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

CH3CH2CH(CH3)CH2OH C(CH3)3CH(OH)CH2CH3 C(CH3)3CH2C(OH)(CH3)2
4-methylpentan-2-ol 3,3-dimethylbutan-2-ol 3-ethyl-2,4-dimethylpentan-2-ol
3-methylpentan-2-ol 3,4-dimethylheptan-4-ol 5-ethyl-
2,5-dimethylheptan-2-ol
4-methylpentan-1-ol 4-ethyl-2-methylhexan-2-ol 5-ethyl-2-methylheptan-
3-ol
2-methylbutan-1-ol 2,2-dimethylpentan-3-ol 2,4,4-trimethylpentan-2-ol

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

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

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
Similar to other organic compounds, the boiling point ……………. with the
number of carbon as the weak Van Der Waals forces increase with
………………………… of the compound.
Though, the hydrogen bonding are weaker when goes down to homologous
series as the polarity of molecules ……………… 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
Alcohol CH3OH C2H5OH C3H7OH C4H9OH C5H11OH C6H13OH C7H15OH C8H17OH
Boiling point oC 46 78 90 115 135 152 169 190
Boiling point
trend BOILING POINT INCREASE DOWN HOMOLOGUS SERIES
increase
molecular mass
decrease

Straight chain molecules have higher boiling point compare to branched
chain because straight chain molecule has a ………… surface area than a
branched chain molecule. The more branches attached to the parent chain,
the ……………. the surface area ; …………… the forces of attraction between
molecules ; ………… the boiling point.
Molecules
Butan-1-ol Butan-2-ol 2-methylpropan-2-ol
Boiling point
(0C)
117 99 82
larger
smaller weaker
lower

The number of the hydroxyl group in an organic compound is also one of the
major factor which contribute to its boiling point
The boiling point of the alcohol increase with the number of –OH. This is a
result caused by more ………………….. bond formed between –OH of the
molecules. So the more the –OH ; stronger the hydrogen bond ; higher the
boiling point.
Molecules
Butan-1-ol Butan-1,2-diol Butan-1,2,3-triol
Boiling point
(0C)
117 208 274
hydrogen

Compare to alkane and haloalkane, alcohol has a higher boiling point
Alcohol has the highest boiling point compare to other organic compound
because it forms strong ……..…………. bond between the molecules.
Fluoroethane has a higher boiling point than propane as fluoroethane is a
……………………. molecules and so, the weak ………………………… forces are
stronger than propane since propane is a …………………..…. molecule.
Compound Ethanol (C2H5OH)
Propane
(C3H8)
Fluoroethane
(C2H5F)
Relative molecular
mass
46 44 48
Boiling point (oC) 78 – 4.2 7
hydrogen
polar
Van Der Waals’
Non-polar

B)Solubility of alcohol in water
Hydrogen bonding occur between alcohol molecules because of the
presence of ……………. 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.
hydroxy

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. (dipole-dipole
interaction between R- and R- are more obvious)
Ethanol is a good solvent for both polar and non polar solute because it
contain non polar (………….) group and a polar group (……………) in it. As a
result, ethanol is used as solvent in many industries
alkyl hydroxyl

(C) Acidity of alcohol
Alcohols are generally a weak acid. Table below shows the pKa value of some
alcohols and water
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.
Alkyl is an electron …………..…….. group whereas phenyl is an electron
…………..……… group.
Compounds
Methanol
(CH3OH)
Ethanol
(C2H5OH)
Propan-1-ol
(C3H7OH)
Phenol
(C6H5OH)
p-methylphenol
CH3-C6H5OH
Water
pKa 15.5 16.0 18.0 10.0 11.0 14.0
donating
withdrawing

Alcohol Explanation
δ+ δ−
CH3CH2–O–H
Ethanol
Ethanol dissociate in water according to the equation
CH3CH2–OH + H2O CH3CH2–O- + H3O+
Alkyl group, which is an electron donating group, donate electron
to O and caused the electron density of O in R–OH increase. As a
result, O is more readily to accept proton, which makes the
equilibrium favours to left.
Phenol
Phenol dissociate in water according to the equation
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
Hydration of
alkene
Steam (H2O)
---------
Phosphoric acid,
(H3PO4 )
At 300oC ; 60
atm
@
Concentrated H-
2SO4 at 800C.
Hydrolysis of
haloalkane
NaOH (aq)
under reflux
1-chloropropane sodium propan-1-ol
Hydroxide
Reaction of
Grignard
reagent
Aldehyde /
ketone with
H2SO4
CH3CH2MgBr + CH3CH=O
CH3CH2CH(OH)CH3
 →
+
OH3

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
• Formation of alkoxide
• Formation of ester
• Oxidation of alcohol
• Dehydration of alcohol
• Reaction with hydrogen halide
• Reaction with phosphorous halide
(PX5) or thionyl chloride, SOCl2
Name of reaction
Reagent used and
condition
Equation
Formation of
alkoxide
Sodium (Na)
2 CH3CH2O–H + 2 Na 2 CH3CH2O–Na+ + H2 (g)
Ethanol sodium sodium ethoxide hydrogen
Esterification
Carboxylic acid
(R–COOH) catalysed
by conc. sulphuric
acid (H2SO4)
Ethanol propanoic acid ethyl propanoate water

Name of
reaction
Reagent used and
condition
Equation
Oxidation of
alcohol
Acidified KMnO4 or
acidified K2Cr2O7 +
heat
propan-1-ol propanal propanoic acid
propan-2-ol propanone
Dehydration
(removal of
water)
from
alcohol
Excess conc.
H2SO4
at 1800C
or
Alumina (Al2O3) at
350oC
Halogenation of
alcohol
Phosphorous
pentachloride
(PCl5)
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
b) C(CH3)2(OH)CH2CH3 + K
c) CH3C(CH3)(OH)CH2CH3 + Na
CH3CH(O–Na+)CH3 + ½ H2
C(CH3)2(O–K+)CH2CH3 + ½ H2
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 Ester Water
CH3CH2OH
Ethanol
CH3COOH
Ethanoic acid
CH3CH2CH2OH
Propan-1-ol
CH3CH2CH2COOH
Butanoic acid
CH3CH2CH2CH2OH
Butan-1-ol
CH3CH2COOH
Propanoic acid
CH3CH2OH
Ethanol
CH3CH2COOH
Propanoic acid
CH3COOCH2CH3
Ethyl ethanoate
H2O
CH3CH2CH2COOCH2CH2CH3
propyl butanoate
H2O
CH3CH2COOCH2CH2CH2CH3
butyl propanoate
H2O
CH3CH2COOCH2CH3
ethyl propanoate
H2O

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 Reaction
10 alcohol
(methanol)
CH3OH
10 alcohol
CH3CH2CH2OH
propan-1-ol
20 alcohol
CH3CH(OH)CH3
propan-2-ol
30 alcohol
CH3C(CH3)(OH)CH3
2-methylpropan-2-ol
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 Alcohol Product
CH3CH2CH2COOH CH3CH2COCH2CH3
CH3CH2COCH3
C(CH3)3COOH
No reaction
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.
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.
Class Example Reaction
10 alcohol
CH3CH2CH2OH
propan-1-ol
20 alcohol
CH3CH(OH)CH3
propan-2-ol
30 alcohol
CH3C(CH3)(OH)CH3
2-methylpropan-2-ol

(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 Result
10 alcohol
CH3CH2CH2OH
propan-1-ol Propan-1-ol propene
20 alcohol
CH3CH2CH(OH)CH3
butan-2-ol
30 alcohol
CH3CH2C(CH3)(OH)CH3
2-methylbutan-2-ol

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
2. CH(CH3)2CH(OH)CH3
3. C(CH3)3OH
CH3CH2CH=CHCH3
CH(CH3)2CH=CH2 + C(CH3)2=CHCH3
C(CH3)2=CH2

(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)
propan-1-ol
CH3CH2CH(OH)CH3 + PCl5
butan-2-ol
CH3C(CH3)(OH)CH3 + SOCl2
2-methylpropan-2-ol
CH3CH2CH2Cl + H2O
CH3CH2CH(Cl)CH3 + HCl + POCl3
C(Cl)(CH3)3 + HCl + SO2

To prepare a bromoalkane, reagent used is concentrated hydrobromic
acid, HBr, catalysed by concentrated sulphuric acid
CH3CH2CH2OH + HBr (conc)
Propan-1-ol CH3CH2CH2Br + H2O

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 cumene
Step 2 : Oxidation of 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 2,4,6-
tribromophenol.
(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 Chemical test Observation
Methanol
(CH3OH) and
other alcohol
Acidified
potassium
manganate (VII)
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
Ethanol
(C2H5OH) and
other alcohol
Iodoform test
Positive test : Ethanol
Add NaOH then iodine and heated gently. Pale yellow crystal of
triiodomethane is formed.
Equation : CH3CH2OH + 4I2 + 6 CHI3 + 5 I- + HCOO-
Alkan-2-ol
(R-CHCH3)
OH
and other
alcohol
Iodoform test
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
→ ]O[→ ]O[

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

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 156
alkene alcohol

Citronellol : optical isomerism
Geraniol : geometrical isomerism
Observation : brown colour of aqueous bromine decolourised
Explanation : due to the presence of unsaturated C=C
Equation :

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
secondary primary tertiary

orange green
Isomer 3

5

(i) sodium metal (ii) Br2 (aq)
(iii) NaOH(aq) (iv) CH3COCl

(v) hot acidified K2Cr2O7 (vi) PCl5

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