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Published by NOR IRWAN SHAH MOHAMED NOOR, 2021-03-09 08:10:24

Tutorial Meka Kumbe (TMK) Chapter 15 Hydroxy Compounds

TMK Chapter 15 Hydroxy Compounds

CHEMISTRY SK025
2020/2021

TUTORIAL

15.1: Introduction & 15.2: Nomenclature

1. Name and classify

Structure IUPAC Name Class

(a) OH

(Br)2CHC(CH 3)2 1,1-dibromo-2-methyl-2-propanol 3 alcohol

(b) CH2I CH3

CH3 CH2 CHCHCHCH3 3-iodomethyl-2-isopropyl -1- 1 alcohol
CH2OH pentanol

(c) CH3

OH 1-methylcyclopentanol 3 alcohol

(d) OH

4-methyl-3-cyclohexen-1-ol 2 alcohol

CH3

(e) H3C Cl

CC

CH3 CH2 CH2OH 2-chloro-3-methyl-2-penten-1-ol 1 alcohol

(f) H

OH

H trans-2-chlorocyclohexanol 2 alcohol

Cl

15.3: Physical Properties

1. Boiling point
Boiling point of n-butane  1-propanol  1,3-propanediol
• n-butane has the lowest boiling point because only van der Waals forces exist between
n-butane molecules.
• Boiling point of 1-propanol is higher than that of n-butane because 1-propanol forms
hydrogen bond between their molecules.
• Hydrogen bond is much more stronger than van der Waals forces.

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• 1,3-propanediol can form more hydrogen bond between their molecules because it has
two –OH group. Therefore boiling point of 1,3-propanediol is higher compare to 1-
propanol.

Solubility
Solubility of n-butane< 1-propanol<1,3-propanediol

• n-butane is insoluble in water because it cannot form hydrogen bond with water
molecule while 1-propanol, and 1,3-propanediol are soluble in water because they can
form hydrogen bond with water molecules.

• 1,3-propanediol is more soluble than 1-propanol because 1,3-propanediol have two –
OH groups. Thus, 1,3-propanediol can formmore hydrogen bond can be formed with
water molecules.

15.4: Preparation & 15.5: Chemical Properties

1. Equation

(a)

OH KMnO4, H+ O
H3CCH CH2CH3 H3C C CH2CH3

Δ

(b)

H3CCH CH3 conc. H2SO4 H3CCH CH2
OH
Δ

(c)

H3CCH CH2OH K2Cr2O7, H+ O
CH3
Δ H3CCHC OH
CH3

(d)

H3C OH PCl5

H3C Cl

SK 2

2. Reagents and conditions CHEMISTRY SK025
2020/2021
I
(a) II H2O, H+
PCl3 or PCl5 or SOCl2 or conc. HCl, ZnCl2,
III
(b) i. CH3CHO , dry ether
ii. H2O, H+
IV
V K2Cr2O7, H+, heat or KMnO4, H+, heat
VI Mg, dry ether
(c) O
CH
VII
K2Cr2O7, H+, heat or KMnO4, H+, heat

3. Synthesis CH3CH CH2 HBr NaOH (aq) CH3CH2CH2OH
(a) H2O2 CH3CH2CH2Br
KOH, ethanol

CH3CH(Br)CH3

reflux

(b)

conc. H2SO4 Br2, CH2Cl2 H2C CH2
H3CCH2OH Δ H2C CH2 Br Br

4. Chemical test

Test Lucas test

Reagent & Concentrated HCl, ZnCl2
Conditions

Compound 2-methyl-1-propanol 2-methyl-2-propanol

Observation A cloudy solution does not formed. A cloudy solution formed
immediately

Cl
+OH conc.HCl, ZnCl2
Equation H3C C CH3 Δ H3C C CH3 H2O

H3C H3C

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(b)

Test Lucas test

Concentrated HCl, ZnCl2
Reagent &
Conditions

Compound 1-butanol 2-propanol

Observatio A cloudy solution does not formed. A cloudy solution formed
n within 5 minutes after
heating.

Cl
+HO conc.HCl, ZnCl2
Equation H3C CHCH3 Δ H3C HC CH3 H2O

5. Structural formulae

Deduction Structure
H3CCH CH2 CH3
P 2 alcohol undergoes oxidation to form a ketone
OH
Q A ketone H3CC CH2CH3

R 2 alcohol reacts with PBr3 to form a 3 haloalkane O
H3CCH CH2 CH3
Haloalkane reacts with Mg in dry ether to form
S Grignard reagent Br
H3CCH CH2 CH3

Mg Br

Equations:

H3CCH CH2CH3 KMnO4, H+

OH Δ H3CC CH2CH3

P O

Q

H3CCH CH2CH3 PBr3
OH
H3CCH CH2CH3
P
Br

R

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H3CCH CH2CH3 Mg, dry ether H3CCH CH2CH3
Br Mg Br

R S

H3CCH CH2CH3 i. Q, dry ether CH3CH3
H3CCH2CH C CH2CH3
Mg Br ii. H2O, H+
OH

S

SK 5

CHEMISTRY SK025
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15.6 : Phenol

1. Name

2. Structure of product O- Na+
ONa CH2CH2 O- Na+

@

(a) CH2CH2ONa

(b) ONa O- Na+

@

CH2CH2OH CH2CH2 OH

(c)
OH

CH2CH2Cl

(d)
OH
O
CH2CH2O C CH2CH3

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3. Explanation
(a) Phenol is soluble in aqueous solution of sodium hydroxide but cyclohexanol is not. This
is due to the properties of phenol which is more acidic than cyclohexanol. That is why
phenol can reacts with NaOH(base) to yield sodium phenoxide which is a water soluble
salt.
But, cyclohexanol is an aliphatic alcohol which is less acidic than phenol. That is why
cyclohexanol does not react with NaOH(base) and it is insoluble in NaOH(aq).

(b) I2 oxidises methyl carbinol group in ethanol to methyl carbonyl, then to iodoform, CHI3
and methanoate ion. Iodoform is a yellow solid observed as yellow precipitate. But, 1-
propanol has no methyl carbinol group to be oxidised. Therefore, it does not form
iodoform, CHI3.

4. Structural formula
(a) Reason:
CH3 - Y: C7H8O, the ratio of C:H = 1:1, contains aromatic ring
- 1 oxygen atom: an alcohol or ether
- Dissolves in dilute NaOH (aq) solution: phenol (aromatic
alcohol is more acidic)
OH - Rapidly forms C7H5OBr3 with bromine water: phenol with a
methyl as a substituent

(b) Equations

OH -+
NaOH(aq)
O Na
+ H2O

CH3 CH3
OH OH
Br Br
Br 2(aq )

CH3 CH3
Br

(c) Confirmatory test

Reagent : FeCl3 (aq)
Colourless solution turns to purple
Observation :

CH3 CH3
FeCl3 (aq) FeCl3

OH OH

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MEKA 15 Classify and name Class
1. IUPAC Name 2° alcohol
1 alcohol
(a) 3-nitrocyclohexanol
(b) 1° alcohol
3-methyl-4-phenyl-1-pentanol
(c)
cis-2-buten-1-ol

2. Structural formulae H3C
(a)
2-methyl-1-propanol H3C CHCH2OH
HO
(b)
5-methyl-3-hexanol H3CHC CH2CHCH2CH3

(c) 2-isopropylcyclopentanol CH3

OH

(d) trans-1,4-cyclohexanediol CH3
H3C

H OH

(e) 4-tert-butyl-2,6- OH H
difluorophenol
OH F
F

H 3C C CH3
CH3

(f) 4-amino-2,5-dichlorophenol OH

3. a)Structural formulae and explain Cl
Structure
Cl
NO 2

Reason

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OH E is a 2 alcohol because cloudy solution is
E formed within 5 minutes when Lucas
reagent is added.
H3C HCCH2CH3
Cl Haloalkane is produced when an alcohol
reacted with PCl5.
F
H3C HCCH2CH3 Grignard reagent is formed when a
MgCl haloalkane is treated with Mg in dry ether.

G Carboxylic acid is formed when Grignard
H3C HCCH2CH3 reagent is treated with CO2 followed by
hydrolysis.
H COOH
H3C HCCH2CH3

b) Name
2-butanol

4. decreasing order of solubility
1,2-ethanediol > 1-hexanol > hexane

Reason:

• Hexane is a non-polar molecule and cannot form a hydrogen bond with water

molecules. It can only dissolve in non-polar solvent such as benzene and CCl4.

Therefore, it is insoluble in water.
• Both 1,2-ethandiol and 1-hexanol are alcohol. They can form hydrogen bond

with water molecules.
• 1,2-ethandiol has more –OH groups than 1-hexanol.
• Thus, 1,2-ethanediol can form more hydrogen bond with water molecules

compared to 1-hexanol.
• Therefore, 1,2-ethanediol is more soluble in water compared to 1-hexanol.

5. Structural formulae

Structure

H3C
A H3C CH2 CHCl

H3C
B H3C CH2 CHMgCl

C H3C
H3C CH2 CHCH2OH
HH

D H3C C C CH3

EO
H3C CH2 C CH3

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O
F H3C CH2 C OHCCH2CH3

CH3

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KUMBE 15

1. Structural formulae & correct name
CH3

a) CH3
3-ethyl-1-methylcyclohexanol

H3C CH2 CH2CHCH2CH2CH2CH2Cl

b) HO

8-chloro-4-octanol

2. equation

OH O-Na+

H2CCH2 Na H2CCH2
a) OH O- Na+

OH

H3CHCCH3 , conc.H2SO4 O
OH
H3C C H3C C O CHCH3
O reflux H3C

b)

c) H3CCH2CH2CH2CH2CH2OH PCC, CH2Cl2 O H
H3CCH2CH2CH2CH2C

OH SOCl2, pyridine Cl
d)

3. descending of solubility ethanol > 1-pentanol > 1-hexanol
Reason:

• Size of alkyl group of ethanol < 1-pentanol < 1-heptanol.
• Size of hydrophobic area of ethanol < 1-pentanol < 1-heptanol.

Solubility of ethanol > 1-pentanol > 1-heptanol.

4. a) name of A
4,6-dimethyl-3-octanol
b) optically active
Yes. Because it has chiral carbon.
c) class
2o alcohol

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d) major product
H3CCH2CH C CH2CH CH2CH3

H3C H3C

5. Explain and structural formulae

Structure Reason

Primary alcohol gives negative result

towards Lucas reagent and forms
B carboxylic acid when oxidized with strong

CH3CH2CH2CH2OH oxidizing agent.

H3C CH CH2CH3 C has methyl carbinol group because it

C gives positive result with I2/NaOH and has

HO chiral carbon.

H3C Tertiary alcohol reacts instantly with Lucas

D H3C C CH3 reagent but does not decolorize acidified

HO solution of potassium dichromate.

6. Predict structural formulae Reason
Structure A is primary alcohol because it turn to
carboxylic acid D when oxidized.
A B is alkene because ozonolysis of B give
CH3CH2CH2CH2OH 2 ethanal.

B H3C C CHCH3 Ozonolysis of B produce aldehyde C.
H
Oxidation of A produce carboxylic acid D.
CO
H3C C H
O

D
H3CCH2CH2C OH

SK 12


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