Download - Organic Chemistry : Alcohol
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
mass46 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 alkanesand 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.
CompoundsMethanol
(CH3OH)
Ethanol
(C2H5OH)
Propan-1-ol
(C3H7OH)
Phenol
(C6H5OH)
p-methylphenol
CH3-C6H5OHWater
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 conditionEquation
Fermentation
of glucoseZymase enzyme
C6H12O6 � 2 CH3CH2OH + 2 CO2 Glucose
ethanol carbon dioxide
Name of
reaction
Reagent used
and conditionEquation
Hydration of
alkene
Steam (H2O)
---------
Phosphoric acid,
(H3PO4 )
At 300oC ; 60
atm
@
Concentrated H-
2SO4 at 800C.
Hydrolysis of
haloalkane
NaOH (aq)
under reflux1-chloropropane sodium propan-1-ol
Hydroxide
Reaction of
Grignard
reagent
Aldehyde /
ketone with
H2SO4
CH3CH2MgBr + CH3CH=O
CH3CH2CH(OH)CH3
→+OH 3
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 reactionReagent used and
conditionEquation
Formation of
alkoxideSodium (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
conditionEquation
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 ethanoateH2O
CH3CH2CH2COOCH2CH2CH3
propyl butanoateH2O
CH3CH2COOCH2CH2CH2CH3
butyl propanoateH2O
CH3CH2COOCH2CH3
ethyl propanoateH2O
� Esterification can also be achieved by replacing carboxylic acid with alkanoylchloride
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 alcoholCH3CH2CH2OH
propan-1-ol
20 alcoholCH3CH(OH)CH3
propan-2-ol
30 alcoholCH3C(CH3)(OH)CH3
2-methylpropan-2-olNo 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 reactionCH(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 alcoholCH3CH2CH2OH
propan-1-ol
20 alcoholCH3CH(OH)CH3
propan-2-ol
30 alcoholCH3C(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 alcoholCH3CH2CH2OH
propan-1-ol Propan-1-ol propene
20 alcoholCH3CH2CH(OH)CH3
butan-2-ol
30 alcoholCH3CH2C(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 hydrobromicacid, 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 dyeingindustries 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 : PCl5Observation : 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