1. ORGANIC CHEMISTRY II
CHM 301
CHAPTER 1
ALCOHOLS

2. ALCOHOLS

Alcohols: Organic compounds containing
hydroxyl (-OH) functional groups.
R

OH
Phenols: Compounds with hydroxyl group bonded
directly to an aromatic (benzene) ring.
OH

3. NOMENCLATURE OF
ALCOHOLS

4. IUPAC RULES
1.
2.
3.
4.
Select the longest continuous chain of carbon atoms
containing the hydroxyl group.
Number the carbon atoms in this chain so that the
one bonded to the –OH group has the lowest
possible number.
Form the parent alcohol name by replacing the final
–e of the corresponding alkane name by –ol. When
isomers are possible, locate the position of the –OH
by placing the number (hyphenated) of the carbon
atom to which the –OH is bonded immediately before
the parent alcohol name.
Name each alkyl branch chain (or other group) and
designate its position by number.

5. This is the longest continuous chain that
contains an hydroxyl group.
Select this chain as the parent compound.

6. 4
3
2
1
This end of the chain is closest to the
OH. Begin numbering here.

7. 4
3
2
1
IUPAC name: 3-methyl-2-butanol
New IUPAC name: 3-methylbutan-2-ol

8. Example:
This is the longest continuous chain that
contains an hydroxyl group.
Select this chain as the parent compound.

9. 5
4
3
2
1
This end of the chain is closest to the
OH. Begin numbering here.

10. 5
4
3
2
1
IUPAC name: 3-methyl-2-pentanol
New IUPAC name: 3-methylpentan-2-ol

11. NOMENCLATURE OF CYCLIC
ALCOHOLS


Using the prefix cycloThe hydroxyl group is assumed to be on C1.
H
5
4
6
3
1
2
Br
IUPAC name:
OH
HO CH2CH3
1
H
trans-2-bromocyclohexanol
new IUPAC name: trans-2-bromocyclohexan-1-ol
3
2
1-ethylcyclopropanol
1-ethylcyclopropan-1-ol

12. NOMENCLATURE OF ALCOHOLS
CONTAINING TWO DIFFERENT
FUNCTIONAL GROUPS

Alcohol containing double and triple bonds:
- use the –ol suffix after the alkene or alkyne name.

The alcohol functional group takes precedence over
double and triple bonds, so the chain is numbered in
order to give the lowest possible number to the carbon
atom bonded to the hydroxyl group.

The position of the –OH group is given by putting its
number before the –ol suffix.

Numbers for the multiple bonds were once given early in
the name.

13. EXAMPLE
5
4
CH2 CH
3
CH2
2
CH
1
CH3
OH
1) Longest carbon chain that contains –OH group
- 5 carbon
2) Position of –OH group
- Carbon-2
3) Position of C=C
- Carbon-4
COMPLETE NAME = 4-penten-2-ol

14. Some consideration:

- OH functional group is named as a hydroxy
substituent when it appears on a structure with a
higher priority functional group such as acids, esters,
aldehydes and ketones.
- Examples:
OH
4
CH3
3
CH
O
2
CH2
1
C
OH
3-hydroxybutanoic acid
4 3 2
5 1
6
OH
O
2-hydroxycyclohexanone

15. MAIN GROUPS
decreasing priority
Acids
Esters
Aldehydes
Ketones
Alcohols
Amines
Alkenes
Alkynes
Alkanes
Ethers
Halides

16. NOMENCLATURE OF DIOLS

Alcohols with two –OH groups

Naming of diols is like other alcohols except that the
suffix diol is used and two numbers are needed to tell
where the two hydroxyl groups are located.
diols or lycols.
5
OH
3
CH3
IUPAC name
2
CH
1
CH2
propane-1,2-diol
1
4
OH
OH
3 2
OH
trans-cyclopentane-1,2-diol

17. NOMENCLATURE OF PHENOLS

The terms ortho (1,2-disubstituted), meta (1,3disubstituted) and para (1,4-disubstituted) are often
used in the common names.
OH
O2N
OH
CH3CH2
Br
IUPAC name:
2-bromophenol
common name: ortho-bromophenol
OH
3-nitrophenol
meta-nitrophenol
4-ethylphenol
para-ethylphenol

18. 
Phenols may be monohydric, dihydric or
trihydric - (number of hydroxyl groups) in the
benzene ring.
OH
OH
OH
OH
OH
benzene-1,3-diol
OH
benzene-1,4-diol
OH
benzene-1,2,3-triol

19. COMMON NAMES


Derived from the common name of the alkyl group
and the word alcohol.
For examples:
CH3
H3C C OH
CH3
IUPAC name: 2-methyl-2-propanol
Common name: tert-butyl alcohol
CH2CHCH3
OH
IUPAC name: 2-propanol
Common name: isopropyl alcohol
CH3CH2OH
IUPAC name: ethanol
Common name: ethyl alcohol
CH3OH
IUPAC name: methanol
Common name:methyl alcohol

20. CLASSIFICATION OF
ALCOHOLS

21. CLASSIFICATION

According to the type of carbinol carbon atom (C bonded to the –
OH group).
C

OH
Classes:
i) Primary alcohol
- -OH group attached to a primary carbon atom
- one alkyl group attached
ii) Secondary alcohol
- -OH group attached to a secondary carbon atom
- two alkyl group attached
iii) Tertiary alcohol
- -OH group attached to a tertiary carbon atom
- three alkyl group attached

22. TYPE
i)
ii)
Primary (1 )
Secondary (2 )
STRUCTURE
H
R C
H
EXAMPLES
CH3
OH
R'
R C OH
H
CH3CH2-OH
CH3CHCH2 OH
ethanol
2-methyl-1-propanol
OH
2-butanol
iii)
Tertiary (3 )
R'
R C OH
R''
OH
H3C CH CH2CH3
cyclohexanol
CH3
H3C C OH
CH3
2-methyl-2-propanol

23. Polyhydroxy Alcohols
• Alcohols that contain more than one OH group polyhydroxy alcohols.
• Monohydroxy: one OH group.
• Dihydroxy: two OH groups.
• Trihydroxy: three OH groups.

24. PHYSICAL PROPERTIES
OF ALCOHOLS

25. PHYSICAL PROPERTIES

PHYSICAL STATES OF ALCOHOLS
- aliphatic alcohols and lower aromatic alcohols
liquids at room temperature.
- highly branched alcohols and alcohols with twelve
or more carbon atoms
solids.

26. 
BOILING POINTS
i) Boiling points of alcohols are higher > alkanes and
chloroalkanes of similar relative molecular mass.
- For example:
C2H5OH
46
78 C
Relative molecular mass:
Boiling point:
CH3CH2CH3
44
-42 C
CH3Cl
50.5
-24 C
- Reason:
* intermolecular hydrogen bonds
R
Ar
δ+
O
H
δ-
H
O
δ-
hydrogen bonding
H
δ+
O
R
δ-
H
Ar
Oδ-
hydrogen bonding

27. 
SOLUBILITY OF ALCOHOLS IN WATER
i) Alcohols with short carbon chains (i.e. methanol, ethanol) dissolve in water.
- dissolve in water (hydrogen bonds are formed).
ii) Solubility decreases sharply with the increasing length of
the carbon chain.
iii) Higher alcohols are insoluble in water.
- alcohol contains a polar end (-OH group) called ‘hydrophilic’
and a non-polar end (the alkyl group) called ‘hydrophobic’.

28. iii) Polyhydroxy alcohols are more soluble than monohydroxy
form more hydrogen bonds with water
molecule.
iv) Branched hydrocarbon increases the solubility of alcohol
in water
branched hydrocarbon cause the
hydrophobic region becomes compact.

29. ACIDITY OF ALCOHOLS AND PHENOLS


Alcohol
weakly acidic.
In aqueous solution, alcohol will donated its proton to
water molecule to give an alkoxide ion (R-O-).
R-O- + H3O+
R-OH + H2O
Ka = ~ 10-16 to 10-18
alkoxide ion
Example
CH3CH2-OH + H2O

CH3CH2-O- + H3O+
The acid-dissociation constant, Ka, of an alcohol is defined
by the equilibrium
R-OH + H2O
Ka
Ka = [H3O+] [RO-]
[ROH]
R-O- + H3O+
pKa = - log (Ka)
* More smaller the pKa
value, the alcohol is
more acidic

30. Acidity OF PHENOLS

Phenol is a stronger acid than alcohols and water.
R-OH + H2O
alcohol
OH
R-O- + H3O+
Ka = ~ 10-16 to 10-18
alkoxide ion
H2O
phenol
H 2 O + H2 O
O-
H3O
+
Ka = 1.2 x 10-10
phenoxide ion
HO- + H3O+
hydroxide ion
Ka = 1.8 x 10-16

31. 
Phenol is more acidic than alcohols by considering
the resonance effect.
i) The alkoxide ion (RO-)
- the negative charge is confined to the oxygen and
is not spread over the alkyl group.
- this makes the RO- ion less stable and more
susceptible to attack by positive ions such as H+
ions.

32. ii) The phenoxide ion
- one of the lone pairs of electrons on the oxygen atom is
delocalised into the benzene ring.
- the phenoxide ion is more stable because the negative
charge is not confined to the oxygen atom but
delocalised into the benzene ring.
- the phenoxide ion is resonance stabilised by the
benzene ring and this decreases the tendency for the
phenoxide ion to react with H3O+.
O
O
O
O

33. EFFECTS OF Acidity

The acidity decreases as the substitution on the alkyl group increase.
- Ethyl group is an electron-donating group
bond
harder to release a proton.
strengthens the –O-H
- i.e: methanol is more acidic than t-butyl alcohol.

The present of electron-withdrawing atoms enhances the acidity of
alcohols.
- The electron withdrawing atom helps to stabilize the alkoxide ion.
- i.e: 2-chloroethanol is more acidic than ethanol because the
electron-withdrawing chlorine atom helps to stabilize the 2chloroethoxide ion.
- Alcohol with more than one electron withdrawing atoms are more
acidic. i.e. 2,2,-dichloroethanol is more acidic than 2-chloroethanol.
- Example of electron-withdrawing atom/groups:
Halogen atoms and NO2.

34. IMPORTANT OF ALCOHOL

Ethanol - solvent for varnishes, perfumes and
flavorings, medium for chemical reactions and in
recrystallization. Also is an important raw material
for synthesis.

Medically, ethanol is classified as a hypnotic (sleep
producer), it is less toxic than other alcohol.

35. 
Ethanol is prepared both by hydration of ethylene and
by fermentation of sugars. It is the alcohol of alcoholic
beverages.
acid
CH2 =CH2 + H2O -----------> CH3CH2OH
yeast
C6H12O6 ----------->
2CH3CH2OH + 2CO2

36. PREPARATION OF ALCOHOLS



Grignard synthesis
Hydrolysis of alkyl halides
Industrial and laboratory preparations of
ethanol

37. GRIGNARD SYNTHESIS

The grignard reagent (RMgX) is prepared by the reaction of
metallic magnesium with the appropriate organic halide.
This reaction is always carried out in an ether solvent,
which is needed to solvate and stabilize the Grignard
reagent as it forms.
R-X + Mg
(X = Cl, Br or I)


CH3CH2OCH2CH3
R-Mg-X
organomagnesium halide
(Grignard reagent)
Grignard reagents may be made from primary, secondary,
and tertiary alkyl halides, as well as from vinyl and aryl
halides.
Alkyl iodides are the most reactive halides, followed by
bromides and chlorides. Alkyl fluorides generally do not
react.

38. EXAMPLES
ether
Mg
CH3I
CH3CH2Br
Br
Mg
Mg
ether
CH3MgI
CH3CH2MgBr
ether
MgBr

39. Grignard reactions of carbonyl compounds

Formaldehyde (H2C=O) reacts with Grignard reagents
giving primary alcohol.
R-MgX
+
H
H
C O
ether
H
R C O
H
MgX
H3O+
R CH2 OH
or
R-MgX +
H
H
i) ether
C O
ii) H3O+
R CH2 OH
Example:
CH3CH2CH2CH2-MgBr +
H
H
butylmagnesium bromide
i) ether
C O
+
ii) H3O
H
CH3CH2CH2CH2-C OH
H
1-pentanol (92%)

40. 
Aldehydes reacts with Grignard reagents giving
secondary alcohols.
R-MgX
R'
+
H
C O
ether
R'
R C O
H
MgX
H3O+
or
R-MgX +
R'
H
i) ether
C O
ii) H3O+
R'
R C OH
H
Example:
CH3CH2-MgBr
+
H3C
H
i) ether
C O
ii) H3O+
CH3
CH3CH2-C OH
H
R'
R C OH
H

41. 
Ketones reacts with Grignard reagents giving
tertiary alcohols.
R-MgX
+
R'
R''
C O
ether
R'
R C O
R''
MgX
H3O+
R'
R C OH
R''
or
R-MgX +
R'
R''
i) ether
C O
ii) H3O+
R'
R C OH
R''
Example:
O
CH3CH2-MgBr + H3C C CH2CH2CH3
i) ether
ii) H3O+
OH
CH3CH2-C CH2CH2CH3
CH3

42. HYDROLYSIS OF ALKYL HALIDES

Hydrolysis of alkyl halides is severely limited as a
method of synthesizing alcohol, since alcohol are
usually more available than the corresponding
halides;indeed, the best general preparation of halides
is from alcohols.

For those halides that can undergo elimination, the
formation of alkene must always be considered a
possible side reaction.

43. Example:
1) Second-order substitution: primary (and some
secondary halides)
KOH
(CH3)2CHCH2CH2-Br
(CH3)2CHCH2CH2-OH

H2O
2) First-order substitution: tertiary (and some secondary)
halides
CH3
H3C C CH3
Cl
acetone/water
heat
CH3
H3C C CH3
OH
CH3
H2C C CH3

44. INDUSTRIAL AND LABORATORY
PREPARATION OF ETHANOL

There are three principle ways to get the simple alcohols that
are the backbone of aliphatic organic synthesis. These
methods are:
a) hydration of alkenes obtained from the cracking of
petroleum
b) the oxo process from alkenes, carbon monoxide and
hydrogen
c) fermentation of carbohydrate

45. REDUCTION OF ALDEHYDES, KETONES AND
CARBOXYLIC ACIDS
Aldehydes and ketones can be reduced to alcohols
using:
a) lithium aluminium hydride (LiAlH4)
b) sodium borohydride (NaBH4)
c) catalytic hydrogenation

O-
O
OH
+
R
C
H
LiAlH4 or NaBH4 or H2, Ni
R
aldehyde
C
H
H
R
C
H
H
H
o
1 alcohol
O
R
C R'
O
LiAlH4 or NaBH4 or H2, Ni
ketone
+
H = diluted acid such as H2SO4
-
OH
+
R
C R'
H
R
C R'
H
H
o
2 alcohol

46. Examples:
O-
O
OH
+
CH3
C
H
LiAlH4
CH3
ethanal
C
H
H
CH3
C
H
H
H
ethanol
O-
O
OH
+
CH3
C
CH3
propanone
H2/Ni
CH3
C CH3
H
H
CH3
C
CH3
H
2-propanol

47. reduced

Carboxylic acids
primary alcohols

Reducing agents: LiAlH4 in dry ether
H
O
R C OH
(1) LiAlH4 / ether
(2) H2O
R C OH
H
carboxylic acids
primary alcohols
examples:
H
O
CH3 C OH
ethanoic acid
(1) LiAlH4 / ether
(2) H2O
CH3 C OH
H
ethanol
- Benzoic acid can be reduced to phenylmethanol by using LiAlH4 in eth

48. CH3 C OH
(1) LiAlH4 / ether
(2) H2O
CH3 C OH
H
 Benzoic acid can be reduced to phenylmethanol by using LiAlH4
ethanol
in ether at low temperatures. to phenylmethanol by using LiAlH4 in ether at
- Benzoic acid can be reduced
 An alkoxide intermediate is formed first.
low temperatures.
 On adding water, hydrolysisformed first.
- An alkoxide intermediate is of the intermediate yields the
- On alcohols.
primary adding water, hydrolysis of the intermediate yields the primary alcohols.
ethanoic acid
O
C OH
(1) LiAlH4 / ether
(2) H2O
H
C OH
H
benzoic acid
phenylmethanol
 LiAlH4 has no effect on the benzene ring or the double bond.
 -COOH is reduced to –CH2OH but the C=C bonds remains
unchanged.
CH3CH2CH=CHCOOH
1) LiAlH4
2) H2O
CH3CH2CH=CHCH2OH

49. Question:
i) Give the structural formulae of L, M and N
OH
A
PBr3
M
Mg
ether
i) L
ii)H3O+
N
ii) How to prepare alcohol A from the
reduction process?
OH

50. Answers
i)
L:
M:
Br
N:
MgBr
O
ii)
O
(1) LiAlH4 / ether
(2) H2O
OH

51. REACTIONS OF ALCOHOLS






Reaction with sodium
Oxidation
Esterification
Halogenation and haloform reactions
Dehydration
Formation of ether (Williamson ether
synthesis)

52. Reaction with sodium

Alcohols reacts with Na at room temperature to
form salts (sodium alkoxides) and hydrogen.
2R-O-H + 2Na → 2R-O- Na+ + H2

For example:
CH3CH2OH + Na → CH3CH2O-Na+ + 1/2H2
alcohol

sodium ethoxide
Reactivity of alcohols towards the reactions
with sodium:
CH3 > 1 > 2 > 3

53. Oxidation
H
1 alcohol
R
C OH
Pyridinium chlorochromate (PCC)
H
CH2Cl2, 25oC
R-C=O
H
1o alcohol
aldehyde
H
R
C OH
Cu or Cr3O/pyridine
H
R-C=O
H
o
1 alcohol
Cr3O/pyridine = Collins reagent
aldehyde
H
R
C OH
H
o
1 alcohol
KMnO4/H+ or K2Cr2O7/H+
or CrO3/H+
O
R-C-OH
carboxylic acid

54. Examples:
1 alcohol
O
CH3(CH2)4-CH2-OH
PCC
1-hexanol
CH3(CH2)4-CH2-OH
1-hexanol
CH3(CH2)4-C-H
hexanal
KMnO4/H+ or K2Cr2O7/H+
or CrO3/H+
O
CH3(CH2)4-C-OH
hexanoic acid

55. H
+
2 alcohol
R
+
KMnO4/H or K2Cr2O7/H
C OH
O
+
or CrO3/H
R'
R-C-R'
ketone
2o alcohol
R"
3 alcohol
+
R
+
KMnO4/H or K2Cr2O7/H
C OH
no reaction
+
or CrO3/H
R'
3o alcohol
Example:
OH
CH3
CH CH2CH3
2-butanol
+
O
+
KMnO4/H or K2Cr2O7/H
+
or CrO3/H
CH3
C
CH2CH3
2-butanone

56. Esterification

Esterification:
- the reaction between an alcohol and a carboxylic acid to
form an ester and H2O.
O
R
C
O
H
H O
carboxylic acid
CH3CH2-O-H
ethanol
R'
O
CH3
C
methanol
O
R'
H2O
O
H+
O
H
ethanoic acid
C OH
benzoic acid
C
ester
O
CH3-O-H
R
alcohol
EXAMPLES
H+ = catalyst
O
H+
CH3
C
OCH2CH3
ethyl ethanoate
H+
O
C OCH3
methyl benzoate
H2O
H2O

57. Esterification also occurs when alcohols
react with derivatives of carboxylic acids
such as acid chlorides
O
O
CH3-O-H
CH3 C Cl
CH3 C OCH3
methanol
ethanoyl chloride
methyl ethanoate
HCl

58. Halogenation and haloform reactions
1) Hydrogen halides (HBr or HCl or HI)
R-OH + H-X → R-X + H2O
Example:
C2H5-OH + H-Br
•
•
H+
C2H5-Br + H2O
Reactivity of hydrogen halides decreases in order HI >
HBr > HCl
Reactivity of alcohols with hydrogen halides:
3 >2 >1

59. 2) Phosphorus trihalides, PX3 or phosphorus
pentahalides, PX5
3R-OH + PX3
3R-X + H3PO3
(PX3 = PCl3 or PBr3 or PI3)
Example:
(CH3)2CHCH2-OH + PBr3 → (CH3)2CHCH2-Br
isobutyl alcohol
isobutyl bromide
3) Thionyl chloride (SOCl2)
R-OH + SOCl2 → R-Cl + SO2 + HCl
Example:
CH3(CH2)5CH2-OH + SOCl2 → CH3(CH2)5CH2-Cl + SO2 + HCl
1-heptanol
1-chloroheptane

60. Dehydration

Dehydration of alcohols will formed alkenes and the
products will followed Saytzeff rules.
R-CH2-CH2-OH

conc. H2SO4
R-CH=CH2 + H2O
Saytzeff rule:
- A reaction that produces an alkene would favour the
formation of an alkene that has the greatest number of
substituents attached to the C=C group.
+
H
CH3CH2-CH-CH3
OH
2-butanol
CH3CH2-CH=CH2 + H2O
+
H
1-butene
CH3CH=CH-CH3 + H2O
2-butene
major product

61. 

Reactivity of alcohols towards dehydration:
3 >2 >1
Reagents for dehydration:
i) Concentrated H2SO4
CH3-CH2-OH
conc. H2SO4
CH2=CH2 + H2O
ii) With phosphoric (v) acid
OH
85% H3PO4, 165-170oC
iii) Vapour phase dehydration of alcohols
CH3CH2OH
Al2O3
heat
CH2=CH2 + H2O
H2O

62. Formation of ether (Williamson ether synthesis)


Involves the SN2 attack of an alkoxide ion on an
unhindered primary alkyl halides.
The alkoxide is made by adding Na, K or NaH to the
alcohol.
R-O- + R’-X → R-O-R’ + Xalkoxide
(R’ must be primary)


The alkyl halides (or tosylate) must be primary, so that
a back-side attack is not hindered.
If the alkyl halides is not primary, elimination usually
occurs to form alkenes.

63. EXAMPLES
CH3CH2-OH
+
Na
CH3CH2-O Na
CH3CH2-O-CH3
CH3I
NaI
or
CH3CH2-OH
OH
1) Na
2) CH3I
1) Na
CH3CH2-O-CH3
OCH2CH3
2) CH3CH2-OTs
cyclohexanol
ethoxycyclohexane
NaI

64. REACTIONS OF PHENOLS




Reaction with sodium
Esterification
Halogenation of the ring
Nitration of the ring

65. REACTION WITH SODIUM
OH
Na
O- Na+
1/2 H2(g)
sodium phenoxide
REACTION WITH AQUEOUS SODIUM HYDROXIDE
OH
NaOH
O- Na+
sodium phenoxide
ROH + NaOH
no reaction
H2O

66. ESTERIFICATION
EXAMPLES
O
OH
NaOH
ONa
H2O
O
CH3CCl
NaCl
OCCH3
NaOH
sodium phenoxide
O
OH
C OH
O
H+
OC
phenyl benzoate
H2O

67. HALOGENATION


More reactive towards electrophilic substitution than benzene.
ortho-para director.
OH
OH
X
3X2
X
room
temperature
3HX
X
EXAMPLES
OH
OH
Br
3Br2
Br
room
temperature
3HBr
Br
2,4,6-tribromophenol (white precipitate)
OH
OH
Cl
3Cl2
Cl
room
temperature
3HCl
Cl
2,4,6-trichlorophenol (white precipitate)

68. • Monobromophenols are obtained if the bromine
is dissolved in a non-polar solvent such as CCl4.
OH
2
OH
OH
Br
2Br2 (CCl4)
2HBr
Br

69. NITRATION

Dilute nitric (v) acids reacts with phenol at room
temperature to give a mixture of 2- and 4-nitrophenols.
OH
2
OH
OH
2HNO3
< 20oC
NO2
2H2O
NO2
2-nitrophenol
4-nitrophenol

70. 
By using concentrated nitric (v) acid, the nitration of
phenol yields 2,4,6-trinitrophenol (picric acid).

Picric acid is a bright yellow crystalline solid. It is used
in the dyeing industry and in manufacture of explosives.
OH
OH
3HNO3
O2 N
NO2
NO2
2,4,6-trinitrophenol
(picric acid)
3H2O

71. Question:
Alcohol W is a secondary alcohol with a molecular formula
of C4H10O.
Compound M
Step 1
CrO3 /
pyrridine
C4H10O
Alcohol W
Step 2
H+ / heat
Reagent A
C4H10ONa
a) Draw and give the IUPAC name for alcohol W.
b) Draw the structural formula for the following
compounds:
i) Compound M
ii)Compound N
iii)Compound O
Compound N
(major)
+
Compound O
(minor)

72. c) Give the correct name for the following:
i) Step 1
ii) Step 2
iii)Reagent A

73. Answers
a) Alcohol W
OH
name: butan-2-ol
b) i) compound M
ii) compound N
O
c) i) Step 1: Oxidation
ii) Step 2: Dehydration (of alcohol)
iii) Reagent A: Na Metal
iii) Compound O

74. TESTS TO DISTINGUISH CLASSES OF
ALCOHOLS
1)
Lucas Test
- The alcohol is shaken with Lucas reagent (a solution
of ZnCl2 in concentrated HCl).
- Tertiary alcohol - Immediate cloudiness (due to the
formation of alkyl chloride).
- Secondary alcohol - Solution turns cloudy within
about 5 minutes.
- Primary alcohol - No cloudiness at room temperature.

75. CH3
CH3
HCl/ZnCl2
CH3 C CH3
OH
room temperature
Cl
o
3 alcohol
CH3 CH CH2CH3
OH
(cloudy solution almost immediately)
HCl/ZnCl2
room temperature
o
(cloudy solution within 5 minutes)
HCl/ZnCl2
room temperature
o
1 alcohol
CH3 CH CH2CH3
Cl
2 alcohol
CH3CH2CH2CH2OH
CH3 C CH3
no reaction

76. 2)
Oxidation of alcohols
- only primary and secondary alcohols are oxidised by
hot acidified KMnO4 or hot acidified K2Cr2O7 solution.
- the alcohol is heated with KMnO4 or K2Cr2O7 in the
presence of dilute H2SO4.
- 1o or 2o alcohol:
→ the purple colour of KMnO4 solution disappears.
→ the colour of the K2Cr2O7 solution changes from
orange to green.
- 3o alcohol do not react with KMnO4 or K2Cr2O7.

77. 3RCH2OH
+ Cr2O
1o alcohol
+
7
+ 8H
27
+
3RCOOH
3+
+ 7H2O
(green)
R'
3 R CH OH + Cr2O2-7 + 8H+
o
+ 2Cr
carboxylic acid
R'
2 alcohol
(green)
aldehyde
+ 8H
(orange)
+ 2Cr3+ + 7H2O
3RCHO
(orange)
3RCHO + Cr2O
aldehyde
2-
(orange)
3R C
ketone
3+
O + 2Cr + 7H2O
(green)

78. HALOFORM TEST TO IDENTIFY METHYL
ALCOHOL GROUP
1) Iodoform:
 Ethanol and secondary alcohols containing the group
methyl alcohol group which react with alkaline
solutions of iodine to form triiodomethane (iodoform,
CHI3).
 Triiodomethane – a pale yellow solid with a
characteristic smell.
H
CH3
C
OH
(methyl alcohol group)

79. H
CH3
C
R + 4I2 + 6NaOH
OH
CHI3 (s) + RCOONa + 5NaI + 5H2O
triiodomethane
(iodoform)
yellow precipitate
where R = hydrogen, alkyl or aryl group
• The iodoform test can distinguish ethanol from methanol
H
CH3
C
O
H + 4I2 + 6OH
-
CHI3 (s) + 5I + 5H2O
iodoform
OH
H
C
O
methanoate
ethanol
positive iodoform test
H
H
C
H + 4I2 + 6OH
OH
methanol
no reaction
negative iodoform test

80. • The iodoform test can distinguish 2-propanol from 1-propanol
CH3
CH3
C
O
H + 4I2 + 6OH
2-propanol
CH3
C
O
ethanoate
iodoform
OH
positive iodoform test
H
H
H
C
C
C
H
H
CHI3 (s) + 5I- + 5H2O
H
OH
H + 4I2 + 6OH
no reaction
negative iodoform test
1-propanol
* TERTIARY ALCOHOLS DO NOT GIVE POSITIVE
IODOFORM TEST

81. 2) BROMOFORM
H
CH3
C
R + 4Br2 + 6NaOH
CHBr3 (s) + RCOONa + 5NaBr + 5H2O
bromoform
OH
where R = hydrogen, alkyl or aryl group
reagent
sample
iodoform

82. Question:
a) Classify each of the following alcohols as
primary, secondary or tertiary.
i) 2-Propanol
ii) 4-methylpentanol
iii)2,3-dimethylbutan-2-ol
b) Name a simple test to distinguish 1°, 2°, 3°
alcohol. State the reagents and conditions
required for the test and write down the
expected observations.

83. Answer:
a) i) 2°
ii) 1°
iii) 3°
b) Test: Lucas test
Reagent and conditions : Lucas reagent /
Mixture of HCl and ZnCl2
Observatios:
- Clear homogenous solution change into 2
layers or cloudiness
- Rate of reaction: 3° > 2° > 1° alcohol