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