CHAPTER 4.0 : AROMATIC COMPOUNDS |CHEMISTRY UNIT KMKt 2021/2022
9. Which of the following compounds is wrongly named?
Compound Name
A NH2 Aniline
BO Benzaldehyde
CH
CO Acetophenone
C CH3
D OH Toluene
10. Which of the following compounds is formed when a mixture of benzene and
chloroethane is reacted together in the presence of anhydrous Aluminium chloride?
A CH3 C Cl
CH2CH3
B CH2 CH3 D Cl
Revised by:
Noor Azida Binti Mohd Yusoff
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CHAPTER 5 : ALCOHOLS
GENERAL OVERVIEW
5.1 Nomenclature a) Draw and name structures of alcohol according
to the IUPAC nomenclature.
* limits parent chain ≤ C10
b) Classify into primary i. within successive
(10), secondary (20), members; and
tertiary (30) alcohols
5.2 a)Compare the ii. with alkanes
Physical boiling points of
Propertie alcohol: *exclude the comparison of boiling
point for different classes of
s alcohols.
b) Compare the solubilty of i.fermentation; and
alcohol in water.
* limit comparison to two
compounds
5.0 5.3 a) Explain the ii. hydration of alkenes.
ALCOHOLS Preparation preparation of alcohols
*exclude the
through: mechanism.
*apply Markovnikov's
rule
Explain the reaction with Group 1 and
reaction of alcohol Group 2 metal:
with references to:
oxidation with hot acidified KMnO4
5.4 or K2Cr2O7:
Chemical
properties oxidation with PCC in CH2Cl2
Explain the chemical esterification
tests to distinguish dehydration
between 10,20 and 30
alcohols using Lucas
reagent
Explain iodoform test to identify
methyl carbinol group,
*include observations and equations
in (b) and (c)
(Experiment 3- Reactions of Alcohol)
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INTRODUCTION
• Hydroxy compounds include aliphatic alcohols, phenols and aromatic alcohols.
Aliphatic alcohols are organic compounds with at least one hydroxyl (-OH) group
attached to the alkyl group.
• The functional group is the hydroxyl (-OH) group.Aliphatic open chain alcohols with
one -OH group have the general formula CnH2n+1OH or CnH2n+2O, where n ≥ 1.
• Aliphatic alcohols can be considered as derivatives of alkanes with hydrogen atoms
replaced by hydroxyl groups.
Structural Formula Chemical Formula
C2H5OH CH3CH2OH
5.1 NOMENCLATURE
In the IUPAC system,alcohols are named by replacing the final ‘e’ of the corresonding
alkane with the suffix ‘ol’ .
Straight chain (unbranched) alcohol
IUPAC names of alcohol have the -ol suffix.
Molecular Structural formula No. of C Name
formula (Condensed Structure) atoms
CH3OH CH3OH 1 Methanol
C2H5OH CH3CH2OH 2 Ethanol
C3H7OH CH3CH2CH2OH 3 1-Propanol
C4H9OH CH3CH2CH2CH2OH 4 1-Butanol
C5H11OH CH3CH2CH2CH2CH2OH 5 1-Pentanol
C6H13OH CH3(CH2)5OH 6 1-Hexanol
C7H15OH CH3(CH2)6OH 7 1-Heptanol
C8H17OH CH3(CH2)7OH 8 1-Octanol
C9H19OH CH3(CH2)8OH 9 1-Nonanol
C10H21OH CH3(CH2)9OH 10 1-Decanol
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NOMENCLATURE OF BRANCHED ALCOHOL
The following steps are used in naming an alcohol.
• Identify the longest carbon • Number the carbon atoms in
chain that includes the carbon
atom joined to the –OH group. the longest chain so that the
carbon atom attached to the –
• Drop the final ‘e’ from the OH group is given the lowest
alkane name and replace it with number possible.
the suffix ‘ol’.
• Name all the subtituents • The position of the hydroxyl
present in the alcohol molecule group is indicated by the
and give their numbers. number of the carbon atom to
which it is attached.
1 23 4
2–methyl–2–butanol 6 54 32 1
5–methyl–3–hexanol
Structure Common name IUPAC name
CH3-OH methyl alcohol methanol
CH3CH2-OH ethyl alcohol
CH3CH(OH)CH3 isopropyl alcohol ethanol
2-propanol
OH cyclohexyl alcohol
cyclohexanol
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Cyclic Alcohol
H OH H OH
C HC CH
HC CH HC CH
HH HH
cyclopropanol cyclobutanol
H OH H OH
H
H CH HC
HC CH C CH
H H
C CH
HH H C C
cyclopentanol HCH
H
H H
cyclohexanol
Ring : Numbering start at C bearing OH. 5
4
6
51 1
42
32
3
2–chlorocyclohexanol
2–bromo–5–methylcyclopentanol
EXAMPLE:
CH —CH —CH —CH —OH
3222
1979 IUPAC recommendation: 1–butanol The first of
1993 IUPAC recommendation: butan–1–ol these
convention
1–butanol
more widely
used
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If higher priority group exist in the same molecule,-OH group is named as prefix hydroxyl.
Examples:
CH Keep in mind!
3
HOOCCHCH CH CH OH Increasing MAIN GROUPS
priority carboxylic acids
1 2 3 24 2 5 2
5–hydroxy-2-methylpentanoic acid esters
Cl OH aldehydes
ketones
1 23 4 5 6 7 alcohol
amine
CH CCHCH CH CHCH alkenes
3 22 3 alkanes / halides
O
3–chloro-6–hydroxy-4-methylheptanone
❑ For unsaturated alcohols (containing a double or triple bond) the hydroxyl group
takes precedence over multiple bonds in the numbering of the carbons atoms.
❑ Thus, the systematic name of allyl alcohol (CH2=CHCH2OH) is 2 –propen-1-ol
and not 1-propen -3-ol.
Example:
6 54 1
2
3
cis–2,3–dimethyl-2–hexene–1-ol 5
1 2 34
CH3
HO CH2 CH CH CH2
pent–2–en–1–ol or 2–penten–1–ol
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1
62
5 3
cyclohex–2–en–1–ol 4
or 2–cyclohexen–1–ol
❑ Alcohols with two –OH groups are called diols or glycols.
❑ They are named like other alcohols except that the suffix diols is used and two
numbers are needed to tell where the two hydroxyl groups are located,
❑ This is the preferred,systematic (IUPAC) method for naming diols.
❑ The term glycol generally means a 1,2–diol,or vicinal diol,with its two hydroxyl groups
on adjacent carbon atoms.
❑ Glycols are usually synthesized by the hydroxylation of alkenes, using peroxy acids,
osmium tetroxide or potassium permanganate.
Example:
12 1
5
HO CH2 CH2 OH
2
1,2–ethanediol @ ethylene glycol
43
3 1,2–cyclopentanediol
54 21
6
4–chloro–1–phenyl–1,5–hexanediol
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AROMATIC ALCOHOLS
❑ In most cases, aromatic alcohols are named with phenol as the parent name except
if higher priority group exist.
1 3
62
14 2
5 43
2 6 1
5
6
3-hydroxybenzoic acid
OH
35 phenol
4
2- nitrophenol
EXERCISE 1:
Give the IUPAC names of the following compounds.
a) b)
c)
HO—(CH ) —OH
e) 2 8
CLASSIFICATION OF ALCOHOLS
❑ Alcohols can be grouped into three classes, namely primary, secondary and tertiary
alcohols according to the number of alkyl groups bonded to the carbon atom that has
the – OH group attached.
❑ In primary alcohols, the –OH group is bonded to a carbon atom with one or no alkyl
groups attached. Ethanol is an example of a primary alcohol.
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❑ In secondary alcohols, the carbon atom attached to the –OH groups has two alkyl
groups attached. For example, 2 – propanol is a secondary alcohol.
❑ In tertiary alcohols, three alkyl groups are attached to the carbon atom bonded to
the –OH group. For example, 2 – methyl- 2 propanol is a tertiary alcohol.
CLASS GENERAL FORMULA EXAMPLE
R–OH CH CH –OH
o R–CH–R
32
1 OH
R CH –CH–CH
o
RC R 33
2
OH OH
o
CH3
3
CH3 C CH3
EXAMPLE 1: OH
Classify the types of alcohol (1o, 2o or 3o) for the following molecules.
(a) CH CH(OH)CH b) (CH ) C(OH)CH CH
33 32 2 3
c)
CH2OH
ANSWERS: b) CH3CH2OH
a) 20 alcohol
b) 30 alcohol 105 | P a g e
c) 10 alcohol
EXERCISE 2:
Classify the following alcohols:
a) CH3CH2CHOH
CH2CH3
c) OH
CHAPTER 5: ALCOHOLS | CHEMISTRY UNIT KMKt 2021/2022
5.2 PHYSICAL PROPERTIES
BOILING POINT
• In liquid form, molecules of alcohol can form hydrogen bond between themselves
(OH group) and also van der Waals forces between alkyl groups.
H Hydrogen bond
O H
CH3CH2 O
Van der Waals forces CH3CH2
• As molecular weight increase (size of molecule increase, number of carbon atoms
increase), the strength of van der Waals forces increase, thus the boiling point will
increase.
Example:
Compound Molecular weight Boiling Point (oC)
(gmol-1) 78
Ethanol 46 118
CH3CH2OH
1-butanol 74
CH3CH2CH2CH2OH
• Alcohols have higher boiling points than alkanes of similar molecular weight.
• Alcohols form hydrogen bond between molecules while alkanes only form weak
van der Waals forces between molecules.
• Hydrogen bond is stronger than van der Waals forces.
Example:
Compound Molecular weight Boiling Point (oC)
(gmol-1)
Ethanol 46 78
CH3CH2OH 44 -42
Propane
CH3CH2CH3
• For compounds of comparable molecular size, the stronger the intermolecular forces,
the higher the boiling point.
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EXAMPLE:
CH CH CH CH CH OCH CH CH CH CH CH OH
3223 3 23 3222
London Forces
Dipole-Dipole Dipole-Dipole Forces ,
Forces and London London Forces,
Forces Hydrogen Bond
Increasing boiling point
• For isomeric alcohol (same molecular weight):
• The boiling point decrease with branching
• Boiling point in decending order: -
• 1° alcohol > 2° alcohol > 3° alcohol
EXAMPLE:
Rank the compounds in each group in order of increasing boiling point.
OH
OH
SOLUTION:
an alkane Intermolecular force:
OH London Forces
an alcohol Hydrogen Bonding, London Forces ,
Dipole – Dipole Forces
lower molar mass
smaller surface area
OH Hydrogen Bonding, London Forces ,
an alcohol Dipole – Dipole Forces
higher molar mass
larger surface area
, OH , OH
Increasing boiling point
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SOLUBITY OF ALCOHOLS IN WATER
• Alcohols with short carbon chains ( such as methanol,ethanol and propanol)
dissolve readily in water.
• When the alcohols dissolve in water , hydrogen bonds are formed between the –OH
group of the alcohol molecule and the –OH group of the water molecule.
• The alcohol molecule contains a polar end ( the –OH group) and a non polar end (the
alkyl group).
• The polar end (hydrophilic) is soluble in water but the non polar end (hydrophobic) is
soluble in non polar solvents and insoluble in water.
• As the number of carbon atoms in the alcohol molecule increases, the influence of
the –OH group becomes less important compared to the influence of the
hydrocarbon chain which becomes more important.
• Consequently the solubility of alcohols in water decreases sharply with the increasing
lenght of the carbon chain.
• For example, 1 – butanol is partially miscible in water, but 1 –pentanol and 1 –
hexanol are only slightly soluble in water.
• Higher alcohols are insoluble in water.
• Alcohols with more than one hydroxyl group (polyhydric alcohols) are more soluble
than monohydric alcohols with the same number of carbon atoms
• This is because they can form more hydrogen bonds with water molecules.
Hydrogen bonding
Example:
Solubility:
HO(CH2)6OH > CH3(CH2)5OH
✓ HO(CH2)6OH has two OH groups.
✓ It can form more hydrogen bonding with water molecule
EXAMPLE :
Rank the compounds in order of increasing water solubility.
CH3CH2CH2 CH2 CH2 CH2OH CH CH CH CH OH
3222
Solution: < CH CH CH CH OH
3222
CH3CH2CH2 CH2 CH2 CH2OH
Solubility in water will decrease with the increase in molecular weight because the hydrophobic
area (alkyl group) becomes bigger.
1-hexanol has the lower solubility because it has the biggest molecular weight (Mr = 102) and
larger hydrophobic area than 1-butanol (Mr = 74).
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EXERCISE 3:
Arrange the following compounds in order of increasing solubility in water:
a) C4H9OH, CH3OH, C4H9Br
b) Pentanol, pentane- 1,2-diol, hexane-1,2,3,4,5,6-hexol
EXERCISE 4:
Rank the compounds in order of increasing boiling point.
CH CH CH CH (CH ) OH CH (CH ) OH
322 3 23 3 27
5.3 PREPARATION OF ALCOHOLS
FERMENTATION
❑ Fermentation is a chemical process in which microorganism such as yeast act on
sugars to produce ethanol and carbon dioxide.
❑ A biological process of enzymatic degradation of sugars by microorganisms
❑ Alcoholic drinks contain ethanol. The fermentation method has been used for
thousands years to make alcoholic drinks.
❑ Ethanol is manufactured industrially by the fermentation of sugars such as sucrose
(from cane sugar) or glucose (from grape juice), followed by distillation.
❑ During fermentation, aqueous glucose is converted into aqueous ethanol and carbon
dioxide .
❑ The reaction is catalyzed by the enzymes in yeast.
❑ The enzymes in yeast are denatured at temperatures above 45 0C and can no longer
function as a catalyst.
❑ Thus, the fermentation process must be carried out at temperature of 25 – 45 0C.
CH O yeast 2CH CH OH + 2CO
6 12 6 32 2
glucose ethanol
HYDRATION OF ALKENES
❑ Method for preparation of simple alcohol.
❑ Reactant : Alkene and water
❑ Condition : Catalyst (H2SO4)
❑ The acid-catalyzed addition of water to the double bond of an alkene (hydration of
an alkene) is a method for the preparation of low molecular weight alcohols.
❑ The most commonly acid used to catalyze the hydration of alkenes are dilute
solution of H2SO4 & H3PO4.
❑ The addition of water to the double bond follows Markovnikov’s rule
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❑ Ethanol can be produced by passing a mixture of ethene and steam over phosphoric
acid as catalayst at 300 0C and 60 atm.
❑ Other alcohols as well as ethanol can be synthesised by passing alkenes into
concentrated sulphuric acid at room temperature or 80 oC to produce alkyl
hydrogensulphate .
❑ The mixture is then diluted with water and distilled. The hydrolysis of alkenes
hydrogensulphate produces an alcohol.
❑ For example, when ethene undergoes indirect hydration in this way, ethanol is
produced.
Example: H2O OH
H2SO4 CH3 CCH2CH2CH2 CH3
CH2 C CH2 CH2CH2CH3
CH3 CH3
2- methyl - 2- hexanol
2- methyl-1-hexene
HO
2
H2SO4
CH CH3 OH
3
1-methylcyclopentanol
1-methylcyclopentene
EXERCISE 5:
Complete the reactions below:
1. CH CH=CH +
32 H O,H O
23
2. CH C(CH )=CH +
3 32 H O,H O
23
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3. H O , H SO4
22
CH3
+
4. (CH CH ) C═CHCH H O,H O
23
3 22 3
5. H O , H SO4
22
6. CH3
CH3 +
H O,H O
23
5.4 : CHEMICAL PROPERTIES OF ALCOHOL
(1) Cleavage of bond between O and H in –OH group (O – H bond is broken in
R – O – H):
(a) Reaction with Group 1 and Group 2 metal
(b) Esterification
(c) Oxidation
(2) Cleavage of bond between C and O (C – O bond is broken in R –O – H):
(a) Dehydration
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REACTION WITH GROUP 1 and GROUP 2 METAL
❑ Alcohols react with sodium at room temperature to form salts (sodium alkoxides) and
hydrogen.
❑ General reaction:
❑ RO-H + Na → RO-Na+ + ½H2
❑ 2RO-H + Mg (RO-)2 Mg2+ + H2
Examples:
CH3CH2OH + Na → CH3CH2O-Na+ + ½H2
❑ For example,ethanol reacts with sodium to form sodium ethoxide.
❑ The reaction is much slower than the reaction of water and sodium.
❑ When sodium is added to ethanol,it sinks (sodium is denser than ethanol) and then a
steady stream of hydrogen is produced.
❑ The evolution of hydrogen on the addition of metallic sodium is used to test the
presence of hydroxyl group in a compound.
❑ The reaction between alcohols and sodium to form hydrogen suggest that ethanol
acts as acids.
❑ The more acidic of alcohols such as methanol and ethanol,react rapidly.
❑ Secondary alcohols react slowly and tertiary alcohols react very slowly.
OXIDATION WITH HOT ACIDIFIED KMnO4 Or K2Cr2O7
❑ Alcohols can be oxidised by hot acidified potassium dichromate(Vl) solution or hot
acidified potassium permanganate(Vl) solution (KMnO4), dilute sulphuric acid is used
for acidification.
❑ Oxidation product depends on the class of alcohol used and types of oxidizing agent.
OXIDATION OF 1O ALCOHOL
❑ 1o alcohols are oxidized to aldehydes or carboxylic acids, depending on the
reagent:
Oxidising agent Product
Hot acidified KMnO4/H+ , Δ or K2Cr2O7/H+, Δ) carboxylic
acid
Both the reagent is strong oxidizing agent
and the product of reaction is carboxylic acid
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R CH H O
10 alcohol [O] R C OH
carboxylic acid
Example: + O
CH3 CH2 CH2 CH2 OH CH3 CH2CH2 C OH
K Cr O / H
2 27
▲
CH3 CH2 OH KMnO4 /H+ OH
CH3 C O
OXIDATION OF 20 ALCOHOL
Secondary alcohols are oxidised to ketones.
OH [O] O
R CR RC R
H ketone
20 alcohol
Example: O
OH
KMnO4 /H+
❑ Ketones are difficult to oxidise further. However, under vigorous oxidising
conditions,ketones are oxidised to form a mixture of carboxylic acids and this process
involves breaking down carbon- carbon linkages.
OXIDATION OF 30 ALCOHOL
❑ Tertiary alcohol are resistant to oxidation and can only be oxidised under severe
oxidising conditions.
❑ The differences in behaviour of alcohols towards oxidising agents may be used to
distinguish between primary,secondary and tertiary alcohols.
❑ Tertiary alcohol have no C–H bond on the C bearing the OH group.
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RC R
[O]
R no oxidation
30 alcohol
OXIDATION WITH PCC IN CH2Cl2
❑ Pyridinium chlorochromate (PCC) is the complex formed when CrO3 is added to
excess pyridine in concentrated HCl.
❑ It has the formula C5H6NCrO3Cl.
❑ Most oxidising agents oxidise primary alcohols to carboxylic acids.
❑ The oxidation of primary alcohols can be easily stopped at the aldehyde stage
without further oxidation to carboxylic acid if PCC in dichloromethane (CH2Cl2) is
used.
PCC O
CH3 CH2 CH2 CH2 OH CH3 CH2 CH2 CH
CH2Cl2
SUMMARY:
Class of alcohol Reaction with Product
Primary K2Cr2O7/ H2Cr2O7/ KMnO4 Carboxylic acid
PCC/CH2Cl2 Aldehyde
Secondary K2Cr2O7/ H2Cr2O7/ KMnO4
PCC/CH2Cl2
Ketone
Tertiary K2Cr2O7/ H2Cr2O7/ KMnO4 No reaction
PCC/CH2Cl2
ESTERIFICATION
❑ Esterification is the reaction between an alcohol and a carboxylic acid to form ester
and water.
❑ For primary alcohols and some secondary alcohols, the hydroxyl group of the
carboxylic acid is replaced by the alkoxyl (R-O) group of the alcohol.
❑ Esterification is an equilibrium reaction and is slow under normal conditions.
❑ The reaction can be speeded up by the addition of a strong acid catalyst.
❑ Esterification is carried out by boiling a mixture of the carboxylic acid and alcohol
(usually in excess) under reflux in the presence of a little concentrated sulphuric acid
or by bubbling gaseous hydrogen chloride through the reaction mixture.
❑ Some typical equations for the formation of esters are as follows.
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Reactant : ESTERIFICATION
Condition : carboxylic acid and alcohol
Product : • Reflux or heat or Δ
• Catalyst
(acid such as H2SO4 or HCl )
• Ester and H2O
O H+ O
R OH + R C OH R C OR + H2O
ester
Examples: H+ O
O
CH3 CH2 OH CH3 C O CH2 CH3 + H2O
CH3 C OH +
O CH3 OH H+ O
C OCH3 + H2O
C OH
+
DEHYDRATION OF AlCOHOLS
❑ Dehydration of alcohols to form alkenes can be carried out by
• Passing the alcohol vapour over an aluminium oxide catalyst ( Lewis Acid)
heated in 350 0C, or
• Heating a mixture of alcohol and Bronsted – Lowry acid (proton donor) such as
concentrated sulphuric acid or concentrated phosphoric(v) acid about 170 – 180
0C.
DEHYDRATION
Reactant : Alcohol
Condition : • Conc . H2SO4
Product : • Heat or reflux or
• Alkene and H2O
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H H H H
RC RC
conc. H2SO4 C H + H2O
H
C OH
heat
H
Examples: conc. H2SO4 CH2 CH2 + H2O
heat
CH3 CH2
OH
conc. H2SO4 + H2O
OH heat
❑ When secondary or tertiary alcohols containing four or more carbon atoms
undergoes dehydration reaction, a mixture of two alkenes is produced .
❑ For example , in the dehydration 2–butanol, 1-butene or 2-butene may be formed.
❑ According to the Saytzeff rule, the isomer with the greatest number of alkyl groups
attached in the double bond predominate.
❑ 1–butene has only one alkyl group attached to the double bond whereas 2–butene
has two alkyl groups attached to the double bond.
❑ Thus ,the dehydration of 2–butanol will produce 2–butene as the predominant
product.
CH3 CH CH2 CH3 conc. H2SO4 CH2 CH CH2 CH3 + CH3 CH CH CH3
OH heat
minor major
EXERCISE 6:
Give the product formed for the following reaction:
a) O H
+ Na
O
b) H+
C OH + CH3OH
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CH3 H2SO4
c)
CH3 C CH2 CH3
OH
EXERCISE 7:
Show how each of the following transformations could be accomplished:
a) CH2 OH ? O
C OH
OH O
b) ?
CHEMICAL TESTS TO DISTINGUISH BETWEEN 10 , 20 and 30 ALCOHOLS USING
LUCAS REAGENT.
❑ Primary, secondary and tertiary alcohols are distinguished by their different rates of
reaction with concentrated hydrochloric acid to form alkyl chloride (chloroalkane).
❑ The alcohol is shaken with Lucas reagent ( a solution of zinc chloride in
concentrated hydrochloric acid).
❑ Immediate cloudiness (due to the formation of alkyl chloride ) indicates a tertiary
alcohol.
❑ If the solution turns cloudy within about five minutes, a secondary alcohol is
indicated.
❑ Primary alcohols show no cloudiness at room temperature.
❑ Lucas reagent: HCl(con.) + ZnCl2
HCl, ZnCl2 R Cl + H2O
R OH
TYPE OF ALCOHOL OBSERVATION
o No cloudy solution form at room
1 temperature even after 15 minutes.
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o Cloudy solution formed within
2 5 – 10 minutes.
o Cloudy solution formed
3 immediately.
Example: HCl no observable changes
ZnCl2 within 15 min
CH CH CH CH CH OH
32222
1–pentanol
o
(1 alcohol)
Observation: No cloudy solution form at room temperature
even after 15 minutes.
OH HCl, ZnCl2 Cl
CH3CH2 CH CH2 CH3 CH3 CH2CHCH2CH3 + H2O
3–chloropentane
3–pentanol
o
(2 alcohol)
Observation: Cloudy solution formed within 5 - 10 minutes.
OH HCl, ZnCl2 Cl
CH3 C CH2 CH3 + H2O
CH3C CH2CH3
CH3
CH3 2–chloro-2-methylbutane
2–methyl-2-butanol
o
(3 alcohol)
Observation: Cloudy solution formed immediately.
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IODOFORM TEST TO IDENTIFY METHYL CARBINOL GROUP
❑ Ethanol and secondary alcohols containing the group
H
CH3 C
OH (methyl alcohol group)
❑ React with alkaline solutions of iodine to form triiodomethane ( iodoform, CHI3 ).
❑ This is called the iodoform reaction .
❑ Triidomethane is a pale yellow solid with a characteristic smell.
❑ To detect methyl carbinol group.
❑ Reagent: excess I2 , NaOH (base)
❑ Methyl carbinol cleavage to give carboxylate ion and iodoform.
H excess l2 O
OH -
R C CH3 R C O- + CHl3
iodoform
OH (yellow precipitate)
Observation: light yellow precipitate formed.
OH excess l2 O
OH -
CH3CH2 CH CH3 CH3 CH2 C O- + CHl3
iodoform
Observation: Light yellow precipitate formed
OH excess l2 O
CH3 CH OH- O- C
+ CHl3
1–phenylethanol iodoform
Observation: Light yellow precipitate formed.
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EXERCISE 8:
Show how you would use a simple test to distinguish between the following pairs of
compounds. Tell what you would observe with each compound:
a) 1-hexanol and cyclohexanol
b) 2–butanol and 3–pentanol
c) 2-butanol and 1-methylcyclohexanol
Prepared by: Revised by:
Suhaina Binti Mohd Yatim Badrisam bin Saindin
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CHAPTER 5.0 : ALCOHOLS |CHEMISTRY UNIT KMKt 2021/2022
1. Classify the following compounds as primary, secondary or tertiary alcohol and give
their IUPAC names.
a) CH3
CH3 CH2 C CH2OH
CH2 CH2 CH3
b) CH3
OH
c) CH2 CH3
CH CH CH2 CH2 OH
d) H3C CH3
CH2 OH
CC
HH
2. Draw the structural formulae of the following alcohol.
a) 1-butanol
b) 5-ethyl-2-methylcyclohexanol
c) 2-methyl-2,5-heptanediol
d) 2-penten-3-ol
e) 3-phenyl-2-butanol
3. The following names are incorrect. Draw the structural formulae and give the correct
IUPAC names.
a) 2-methyl-3-butanol
b) 1,1-dimethyl-2-cyclopentanol
c) 1-chloro-2-phenol
4. Arrange the following compounds in order of increasing boiling points. Explain your
answer
a) (CH3)2CHCH2OH, (CH3)2CH(CH2)2CH2OH, (CH3)2CH(CH2)3CH2OH
b) 1,2-ethanediol, butane, 1-propanol, 1,3-propanediol
5. Arrange the compounds in each set in decreasing order of solubility in water. Explain your
answer.
a) ethanol, 1-pentanol, 1-hexanol
b) hexane, 1-hexanol, 1,2-ethanediol
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CHAPTER 5.0 : ALCOHOLS |CHEMISTRY UNIT KMKt 2021/2022
6. The structural formula of alcohol A is given below
OH CH2 CH3
CH3 CH2 CHCHCH2 CHCH3
CH3
a) Give the IUPAC name of A.
b) Classify alcohol A as 1, 2 or 3.
c) Give the major product of the dehydration of A.
7. Draw the structural formulae of A to D in the following equations
a) +
H2O, H A
b) conc H2SO4 B
CH3CH2CH2OH
8. Write the complete equation for each of the following reactions.
a) 2-butanol heated with acidified KMnO4 solution
b) 2-propanol heated with concentrated H2SO4
c) 2-methyl-1-propanol heated with acidified K2Cr2O7 solution
d) 1,2-ethanediol with potassium
e) glacial ethanoic acid heated with 2-propanol in the presence of concentrated H2SO4
9. Give a chemical test to distinguish between
a) 2-methyl-1-propanol and 2-methyl-2-propanol
b) 1-butanol and 2-propanol
c) 2-methyl-2-butanol and 2-butanol
10. Outline the step(s) involved in the synthesis of
a) 1,2-dibromoethane from ethanol
b) 2,3-pentanediol from 2-pentanol
11. A and B are two isomeric alcohols
CH3CH(OH)CH2CH3 (CH3)3COH
A B
a) State the reagent used to dehydrate A and B to alkenes.
b) Draw the structural formulae of TWO alkenes obtained by dehydrating A.
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CHAPTER 5.0 : ALCOHOLS |CHEMISTRY UNIT KMKt 2021/2022
12. Based on the following reaction scheme;
OH
K2Cr2O7 / H+ CH3COOH / H +
S
U
R
Na
T
Draw the structure of S, T and U.
13. Compounds P and Q are the major and minor products obtained from the dehydration
of alcohol R , CH3CH(OH)CH(CH3)2 with concentrated sulphuric acid.
a) Draw the structural formulae and give the IUPAC name of P and Q.
b) State the rule that is used to determine the major products.
14. Alcohol R, C3H8O does not show any changes at room temperature when Lucas reagent
is added. After addition with hot ascidified potassium permanganate to alcohol R ,
propanoic acid is formed.
a) Draw the structure of alcohol R. Explain your answer.
b) Write the chemical equation and observation involved
15. The structural formula of menthol, the compound responsible for odour of peppermint, is
given below:
CH3 CH3 CH3
O
CH3CHCH3 OH OCCH3
CH3CHCH3 CH3CHCH3
X menthol Y
a) Give the IUPAC name for menthol.
b) Suggest the reagents and conditions necessary to convert menthol into compound
X and Y.
c) Draw the chemical equation reaction between menthol with acidified potassium
dichromate(VI).
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CHAPTER 5.0 : ALCOHOLS |CHEMISTRY UNIT KMKt 2021/2022
OBJECTIVE QUESTIONS
1. Which of the following alcohol gives 2-methylpropanoic acid when it is oxidized?
A. 1-Butanol
B. 2-Butanol
C. 2-Methyl-1-propanol
D. 2-Methyl-2-propanol
For questions 2and 3 which followed by three answer options with one or more
possible answers, choose the most appropriate answer from the options given
below;
A. I only
B. I and III only
C. II and III only
D. I, II, and III
2. Which of the following compound(s) could be formed by the dehydration of 1-butanol?
I. 2-Butene
II. 1,3-Butadiene
III. 2-Methyl-1-propene
3. Compound X, C4H10O, can be oxidized to C4H8O2 by acidified potassium permanganate
solution. X could be
I. (CH3)2CHCH2OH
II. CH3CH2CH2CH2OH
III. CH3CH(OH)CH2CH3
4. The reagent that can be used to differentiate 2-propanol from 1-propanol is
A. Br2/CCl4
B. Aqueous NaOH
C. Hot acidified KMnO4
D. ZnCl2/ concentrated HCl
5. Which compounds is a tertiary alcohols?
A. (CH3CH2)2CHOH
B. (CH3)3COH
C. CH3CH2CH(OH)CH3
D. CH3CH2OCH3
6. 2-methylpropan-2-ol is an example of
A. primary alcohol
B. secondary alcohol
C. tertiary alcohol
D. quarternary alcohol
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CHAPTER 5.0 : ALCOHOLS |CHEMISTRY UNIT KMKt 2021/2022
7. Which of the following alcohols would react fastest with Lucas reagent?
A. (CH3)3CCH2OH
B. (CH3)2CHCH2CH2OH
C. (CH3)2C(OH)CH2CH3
D. CH3CH2CH2CH(OH)CH3
8. When an alcohol Q is reacted with acidified hot potassium manganate (VII) solution ,the
end product obtained gave a positive result to the iodoform test. Which of the
following formula corresponds to Q?
A. C2H5OH
B. C6H5CH2OH
C. CH3CH(OH)CH2CH3
D. CH3CH2CH2CH2OH
9. Which of the following is/are tertiary alcohol ?
CH3 OH OH OH
I II III
A. I only
B. I and II
C. II and III
D. I, II, and III
10. Which of the following compounds is expected to have the highest boiling point ?
A. CH3CH2CH2OH
B. CH3CH(OH)CH3
C. CH3OCH2CH3
D. CH3CH(OH)CH2OH
Revised by:
Badrisam bin Saindin
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
CHAPTER 6: CARBONYL COMPOUNDS
GENERAL OVERVIEW a) General structural formula
6.1 Structures
6.2 Nomenclature a) IUPAC nomenclature for aldehydes
(C1-C10) and ketones (C3-C10)
*include common names
6.3 Preparation a) Preparation through oxidation of
alcohols
6.0 CARBONYL a) Reactive sites for
COMPOUNDS addition reaction
6.4 Chemical
Properties
b) Chemical i. addition
Properties ii. reduction
v. iodoform iii. condensation
iv. oxidation
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
6.1 STRUCTURE
Functional group : Carbonyl, C=O
General formula: CnH2nO
Aldehyde & Ketone are isomeric.
Carbonyl compounds are polar compound since oxygen is more electronegative than
carbon and forms a partially charged dipole.
In general, an aldehyde is more reactive than ketone towards nucleophilic addition
reaction.
6.2 NOMENCLATURE
Aldehyde is named by substituting the letter –e of the corresponding alkane with –al.
Basic name depends on the longest chain with –CHO group.
The chain is numbered by starting with –CHO group as C1.
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXAMPLE 1:
Structural Formula of Decanal
Condensed structure
CH3CH2CH2CH2CH2CH2CH2CH2CH2CHO
Expanded structure
1
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
Nomenclature Aliphatic Aldehydes
The chain is numbered by starting with –CHO group as C1.
2-ethyl-3-hydroxylbutanal
2-phenyl ethanal
3-butenal
4-bromo -3-cyclopentylhex-5-enal
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXAMPLE 2:
Cyclic aldehyde
Common names of Aldehyde
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
KETONE is named by substituting the letter –e of the corresponding alkane with –one.
Basic name depends on the longest chain with –COR group.
EXAMPLE 3:
Structural Formula of decanone
Condensed structure
CH3CH2CH2CH2CH2CH2CH2CH2COCH3
Expanded structure
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXAMPLE 4:
Cyclic ketones
The longest chain with carbonyl group is numbered so that C in carbonyl group gets the
smallest number.
pentan-2-one or 2-pentanone
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
1-cyclopentylpropan-2-one
1-cyclopentylpropan-1-one
Common names of Ketone
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXERCISE 1:
3-...................-5-methyl........................
1-...................propan-.......-one
hex-......-.........-3-one
1,5-...............pentan-2,4-...........
EXERCISE 2:
Draw all structural formulae for compounds with molecular formula C5H10O and give the
IUPAC name. (Hint : 7 possible structures)
Answer :
Possible structural formulae:
Pentanal
3-methylbutanal
2-methylbutanal
2,2-dimethylpropanal
2-pentanone
3-pentanone
3-methyl-2-butanone
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
6.3 PREPARATION
Carbonyl compounds can be prepared through Oxidation of Alcohols
Reagents are limited to
PCC, CH2Cl2
KMnO4, H+, Δ
K2Cr2O7, H+, Δ
Oxidation of primary alcohol yield aldehyde.
Oxidation of secondary alcohol yield ketone.
Tertiary alcohol does not undergoes oxidation.
-Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries
the OH group does not have a hydrogen atom attached.
Oxidation of Primary Alcohols
Aldehyde can be prepared by oxidation of primary alcohol using PCC.
General reaction :
Oxidation of Alcohols Secondary
Oxidation of 2o alcohols using any oxidizing agent will yield a ketone.
General reaction :
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXAMPLE 5:
6.4 CHEMICAL PROPERTIES
WHY C=O is susceptible to nucleophilic addition ?
In C=O, the electron density is drawn more towards the O atom making the C atom
deficient in electrons.
Thus, the C atom becomes the site for Nu- attack.
When carbonyl C is attacked by Nu-, the carbonyl bond is broken and a tetrahedral
intermediate is formed.
Hybridization of sp2 C atom changes to sp3.
Aldehydes are more reactive than ketone in Nu- addition reaction. Because ....
1. Steric factor
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
2. Electronic factor
Hydrogen is an electron withdrawing group (EWG)
Alkyl group is an electron donating group(EDG).
Therefore, aldehyde has a greater partial positive charge on its carbonyl carbon than a
ketone.
Chemical Properties Carbonyl compounds
i) Nucleophilic addition reaction with KCN,H3O+ or NaCN,H3O+
ii) Reduction to alcohol using LiAlH4 followed by H3O+ or NABH4 in methanol or H2
in the presence of catalyst
iii) Condensation with 2,4-dinitrophenylhydrazine(2,4-DNPH) as identification test
iv) Oxidation with Tollen’s reagents to differentiate aldehyde from ketone.
iv) Iodoform test to identify methyl carbonyl group
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
i) Addition of KCN, H3O+ or NaCN, H3O+ (Nucleophilic Addition)
Reagents : KCN, H3O+ or NaCN, H3O+
The HCN is prepared ‘in situ’ by the reaction of a mixture of KCN or NaCN with H2SO4 and
the product is Cyanohydrin
General reaction :
EXAMPLE 6:
ii) Reduction to alcohol using LiAlH4 followed by H3O+ or NABH4 in methanol or H2
in the presence of catalyst
Carbonyl compounds undergo reduction reaction with :
Reducing agents:
a) LiAlH4 followed by H3O+
b) NaBH4 in methanol @ ethanol
c) catalytic hydrogenation, H2 in catalyst (Pt or Ni)
The product of reduction reaction is:
i. Aldehyde – 10 alcohol
ii. Ketone – 2o alcohol
When a compound contains both C=O and C=C, selective reduction of one functional group
by proper choice of reducing agent :
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
Reduction of C=O
Selectivity of Reducing Agents
LiAlH4 - does not reduce C=C and C≡C
NaBH4 - does not reduce C=C, C≡C and C=O in acid carboxylic and esters
H2/Pt - reduce C=C, C≡C and C=O
Hence, NaBH4 is the best reducing agent for reduction of simple ketone and aldehyde.
EXAMPLE 7:
1.
2.
iii. Condensation with 2,4-dinitrophenylhydrazine(2,4-DNPH) as identification test
A condensation reaction is a chemical reaction in which two molecules combine to form
a larger molecule with the elimination of a smaller molecule such as water.
Reaction of 2,4-DNPH is used as an identification test for the; presence of carbonyl
group in a compound.
Also known as the Brady’s Test.
Observation : Formation of yellow or orange precipitate.
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXAMPLE 8
iv) Oxidation with Tollen’s reagents to differentiate aldehyde from ketone.
Oxidizing agents [O] :
KMnO4, K2Cr2O7,CrO3
Tollens ---- These oxidation reactions can be used as differentiation test for aldehydes
and ketones
Ketones are resistant to oxidation and No reaction with these oxidising agents !
Aldehydes are easily oxidised to carboxylic acid by strong oxidising agents.
Mild oxidising agents such as Tollen’s solutions can also oxidise aldehydes.
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
General reaction :
Silver mirror
Observation: Silver mirror is deposited on the wall of the test tube
EXAMPLE 9:
v) Iodoform test to identify methyl carbonyl group
To detect the presence of methyl carbonyl group.
Reagent : I2 in NaOH(aq)
OBSERVATION: light yellow precipitate formed
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CHAPTER 6: CARBONYL COMPOUNDS | CHEMISTRY UNIT KMKt 2021/2022
EXAMPLE 10:
EXERCISE 3 :
Chooos the carbonyl structure can react with Iodoform test:
Answer:
Prepared by: Revised by:
Khairul Aswari Bin Abdul Rahman Melani Sari Binti Wan Ismail
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CHAPTER 6 : CARBONYL COMPOUNDS |CHEMISTRY UNIT KMKt 2021/2022
1. Give the IUPAC name for each of the following aldehydes:
(a) CH3 CH2 CH2 C O
H
(b) CH3
CH3 CH2 CHC O
H
(c) Cl CH3
H3C C CH2 C C O
CH2 CH3 H H
2. Give the IUPAC name for each of the following ketones:
(a) Br
CH3CHCHC O
CH3 CH3
(b) O
CH3 C CHCH3
CH3
(c) CH3 CH2 CH2 CH2 C O
CH3
(d)
O
(e)
CO
CH3
(f)
CH2 C O
CH3
3. Draw the structural formula for each aldehydes:
(a) methanal @ formaldehyde
(b) 2-ethyl-3-methylpentanal
(c) Acetaldehyde @ ethanal
4. Draw the structural formula for each ketones:
(a) 4-methyl-2-pentanone
(b) Acetone @ propanone
(c) 4-chloro-2-methyl-3-hexanone
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5. Draw the structural formulae and give IUPAC names for:
(a) The seven isomeric aldehydes and ketones with the molecular formula of
C5H10O.
(b) Aldehydes and ketones with molecular formula of C8H8O containing a
benzene ring.
6. Complete the following reaction and write the structural formulae of A to C. Specify the
reagent D
(a) KMnO4/ H+ A
CH3 CH2 CH2OH
(b) K2Cr2O7/H+ B
CH3CH2CH(OH)CH2
(c) OH
PCC / CH2Cl2
(d) D O
CH2OH CH
7. Aldehyde is reactice than ketone by nucleophile addition. Why?
8. (a) Draw structural formula for the products formed when propanal reacts with
(i) NaBH4 in methanol
(ii) LiAlH4 followed by hydrolysis
(b) What type of reaction?
9. Propanone can be converted to various compounds as shown in the following reaction
scheme. Write the structural formulae of A to C.
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O KCN A H3O+ B
CH3 C CH3
LiAlH4 /H3O+
C
10. Compound X has the molecular formula of C5H10O. It does not react with Tollen’s
reagent but can be reduced to an alcohol with LiAlH4. Draw all the possible structural
formula of X.
11. Show how the following conversions can be carried out:
(a) propanone to propene
(b) 3-pentanone to 3-pentanol
12. Complete the following reactions:
(a) i. LiAlH4
HO CH2 CH2 CH2 CHO A
+
ii. H3O
(b) NaCN CN
B HCl C OH
CH3
(c) C2H5COCl O
AlCl3 C C2H5
E
(d) O + F
CH3 CH2 CH2 C H
K2Cr2O7, H
(e)
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13. Specify the reagent C and D
(a)
CH3 CH2 CH2 CH2 C O C O- + Ag(s)
(b) CH3 CH2 CH2 CH2 C O silver mirror
H
CH3 D CH3 H
CH3 C CH2 C O
CH3 C CH2 CH2OH
OH
OH
14. Give a chemical test to distinguish between:
(a) hexanal from 2-hexanone
(b) 2-pentanone from 3-pentanone
15. A list of carbonyl compounds is given below :
OO
CH3CHO CH3C CH3 C2H5C C2H5
R S T
Give a chemical test in each case to distinguish between:
(a) R and S
(b) S and T
OBJECTIVE QUESTIONS
1. Carbonyl group in aldehyde is
A. C=O
B. C-O
C. CO
D. CHO
2. CH3CH2CH2CH2CHO is formula for
A. methanol
B. propanal
C. butanal
D. pentanal
3. Oxidation of primary alcohols give
A. aldehydes
B. ketones
C. both A and B
D. alcohols
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