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| Chapter | 11. Alcohols, Phenols and Ethers |
| Subject | Chemistry |
| Textbook | Chemistry II |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 11: Alcohols, Phenols and Ethers.
NCERT Solutions for Class 12 Chemistry Chapter 11
Alcohols, Phenols and Ethers Solutions
Q1) Write IUPAC names of the following compounds:
Answer)
(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
Q2) 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.
Answer)
Q3)
(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.
Answer i) The structures of all isomeric alcohols of molecular formula, C5H12O are shown below:
(a) CH3-CH2-CH2-CH2-CH2-OH
Pentan – 1 – ol (1 °)
2 – Methylbutan – 2 – ol ( 3 ° )
Answer 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
Q4) Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer) Propanol undergoes intermolecular H-bonding because of the presence of −OH group. On the other hand, butane does not
Hence, additional energy would be required to break the intermolecular hydrogen bonds. This is the reason why hydrocarbon butane has a lower boiling point than propanol.
Q5) Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer) 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.
Q6) What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer) 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 addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
Q7) Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Answer)
Q8) While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer) 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.
Q9) Give the equations of reactions for the preparation of phenol from cumene.
Answer) 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.
Q10) Write the mechanism of hydration of ethene to yield ethanol.
Answer) The mechanism of hydration of ethene to form ethanol involves three steps.
Q11) Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer) Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.
Q12) You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Q13) Show how will you synthesize:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
Answer)
Answer iii) When 1-chloropentane is treated with NaOH, pentan-1-ol is produced.
Q14) Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer) The acidic nature of phenol can be represented by the following two reactions:
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 ethoxide ion does not.
Q15) Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Answer) The nitro – group is an electron-withdrawing group. The existence of this group in the ortho position decreases the electron density in the O – H bond. Consequently, it is easier to give away a proton. Furthermore, the o -nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Therefore, ortho nitrophenol is a stronger acid. In contrast, a methoxy group is an electron – releasing group. Hence, it increases the electron density in the O – H bond and thus, losing proton is not possible easily. Therefore, ortho – nitrophenol is more acidic than ortho – methoxyphenol.
Q16) Explain how does the −OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Answer) The density of the electron increases in the benzene ring as the – OH group acts as an electron-donating group. This is clearly shown in the resonance structure of phenol given here-
As a result, the benzene ring is activated towards electrophilic substitution.
Q17) 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.
Answer i)
Answer ii)
Answer iii)
Answer iv)
Q18) Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether.
Answer i) 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.
Answer ii) 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.
Answer iii) 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 SN2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in case of primary alkyl halides.
If the alkyl halide is secondary or tertiary, then elimination competes over substitution.
Answer iv) 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).
Q19) Write the mechanism of acid-catalysed dehydration of ethanol to yield ethene.
Answer) The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:
Step 1-
Step 3-
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.
Q20) 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.
Answer i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.
Answer ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.
Answer iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.
Answer iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.’
Q21) Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a 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.
Answer i) Acidified potassium permanganate
Answer ii) Pyridinium chlorochromate (PCC)
Answer iii) Bromine water
Answer iv) Acidified potassium permanganate
Answer v) 85% phosphoric acid
Answer vi) NaBH4 or LiAlH4
Q22) Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer) Ethanol experiences intermolecular H – bonding because of the presence of – OH group, which results in the association of molecules. additional energy is necessary to break these hydrogen bonds. Conversely, methoxymethane does not experience H – bonding. Therefore, ethanol has a higher boiling point when compared to methoxymethane.
Q23) Give IUPAC names of the following ethers:
Answer)
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 1-Ethoxy-4, 4-dimethylcyclohexane
(vi) Ethoxybenzene
Q24) 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
Answer i)
Answer ii)
Answer iii)
Answer iv)
Q25) Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer) The reaction of Williamson synthesis involves SN2 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.
Q26) How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Answer) 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
Q27) Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer) The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN2) 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.
Q28) Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether
Answer i)
Answer ii)
Answer iii)
Q29) 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.
Answer 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.
Answer 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.
Q30) Write the mechanism of the reaction of HI with methoxymethane.
Answer) The mechanism of the reaction of HI with methoxymethane involves the following steps:
Step1: Protonation of methoxymethane:
Step2: Nucleophilic attack of I−:
Step 3: The methanol synthesized in the above step is made to react with another HI molecule which then converts to methyl iodide. This is done at a high temperature & HI is present in excess.
Q31) 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.
Answer i)
Answer ii)
Answer iii)
Answer iv)
Q32) Show how would you synthesise the following alcohols from appropriate alkenes?
Answer) The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed hydration of appropriate alkenes.
Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan-3-ol.
Q33) When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
Give a mechanism for this reaction.
Answer) The mechanism of the given reaction involves the following steps:
That’s it. These were the solutions of NCERT Class 12 Chemistry Chapter 11 – Alcohols, Phenols and Ethers. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.