NCERT Solutions Class 12 Chemistry Chapter 7 - Alcohols, Phenols & Ethers

Page No. 195 - Intext Questions

Q7.1: Classify the following as primary, secondary and tertiary alcohols:

Ans: Primary alcohol → (i), (ii), (iii)

Secondary alcohol → (iv), (v)

Tertiary alcohol → (vi)

Q7.2: Identify allylic alcohols in the above examples.

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

Page No. 198 - Intext Questions

Q7.3: Name the following compounds according to IUPAC system.

Exercises

Q7.1: Write IUPAC names of the following compounds:

Ans: (i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2, 4-diol
(iii) Butane-2, 3-diol
(iv) Propane-1, 2, 3-triol
(v) 2-Methylphenol  
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol  
(viii) 2, 6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane    
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane      
(xii) 2-Ethoxybutane

Q7.2: Write structures of the compounds whose IUPAC names are as follows:

(i) 2-Methylbutan-2-ol  
 (ii) 1-Phenylpropan-2-ol
 (iii) 3,5-Dimethylhexane −1, 3, 5-triol
 (iv) 2,3 − Diethylphenol
 (v) 1 – Ethoxypropane    
 (vi) 2-Ethoxy-3-methylpentane
 (vii) Cyclohexylmethanol
 (viii) 3-Cyclohexylpentan-3-ol
 (ix) Cyclopent-3-en-1-ol  
 (x) 3-Chloromethylpentan-1-ol.
Ans:

Q7.3: (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.
Ans:
(i) The structures of all isomeric alcohols of molecular formula, C₅H₁₂O are shown below:

(a)

Pentan-1-ol (1°)

(b)

2-Methylbutan-1-ol (1°)

(c)

3-Methylbutan-1-ol (1°)

(d)

2, 2-Dimethylpropan-1-ol (1°)

(e)

Pentan-2-ol (2°)

(f)

3-Methylbutan-2-ol (2°)

(g)

Pentan-3-ol (2°)

(h)

2-Methylbutan-2-ol (3°)

(ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol; 3-Methylbutan-1-ol; 

2, 2−Dimethylpropan-1-ol

Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol; Pentan-3-ol

Tertiary alcohol: 2-methylbutan-2-ol

Q7.4: Explain why propanol has higher boiling point than that of hydrocarbon, butane?
Ans: Propanol undergoes intermolecular H-bonding because of the presence of the −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.

Q7.5: 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.

Q7.6: What is meant by the 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 (BH₃)₂ to form trialkyl borane as an additional product. This additional product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Q7.7: Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Ans:

Q7.8: 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.

Q7.9: 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.

Q7.10: Write a 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. 

Q7.11: Write the mechanism of hydration of ethene to yield ethanol.
Ans: The mechanism of hydration of ethene to form ethanol involves three steps.

Step 1:

Protonation of ethene to form carbocation by the electrophilic attack of H₃O :

Step 2:

Nucleophilic attack of water on carbocation:

Step 3:

Deprotonation to form ethanol:

Q7.12: You are given benzene, conc. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Ans:

Q7.13: Show how will you synthesize:

(i) 1-phenylethanol from a suitable alkene.

(ii) cyclohexylmethanol using an alkyl halide by an S_N2 reaction.

(iii) pentan-1-ol using a suitable alkyl halide?

Ans: (i) By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenylethanol can be synthesized.

 (ii) When chloromethylcyclohexane is treated with sodium hydroxide, cyclohexyl methanol is obtained.

 (iii) When 1-chloropentane is treated with NaOH, pentan-1-ol is produced.

Q7.14: Give two reactions that show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Ans: The acidic nature of phenol can be represented by the following two reactions:

(i) Phenol reacts with sodium to give sodium phenoxide, liberating H₂.

(ii) Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products.

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas the ethoxide ion does not.

Q7.15: Explain why is ortho nitrophenol more acidic than ortho methoxy phenol?
Ans:

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, the 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-methoxy phenol.

Q7.16: 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.

Q7.17: Give equations of the following reactions:

(i) Oxidation of propan-1-ol with alkaline KMnO₄ solution.

(ii) Bromine in CS₂ with phenol.

(iii) Dilute HNO₃ with phenol.

(iv) Treating phenol with chloroform in presence of aqueous NaOH.

Ans:

Q7.18: Explain the following with an example.

(i) Kolbe’s reaction.
 (ii) Reimer-Tiemann reaction.
 (iii) Williamson ether synthesis.  
 (iv) Unsymmetrical ether.

Ans: (i) Kolbe’s reaction:

When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe’s reaction.

(ii) Reimer-Tiemann reaction:

When phenol is treated with chloroform (CHCl₃) in the presence of sodium hydroxide, a −CHO group is introduced at the ortho position of the benzene ring.

This reaction is known as the Reimer-Tiemann reaction.

The intermediate is hydrolyzed in the presence of alkalis to produce salicylaldehyde.

(iii) Williamson ether synthesis:

Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.

This reaction involves an S_N2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in the case of primary alkyl halides.

If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

(iv) Unsymmetrical ether:

An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For example ethyl methyl ether (CH₃−O−CH₂CH₃).

Q 7.19: Write the mechanism of acid-catalysed dehydration of ethanol to yield ethene.

Ans: The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:

Step 1:

Protonation of ethanol to form ethyl oxonium ion:

Step 2:

Formation of carbocation (rate-determining step):

Step 3:

Elimination of a proton to form ethene:

The acid consumed in step 1 is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

Q7.20: How are the following conversions carried out?

(i) Propene → Propan-2-ol

(ii) Benzyl chloride → Benzyl alcohol

(iii) Ethyl magnesium chloride → Propan-1-ol.

(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.

Ans: (i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

(iii) When ethyl magnesium chloride is treated with methanal, an adduct is produced which gives propan-1-ol on hydrolysis.

(iv) When methyl magnesium bromide is treated with propane, an adduct is a product that gives 2-methylpropane-2-ol on hydrolysis.

Q7.21: Name the reagents used in the following reactions:

(i) Oxidation of primary alcohol to carboxylic acid.

(ii) Oxidation of primary alcohol to aldehyde.

(iii) Bromination of phenol to 2,4,6-tribromophenol.

(iv) Benzyl alcohol to benzoic acid.

(v) Dehydration of propan-2-ol to propene.

(vi) Butan-2-one to butan-2-ol.

Ans: (i) Acidified potassium permanganate

(ii) Pyridinium chlorochromate (PCC)

(iii) Bromine water

(iv) Acidified potassium permanganate

(v) 85% phosphoric acid

(vi) NaBH₄ or LiAlH₄

Q7.22: Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.
Ans: Ethanol undergoes intermolecular H-bonding due to the presence of the −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.

Q7.23: Give IUPAC names of the following ethers:

Ans: (i) 1-Ethoxy-2-methylpropane

(ii) 2-Chloro-1-methoxyethane

(iii) 4-Nitroanisole

(iv) 1-Methoxypropane

(v) 1-Ethoxy-4, 4-dimethylcyclohexane

(vi) Ethoxybenzene

Q7.24: Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:

(i) 1-Propoxypropane                                
 (ii) Ethoxybenzene
 (iii) 2-Methoxy-2-methylpropane            
 (iv) 1-Methoxyethane

Ans:

(i)

(ii)

(iii)

(iv)

Q7.25: Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Ans: The reaction of Williamson synthesis involves the S_N2 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.

Q7.26: How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Ans: 1-propoxypropane can be synthesized from propan-1-ol by dehydration.

Propan-1-ol undergoes dehydration in the presence of protic acids (such as H₂SO₄, H₃PO₄) to give 1-propoxypropane.

The mechanism of this reaction involves the following three steps:

Step 1: Protonation

Step 2: Nucleophilic attack

Step 3: Deprotonation

Q7.27: 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 (S_N2) involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In the 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.

Q7.28: Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane                                
(ii) Methoxybenzene and
(iii) Benzyl ethyl ether
Ans: (i)

(ii)

(iii)

Q7.29: 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 the 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.

Q7.30: Write the mechanism of the reaction of HI with methoxymethane.

Ans: The mechanism of the reaction of HI with methoxymethane involves the following steps:

Step1: Protonation of methoxymethane:

Step2: Nucleophilic attack of I^−:

Step3:

When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide

Q7.31: Write equations of the following reactions:

(i) Friedel-Crafts reaction−alkylation of anisole.

(ii) Nitration of anisole.

(iii) Bromination of anisole in ethanoic acid medium.

(iv) Friedel-Craft’s acetylation of anisole.

Ans: (i)

(ii)

(iii)

(iv)

Q7.32: Show how would you synthesise the following alcohols from appropriate alkenes?

Ans: The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed hydration of appropriate alkenes.

(i)

(ii)

(iii)

Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan-3-ol.

Thus, the first reaction is preferred over the second one to get pentan-2-ol.

(iv)

Q7.33: 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 the 3rd carbon atom.

Ans: The mechanism of the given reaction involves the following steps:

Step 1: Protonation

Step 2: Formation of 2° carbocation by the elimination of a water molecule

Step 3: Re-arrangement by the hydride-ion shift

Step 4: Nucleophilic attack

Old NCERT Solution