Alcohols, Phenols and Ethers
1) Write IUPAC names of the following compounds:
Solution:
(1) 2, 2, 4-TRimethylepentan- 3-ol
(2) 5-Ethyleheptane-2, 4-diol
(3) Butane-2, 3-diol
(4) Propane-1, 2, 3-triol
(5) 2-Methyle phenol
(6) 4-Methyl phenol
2) Write IUPAC names of the following compounds:
Solution:
(7) 2, 5-Dimethyl phenol
(8) 2, 6-Dimethyl phenol
(9) 1-Methoxy-2-methyle propane
(10) Ethoxybenzene
(11) 1-Phenoxyheptane
(12) 2-Ethoxybutane
3) Write structures of the compounds whose IUPAC name is as follows:
(1) 2-Methylbutan-2-ol
(2) 1-Phenylpropan-2-ol
(3) 3, 5-Dimethylhexane-1, 3, 5-triol
(4) 2, 3-Diethylphenol
(5) 1-Ethoxypropane
(6) 2-Ethoxy-3-methylpentane
(7) Cyclohexylmethanol
(8) 3-Cyclohexylpentan-3-ol
(9) Cyclopent-3-en-1-ol
(10) 3-Chloromethylpentan-1-ol
Solution:
4) (1) Draw the structures of all isomeric alcohols of molecular formula
C5H12O and give their IUPAC names.
(2) Classify the isomers of alcohols in question 11.3 (1) as primary,
secondary and tertiary alcohols.
Solution:
The possible isomers of alcohols of molecular formulas, C5H12O are:
5) Explain why propanol has higher boiling points than that of the
hydrocarbon, butane?
Solution:
Higher boiling point of propanol in comparison to that of butane is due to the
presence of intermolecular H-bonding among molecules of propanol while butane
lacks the ability to form hydrogen bond, due to the absence of –OH group. In
butane only weak van der Waal’s forces exist as intermolecular forces between the
molecules.
6) Alcohols are comparatively more soluble in water than hydrocarbons of
comparable molecular masse. Explain this fact.
Solution:
Higher solubility of alcohols in water is due to presence of polar O-H group by
which molecules of alcohols are able to form H-bonds with molecules of water.
Hydrocarbon lacks the ability to form hydrogen bonds with water and hence are
insoluble in water.
Intermolecular hydrogen bonding between water and alcohol molecules
7) What is meant by hydroboration oxidation reaction? Illustrate it with the
example.
Solution:
Conversion of alkenes into alcohols by addition of diborane followed by oxidation
of the intermediate trialkyl boron by hydrogen peroxide is called hydroboration
oxidation reaction.
The addition of water to alkenes takes place in an opposite manner to
Markownikov’s rule.
8) Give the structures and IUPAC names of monohydric phenols of
molecular formula, C7H8O.
Solution:
The possible monohydric phenols are:
2-Methyl Phenol 3 – Methyl Phenol 4-Methyl Phenol
9) While separating a mixture of ortho and Para nitro phenols by steam
distillation, name the isomer which will be steam volatile. Give reason.
Solution:
O-Nitrophenol is more steam volatile than p-nitrophenol as it has intermolecular
H-boding within single molecular structure while p-nitrophenol has stronger
intermolecular H-bonding among its several molecules.
O-Nitrophenol p-Nitrophenol
10) Give the equations of reactions for the preparation of phenol from
Cumene.
Solution:
+ O2
Cumene Cumene hydroxide
Phenol Propanone
11) Write chemical reaction for the preparation of phenol from
chlorobenzene.
Solution:
Chlorobenene Sodium phenoxide Phenol
12) Write the mechanism of hydration of ethane to yield ethanol.
Solution:
Step-I
Ionisation of water
Step-II
Attack of H+ (proton) on ethane to form carbonium ion.
Ethene Proton Carbonium ion
Step-III
Attack of OH- ion on carbonium ion to give ethanol
Carbonium ion Hydroxide ion Ethanol
13) You are given benzene, conc. H2SO4 and NaOH. Write the equations for
the preparation of phenol using these reagents.
Solution:
Benzene Benzene sulphonic acid sodium phenoxide
Phenol
14) Show how you will synthesize:
(1) 1-phenylethanol from suitable alkenes
(2) Cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(3) pentan-1-ol using a suitable alkyl halide?
Solution:
Phenyl ethane 1-Phenyl ethanol
Cyclohexylbromoethane Cyclohexylmethanol
1-Bromopentane Penton-1-ol
15) Give two reactions that show the acidic nature of phenol. Compare
acidity of phenol with that of ethanol.
Solution:
Two reactions which show acidic nature of phenol are:
(1) Phenol reacts to give hydrogen with highly reactive metals like Na and K.
Phenol Sodium phenoxide
(2) Phenol forms salt and water on reaction with strong alkalies like NaOH and
KOH.
Phenol Sodium Phenoxide
Phenol is more acidic than alcohol because it has tendency to lose proton (H+
ion)
to form more stable resonance stabilized phenoxide ion. On the other hand
formation of ethoxide ion from ethanol is less favoured as it is not resonance
stabilized like phenoxide ion.
16) Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Solution:
Ortho nitrophenol is more acidic than ortho methoxyphenol because nitro group
being electron withdrawing increases the extent of polarization in O-H bond while
methoxy group being electron releasing decreases the extent of polarization in O-H
bond.
Ortho nitrophenol ortho methoxyphenol
More acidic less acidic
17) Explain how does the –OH group attached to a carbon of benzene ring
activate it towards eletrophilic substitution?
Solution:
-OH group releases its electron pair to the benzene ring and hence the electron
density of benzene ring is increased. As a result electrophiles attack the benzene
ring more readily. Hence it is said that –OH group has activated the benzene ring
towards eletrophilic substitutions. Such groups that increase the electron density on
the benzene ring are also called ring activating groups.
The resonance in phenol can be shown as:
18) Give equations of the following reactions:
(1) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(2) Bromine in CS2 with phenol.
(3) Dilute HNO3 with phenol.
(4) Treating phenol with chloform in presence of aqueous NaOH.
Solution:
(1)
1- Propanol Propanoic acid
(2)
Phenol o-Bromophenol p-bromophenol
(minor product) (major product)
Phenol o-nitrophenol p-nitrophenol
Phenol Intermediate Salicylaldehyde
19) Explain the following with an example.
(1) Kolbe’s reaction (2) Reimer –Tiemann reaction
Solution:
(1) Kolbe’s reaction
This reaction takes place by heating sodium phenoxide (obtained by reacting
phenols with sodium hydroxide) with carbon dioxide at 4-7 atm pressure followed
by acidification with acids to given salicylic acid.
Phenol Sodium phenoxide sodium salicylate
Salicylic acid
(2) Reimer –Tiemann Reaction
When phenol is treated with chloroform in the
presence of an alkali (NaOH), Salicylaldehyde is formed.
Intermediate Salicylaldehyde
20) Explain the following with an example.
(3) Williamson ether synthesis (4) Unsymmetrical ether
Solution:
(3) Williamson ether synthesis
The reaction of an alkyl halide with sodium alkoxide to form ether is Williamson
ether synthesis.
1, 1-Dimethyl sodium ethoxide Ethyl chloride 2, 2- Dimethyl ethoxy ethane
Ethers containing substituted alkyl groups are prepared by this method.
(4) Unsymmetrical ethers
Ethers in which different alkyl groups are attached to the ethereal oxygen
atom are called unsymmetrical ethers.
Methoxybenzene Methoxybenzene
21) Write the mechanism of acid catalysed dehydration of ethanol to yield
ethane.
Solution:
Ethanol undergoes dehydration in presence of concentrated sulphuric acid to give
ethene.
Ethanol Ethene
Step I: Ionisation of H2SO4
H2SO4 → H+ + H2S04
Step II: Attack of proton on ethanol to form protonated alcohol.
Protonated alcohol
Step III: Loss of water from protonated alcohol to form carbocation. This is slow
step.
Step IV: Loss of proton (H+) from carbocation to form ethene.
22) How are the following conversions carried out?
(1) Propene → Propan – 2-ol
(2) Benzyl chloride → Benzyl alcohol
(3) Ethyl magnesium chloride → Propan-1-ol
(4) Methyl magnesium bromide → 2- Methylpropan-2-ol
Solution:
(1)
Propene Propan 2-ol
Benzyl chloride Benzyl alcohol
Methanal Ethyl magnesium chloride Propan-1-ol
Propanone Methyl magnesium bromide
2-Methylpropan-2-ol
23) Name the reagents used in the following reactions:
(1) Oxidation of a primary alcohol to carboxylic acid.
(2) Oxidation of a primary alcohol to aldehyde.
(3) Bromination of phenol to 2, 4, 6-tribromophenol.
(4) Benzyl alcohol to benzoic acid.
(5) Dehydration of propa-2-ol to propene.
(6) Butan-2-one to butan-2-ol.
Solution:
(1) Acidified KMnO4
(2) Pyridinium chlorochromate (PCC) using CH2Cl2 as solvent
(3) Aqueous solution of bromine
(4) Acidified KMnO4
(5) Conc. H2SO4 at 443 K or 85% H3PO4 at 440K
(6) NaBH4 or LiAlH4 using dry ether as solvent
24) Give reason for the higher boiling point of ethanol in comparison to
methoxymethane.
Solution:
The high boiling point of ethanol in comparison to methoxy methane is due to its
ability to form intermolecular H-bonding. While methoxy methane does not form
intermolecular H-bonds.
25) Give IUPAC names of the following ethers:
Solution:
(1) 1-Ethoxy-2-methyl propane
(2) 2-Chloro-1-methoxymethane
(3) 1-methoxy-4-nitrobenzene
(4) 1-Methoxypropane
(5) 1-Ethoxy-4, 4—dimethyl cyclohexane
(6) Ethoxybenzene
26) Write the names of reagents and equations for the preparation of the
following ethers by Williamson’s synthesis:
(1) 1-Propoxy propane
(2) Ethoxybenzene
(3) 2-Methoxy-2-methylpropane
(4) 1-Methoxyethane
Solution:
Sodium propoxide Bromo propane
1-Propoxypropane
Sodium phenoxide Bromoethane Ethoxybenzene
Sodium-2-methyl-2-propoxide Bromomethane 2-methoxy-2-methoxypropane
Sodium ethoxide Chloromethane 1-methoxy ethane
27) Illustrate with examples the limitations of Williamson synthesis for the
preparation of certain types of ethers.
Solution:
Limitation of Williamson’ synthesis is:
(1) Tertiary alkyl halide does not form ethers by Williamson’s synthesis as in the
presence of strong alkoxide bases they prefer elimination rather than substitution.
Tertiary butyl iodide 2-methyl propene
(2) Aryl halides also do not undergo Williamson’s synthesis as they are less
reactive in nucleophilic substitution reactions due to partial double bond character
of C-X bond as a result of resonance.
Bromobenzene
28) How is 1-propoxypropane synthesized from propan-1-ol? Write
mechanism of this reaction.
Solution:
1-Propaxypropane is obtained by the acid catalysed dehydration of Propan-1-ol at
415 K.
Propan-1-ol 1-Propoxy propane
Mechanism
Step-I
Ionisation of H2SO4
H2SO4 →H+ + HSO4
-
Step-II
Protonation of alcohol by H+
Protonated -1-Propanol
Step-III
Attack of 2nd
molecule o alcohol on protonated alcohol to form protonated ether,
which forms ether with the loss of proton (H+)
1-Propoxy propane
29) Preparation of ethers by acid dehydration of secondary or tertiary
alcohols is not a suitable method. Give reason.
Solution:
Preparation of ethers by acidic dehydration of secondary and tertiary alcohols is
not a suitable method as elimination competes over substitution and alkenes are
formed more readily in preference to ethers.
30) Write the equation of the reaction of hydrogen iodide with:
(1) 1-propoxypropane
(2) Methoxybenzene and
(3) Benzyl ethyl ether
Solution:
1-Propoxypropane Propane-1-ol Iodopropane
Methoxybenzene Phenol Iodomethane
Benzyl ethyl ether Benzyl iodide
31) Explain the fact that in aryl alkyl ethers
(1) The alkoxy group activates the benzene ring towards eletrophilic
substitution and
(2) It directs the incoming substituent to ortho and Para positions in benzene
ring.
Solution:
(1) The resonating structures of aryl alkyl ethers are given as
Since, alkoxy group releases its electrons to the benzene ring, the electron density
of the ring is increased and hence the benzene ring is activated towards the attack
of the electrophiles.
(2) The resonance hybrid of the resonating structures of aryl alhyl ether can be
given as
Since, alkoxy group by releasing its electron to benzene ring causes high electron
density areas at ortho and Para positions of the benzene ring, the incoming
electrophiles (E+) attacks at these positions giving ortho and Para substituted
products.
o – Product p-product
Hence, alkoxy group directs the incoming substituent to ortho and Para positions in
benzene ring.
32) Write the mechanism of the reaction of HI with methoxymethane.
Solution:
The reaction follows SN2 mechanism and proceeds through following steps
Step-I:
Formation of protonated ether by attack of HI.
Step-II:
Attack of the nucleophile.
Step-III:
Dissociation of the transitory compound into products.
33) Write equations of the following reactions:
(1) Friedel Crafts reaction-alkylation of anisole.
(2) Nitration of anisole
(3) Bromination of anisole in ethanoic acid medium.
(4) Friedel-Craft’s acetylation of anisole.
Solution:
Anisole 2-Methoxy toluene 4-methoxy toluene
(Minor product (major product)
Anisole o-nitroanisole p-nitroanisole
(Minor product) (Major product)
Anisole o-Bromoanisole p-Bromoanisole
(Minor product) (major product)
Anisole o-methoxy acetophenone p-methoxy acetophenone
(Minor product) (Major product)
34) Show how would you synthesize the following alcohols from appropriate
alkenes?
Solution:
1-Methylcyclohexene 1-Methylcyclohexan-1-ol
(b)
4-Methylhept-3-ene 2-Propyl pentan-2-ol
Pent-1-ene Pentan-2-ol
2-Cyclohexyl but-1-ene 2-Cyclohexyl butan-2-ol
2-cyclohexylbut-1-ene will give the required alcohol whereas others will give a
mixture of alcohols.
35) When a 3-methylbutan-2-ol is treated with HBr, the following reaction
takes place:
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more
stable tertiary carbocation by a hydride ion shift from 3rd
carbon atom.)
Solution:
The reaction takes place in the following steps
Step-I: Ionisation of HBr
Step-II: Protonation of alcohol
Step-III: Loss of water from protonated alcohol to give carbocation
Step-IV: Rearrangement of 20 carbocation into more stable carbocation by 1, 2
hydride shift
30 carbocation
Step-V: Attack of nucleophile, Br- ion on carbocation to form final product
2-Bromo-2-methyle butane