This presentation is useful for XII CBSE chemistry teachers.

1. ALCOHOLS, PHENOLS,
ETHERS
Mr. Pranav A.
Galgale

2. Chemical Reactions of alcohols and Phenols
Alcohol acts as
Electrophile
Nucleophile Attraction towards positive charge
Attraction towards negative charge
1. Alcohols as nucleophiles
R – O C
H
+
2. Alcohols as electrophiles
C
+
R – O – H +
R – CH2 –O – H + H+ R – CH2 –O
H
H
+
Br - R – CH2 – BrR – CH2 – O
H
H
+
+
The bond between C – O is broken when alcohols react as nucleophiles.
R – O C H++
Protonated alcohols
The bond between O H is broken when alcohols react as nucleophiles.

3. Reactions are classified into two types
Reaction involving cleavage of O – H bond Reaction involving cleavage of C – O bond
Acidity of alcohol
1. Reaction with metals: Alcohols and phenols react with active metals such as sodium, potassium and aluminium to yield
corresponding alkoxides and hydrogen.
2 R – O – H + 2Na 2 R – O – Na + H2
Which is characteristics of acids
Alcohols and phenols are Brönsted acids i.e., they can donate a proton to a stronger base (B:).
R – O – H B:+ B – H R – O+
BaseAcid Conjugate
Acid
Conjugate
Base
Base + H+ = Conjugate acid
Acid – H+ = Conjugate base

4. Acidity of alcohols
The acidic character of alcohols is due to the polar nature of O–H bond
R – groups (i.e.) alky group is electron donating group (EDG). So it shows +I effect.
e. g. –CH3, –C2H5
R – O – H
d- d+
polar nature due to partial separation of charges
R
CHOH
R
R CH2OH
R
C – OH
R
R
Primary alcohol Secondary alcohol Tertiary alcohol
So, as R – group increases;
polarity of alcohols decreases
and as polarity decreases
acidity of alcohol decreases
1o > 2o > 3o
Acidity decrease

5. Alcohols are, however, weaker acids than water. This can be explained by the reaction of water with an alkoxide.
R – O H – O – H+
Base Acid
Conjugate
Acid
Conjugate
Base
R – O – H + : O – H –
Here water
acts as acid
This reaction shows that water is a better proton donor (i.e.,stronger acid) than alcohol.
Alcohols act as Bronsted bases as well. It is due to the presence of lone electron pairs on oxygen, which makes them
proton acceptors. (Protonated alcohols)

6. Acidity of phenols
1. phenols react with active metals such as sodium, potassium and aluminium to yield corresponding phenoxides and
hydrogen.
OH
2Na+
O Na+
H2+
2. phenols react phenols react with aqueous sodium hydroxide to form sodium phenoxides.
OH
+ NaOH
O Na+
H2O+

7. The hydroxyl group, in phenol is directly attached to the sp2 hybridised carbon of benzene ring which acts as an electron withdrawing
group.
1. Due to resonance, oxygen of –OH group gets positive charge. So, phenols are more acidic
than alcohols
50 % s characters
In phenol ‘C’ to which –OH is attached is sp2 hybridised which is more electronegative than sp3
hybridised ‘C’ in alcohols This increases the polarity of O–H bond
2. We know that,
sp hybridised carbon sp2 hybridised carbon sp3 hybridised carbon
Electronegativity decreases because ‘s’ characters of hybrid orbitals decrease
33.33% s characters 25% s characters Due to all these reasons
phenol is more acidic than
alcohol
phenol is more acidic than alcohol….. Why???
> >

8. O O
+
O
+
O
+
Resonance structures of phenoxide ion
OH OH
+
OH
+
OH
+
OH
Resonance structures of phenol
O
-

9. If electron withdrawing groups (EWG) are present on phenol ring (e.g. – NO2, –CN, –CHO) ; acidic strength of phenol
increases. Because EWG favour formation of phenoxide ion.
If electron donating groups (EDG) are present on phenol ring (e.g. alkyl groups) ; acidic strength of phenol decrease.
Because these groups do not favour formation of phenoxide ion.
Same you can apply for alcohols as well.

10. Do it yourself…..

11. Esterification
Alcohols and phenols heated with carboxylic acid, acid chlorides and acid anhydrides to form esters in presence of H2SO4
R – OH + R’ COOH
H+
R COOR’ + H2O
OH
+ R’COOH
H+
O – C – R’
+ H2O
||
O
H+
R COOR’ + H2O
R’ C – O
O
R’ C – O
R – OH +
Also represented as (R’CO)2O
OH
+ R’COCl Pyridine used in this reaction neutralises HCl so equilibrium shifts
towards RHS
O – C – R’
+ HCl
||
O
The introduction of acetyl (CH3CO) group in alcohols or phenols
is known as acetylation. Acetylation of salicylic acid
produces aspirin.
H+
COOH
CH3COOH
COOCH3
+
COOH
(CH3CO)2O+
OH
Aspirin
Aspirin has analgesic, anti inflammatory & anti pyretic
properties
IMPORTANT
pyridine
OH – is removed from acid
H+ is removed from acid

12. + R – O – H
Step 2 Attack of alcohol on carbonyl carbon
R’ – C – O – H
||
O
H++ R’ – C – O – H
|
O
|
– H
R – O – H
+
Step 1 Protonation
R’ – C – O – H
|
O
|
– H
R – O – H
+
H+
R’ – C – O – H
|
O
|
– HH –
R – O – H
+
+
Step 3 Protonation
R’ – C – O – H
|
R – O
+
Step 4 Dehydration and deprotonation
– H2O
– H+
Step 5 Deprotonation
R’ – C – O
|
R – O
+ –
R’ – C O
|
R – O
––
Role of H2SO4 – 1. Protonation
2. Dehydration
Mechanism
R’ – C – O – H
|
O – H
+

13. Reactions involving braking of C – O bond
1.Reactions with HX
CH3OH HCl+
ZnCl2
CH3Cl + H2O
CH3CH2OH HBr+ CH3CH2Br + H2O
2.Reactions with PX3 and PX5
3R – O – H + PCl3
3R – Cl + H3PO3
R – O – H + PCl5
R – Cl + POCl3 + HCl

14. 3.Reactions with SOCl2 (Darzen’s method)
R – O – H + SOCl2
R – Cl + SO2 + HCl
R – Cl + SO2 + HCl
N
Follows SNi mechanism
Follows SN2 mechanism
Without pyridine

15. 4. Dehydration of alcohols
1o alcohols – Conc. H2SO4 at 443 K
2o alcohols – 85% H3PO4 at 440 K
3o alcohols – 20% H3PO4 at 358 K
Ease of reaction:
1o alcohols < 2o alcohols < 3o alcohols
Ease of reaction increases
Conc. H2SO4
443 K
CH2 = CH2 + H2O
CH3 – CH – CH3
|
OH
85% H3PO4
440 K
CH2 = CH – CH3 + H2O
CH3 – C – CH3
|
OH
|
CH3
20% H3PO4
358K
CH3 – C = CH2 + H2O
|
CH3
CH2 – CH2
|
H
|
OH
ab
Because in this reaction carbocation
formation takes place Tertiary carbocation is
most stable that’s why tertiary alcohols can
undergo dehydration easily

16. Mechanism
CH3 – CH – CH3 + H +
|
O – H
CH3 – CH – CH3
|
O
HH
+
CH3 – CH – CH3
|
O
HH
+
CH3 – CH – CH2
+
|
H
Step 1: Protonation
Step 2: Dehydration (formation of carbocation)
Step 3: Deprotonation
CH3 – CH – CH3
+
- H2O
CH3 – CH = CH2 + H+
Here H+ (acid) used in step 1 forms again
in step 3
To shift equilibrium to RHS alknene is removed

17. 5. Oxidation
H – C – O – H
|
|
In oxidation these two bonds break
C = O
[O]
Following oxidation agents are used,
1. KMnO4 / OH – followed by hydrolysis
2. KMnO4 / H+
3. K2Cr2O7 / H+
4. Conc. HNO3
Strong oxidising agents
Alcohols directly
converted into
carboxylic acid
PCC (Pyridinium chloro chromate)
N
H
+
Cl -
+ CrO3
CrO3
Mild oxidising agents
Specifically oxidise
1o Alcohols into aldehydes
2o alcohols into ketones
R – CH2 – OH
PCC
R – CHO
CH – OH
R
R
CrO3
C = O
R
R
Tertiary alcohols do not oxidise easily. If vigorous oxidation is carried out they undergo elimination i.e. they form alkenes.

18. When the vapours of a primary or a secondary alcohol are passed over heated copper at 573 K, dehydrogenation takes place and an
aldehyde or a ketone is formed while tertiary alcohols undergo dehydration.
6. Dehydrogenation reaction
CH3 – CH2 – OH
Cu
573K
CH3 – CHO
CH – OH
CH3
CH3
Cu
573K
C = O
CH3
CH3
CH3 – C – CH3
|
OH
|
CH3
Cu
573K
CH3 – C = CH2 + H2O
|
CH3

19. Reactions of phenols
1. Electrophilic aromatic substitution
The –OH group attached to the benzene ring. Due to this p electrons
delocalise on ortho and para position. So, electron density at ortho
and para position increases.
Therefore, phenols show electrophilic substitution reaction. –OH
group directs the incoming group to ortho and para positions in the
ring.
OH
a. Nitration
O
H
O
d+
d-
d-
N = O
H – O N
OO
d+ d-
OH N
OO
d-
Intermolecular H – Bonding
Intramolecular H – Bonding
Steam volatile
Steam on – volatile
So, both ortho and para products are separated by
steam distillation
Dilute HNO3
OH
NO2
OH
NO2
o – Nitrophenol p – Nitrophenol
+

20. With concentrated nitric acid, phenol is converted to 2,4,6-trinitrophenol. The product is commonly known as picric acid. The
yield of the reaction product is poor.
OH
concentrated HNO3
OH
NO2O2N
NO2
Nowadays picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2,4, 6-
triisulphonic acid, and then with concentrated nitric acid to get 2,4,6-trinitrophenol.
OH
concentrated H2SO4
OH
SO3HHO3S
SO3H
Poor yield
concentrated HNO3
OH
NO2O2N
NO2
Maximum yield

21. Phenol is heated with acyl cholride in presence of anhydrous AlCl3(Lewis acid) to give hydroxy phenyl acetate (para and ortho
product). Out of two; para product is major.
Friedel Craft’s acylation and alkylation
OH
Anhydrous AlCl3
+ CH3COCl
OH
+
COCH3
OH
COCH3
Major
Phenol is heated with alkyl cholride in presence of anhydrous AlCl3(Lewis acid) to give p and o alkyl phenol. Out of two; para
product is major.
OH
Anhydrous AlCl3
+ CH3Cl
OH
+
CH3
OH
CH3
Major

22. b. Halogenation
When the reaction is carried out in solvents of low polarity such as CHCl3 or CS2 and at low temperature, monobromophenols are
formed.
OH
Br2 in CS2
273 K
OH
Br
The usual halogenation of benzene takes place in the presence of a Lewis acid, such as FeBr3 which polarises the halogen
molecule.
In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly
activating effect of –OH group attached to the benzene ring.
When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.
OH
3Br2
Bromine water
OH
BrBr
Br

23. Write the structures of the major products expected from the following
reactions:
(a) Mononitration of 3-methylphenol
(b) Dinitration of 3-methylphenol
(c) Mononitration of phenyl methanoate.
Do it yourself…..
OH
CH3
NO2
OH
CH3
NO2
(a) (b)
OH
CH3
NO2
NO2
(c)
OCOCH3
NO2

24. 2. Kolbe’s reaction
Phenoxide ion generated by treating phenol with sodium hydroxide is even more reactive than phenol towards electrophilic
aromatic substitution. Hence, it undergoes electrophilic substitution with carbon dioxide, a weak electrophile. Ortho
hydroxybenzoic acid is formed as the main reaction product.
OH
NaOH i. CO2
ii. H+
OH
COOH
3. Reimer-Tiemann reaction
OH
CHCl3 + NaOH
O Na+
–
O Na+
CH2Cl
– O Na+
CHO
–
NaOH H+
OH
CHO
Slicyladehyde
Slicylic acid

25. OH
Zn
+ ZnO
4. Reduction
OH
Na2Cr2O7
O
O
5. Oxidation
Benzoquinone

26. ROH
2Na
R’COOH
H2SO4
2 R – O – Na + H2
RCOOR’ + H2O
1o alcohols – Conc. H2SO4 at 443 K
2o alcohols – 85% H3PO4 at 440 K
3o alcohols – 20% H3PO4 at 358 K
Dehydration
C = C
Oxidation
C = O
Cu / 573 K
C = O 1o alcohols – Aldehyde 2o alcohols – Ketone 3o alcohols form alkenes
HCl ZnCl2
RCl
SOCl2
R – Cl + SO2 + HCl
KMnO4 / OH – followed by hydrolysis
KMnO4 / H+
K2Cr2O7 / H+
Alcohols Carboxylic acids1o alcohols – Aldehyde
2o alcohols – Ketone
PCC or CrO3
3o alcohols undergo elimination and form alkenes

27. OH
Dilute HNO3
OH
NO2
OH
NO2
Conc HNO3
OH
NO2O2N
NO2
Poor yield
concentrated H2SO4
OH
SO3HHO3S
SO3H
concentrated HNO3
OH
NO2O2N
NO2
Maximum yield
Br2 in CS2
273 K
OH
Br
3Br2
Bromine water
OH
BrBr
Br
OH
Zn
+ ZnO
Reduction
OH
Na2Cr2O7
O
O
Oxidation

28. OH
NaOH i. CO2
ii. H+
OH
COOH
O Na+
–
Kolbe’s reaction
OH
CHCl3 + NaOH
O Na+
CH2Cl
– O Na+
CHO
–
NaOH H+
Slicyladehyde
OH
CHO
Reimer-Tiemann reaction
IMPRTANT NAME REACTIONS
OH
Anhydrous AlCl3
+ CH3COCl
OH
COCH3
Major
OH
Anhydrous AlCl3
+ CH3Cl
OH
CH3
Major
Friedel Craft’s acylation and alkylation

29. Ethers
Ethers are organic compounds in which two alkyl groups attached with oxygen this oxygen is known as ethereal oxygen.
Classification of ethers
Simple ethers / Symmetrical ethers Mixed ethers / Asymmetrical ethers
CH3 – O – CH3
C2H5 – O – C2H5
O
CH3 – O – C2H5
CH3 – O – CH
CH3
CH3
O CH3

30. Classification of ethers
Aliphatic ethers Aromatic ethers Cyclic ethers
C2H5 – O – C3H7
CH3 – O – C4H9
O
CH2
–
O CH3
Oxirane
Oxetane
O
CH2
CH2
H2C
H2C CH2
O
Benzyl phenyl ether
Methyl phenyl ether

31. Nomenclature of ethers
CH3 – O – CH3
C2H5 – O – C2H5
C2H5 – O – CH3
Dimethyl ether
Common name IUPAC name
Methoxymethane
Diethyl ether Ethoxyethane
Ethyl methyl ether Methoxyethane
Methyl isopropyl ether 2 – methoxypropaneCH3 – O – CH
CH3
CH3
1
2
3
CH3 – O – C – CH3
CH3
CH3
1
2
3
tertbutyl methyl ether 2–methyl–2–methoxypropane

32. O CH3
O C2H5
O
CH3 – CH – O – CH – CH3
CH3CH3
1
2 3
1
23
CH3 – CH – CH2 – CH2 – O – CH3
OH
1 2 3 4
methyl phenyl ether
Anisol
Methoxybenzene
Phenetole Ethoxybenzene
Diphenyl ether Phenoxybenzene
2 – (2–propoxy)propane--------
--------
2 – methoxybutan–2–ol
Common name IUPAC name

33. – O – CH – CH2 – CH3
CH3
1
2 3 4
O C2H5
NO2
1
2
2 – cylcopropoxybutane
1– ethoxy–2–nitrocylcohexane
--------
--------
Common name IUPAC name

34. Structure of functional group
O
C C
111.7o
Two Sp3 hybridised orbital with lone pair of electrons
Bond angle greater than usual tetrahedral
angle (109.5o)because stearic hindrance of
two alkyl groups.
Sp3 hybridisation

35. Methods of preparation of ethers
1. By dehydration of alcohols
Ethanol protic acids (H2SO4, H3PO4)
At 443 K
Intermolecular dehydration (Substitution)
Intramolecular dehydration (Alkenes are formed)
(Elimination)
protic acids (H2SO4, H3PO4)
At 413 K
Intermolecular dehydration (Ethers are formed)
(Substitution)
CH3 – CH2 – OH
H2SO4
443 K
H2SO4
413 K
CH2 = CH2
CH3 – CH2 – O – CH2 – CH2

36. Mechanism
Intermolecular dehydration of alcohols follows SN2 mechanism
CH3 – CH2 – O – H + H+
CH3 – CH2 – O
H
H
+
CH3 – CH2 – O
H
H
+
+CH3 – CH2 – O – H CH3 – CH2 – O – CH2 – CH3
|
H
+
CH3 – CH2 – O – CH2 – CH3
|
H
+
CH3 – CH2 – O – CH2 – CH3
First molecule of alcohol
Second
molecule
The reaction follows SN1 pathway when the
alcohol is secondary or tertiary
However, in case of secondary and tertiary
alcohols there is competition between
substitution reaction and elimination reaction.
In this elimination reaction dominates over
the substitution.
This method is inappropriate because asymmetrical ethers
cannot be prepared by this process

37. It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an
alkyl halide is allowed to react with sodium alkoxide.
2. Williamson synthesis
R – X R’ – O Na
– +
+ R’ – O – R + NaX
Ethers containing secondary or tertiary alkyl groups may also be prepared by this method.
This reaction follows SN2 mechanism
CH3
CH3 – C – O Na
|
|
CH3
+–
+ CH3 – Br
CH3
CH3 – C – O – CH3
|
|
CH3
+ NaBr
1. Better results are obtained if the alkyl halide is primary.
2. In case of secondary and tertiary alkyl halides, elimination competes over substitution.
3. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed.
Alexander William Williamson

38. CH3 – C = CH2
|
CH3
+ NaBr + CH3OHCH3 – O Na
+–
CH3
CH3 – C – Br
|
|
CH3
+
Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide ion.
OH
NaOH
O Na
– +
R’– X
O – R’

39. Due to this polar nature ethers have higher boiling point than alkanes of comparable masses. But they have lower boiling points
than alcohols because absence of intermolecular H – bonding
Physical properties of ethers
O
C C
|
|
d+
d-
The C-O bonds in ethers are polar and thus, ethers have a net dipole moment. But they are weakly polar than alcohols
1. Boiling point
2. Solubility in water
Ethers form intermolecular H – bonding with water just like alcohols so, they are soluble in water.
O
R
R
d-
O
HH
O
H Hd+
d+
d+

40. Chemical reactions of ethers
Ethers are the least reactive of the functional groups. The cleavage of C-O bond in ethers takes place under drastic conditions
with excess of hydrogen halides. The reaction of dialkyl ether gives two alkyl halide molecules.
1. Cleavage of C–O bond in ethers
R – O – R + HX RX + R – OH
R – O – R
H + X
–
Similarly, phenols also can be formed by this method.
O – R
+ HX
OH
+ R – OH

41. Ethers with two different alkyl groups are also cleaved in the same manner.
R – O – R’ + HX RX + R’ – OH
The order of reactivity of hydrogen halides is as follows:
HI > HBr > HCl.
Mechanism
CH3 – O – C2H5 + H – I CH3 – O – C2H5
H
| +
+ I
–
Oxonium ion
CH3 – O – C2H5
H
| +
I +
–
CH3 O – C2H5
H
|+
I CH3 – I + C2H5 – OH

42. If HI is used in excess the molecule of alcohol also get converted into alkyl iodide
+ H – I + I
–
CH3 – CH2 – O
H
H
+
I +
– CH3 – CH2 – I + H2O
CH3 – CH2 – O
H
H
+
CH3 – CH2 – O – H
If aromatic ethers are used, then phenols get formed. Phenols cannot be attacked by I- again because C – O bond in
phenol cannot be broken easily as this bond is strong bond. This bond is strong because,
1. it shows partial double bond character because of resonance
2. Due to higher electronegativity of sp2 hybridised carbon.

43. However, when one of the alkyl group is a tertiary group, the halide formed is a tertiary halide.
CH3
CH3 – C – O – CH3
|
|
CH3
+ HI
CH3
CH3 – C – I
|
|
CH3
CH3 – OH +
But in case of aromatic ethers, always phenol gets formed because C – O bond cannot be broken easily as it shows
partial double bond characters.

44. The alkoxy group (-OR) is ortho, para directing and activates the aromatic ring towards electrophilic substitution in the same way
as in phenol.
2. Electrophilic substitution
O – R O – R
+
O – R
+
O – R
+
O – R

45. a. Halogenation:
Anisole undergoes bromination withbromine in ethanoic acid even in the absence of iron (III) bromide catalyst. It is due to the
activation of benzene ring by the methoxy group. Para isomer is obtained in 90% yield.
O – CH3
Br2
Bromine water +
O – CH3
Br
O – CH3
Br
p – bromoanisol
Major
o – bromoanisol
Minor

46. Phenol is heated with acyl cholride in presence of anhydrous AlCl3(Lewis acid) to give (para and ortho product). Out of two; para
product is major.
b. Friedel Craft’s acylation and alkylation
O CH3
Anhydrous AlCl3
+ CH3COCl
O CH3
+
COCH3
O CH3
COCH3
Major
Phenol is heated with alkyl cholride in presence of anhydrous AlCl3(Lewis acid) to give p and o productsl. Out of two; para
product is major.
OH
+ CH3Cl
O CH3
+
CH3
O CH3
CH3
Major
4 – methoxyacetophenone 2 – methoxyacetophenone
Anhydrous AlCl3
CS2
4 – methoxytoulene 2– methoxytoulene

47. O CH3
c. Nitration
H2SO4 + HNO3
O CH3
NO2
O CH3
NO2
o– Nitroanisol
Minor
p – Nitroanisol
Major
+

48. Some Commercially Important Alcohols
Methanol, CH3OH, also known as ‘wood spirit’, was produced by destructive distillation of wood. Today, most of the methanol is
produced by catalytic hydrogenation of carbon monoxide at high pressure and temperature and in the presence of ZnO – Cr2O3
catalyst.
CO + 2H2
ZnO + CrO3
200 – 300 atm
573 – 673 K
CH3 OH
Methanol is a colourless liquid and boils at 337 K.
It is highly poisonous in nature.
Ingestion of even small quantities of methanol can cause blindness and large quantities causes even death.
Methanol is used as a solvent in paints, varnishes and chiefly for making formaldehyde.
1. Methanol

49. Ethanol, C2H5OH, is obtained commercially by fermentation, the oldest method is from sugars. The sugar in molasses,
sugarcane or fruits such as grapes is converted to glucose and fructose, (both of which have the formula C6H12O6), in the
presence of an enzyme, invertase. Glucose and fructose undergo fermentation in the presence of another enzyme, zymase,
which is found in yeast.
2. Ethanol
C12H22O11 + H2O
Invertase
C6H12O6 + C6H12O6
C6H12O6
Zymase
2C2H5OH + 2CO2

50. Wine making
grapes are the source of sugars and yeast As grapes ripen, the quantity of sugar increases and yeast grows
on the outer skin.
When grapes are crushed, sugar and the enzyme
come in contact and fermentation starts
Fermentation takes place in anaerobic
conditions i.e. in absence of air. Carbon
dioxide is released during fermentation
The action of zymase is inhibited
once the percentage of alcohol
formed exceeds 14 percent. If air gets into fermentation mixture, the oxygen of air oxidises ethanol to ethanoic acid which
in turn destroys the taste of alcoholic drinks.

51. Ethanol is a colourless liquid with boiling point 351 K.
It is used as a solvent in paint industry and in the preparation of a number of carbon compounds. The commercial alcohol is
made unfit for drinking by mixing in it some copper sulphate (to give it a colour) and pyridine (a foul smelling liquid). It is
known as denaturation of alcohol.
Properties of ethanol