Ncert Chemistry Class 12 intext questions solutions chapter 7 Alcohols Phenols and Ethers

Alcohols Phenols and Ethers intext questions solutions: Class 12 Chemistry Chapter 7 intext questions solutions

Textbook	Ncert
Class	Class 12
Subject	Chemistry
Chapter	Chapter 7
Chapter Name	Alcohols Phenols and Ethers Class 12 intext questions solutions
Category	Intext questions
Medium	English

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Question 7.1: Classify the following as primary, secondary and tertiary alcohols.

(i) 

(ii)

(iii)

(iv) 

(v) 

(vi) 

Solution 7.1:

• Primary alcohol → (i), (ii), (iii)
• Secondary alcohol → (iv), (v)
• Tertiary alcohol → (vi)

Question 7.2: Identify allylic alcohols in the above examples.

Solution 7.2: The alcohols given in (ii) and (vi) are allylic alcohols.

Question 7.3: Name the following compounds according to IUPAC system.

(i) 

(ii) 

(iii) 

(iv) 

(v) 

Solution 7.3:

• (i) 3-Chloromethyl-2-isopropylpentan-1-ol
• (ii) 2, 5-Dimethylhexane-1, 3-diol
• (iii) 3-Bromocyclohexanol
• (iv) Hex-1-en-3-ol
• (v) 2-Bromo-3-methylbut-2-en-1-ol

Question 7.4: Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?

(i) 

(ii) 

Solution 7.4: (i)

(ii)

Question 7.5: Write structures of the products of the following reactions:

Solution 7.5: (i) Propene to Propene-2-ol

(ii) Methyl (2-oxocyclohexyl) ethanoate to Methyl (2-hydroxycyclohexyl) ethanoate

(iii) 2-Methylbutanal to 2-Methylbutan-1-ol

Question 7.6: Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl₂ (b) HBr and (c) SOCl₂.

• (i) Butan-1-ol
• (ii) 2-Methylbutan-2-ol

Solution 7.6: (a) Reaction with HCl-ZnCl₂:

(i) Primary alcohols do not react appreciably with Lucas’ reagent (HCl-ZnCl₂) at room temperature.

(ii) Tertiary alcohols react immediately with Lucas’ reagent.

(b) Reaction with HBr :

(i)

(ii)

(c) Reaction with SOCl₂:

(i)

(ii)

Question 7.7: Predict the major product of acid catalysed dehydration of

• (i) 1-Methylcyclohexanol
• (ii) Butan-1-ol

Solution 7.7: (i)

(ii)

Question 7.8: Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

Solution 7.8: Resonance structure of the phenoxide ion

Resonance structures of p-nitrophenoxide ion

Resonance structures of o-nitrophenoxide ion

It can be observed that the presence of nitro groups increases the stability of phenoxide ion.

Question 7.9: Write the equations involved in the following reactions:

• (i) Reimer-Tiemann reaction
• (ii) Kolbe’s reaction

Solution 7.9: (i) 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 salicyclaldehyde.

(ii) 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.

Question 7.10: Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Solution 7.10: In Williamson synthesis, an alkyl halide reacts with an alkoxide ion. Also, it is an S_N2 reaction. In the reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from 3-methylpentan-2-ol.

Question 7.11: Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

(i) 

(ii) 

Solution 7.11: Set (ii) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene.

In set (i), sodium methoxide (CH₃ONa) is a strong nucleophile as well as a strong base. Hence, an elimination reaction predominates over a substitution reaction.

Question 7.12: Predict the products of the following reactions:

(i) CH₃ −CH₂ −CH₂ −O – CH₃ +HBr →

Solution 7.12:

(i)

(ii)

(iii)

(iv)