This document discusses the nomenclature, properties and reactions of alcohols, phenols, and ethers. It defines each compound group and provides IUPAC names for examples. Alcohols are formed by replacing hydrogen in hydrocarbons with hydroxyl groups. Phenols have hydroxyl groups attached to aromatic systems. Ethers have an alkoxy or aryloxy group in place of hydrogen. The document outlines common preparation methods for each group and describes physical properties like boiling points. It also explains reactions like dehydration, esterification, and oxidation of alcohols.

2. After studying this topic, you will be able to:
•name alcohols, phenols and ethers according to
the IUPAC system of nomenclature

3. If H of hydrocarbon is replaced by:
• One or more hydroxyl (OH) group directly
attached to carbon atom(s), of an aliphatic
system we get an ALCOHOL eg: C2H5OH
• C6H5CH2OH ALCOHOL AROMATIC
• If (–OH) group(s) directly attached to carbon
atom(s) of an aromatic system, we get a
PHENOL (C6H5OH).
• an alkoxy or aryloxy group (R–O/Ar–O), we get
ETHER eg: CH3OCH3 .

5. Structure Name (common) IUPAC Name Type of phenol
Phenol Phenol Monohydric
o- cresol 2-Methylphenol Monohydric
Catechol Benzene-1,2-diol
Dihydric
Resorcinol Benzene-1,3-diol
Dihydric
Hydroquinone Benzene-1,4-diol
Dihydric

6. Structure Name (common) IUPAC Name Type of ethers
CH3OCH3 Dimethyl ether Methoxymethane Symmetrical
CH3CH2OCH2CH3 Diethyl ether Ethoxyethane Symmetrical
CH3CH2CH2OCH3 Methyl n- propyl
ether
Methoxypropane Unsymmetrical
C6H5OCH3 Methyl phenyl
ether (Anisole)
Methoxybenzene
(Anisole)
Unsymmetrical
C6H5OC2H5 Ethyl phenyl ether Ethoxybenzene Unsymmetrical
(i)4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol (ii) 2, 5-Dimethylhexane-1,3-diol
(iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol

7. Preparation of Alcohols
• 1. Hydration of alkenes (Markovnikov addition of
water)
CH3CH=CH2 H2O, H2SO4 CH3CH(OH)CH3
(Markovnikov rule, OH- goes least hydrogenated carbon)

8. CH3C**H=C*(CH3)2 H2O, H2SO4 CH3CH2C(OH)(CH3)2
• 2. Hydration of alkenes (Anti Markovnikov addition of
water) Hydroboration oxidation
CH3CH=CH2 (i)B2H6 (ii)H2O, H2O2,OH- CH3CH2CH2OH
CH3CH=CH2 (HBH2)2 CH3CH(H)CH2(BH2)
CH3CH(H)CH2(BHH) + CH3CH=CH2 (CH3CH2CH2)2BH
(CH3CH2CH2)2BH + CH3CH=CH2 (CH3CH2CH2)3B
(CH3CH2CH2)3B H2O, H2O2,OH- 3CH3CH2CH2OH + B(OH)3
CH3C*H=C**(CH3)2 (i)B2H6 (ii)H2O, H2O2,OH-

9. • 3.Reduction of aldehydes and ketones
• Aldehyde / ketone on reduction give alcohol
• Any one of the following Reducing agents:
(i)H2 Pt or Pd or Ni (ii)NaBH4 (iii)LiAlH4
• RCHO NaBH4 RCH2OH (PRIMARY ALCOHOL)
• CH3CH2CH2CHO H2 / Pt CH3CH2CH2CH2OH
• RCOR LiAlH4 RCH(OH)R (SECONDARY ALCOHOL)
• CH3CH2COCH3 H2 / Pt CH3CH2 CH CH3
OH
Acetone on reduction with NaBH4 gives? CH3CH(OH)CH3
Ethanal on reduction with LiAlH4 gives? CH3CH2OH
Alcohol ----oxidation -------ald/ketone -------oxidation -----carboxylic acid

10. • 4.Grignard reagent and aldehydes / ketones
• >Cδ+=Oδ- + Rδ-Mgδ+X >C-O- Mg+X H2O >C-OH
R R
HCHO + CH3MgCl  H-C-OMgCl H2O H-C-OH
• H CH3 H CH3
• CH3CHO + CH3MgCl 
• CH3COCH3 + CH3MgBr ?
Formaldehyde on reaction with Grignard reagent followed by hydrolysis gives 1o
alcohol
ANY OTHER aldehyde on reaction with Grignard reagent followed by hydrolysis gives
2o alcohol
KETONE on reaction with Grignard reagent followed by hydrolysis gives 3o alcohol

11. • 5. Carboxylic acid on reduction
• LiAlH4 is an expensive reagent, so is not used
commercially
• Commercial method
• RCOOH R’OH, H+ RCOOR’ H2, Pt RCH2OH + R’OH
RCOOH (i) LiAlH4 (ii) H2O RCH2OH

12. Preparation of Phenols
• From Haloalkanes (Dow’s Process)
• From Benzene sulphonic acid
• Oleum conc H2SO4 : H2S2O7

13. Preparation of Phenols
From Diazonium salt
273K
From cumene COMMERCIAL

14. Preparation of Phenols
HCHO +

15. 11.4 HCHO + ?  CH3CHCH2OH
CH3
H2C=O + CH3CHMgCl 
CH3
MgBr CH2OH
HCHO + 
11.5 (i)CH3CH=CH2 H2O, H+ CH3CH(OH)CH3
(ii) O OH
CH2COOCH3 NaBH4 CH2COOCH3
(iii) CH3CH2CHCHO NaBH4 CH3CH2CHCH2OH
CH3 CH3

16. Physical Properties
• Boiling point
 Increase with increase in number of carbon atoms as (van
der waal forces increase)
 decrease with increase of branching in carbon
chain(decrease in van der Waals forces with decrease in
surface area)
 CH3CH2CH2CH2OH CH3CH2CHCH3
 OH
 Alcohols and phenols have higher boiling point than
hydrocarbons, ethers, haloalkanes and haloarenes of
comparable molar masses (alcohols have intermolecular
hydrogen bonding, hydrocarbons van der waal forces,
ethers and haloalkanes have dipole -dipole interaction)

17. Alkane bp oC Alkyl
halide
bp oC alcohol bp oC ether bp oC
propane -42 Methyl
chloride
-24 ethanol 79 Dimethyl
ether
-25

18. Solubility
• alcohols and phenols are soluble in water is
due to their ability to form hydrogen bonds
with water molecules.
• The solubility decreases with increase in size
of alkyl/aryl (hydrophobic) groups.
• Several of the lower molecular mass alcohols
are miscible with water in all proportions.

19. Back Exercise
11.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O
and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary
and tertiary alcohols.
11.4 Explain why propanol has higher boiling point than that of the hydrocarbon,
butane?
11.5 Alcohols are comparatively more soluble in water than hydrocarbons of
molecular masses. Explain this fact.
11.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
11.7 Give the structures and IUPAC names of monohydric phenols of molecular
formula, C7H8O.
11.9 Give the equations of reactions for the preparation of phenol from cumene.
11.10 Write chemical reaction for the preparation of phenol from chlorobenzene.
11.11 Write the mechanism of hydration of ethene to yield ethanol.
11.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the
preparation of phenol using these reagents.

20. • CH3CH2CH2CH2CH2OH
• CH3CH2CH2CH2OH
CH3

21. 11.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by
an SN
2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
(i) CH=CH2 H2SO4/H2O CH(OH)CH3
CH2Br aq NaOH

22. The bond between O–H is broken(H+ goes) when alcohols react
as nucleophiles.
Breaking of RO-H
1. Acidity of alcohols and phenols
2ROH + 2Na  2RONa + H2
ROH + NaOH  no reaction
2C6H5OH + 2Na  2C6H5ONa + H2
C6H5OH + NaOH  C6H5ONa + H2O

23. • The acidic character of alcohols is due to the
polar nature of O–H bond.
• An electron-releasing group (–CH3, –C2H5)
increases electron density on oxygen tending
to decrease the polarity of O-H bond.
• Alcohols are, however, weaker acids than
water.

24. The acid strength of alcohols decreases in the following
order:
• Primary> secondary>tertiary
•
• (O is more electronegative than H so attracts shared pair towards
itself thus making the bond polar. When e- density on O is
increased, its tendency to attract electrons is decreased making bond
less polar)

25. Acidity of phenols
• Phenol is a stronger acid than alcohols and water.
• WHY?
• Which is more polar: alcohol or phenol? C3-O C2-O
• Due to high electronegativity of sp2 hybrid carbon attached to –OH in
phenol, electron density decreases on O making O-H bond in phenols more
polar than alcohols (which have sp3 hybrid carbon)
25

26. Acidity of phenols
• Which ion is more stable: phenoxide or
alkoxide?
• When H+ is released from alcohol, it forms alkoxide
ion, which is less stable.
• When H+ is released from phenol, it forms phenoxide
ion, which is stable due to resonance.
26

27. Resonance Recap
• 1. More the number of resonance structures
higher is the stability.
• 2.Structures which involve separation of
charges are less stable so contribute less
towards resonance hybrid.
• 3.Structures in which –ve charge is delocalised
on more electronegative atom are more stable
than those where there is localisation of
charge.

28. Compare resonance structures of phenol and phenoxide ion
Structures II,III and IV of phenol involve separation of charges thus they
contribute less towards resonance hybrid.
Phenoxide ion involves delocalisation of negative charge so is more
stable. The reaction shifts in forward direction, towards more stable
phenoxide ion making phenols more acidic than alcohols.
In Alkoxide ion negative charge
is localised on oxygen so it is
less stable than alcohol

29. resonance structures of phenol

30. Effect of substituent on acidity
• Higher the pKa = - log Ka value weaker is the acid
• Cresols are weaker acids than phenol while nitrophenols are stronger acids than
phenol.
• Nitro group is e- withdrawing group, it withdraws electrons from phenoxide ion
making it more stable thus increasing acidity of phenols.
• Methyl group is e- donating group, it destabilizes phenoxide ion making phenols
less acidic.

31. • 11.14, 11.15 11.16 Back exercise

32. Esterification
• Alcohols and phenols react with carboxylic acids,
acid chlorides and acid anhydrides to form
esters.
• ROH + R’COOH H+ R’COOR + H2O
CH3.CH2.OH + HOOC.CH2.CH3 H+ CH3.CH2.OOC.CH2.CH3 + H2O
ROH + ( R’CO)2O H+ R’COOR + R’COOH
Water/ carboxylic acid formed in the reactions above is removed as soon as it is
formed so that reaction moves in forward direction. (Le Chatlier’s principle)
32

33. RCO
O + R’O-H RCOOR’ + RCOOH
RCO

34. Esterification
ROH + R’COCl pyridine R’COOR + HCl
CH3CH2OH + ClOCCH2CH3 pyridine CH3CH2OOCCH2CH3 + HCl
• Pyridine is a weak base. It is added to neutralize
HCl formed during the reaction so that reaction
moves in forward direction. (Le Chatlier’s
principle)
34

35. Reactions involving breaking of carbon – oxygen (C–O) bond
1. Reaction with hydrogen halides
Alcohols react with hydrogen halides to form alkyl halides.
ROH + HX (anhy ZnCl2) → R–X + H2O
LUCAS REAGENT (conc. HCl and ZnCl2) TEST
Alcohols are soluble in Lucas reagent while their halides are immiscible
and produce turbidity in solution.
Tertiary alcohols, turbidity is produced immediately
Secondary alcohols, turbidity is produced within 5 minutes
Primary alcohols do not produce turbidity at room temperature.
2. Reaction with SOCl2, PCl3 and PCl5
ROH + PCl3 → RCl +H3PO3
ROH + PCl5 →RCl + POCl3 + HCl
ROH + SOCl2 → RCl + SO2 + HCl

36. 2. Dehydration of alcohols
CH3CHCH2CH3
conc H2SO4 CH3CH=CHCH3 + CH2=CHCH2CH3
Major Minor
Saytzeff or Zaitsev Rule states that the more substituted alkene will be the
major product. (H will be removed from least hydrogenated carbon during
elimination reaction)
OH

37. • Ques: Why does the relative ease of dehydration
of alcohols follow the following order:
Tertiary > Secondary > Primary
• Ans: Mechanism of dehydartion involves
carbocation intermediates. Tertiary carbocations
are more stable and therefore are easier to form
than secondary and primary carbocations thus
tertiary alcohols are the easiest to dehydrate.

38. Oxidation
• Oxidation of alcohols involves the formation of a carbon-oxygen
double bond with cleavage of an O-H and C-H bonds. These are also
known as dehydrogenation reactions as these involve loss of
dihydrogen from an alcohol molecule.
•
• RCH2OH [O] RCHO [O] RCOOH
Primary alcohol on oxidation gives aldehyde same number of C atoms which
on further oxidation gives carboxylic acid with same number of carbon
atoms
RCH(OH )CH3
[O] RCOCH3
[O] RCOOH
Secondary alcohol on oxidation gives ketone same number of C atoms which
on further oxidation gives carboxylic acid with lesser number of carbon
atoms
38

39. Alcohol to aldehyde/ ketone:Weak oxidising agent like PCC (pyridium chloro chromate)
and CrO3
Alcohol to Carboxylic acid: strong oxidising agent like acidified/ alkaline K2Cr2O7 or
acidified/ alkaline KMnO4
CH3CH2OH CrO3 or PCC CH3CHO
CH3CH(OH )CH3 CrO3 CH3COCH3
PCC does not effect double bond
CH2=CHCH2CH2OH PCC CH2=CHCH2CH2CHO
CH3CH2OH acidified KMnO4 CH3COOH
Tertiary alcohols do not undergo oxidation reaction. Under
strong reaction conditions such as strong oxidising agents
(KMnO4) and elevated temperatures, cleavage of various C-C
bonds takes place and a mixture of carboxylic acids
containing lesser number of carbon atoms is formed.

40. Reaction with Cu at 573K
CH3CH2OH Cu/573K CH3CHO
CH3CH(OH )CH3 Cu/573K CH3COCH3
CH3CH(OH )CH3 Cu/573K CH2=CHCH3
CH3 CH3

41. Reactions of phenols
• Electrophilic aromatic substitution
• The –OH group attached to the benzene ring
activates it towards electrophilic substitution.
Also, it directs the incoming group to ortho and
para positions in the ring.
41

42. Nitration

43. Nitration
43
With dil. HNO3, a mixture of ortho and para
Nitrophenols are formed

44. o-Nitrophenol is steam volatile due to intramolecular
hydrogen bonding while p-nitrophenol is less volatile due to
intermolecular hydrogen bonding which causes the association
of molecules.
O and p nitro phenol can be separated by distillation

45. With conc HNO3, picric acid is produced, but yield is low
Phenol + sulphuric acid  o and p hydroxysulphonic acid
Nitration Picric (good yield)

46. Halogenation
1. Halogenation does not require a Lewis acid like FeCl3 or AlCl3
because 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.

47. Halogenation
In presence of CS2 the product was monobromophenol while
bromine water gives tribromo derivative (a white ppt) Why?
In aq medium, phenoxide ion is formed which further
activates benzene ring while in non polar solvents it is not
formed

48. Conversion to benzene
OH
+ Zn (dust) 
Oxidation

50. --CHCl2 + 2NaOH 2 NaCl + CH(OH)2

51. CHCl2 + NaOH 2 NaCl +
CH(OH)2

52. CH3CH2CH2CH2OH  CH3CH2CH=CH2
CH3CH2C(CH3)(OH)CH3+ 

53. CH3 OH

54. C6H5CH2OH

55. Some commercially important alcohols
• Methanol (wood spirit)
Poisonous, causes blindness and death
• Used as solvent for paint and varnishes and in
preparing formaldehyde
• Ethanol