CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 11- Alcohols, Phenols and Ethers

Check important questions and answers for the chapter Alcohols, Phenols and Ethers here for the upcoming CBSE Class 12th Chemistry Examination 2020. 

Created On: Mar 5, 2020 18:17 IST
chapter 11
chapter 11

The students who are appearing CBSE Class 12th Chemistry Examination 2020 can go through the below-mentioned important questions for the chapter Alcohols, Phenols and Ethers. The questions and answers mentioned in this article are strictly based on the latest CBSE pattern. 

Key Points to be mentioned while writing the answers to the below mentioned important questions: 

Question 1- Write the products for the following: 

(i) 

(ii) 

(iii)

Answer: (i) 

(ii) 

(iii)

 

Question 2- Define: 

(i) Kolbe’s reaction

(ii) Williamson synthesis 

(iii) Reimer-Tiemann reaction

(iv) Friedel-Crafts reaction

Answer: (i) Kolbe’s reaction: Phenoxide ion generated by treating phenol with sodium hydroxide is even more reactive than phenol towards electrophilic aromatic substitution. Hence, it undergoes electrophilic substitution with carbon dioxide, a weak electrophile. Ortho hydroxybenzoic acid is formed as the main reaction product.

(ii) Williamson synthesis: It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide. Ethers containing substituted alkyl groups (secondary or tertiary) may also be prepared by this method. The reaction involves the SN2 attack of an alkoxide ion on primary alkyl halide.  Better results are obtained if the alkyl halide is primary. In case of secondary and tertiary alkyl halides, elimination competes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed. Example, the reaction of CH3ONa with (CH3)3C–Br gives exclusively 2-methylpropene. It is because alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to elimination reactions. Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.

(iii) Reimer-Tiemann reaction: On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at the ortho position of the benzene ring. This reaction is known as Reimer - Tiemann reaction. The intermediate substituted benzal chloride is hydrolysed in the presence of alkali to produce salicylaldehyde.

(iv) Friedel-Crafts reaction: Anisole undergoes Friedel-Crafts reaction, i.e., the alkyl and acyl groups are introduced at ortho and para positions by reaction with an alkyl halide and acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as a catalyst.



Question 3- How can the following conversions be carried out? 

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

(ii) Methyl magnesium bromide → 2-Methylpropan-2-ol

(iii) Phenol → benzoquinone

(iv) Propene → propan-2-ol

Answer: (i) 

(ii)

(iii)

(iv)

 

Question 4- State Reasons: 

(i) Alcohol is more soluble in water than the hydrocarbons of comparable molecular masses.

(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol.

(iii) The boiling point of ethanol is higher than that of methoxymethane. 

(iv) Phenol is more acidic than ethanol.

Answer: (i)  Alcohol can form H-bonds with water and break the existing H-bonds between water molecules whereas hydrocarbons cannot form H-bonds with water. Therefore, alcohol is more soluble in water than the hydrocarbons of comparable molecular masses.

(ii) Due to the strong – R and -1 effect of the – NO2 group, electron density in the – OH bond decreases and proton loss is easier. In addition to this,  O-nitrophenoxide ion is stabilized by resonance, 1 thereby making O-nitrophenol a stronger acid.

In O-methoxyphenol, due to + R effect of the – OCH3 group the electron density in the O – H bond increases thereby making the loss of proton difficult. Additionally, the O-methoxyphenoxide ion left after the loss of a proton is destabilized by resonance as negative charges repel each other. Therefore, O-methoxyphenol is a weaker acid than O-nitrophenol.  

(iii) The boiling point of ethanol is higher than that of methoxymethane because of the presence of intermolecular H-bonding, associated molecules are formed, hence ethanol has a high boiling point while methoxymethane does not have intermolecular H-bonding.

(iv) Phenol is more acidic than ethanol because on losing H+ ions, phenol forms phenoxide ion and ethanol on losing H+ ion forms ethoxide ion. Also, phenoxide ion exists in resonance structure. 

Question 5- Explain the mechanism of : 

(i)

(ii) acid dehydration of ethanol to yield ethene.

Answer: (i) 

(ii) The following steps are carried out in the dehydration of ethanol to yield ethene:

Step 1:  Formation of protonated alcohol: 

Step 2:  Formation of Carbocation: 

Step 3:  Formation of ethene by the elimination of a proton: 



Question 6-  Arrange the following in the increasing order of their acidic strength:

(i) p-cresol, p-nitrophenol, phenol

(ii) Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.

Answer: (i) p-cresol < phenol < p-nitrophenol

(ii) Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 3,5-dinitrophenol < 2,4, 6-trinitrophenol.

 

Question 7- (i) What happens when CH3—O—CH<sub3 is heated with HI?

(ii) Explain mechanism for hydration of acid-catalyzed ethene :

Answer: (i) When CH3—O—CH<sub3 is heated with HI then Methyl Iodide (CH3l) and Methanol (CH3OH) are formed. 

(ii) The mechanism for hydration of acid-catalyzed ethene is as follows:

Step 1: Protonation of the alkene to form carbocation by an electrophilic attack of H3O+.

Step 2: Nucleophilic attack of water on carbocation.

Step 3: Deprotonation to form alcohol. 

Question 8- Give simple chemical tests to distinguish between the following pairs of compounds:

(i) Ethanol and Phenol

(ii) Propanol and 2-methylpropan-2-ol

Answer: (i) Ethanol gives +ve Iodoform test whereas Phenol gives a -ve Iodoform test. 

(ii) Through Lucas Test:

Turbidity appears after heating in Propanol while Turbidity appears immediately in 2-methylpropan-2-ol. 

Question 9-Classify the following as primary, secondary and tertiary alcohols:

Answer:  (i) Primary alcohols

(ii) Primary alcohols

(iii) Primary alcohols

(iv) Secondary alcohols

(v) Secondary alcohols

(vi) Tertiary alcohols

 

Question 10- Write IUPAC names along with the structures of the products: 

(a) Catalytic reduction of butanal. 

(b) Hydration of propene in the presence of dilute sulphuric acid. 

(c) The reaction of propanone with methylmagnesium bromide followed by hydrolysis.

Answer: (a) Butan-1-ol

(b) Propan-2-ol

(c) 2-Methylpropan-2-ol 

The above-mentioned questions and answers are based on the NCERT textbook, previous year question papers and sample papers. The students are advised to go through the links mentioned below: 

CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 10 - Haloalkanes and Haloarenes

CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from All Chapters of NCERT Textbooks - Part I & II

CBSE Board Exam 2020: Check Important Questions of Class 12th Chemistry Subject