Class 12th Chemistry Chapter 11 Alcohols, Phenols and Ethers Notes & Important Question is a platform where you can get pdf notes from 6th to 12th class notes, General Knowledge post, Engineering post, Career Guidelines , English Speaking Trick , How to crack interview and lots more. ( Alcohol Phenols and Ethers )

Alcohols, Phenols and Ethers Notes

Isomerism  in alcohols                                                

Isomerism in alcohols: They basically show

  • Position isomerism
  • Chain isomerism
  • Functional isomerism
  • Optical isomerism

Position: In this the position of –OH group changes.



    (2-propanol)          (Propanol)                                        

CHAIN ISOMERISM: In this the number of carbon atom in parent chain is going to be different. 



Functional isomerism: they are functional isomer of alcohol & ether   

Example:   CH3CH2CH2OH and CH3CH2O CH3 are functional isomers

                       (Propanol)             (Methoxy ethane)

Optical Isomerism: they also show optical isomerism but there should be a chiral centre

Like: in the given molecule of alcohol



   (In this there is the chiral centre marked as C* as it has different groups attached to it).

Preparation of alcohols:

It can be prepared by following methods:

  1. From Haloalkane: When any haloalkane reacts with aqueous KOH it undergoes nucleophilic substitution and leads to the formation of alcohol and potassium halide .

   If we take secondary alcohol than secondary alcohol potassium halide and    traces of alkenes is formed.

Example: If we take tertiary alkyl halide then on reacting with KOH we get alkene as major product and potassium halide.


                                                                             ( 2-methyl butene)

  1. By reduction of aldehyde and ketones: Whenever we reduce aldehyde or ketones in presence of reducing agent like Ni, Pd ,Pt, LiAlH4 etc we get alcohol .

  We can’t produce tertiary alcohol by this method.

  1. By reduction of carboxylic acid: Whenever we reduce carboxylic acid in presence of reducing agent it forms respective alcohol as shown
  1. From Grignard reagent i.e RMgX
  • Please remember whenever Grignard reagent reacts with formaldehyde it forms primary alcohol.
  • When Grignard reagent reacts with any other alcohol it forms secondary alcohol.
  • When Grignard reagent reacts with ketone it forms tertiary alcohol.
  1. By hydrolysis of ester: Whenever ester reacts with caustic soda or water it forms alcohol and sodium salt of carboxylic acid.

6.    From alkenes: by their acidic hydration


Another method employed for alkenes is: Hydroboration oxidation reaction

In this we add BH3 to unsymmetrical alkenes:

  1. From aliphatic primary amines: Whenever amine reacts with nitrous acid in presence of NaNO2 and HCl it forms respective alcohol and nitrogen gas and water.



Industrial preparation of alcohol is by two methods: Hydration of alkene and fermentation of sugar (used in breweries)

  1. Fermentation – Breakdown of complex substance into simple substance by action of enzymes released by anaerobic bacteria.
  • By this method we can prepare only ethanol.
  • In this firstly the sugarcane juice and water are mixed and heated to form sweet liquid called molasses.
  • Then in presence of yeast and enzyme invertase it breaks in to glucose and fructose and then further glucose and fructose break into further simpler substances called ethanol and carbon dioxide.
  • This method is generally carried out in breweries.
  1. Hydration of alkene: This is a method by which any alcohol can be prepared and accordingly we take member of alkene. Suppose we need to prepare ethanol than we need to take ethene. When water is added to ethene in presence of some weak acid like phosphoric acid then ethene undergoes addition reaction and leads to formation of ethanol.


  1. State of existence: Lower members are colorless liquids with characterize smell. Higher members with more than 12 carbons are solid
  2. Boiling point: They have high boiling points due to hydrogen bonding.
  • If we compare their boiling points with hydrocarbons they will possess higher boiling point because of existence of hydrogen bonds in them due to which they occur at associated molecules.
  • It is noted that boiling point increase with increase in carbons in chain: As no. of carbon atoms increases Vander wall force increases hence boiling point increases.


  • If we look for the molecule of alcohol out of them with higher boiling point than :

              CH3CH2CH2OHAND   CH3CH2CH2 CH2CH2OH

       The Second compound will definitely have the higher boiling point as it has more number of carbon atoms.

  • Boiling point decreases with branching: as surface area gets reduced therefore Vander wall force decrease and boiling point also decreases.

Example:  If we look for the molecule of alcohol out of them with higher boiling point than: 


(Primary)                      (Secondary)          (Tertiary)

The one degree will have higher boiling point as it has more surface area.

  1. Solubility: They are soluble in water due to formation of hydrogen bonds in them .However solubility decreases with increase in number of carbon atoms in chain as non polar part increases therefore solubility decreases.

Example: If we look for the molecule of alcohol out of them with higher solubility than:


The second one will have lower solubility because the non polar part is more in it.

Chemical properties of alcohols

Alcohol is most versatile organic compound. They can behave in different manner to give different reaction. Three type of reaction we will study

  1. Alcohol behaves as nucleophiles

They are electron rich species


 2. Alcohols behave as electrophile

Electron deficient

Carbocation is formed   

  R-OH   –>       R      + OH

  alcohol       carbocation   Hydroxide ion

                        Mixed reaction

Involving alkyl and OH group

  1. Reaction as nucleophiles

It shows their acidic character.

  1. Reaction with active metals

Comparison of acidic character of H2O with alcohol. 


In alcohols R is electron releasing group which increases electron density on oxygen and pass of electron from O-H bond to O becomes difficult due to +inductive  effect. Therefore water is more acidic than alcohols.

Order of reactivity as nucleophile /acidic strength

  1. If we compare acidic strength of primary, secondary and tertiary alcohols we found that more the alkyl groups attached, more they release electrons and more is the difficulty in loosing H+

Primary Alcohols (Less Acidic), Secondary Alcohols(Acidic) and Teritary Alcohols(Most Acidic)

Here are some reactions showing acidic strength of alcohols:

  • Reaction with metal hydrates
  • Reaction with carboxylic acid : Esterification
  • Order of reacting of alcohols towards esterification :

                            CH3 OH   >  1>   20>   30           (alcohol)

  • Order of reactivity of carboxylic acid towards esterification:

                     H COOH   >  CH3COOH  > (CH3)2  CH COOH  >  (CH3)3 COOH

Less number of alkyl groups attached, more is the acidic strength that is release of hydrogen ion becomes easy.

2.Alcohol behave as electrophiles

In this the bond break in such a way that electrophile is generated

            R -OH   –>     R+     +   OH

  • Reaction with HX: This reaction takes place in presence of zinc chloride and the equation involved is given below.

           The order of reactivity of different halogen acids towards this reaction is :

      HI  >  HBr   >   HCl   >  HF

In this HI is most reactive towards this reaction as the bond that exist between H and I is weak bond whereas HF do not participate in this reaction as the bond between them is strong enough due to comparable sizes .

  • Reaction With PCl5 and PCl3

                ROH     +    PCl5                                    —> RCl +       POCl3  +     H2O

                Alcohol  PhosphorousPentachloride      ChloroAlkane  PhosphorylChoride

              ROH +        PCl3                        —>   RCl     +    H3PO3

Alcohol  Phosphoroustrichloride            AlkylHalide   Phosphoric Acid

  •  Reaction With SOCl2
  1. Clevage of both alkyl group and hydroxyl group
  • Acidic dehydration of alcohols

(Ethanol)                           (ethene)


  • Dehydrogenation – It is also called as oxidation reaction.

In this primary alcohols give aldehydes in presence of oxidizing agent acid

  • Acidified K2Cr2O7,CrO3, Collins-reagent, CrO3-pyridine,PCC (pyridinium chlorochromate), PDC (Pyridium dichromate).

  Please note: If we want to stop the reaction at stage of aldehyde, we can use mild oxidizing agent. It is to be noted that primary alcohols give aldehyde and secondary alcohols give ketones.


We have reactions where you will see that secondary alcohols gives ketones  but under drastic conditions like use of potassium dichromate it gives carboxylic acid with less number of carbon atom than in alcohol.


Tertiary alcohols – Gives alkenes because they do not have H atom so oxidation is not possible and they show elimination reaction.

  •  Dehydrogenation – removal of Hydrogen.
  • Please note that C2H5OH is used for drinking purpose are beverage but

CH3 OH is not good to consume. When in C2H5OH, CH3OH is added it becomes denatured alcohol if it is consumed by anyone it causes death. It hits the nervous system and can cause blindness.

Partial Oxidation


 Methanol    Methanal

Chemical properties of phenols

  1. Acidic nature of phenols –
  • It is sp2 hybridized that is it has 33% s character and 67% p character that shows it is more electronegative . As a result electron density on oxygen decreases and Polarity of O – H bond increase therefore Hydrogen is not easily removed.
  • Although phenol is acidic but it is less acidic than carboxylic acid therefore It doesn’t react with Na2CO3 and NaHCObut reacts with sodium metal as shown below.
  • As compared with alcohol, phenol is more acidic than alcohol. Because in alcohol it is sphybridised in phenol and Sp3 hybridised C atom in alcohol.
  • The phenoxide ion is resonantly stabilized due to resonance as shown :

In this only one negative charge is getting dispersed.


On the other hand:

            C2H5OH  –>     C2H5 O   + H+

                 ethanol                  ethoxide ion

No such resonance is possible therefore stability of alkoxide is less than phenoxide as a result phenol has more tendency to form phenoxide ion due to which phenol is more acidic than alcohols.

 If electron withdrawing group is present at ortho or para position like  NO2 , CN , Cl etc then  they disperse  this negative charge and will enhance acidic strength, whereas if electron donating group’s like CH3, alkyl group, NH2 (amine) or OR etc are present they decrease the acidic strength.

 1. Reacting with zinc dust: Benzene is formed .


          (phenol)     (Zinc)           (benzene)

  1. Reaction with NH3: aniline is formed
  1. Reaction with Acid chloride: Phenyl ethanoate is formed .The reaction is called as acetylation. If we react this phenyl ethanoate with ethyl anhydride then we get aspirin :
  1. Fries rearrangement

Phenyl ethanoate                           O – (hydroxy acetophenone)

  1. Shorten baumem reaction

         (Phenol)    ( benzoyl chloride)        (phenyl benzoate)

In all these reaction cleavage takes place between O and H.

Reactions of benzene ring

            In this Electrophile substitution reaction takes place .As we know OH- group is Ortho and Para directing so according to resonating structures the negative charge is on the Ortho and Para position. Therefore Electrophile attacks at O and P position.

Reactions involved:

  1. Bromination: Substitution by bromine

If it occurs in presence of CS2 then mixture of ortho and para isomers are formed:

  1. Nitration: Substitution by nitro group by adding nitric acid. If we use Concentrated HNO3 we get tri-substituted  
  1. Friedel craft alkylation reaction
  1. Sulphonation: Substitution by SO3H group.

There are certain reactions that phenol undergoes –

  • Kolbe’s reaction

By this Aspirin and salol can be formed a shown :

  • Aspirin :
  • Salol – phenyl salicylate
  • Reimer – tieman reaction –
  • Reaction with carbon tetra chloride               
  • Coupling reaction – in this we get dyes of different colours.
  • Hydrogenation reaction –
  • With Zn dust
  • Oxidation reaction

Test to distinguish between 10  and  20  and  30alcohol

  • First test :Lucas test
  1. Primary alcohol + HCl( ZnCl2  / warm) —-> Turbidity appears on warm

              Less reactive                              

  1. Secondary alcohol + HCl (ZnCl2  / NO warming) –>Turbid on standing                                                                                          On its own.

 3.  Tertiary alcohol + HCl (ZnCl2 )  —> immediate turbid                  

  • Second test

             Victor Mayer’s test


Test to distinguish between alcohol and phenol

  1. Litmus test: Phenol turns blue litmus red, whereas alcohols have no effect.
  1. Coupling reaction – azo dye test

Phenol give orange colour, alcohol do not give any reaction.

  1. Bromine water test

When Phenol is added to Br2water they give white ppt. Due to formation of 2,4,5 tri nitro phenol But  alcohol do not give any such test.



            They are functional isomers of alcohols .The functional group in them is ROR and are represented as ROR where R is alkyl group.             

Depending upon the type of alkyl group on both sides we can classify ethers as:

  1. Symmetrical ethers: When it has same R on both sides.

                    Like:       CH3 OCH3                               C2 H5OC2 H5

                 Di methyl ether                  Diethyl ether

  1. Unsymmetrical ethers: They have different R on both sides.

                   Like:        CH3OC2 H5

               Ethyl methyl ether

Depending upon the number of alkyl groups present we can classify ethers as:

  1. Primary ,secondary and tertiary ethers


The common name used for them is alkoxy alkane. In this smaller group is written as alkoxy group and bigger group is written as Alkane.


Isomerism in ethers

  1. Chain isomerism: In this chain is going to be different.
  1. Functional Isomerism: In this the functional group changes.

 3. Metamerism: it is shown by only ethers , ester and ketones.


Preparation of ethers  

  1. By dehydration of alcohol : by this method only primary ethers can be prepared but not tertiary or unsymmetrical ethers .the reaction involved is shown below .
  1. The Conditions required to form ether from alcohol
  • Temperature must be maintained at 413k.
  • Large amount of alcohol should be used as we will see in the mechanism we need more alcohol as shown below in second step of mechanism.
  • Alkyl group should be unhindered.

Mechanism: Protonation and deprotonation

(a)From of protonated alcohol


Ethanol          hydrogen ion    (protonated alcohol)

        (b)Another molecule of alcohol also react


 (c)De protonation: removal of proton

  1. Williamson’s synthesis – By this method tertiary ether ,unsymmetrical ether can be prepared .The reaction involved is given .

            Key points –

  • If secondary or tertiary ether have to formed than the RX should be
  • From secondary or tertiary RX we get alkenes instead of ether.
  1. From alkyl halides by treating with dry silver oxide

 5. By action of diazomethane on alcohol


Physical properties of ethers

  1. Existence: Lower members are gases and higher members may be liquids or solid due to increase in the Vander wall force.
  2. Dipole moment: Ethers have higher dipole moments because of electro negativity difference and structure is bent.
  • They have larger bond angles than alcohol. Alcohol have one hydrogen and ether there are two bulky groups so, repulsion between bulky groups are more therefore Bond angle will increase.
  • More the bulky group’s, lower is the value of dipole moments.

            CH3  – O  –  CH3                     C2 H5OC2  H5

                        1.3D                                       1.18D

  1. Boiling point: Ethers have low boiling point because of less polarity.

CH3 CH2 O CH2 CH3             CH3 CH2CH2CH2 OH

  Boiling point – 307.6k           boiling point – 390k

                                               “more polarity”

  1. Solubility: lower members are more soluble .

Example – C2 H5 OH                       CH3OCH3       ] same solubility

  • But as number of Carbon atoms increase, solubility decrease because non polar part increase.

Chemical properties of ethers

Ethers are very stable compounds –therefore they are least reactive among all functional groups .still we have few reactions lets study them.

  1. [ Reaction of ethereal oxygen]

ROR  (ether )

CH3OCH(methoxy methane )

  1. Action of air on ether
  1. With Lewis acid
  1. Cleavage of C = O bond of ether

(a)Cleavage by halogen acids: The order of reactivity of halogen acids towards this reaction is:

HI > HBr  >  HCl  >  Hl

The HI bond is broken easily but HF bond can’t be broken .The reaction involved is


                        The Mechanism involved:

  • In case of unsymmetrical ether :

          Another example:   

  • If we take tertiary than Iodide goes with tertiary group
  • More positive is the carbocation more is the tendency of iodine to attach.

 2. Reaction of alkyl groups

Reaction taking place in dark in which chlorine molecule is added


Reaction taking place in light in which chlorine molecule is added

CH3CH2OCH2CH3Cl2  — > CCl3CCl2OCCl2CCl3

                                ( Light )      Perchloro diethyl ether 

  1. Ring substitution reaction in ether

If we look at resonance structures of ether ,we see that attack occur at ortho and para position because on those position electron density is high so attack of electrophile occur at ortho and para .

  • Halogenations: Attack of halogen.
  • Chlorination

            (B)       Nitration: Reaction with nitric acid  

  •  Friedel craft alkylation and acylation

Uses of ethers

  1. Ethers are used as a refrigerant.
  2. They are used as industrial solvent for fat and oils.
  3. When mixed with alcohol they make alcohol denatured.
  4. They are used in Perfumery like:
  • In aniseeds – Anethol is present which is a phenyl ether.
  • In clove oil – Eugenol is present.
  • In vanilla extract vanillin is present.
  • In mint – Thymol is present.

Introduction of alcohol

It belongs to category of groups which when attached to a carbon chain governs the properties of organic compound or in other words we can say it is a functional groups. They are indicated by presence of—OH group in a chain .This -OH is regarded as hydroxyl group.

Functional group – OH

Suffix used: “ol”

Examples: Its  homologous series is :

Methanol  (CH3OH)

Ethanol (C2H5OH)

Propanol  (C3H7OH)

Butanol  (C4H9OH)

Pentanol (CH11 OH) and so on.

Classification of alcohols

We can classify alcohols on the basis of different factors:

  • Number of hydroxyl groups attached , hybridization ,number of alkyl groups attached to alpha carbon (alpha carbon is that which has functional groups attached to it )
  1. On the basis of number of -OH(hydroxyl ) group attached we have :
  • Monohydric alcohols: They are those that have only one hydroxyl group attached.

 Example: methanol (CH3OH) ethanol (C2H5OH) and more

  • Dihydric alcohols: They have two hydroxyl groups attached.

               Example: ethylene-glycol 

  • Polyhydric alcohol: That have three or more hydroxyl groups in it.

        Example glycerol 


(2). Based on the type of hybridization:

  • It can be sp3 like ethanol – CH3-CH2-OH  
  • Another example :

It can be sp2 like Benzyl alcohol-

In this second carbon is alpha carbon and that is sp3


(Benzyl alcohol)

Allyl alcohol CH2=CH- CH2OH  (In it also  alpha carbon is sp3  hybridised ).

Another type of hybridization they possess is sp2

Example: Phenol  C6H5OH, Vinyl alcohol CH3-CH=CH-OH etc

(c) Classification is on the basis of primary secondary or tertiary carbon atom (alpha carbon)


              10           20                             30 


CBSE Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

1 Mark Questions

1. Write IUPAC names of :-


Ans. 5 – Methyloctan-3-ol


Ans. 1-Chloro propan-2-ol


Ans. Pentan – 1,3,4 – triol


Ans. 2 – Methylpropan -2-ol


Ans. 1,3 – Dibromo – 4- chloro – 2- butanol


Ans. 5 – Chloro – 4 – ethyl – 5- methyl hexanol.


Ans. 2 – Phenyl ethanol


Ans. 2- Methyl phenol.


Ans. 4- Bromo -3, 3,5 – trimethyl – hex -1-ene- 2,5- diol


Ans. 2,3 – Dimethylbutan – 2,3 –diol

2. Write structural formula and give IUPAC names :-

i. Benzyl Alcohol


ii. Ethylene Glycol


iii. Glycerol


iv. m- cresol




vi. Ethylphenylether


vii. Methylpropylether


viii. Anisole


ix. Isopropyl methylether


x. Phenetole


3. What is denatured alcohol?

Ans. Industrial alcohol (ethyl alcohol) is made unfit for drinking purpose by addition of a small quantity of poisonous substance, methanol. This is called denatured alcohol. For denaturation, copper sulphate or pyridine can also be added.

4. Mention two important uses of methanol.

Ans. Uses of methanol:-

(i) It is used for denaturing alcohol.

(ii) It is used as a solvent for paints & varnishes.

5. Arrange in the increasing order of acid strength.

Ans. The order is.

6. Give the structure of main product of action of excess of on phenol.

Ans. The main product will be 2,4,6-tribromophenol.

7. What is wood sprit? Why is it so called?

Ans. Wood spirit is methanol. It is so called because earlier it was obtained from destructive distillation of wood only.

8. Identify allylic alcohols in the above examples.

Ans. The alcohols given in (ii) and (vi) are allylic alcohols.

2 Marks Questions

1. Ethene to 1,2 -ethanediol


2. Phenol to Salicyldehyde


3. Butanol to Butanoic acid


4. Ethanol to propanone


5. Phenol to salicylic acid


6. Methanol to Ethanol


7. Ethanol to propanol


8. Phenol to Benzyl Alcohol


9. Ethanal to propan -2- ol


0.   l – propanol to 2 – bromo propane


11.  How is the presence of peroxides in ethers detected? How are peroxides removed from Ethers?

Ans. Presence of peroxides in ethers is detected by addition of freshly prepared and KCNS. Appearance of blood red colour confirms the presence of peroxide. They can be removed by shaking ethers well with  solution.

12.  Explain a chemical test to distinguish between primary, secondary and tertiary alcohols.

Ans. Primary tertiary and secondary alcohols can be distinguished by oxidation reaction. Primary alcohols give aldehyde with 

Secondary alcohols give ketone with CrO3.

Tertiary alcohol do not get oxidized with CrO3

13.  What is Lucas test?

Ans. Lucas test is used for distinguishing between primary secondary and tertiary alcohols. When a tertiary alcohol is dissolved in Lucas reagent  it produces turbidity immediately where as secondary alcohols produce turbidity after some time and primary alcohols do not react at all and no turbidity is product.

14. What is Picric acid? How is it prepared from phenol?

Ans. Picric acid is 2,4,6 – trinitrophenol

It is prepared from phenol by nitration with conc. HNO3.

15.  Give equations for preparation of ethanol by fermentation.

Ans. Ethanol can be prepared by fermentation of sugar –

16. Show how are the following alcohols prepared by the reaction of a suitable

Grignard reagent on methanal?



Ans. (i)


17. Predict the major product of acid catalysed dehydration of

(i) 1-Methylcyclohexanol and

(ii) Butan-1-ol

Ans. (i)


18. Write the equations involved in the following reactions:

(i) Reimer-Tiemann reaction

(ii) Kolbe’s reaction

Ans. (i) ReimerTiemann reaction

(ii) Kolbe’s reaction

19. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?



Ans. Set (ii) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene.

In set (i), sodium methoxide is a strong nucleophile as well as a strong base. Hence, an elimination reaction predominates over a substitution reaction.

20. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Ans. Propanol undergoes intermolecular H-bonding because of the presence of -OH group. On the other hand, butane does not

Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane.

21.   Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Ans. Alcohols form H-bonds with water due to the presence of -OH group. However, hydrocarbons cannot form H-bonds with water.

As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

22. Give the structures and IUPAC names of monohydric phenols of molecular formula,.


23. Write chemical reaction for the preparation of phenol from chlorobenzene.

Ans. Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.

24. You are given benzene, conc. and NaOH. Write the equations for the preparation of phenol using these reagents.


25. Explain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Ans. The -OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.

As a result, the benzene ring is activated towards electrophilic substitution.

26. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Ans. Ethanol undergoes intermolecular H-bonding due to the presence of -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxymethane.

27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Ans. The formation of ethers by dehydration of alcohol is a bimolecular reaction () involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alkenes are formed.

28. Explain the fact that in aryl alkyl ethers

(i) The alkoxy group activates the benzene ring towards electrophilic substitution and

(ii) It directs the incoming substituents to ortho and para positions in benzene ring.

Ans. (i)In aryl alkyl ethers, due to the +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

(ii) It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

3 Marks Questions

1. Phenol is acidic in nature.

Ans. Phenol is acidic in nature because

(a) phenol, due to resonance, the positive charge rests on oxygen making the shared pair of electrons more towards oxygen and hydrogen as 

(b) The carbon attached to OH is hybridize and is more electronegative, this decreases the electron density on oxygen, increasing the polarity of O-H bond and ionization of phenol.

The phenoxide ion formed by loss of  is more resonance stabilized than phenol itself.

2. Phenol has a smaller dipole moment than methanol.

Ans. In phenol due to electron rich benzene ring the C-O bond is less polar whereas in methanol the C-O bond is highly polar. Therefore the dipole moment of methanol is higher than phenol.

3. o- nitrophenol has lower boiling point (is more volatile) than p – nitrophenol.

Ans. P- nitrophenol has intermolecular hydrogen bonding which increases the boiling point while in o- nitro phenol due to presence of intra molecular hydrogen bonding, there is a decrease in boiling point and increase in volatility.

4. Methanol is miscible with water while iodomethane is not.

Ans. Methanol can form intermolecular hydrogen bonding with water but there is no hydrogen bonding in iodomethane and water. Therefore methanol in miscible in water.

5. Alcohols have higher boiling points than isomeric ethers.

Ans. Alcohols can form intermolecular hydrogen bonds due to their high polarity whereas, ether cannot. Therefore alcohols have higher boiling points than isomeric ethers.

6. Ethers are soluble in water alkanes are not.

Ans. Ethers can form H- bonding with water molecule whereas alkenes cannot. Therefore ethers are soluble in water and alkanes are not.

7. The order of acidic strength in alcohols is R 

Ans. In alcohols, the acidic strength is due to polar nature of O-H bond. An electron releasing group e.g., alkyl groups, increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid strength. Therefore the order of acid strength is .

8. During preparation of ester from alcohol and acid, water has to be removed as soon as it is formed.

Ans. The reaction between alcohol and carboxylic acid is reversible and goes in backward direction if water is not removed as soon as it is formed.

9. Ethers can not be prepared by dehydration of secondary or tertiary alcohols.

Ans. For secondary and tertiary alcohols, elimination competes over substitution and alkenes are formed on acidic dehydration as the reaction follows Sn1 mechanism. Therefore the acidic dehydration of secondary or tertiary alcohols does not give ethers.

10. Reaction of anisole with HI gives methyl iodide and phenol.

Ans. In case of anisole, methyl phenyl oxonium ion,  is formed by protonation of ethers during reaction with HI. The bond between O- CH3 is weaker than the bond between  because carbon of phenyl group is hybridised and there is a partial double bond character. Therefore the attack by I ion breaks bond to form CH3I.

11. An organic compound ‘ A ‘ having molecular formula on treatment with aq.  give ‘B’ which on treatment with Lucas reagent gives ‘C’. The compound ‘C’ on treatment with ethanolic KOH gives back ‘A’ .Identify A, B, C .


12.  An organic compound gives a characteristic colour with aq. solution. (A) On reacting with and NaOH at 400k under pressure gives (B) which on acidification gives a compound (C) .The compound (C) reacts with acetyl chloride to give (D) which is a popular pain killer. Deduce the structure of A,B,C & D.


13.  An organic compound (X) when dissolved in ether and treated with magnesium metal forms a compound Y. The compound, Y, on treatment with acetaldehyde and the product on acid hydrolysis gives isopropyl alcohol. Identify the compound X. What is the general name of the compounds of the type Y.

Ans. The compound X is and Y is The compounds of the type ‘Y’ are called Grignard reagent.

14.  A compound ‘A’ with molecular formula C4H10O on oxidation forms compound ‘B’ gives positive iodoform test and on reaction with CH3MgBr followed by hydrolysis gives (c). Identify A, B & C.

Ans. The compound ‘B’ is obtained by oxidation of and gives positive iodoform test and also reacts with , it must be methyl Ketone  , it must be methyl ketone having four carbon atoms i.e, .

This can be obtained by oxidation of 2 – butanol i.e , Therefore , the reactions are.

15.  An aromatic compound (A) having molecular formula  on treatment with CHCl3 and KOH gives a mixture two isomers ‘B’ and  ‘C’ both of  ‘B’ & ‘C’ give same product  ‘D’ when distilled with Zn dust. Oxidation of ‘D’ gives ‘E’ of formula. The sodium salt of ‘E’ on heating with soda lime gives ‘F’ which may also be obtained by distilling ‘A’ with zinc dust. Identify compounds ‘A’ to ‘F’ giving sequence of reactions.


The aromatic compound having molecular formula   and which gives a mixture of two isomers on reacting with and KOH is phenol i.e.

16.  Compound ‘A’ of molecular formula gives a compound ‘B’ of molecular formula when treated with aq. NaOH. On oxidation the compound yields a mixture of acetic acid & propionic acid. Deduce the structure of A, B & C.


Since acetic acid & propionic acid are the products of oxidation of C which is a ketone, C is   . Since it is the oxidation product of B, therefore

The reactions are





19. Write structures of the products of the following reactions:







Ans. (i)






20.What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Ans. The addition of borane followed by oxidation is known as the hydroboration-oxidation reaction. For example, propan-1-ol is produced by the hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.


21. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Ans. Intramolecular H-bonding is present in o-nitrophenol. In p-nitrophenol, the molecules are strongly associated due to the presence of intermolecular bonding. Hence, o-nitrophenol is steam volatile.


22.   Give the equations of reactions for the preparation of phenol from cumene.

Ans. To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-peroxide.


Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-products.


23. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?



The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. As a result, it is easier to lose a proton. Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.

On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily.

For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

24. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.

Ans. The reaction of Williamson synthesis involves attack of an alkoxide ion on a primary alkyl halide.


But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.


25. Write the equation of the reaction of hydrogen iodide with:

(i) 1-propoxypropane

(ii) Methoxybenzene and

(iii) Benzyl ethyl ether

Ans. (i)






NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols Phenols and Ethers

NTEXT Questions

Question 1. ( Alcohol Phenols and Ethers )
Classify the following as primary, secondary and tertiary alcohols :

Primary alcohols : (i), (ii) and (iii)
Secondary alcohols : (iv) and (v)
Tertiary alcohols : (vi)

Question 2.
Identify allylic alcohols in the above examples.
Allvlir alcohols : (ii) and (vi).

Question 3. ( Alcohol Phenols and Ethers )
Name the following compounds according to IUPAC system.


  1. 3-Chloromethyl-2-isopropylpentan-l-ol
  2. 2, 5-Dimethylhexane-1,3-diol
  3. 3-Bromocyclohexanol
  4. Hex-l-en-3-ol
  5. 2-Bromo-3-methylbut-2-en-1 -ol

Question 4. ( Alcohol Phenols and Ethers )
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.


Question 5. ( Alcohol Phenols and Ethers )
Write structures of the products of the following reactions :


Question 6. ( Alcohol Phenols and Ethers )
Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl – ZnCl2 (b) HBr and (c) SOCl2.

  1.  Butan-1-ol
  2. 2-Methylbutan-2-ol


Question 7. ( Alcohol Phenols and Ethers )
Predict the major product of acid catalysed dehydration of

  1. 1-methylcyclohexanol and
  2. butan-1-ol


Question 8. ( Alcohol Phenols and Ethers )
Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
The attachment of the -NO2 group to the phenol molecule at 0- and p-positions decreases the electron density on oxygen atom. This causes the oxygen atom to pull the bond pair of electrons of the O – H bond towards itself thereby facilitating the release of H as H+.

The resonance structures of the phenoxide ions are :

Question 9. ( Alcohol Phenols and Ethers )
Write the equations involved in the following reactions :

  1. Reimer-Tiemann reaction
  2. Kolbe’s reaction


Question 10. ( Alcohol Phenols and Ethers )
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Question 11. ( Alcohol Phenols and Ethers )
Which of the following is an appropriate set of reactants for the preparation of 1 -methoxy-4- nitrobenzene and why?


Question 12. ( Alcohol Phenols and Ethers )
Predict the products of the following reactions :


NCERT Exercises

Question 1. ( Alcohol Phenols and Ethers )
Write IUPAC names of the following compounds :


  1. 2,2,4-Trimethylpentan-3-ol
  2. 5-Ethylheptan-2,4-diol
  3. Butane-2,3-diol
  4. Propane-1,2,3-triol
  5. 2-Methylphenol
  6. 4-Methylphenol
  7. 2,5-Dimethylphenol
  8. 2,6-Dimethylphenol
  9. l-Methoxy-2-methylpropane
  10. Ethoxybenzene
  11. 1-Phenoxyheptane
  12. 2-Ethoxybutane

Question 2. ( Alcohol Phenols and Ethers )
Write structures of the compounds whose IUPAC names are 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-Cydohexylpentan-3-ol
  9. Cyclopent-3-en-1 -ol
  10. 3-Chloromethylpentan-1-ol


Question 3. ( Alcohol Phenols and Ethers )

  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 3 (i) as primary, secondary and tertiary alcohols.

The isomeric alcohols with molecular formula C5H12O are :

Question 4. ( Alcohol Phenols and Ethers )
Explain why propanol has higher boiling point than that of the hydrocarbon, butane.
The boiling point of any compound depends on the strength of inter-molecular forces. Stronger is the inter-molecular attraction, higher is the boiling point.

In butane, the molecules interact with each other through weak van der Waals forces. These weak forces can be easily overcome by supplying small amount of heat energy. Thus, they have low boiling point.

In propanol, the molecules are held together by strong hydrogen bonding. These attractive forces operating between molecules are more difficult to break and therefore higher amount of heat needs to be supplied, therefore, the higher boiling point.

Question 5. ( Alcohol Phenols and Ethers )
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Organic compounds are soluble in water if they are able to form hydrogen bonds with it. Alcohols are able to establish this interaction by the virtue of their OH group and are therefore soluble in water. On the other hand, other hydrocarbons of comparable mass do not dissolve in water since they cannot form hydrogen bonds.

Question 6. ( Alcohol Phenols and Ethers )
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Hydroboration-oxidation is a method of preparation of alcohols from alkenes. The main advantage of this method is the high yield of alcohol obtained. During hydroboration, diborane (BH3)2 is made to react with an alkene to form an addition product. This product is then treated with hydrogen peroxide in the presence of sodium hydroxide to give alcohol.
For example,

Question 7. ( Alcohol Phenols and Ethers )
Give the structure and IUPAC names of monohydric phenols of molecular formula, C7H8O.

Question 8. ( Alcohol Phenols and Ethers )
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
During steam distillation, it is the lower boiling compound which distills out first. Between ortho- and para-nitrophenol it is the ortho-isomer which will be steam volatile since it has a lower boiling point.

The difference in boiling point between the two isomers can be understood based on the structural difference. In ortho-isomer intra¬molecular hydrogen bonding takes place while in the pttra-isomer, inter-molecular hydrogen bonding takes place.

As a result of the strong forces operating between the molecules of p-isomer, the boiling point is higher and it is not steam volatile.

Question 9. ( Alcohol Phenols and Ethers )
Give the equations of reactions for the preparation of phenol from cumene.

Question 10. ( Alcohol Phenols and Ethers )
Write chemical reaction for the preparation of phenol from chlorobenzene.

Question 11. ( Alcohol Phenols and Ethers )
Write the mechanism of hydration of ethene to yield ethanol.
The acid catalysed hydration of ethene may be represented as :

Question 12. ( Alcohol Phenols and Ethers )
You are given benzene, cone. H2SOand NaOH. Write the equations for the preparation of phenol using these reagents.
Using the given reagents, phenol may be prepared as :

Question 13. ( Alcohol Phenols and Ethers )
Show how will you synthesise :

  1. 1 -Phenylethanol from a suitable alkene.
  2. Cyclohexylmethanol using an alkyl halide by an SNreaction.
  3. Pentan-1-ol using a suitable alkyl halide.


Question 14. ( Alcohol Phenols and Ethers )
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

The resonance stabilization provided by the contributing structures (I)-(V) more than compensates for the bond breakage energy of O – H bond and thus causes phenol to be acidic in nature.

No such resonance structures are possible for ethoxide ion and therefore the conversion of ethanol to ethoxide is not favoured under normal conditions. Therefore, ethanol is less acidic than phenol.

Question 15. ( Alcohol Phenols and Ethers )
Explain why is ortho nitrophenol more acidic than methoxyphenol?

Question 16. ( Alcohol Phenols and Ethers )
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
In an electrophilic substitution reaction, an electron deficient species attacks the benzene ring which is e rich.

When an -OH group is attached to the benzene ring, by the virtue of its electron releasing nature increases the e density of the ring and thus activates it, i.e., makes it a welcome site for electrophiles.

The increase in electron density can be visualised as :

From structures (I) – (V), we find that the attachment of hydroxyl group to benzene has increased the electron density (-ve charge) on the ring carbon atoms (especially C-2, C-4 and C-6). It is therefore said to have activated the ring towards electrophiles which are attracted to the increased electron density.

Question 17. ( Alcohol Phenols and Ethers )
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 chloroform in presence of aqueous NaOH.


Question 18. ( Alcohol Phenols and Ethers )
Explain the following with an example.
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis
(iv) Unsymmetrical ether.
(i) Kolbe’s reaction :
The fact that phenoxide ion is even more reactive than phenol towards incoming electrophiles is made use of in this reaction. Sodium phenoxide is reacted with CO2 followed by acid treatment to yield o-hydroxybenzoic acid as the major product.

(ii) Reimer-Tiemann reaction :
Treatment of phenol with chloroform in the presence of aqueous alkali introduces a CHO group at the ortho position. Acidification yields salicylaldehyde.

(iii) Williamson synthesis :
In this method, an alkyl halides is reacted with sodium alkoxide.
R-X + R’ – ONa → R-O-R’ + NaX
The reaction involves SN2 attack of an alkoxide ion on 1° RX.

(iv) Unsymmetrical ethers :
Unsymmetrical ethers are organic compounds where the ethereal oxygen atom is attached to two different alkyl or aryl groups, e.g.,
C2H5 – O – CH3, C6H5O – C2H5, etc.

Question 19. ( Alcohol Phenols and Ethers )
Write the mechanism of acid dehydration of ethanol to yield ethene.

Question 20. ( Alcohol Phenols and Ethers )
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.


Question 21. ( Alcohol Phenols and Ethers )
Name the reagents used in the following reactions :

  1. Oxidation of primary alcohol to carboxylic acid.
  2. Oxidation of primary alcohol to aldehyde.
  3. Bromination of phenol to 2,4,6-tribromophenol.
  4. Benzyl alcohol to benzoic acid.
  5. Dehydration of propan-2-ol to propene.
  6. Butan-2-one to butan-2-ol.


  1. Alkaline KMnO4
  2. Pyridinium chlorochromate in chloromethane (CH2Cl2)
  3. Br2/H2O
  4. Alkaline KMnO4
  5. Cone. H2SO4 or H3PO4 at 433-443 K
  6. H2/Ni or NaBH4 or LiAlH4

Question 22. ( Alcohol Phenols and Ethers )
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
The higher boiling point of ethanol may be attributed to the presence of intermolecular hydrogen bonding in it.

Due to such extensive bonding, more energy needs to be supplied to ethanol to break these bonds and move it into the vapour phase.

Question 23. ( Alcohol Phenols and Ethers )
Give IUPAC names of the following ethers :


  1. 1 -Ethoxy-2-methylpropane
  2. 2-Chloro-l-methoxyethane
  3. 4-Nitroanisole
  4. 1-Methoxypropane
  5. 1-Ethoxy-4,4-dimethylcyclohexane
  6. Ethoxybenzene

Question 24. ( Alcohol Phenols and Ethers )
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis :

  1. 1-Propoxypropane
  2. Ethoxybenzene
  3. 2-Methyl-2-methoxypropane
  4. 1-Methoxyethane


Question 25. ( Alcohol Phenols and Ethers )
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
The main limitation of Williamson’s ether synthesis lies in its unemployability for preparation of unsymmetrical ethers where the compound contains secondary or tertiary alkyl groups.

e.g., reaction between tert-butyl bromide and sodium methoxide yields an alkene.

This is because the competing elimination reaction predominates over SN2 and alkene is formed.

Question 26. ( Alcohol Phenols and Ethers )
How is 1-propoxypropane synthesised from propan-1 -ol? Write mechanism of this reaction.

Question 27. ( Alcohol Phenols and Ethers )
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Consider the reaction between propan- 2-ol molecules in the presence of acid.

If an ether is to be formed, another alcohol molecule must carry out a nucleophilic attack on the carbocation as

However, this does not happen because of
(a) the steric hindrance around the carbocation, and
(b) bulky size of the nucleophile which would further cause crowding.
As a result, the carbocation prefers to lose a proton and forms an alkene.

For the same reason 3° alcohols in the presence of acid do not form ethers since 3° alcohols are even more sterically hindered than 2° alcohols.

Question 28. ( Alcohol Phenols and Ethers )
Write the equation of the reaction of hydrogen iodide with :

  1. 1-propoxypropane
  2. methoxybenzene; and
  3. benzyl ethyl ether.


Question 29. ( Alcohol Phenols and Ethers )
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Consider the following resonance structures of aryl alkyl ethers :

(i) From the above structures we find that the presence of the OR group has increased the electron density on the benzene ring and therefore the ring is said to have been activated towards incoming electrophiles.

(ii) From structures (II), (III) and (IV) we find that and the electron density has increased on C-2, C-4 and C-6, i.e., at the ortho and para positions. As a result the electrophile (E ) attaches itself to these e- rich sites and the -OR group is said to have directed the E® to ortho and para positions

Question 30. ( Alcohol Phenols and Ethers )
Write the mechanism of the reaction of Hl with methoxymethane.
The reaction between methoxymethane and Hl is :

Question 31. ( Alcohol Phenols and Ethers )
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.


Question 32. ( Alcohol Phenols and Ethers )
Show how would you synthesise the following alcohols from appropriate alkenes.


Question 33. ( Alcohol Phenols and Ethers )
When 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.]

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