colour of bromine water decolourised and formed a white precipitate of 2,4,6tribromophenol.

(2) Nitration of phenol
When concentrated nitric acid (HNO3) is used to react with phenol, it formed a yellow crystalline solid of 2,4,6-trinitrophenol. This yellow crystal is used in dyeing industries and make explosive

(3) Reaction with iron (III) chloride, FeCl3
When a few drops of iron (III) chloride solution is added to phenol, a violet-blue colouration produced. A methylphenol produce blue colour.

5.5

Chemical Test to distinguish between alcohols

Differentiate Methanol (CH3OH) and other alcohol Ethanol (C2H5OH) and other alcohol Alkan-2-ol (RCHCH3) OH and other alcohol

Chemical test

Observation

Positive test : Methanol Purple colour of potassium manganate is decolourised. Effervescence (Bubbling) occurs. Gas released turn lime water chalky Equation : CH3OH → H2C=O → CO2 + H2O Positive test : Ethanol Add NaOH then iodine and heated gently. Pale yellow Iodoform test crystal of triiodomethane is formed. Equation : CH3CH2OH + 4I2 + 6
CHI3 + 5 I- + HCOOAcidified potassium manganate (VII)

Iodoform test

[O]

[O]

Positive test : alkan-2-ol Add NaOH then iodine and heated gently. Pale yellow crystal of triiodomethane is formed. Equation : R–CH(OH)CH3 + 4 I2 + 6
CHI3 (s) + RCOO– + 5 I– + 5 H2O

Phenol with other alkylalcohol

Positive test : Phenol Add bromine water directly to phenol. The brown colour of Bromine water bromine water is bleached instantly and a white precipitate is formed. Equation : refer above Positive test : Phenol Iron (III) Add iron (III) chloride solution to phenol and a violet-blue chloride solution formed instantly.

H

H

H

CH3 H

C

C

C

C

H

H

H

H

OH

H

H

CH3 H

H

C

C

C

C

H

H

H

H

OH

CH3CH2CH2CH2CH2OH + HBr
CH3CH2CH2CH2CH2Br + H2O

Mol of 1-bromopentane formed = 15.0 / 5(12) + 11(1) + 80 = 0.09934 mol Since only 60% ; mol of 1-bromopentane should be formed = 0.09934 x 100 / 60 mol = 0.1657 mol Mass of pentan-1-ol required = 0.1657 x [ (5(12) + 12(1) + 16) ] = 14.6 g (3.s.f. with unit)

Excess concentrated H2SO4 under reflux/ Al2O3 heated strongly CH3CH2CH2CH2CH2OH
CH3CH2CH2CH=CH2 + H2O

KMnO4 / H+ or K2Cr2O7 / H+ under reflux CH3CH2CH2CH2CH2OH + KMnO4 / H+
CH3CH2CH2CH2COOH + H2O CH3COOH catalysed by conc. H2SO4 under reflux / CH3COCl CH3CH2CH2CH2CH2OH + CH3COOH
CH3CH2CH2CH2OCOCH3 + H2O

100

Type of reaction : elimination reaction

Reagent : PCl5 Observation : White fume released by leaf alcohol while the other does not Equation : CH3CH2CH=CHCH2CH2OH + PCl5
CH3CH2CH=CHCH2CH2Cl + POCl3 + HCl

C10H20O alkene

156 alcohol

Citronellol : optical isomerism

Geraniol : geometrical isomerism

Observation : brown colour of aqueous bromine decolourised Explanation : due to the presence of unsaturated C=C Equation :

O CH(CH3)2CH2CH2CH2

ii.

CH3

CH2O C

C

H

C

OSO3H H CH(CH3)2CH2CH2CH2 CH2Br

iii.

CH3

C

C

Br

H

H

CH3 + H2O

chlorine gas Electrophilic aromatic substitution

neutralisation

Bromine water white precipitate is formed brown colour of bromine remain unchanged

Carbon attached with –OH, that was surrounded by 1 carbon

H

H

H

OH H

C

C

C

C

H

H

H

H

secondary

H H

H

CH3 H

C

C

C

H

H

H

primary

OH H

H

CH3 H

C

C

H

OH H

tertiary

C

H

orange Isomer 3

green

5

(i) sodium metal

(ii) Br2 (aq) OH CH3

Br

HO Br

(iii) NaOH(aq)

(iv) CH3COCl

(v)

hot acidified K2Cr2O7

(vi)

PCl5

H

H

H

H

H

H

C

C

C

C

C

H

H

H

H

H

OH

H

H

H

OH H

H

C

C

C

C

C

H

H

H

H

H

B (pentan-2-ol) B (pentan-2-ol) A yellow precipitate is formed H H C H

H H

O

C C C H H

+ CHI3 O

H