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Chemistry is something that gives us the reason behind things we observe around us whether it is the fizz of soft drinks or the working of medicines. Alcohols, Phenols, and Ethers are three important classes of compounds that feature oxygen-containing organic compounds that are widely used in the medical industry and daily life. Have you ever wondered the things like soft drinks, perfumes, antiseptics, and disinfectants that we use in our daily lives come from? So, all of these come from oxygen-containing organic compounds like alcohol such as ethanol used in soft drinks, beverages, and sanitizers while Phenols are used in antiseptics and disinfectants and Ether is used in the manufacturing of perfumes.
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Class 12 Chemistry chapter 7 explains various concepts that we observe in our daily lives. When the OH group replaces the hydrogen atom in aliphatic hydrocarbons Alcohols are formed same goes for phenols when the OH group replaces the hydrogen of aromatic hydrocarbons then Phenols are formed while Ethers are formed when alkoxy (R-O) or aryloxy (Ar-O) group substitutes hydrogen atom in hydrocarbons. NCERT Chapter 7 Class 12 Chemistry Solutions provides a structured approach to understanding the classification, properties, preparation, and reactions. Students will learn about important concepts like nomenclature, structure and uses of organic compounds containing oxygen. The chapter also covers essential reaction mechanisms like dehydration of alcohol, electrophilic substitution in phenols, and Williamson’s synthesis of ethers.
NCERT Solutions of class 12 chemistry Alcohols, Phenols, and Ether is one of the essential sources of preparation for organic chemistry whether students are preparing for class 12 CBSE boards or other competitive exams like NEET, JEE, BITSAT, etc. Solutions are prepared by our subject experts in a very comprehensive way that helps students understand the concepts used in solving questions in a very simple and easy way. Scroll down to learn more details about the NCERT solutions for Class 12 Chemistry Chapter 7.
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Question 7.1 (1) Classify the following as primary, secondary and tertiary alcohols:
Answer :
To classify we look at the OH bonded carbon.
Here, only 1 carbon is attached to it, hence it is primary alcohol .
Question 7.1 (2) Classify the following as primary, secondary and tertiary alcohols:
Answer :
To classify we look at the OH bonded carbon.
Here, only 1 carbon is attached to it, hence it is primary alcohol.
Question 7.1 (3) Classify the following as primary, secondary and tertiary alcohols:
Answer :
To classify we look at the OH bonded carbon.
Here, only 1 carbon is attached to it, hence it is primary alcohol.
Question 7.1 (4) Classify the following as primary, secondary and tertiary alcohols:
Answer :
To classify we look at the OH bonded carbon.
Here, 2 carbons are attached to it, hence it is secondary alcohol.
Question 7.1 (5) Classify the following as primary, secondary and tertiary alcohols:
Answer :
To classify we look at the OH bonded carbon.
Here, 2 carbons are attached to it, hence it is secondary alcohol.
Question 7.1 (6) Classify the following as primary, secondary and tertiary alcohols :
Answer:
To classify we look at the OH bonded carbon.
Here, 3 carbons are attached to it, hence it is tertiary alcohol.
Question 7.2 Identify allylic alcohols in the above examples.
Answer :
The alcohols (ii) and (vi) are allylic alcohols. Because -C=C-C-OH is the skeleton of allylic alcohol.
Question 7.3 (1) Name the following compounds according to IUPAC system.
Answer :
3-Chloromethyl-2-isopropylpentan-1-ol
Question 7.3 (2) Name the following compounds according to IUPAC system.
Answer :
2, 5-Dimethylhexane-1, 3-diol
Question 7.3 (3) Name the following compounds according to IUPAC system.
Answer :
3-Bromocyclohexanol
Question 7.3 (5) Name the following compounds according to IUPAC system.
Answer :
2-Bromo-3-methylbut-2-en-1-ol
Question 7.4 (1) Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?
Answer :
The reaction of a suitable Grignard reagent on the methanal is mentioned below:
Question 7.4 (2) Show how are the following alcohols prepared by the reaction of a suitable
Grignard reagent on methanal ?
Answer :
Question 7.5 (2) Write structures of the products of the following reactions:
Answer :
Product of the given reaction is-
Question 7.5 (3) Write structures of the products of the following reactions:
Answer :
Product of the given reaction is
Question 7.6 (1) Give structures of the products you would expect when each of the following alcohol reacts with
Question 7.6 (2) Give structures of the products you would expect when each of the following alcohol reacts with
Answer :
Reaction of
Question7.6 (3) Give structures of the products you would expect when each of the following alcohol reacts with HBr with Butan-1-ol
Answer :
Reaction of HBr with Butan-1-ol
Question 7.6 (4) Give structures of the products you would expect when each of the following alcohol reacts with HBr with 2-Methylbutan-2-ol
Answer :
Reaction of HBr with 2-Methylbutan-2-ol
Question 7.6 (5) Give structures of the products you would expect when each of the following alcohol reacts with
Answer :
Reaction of
Question 7.6 (6) Give structures of the products you would expect when each of the following alcohol reacts with
Answer :
Reaction of
Question 7.7(1) Predict the major product of acid catalysed dehydration of 1-methylcyclohexanol
Answer :
Dehydration of 1-methylcyclohexanol
1-Methylcyclohexene is the major product.
Question 7.7 (2) Predict the major product of acid catalysed dehydration of butan-1-ol
Answer :
Dehydration of butan-1-ol
But-2-ene is the major product.
Answer :
Resonance structure of ortho-nitrophenol
Resonance structure of para-nitrophenol
Question 7.9 (1) Write the equations involved in the following reactions: Reimer - Tiemann reaction
Answer :
Reimer - Tiemann reaction
Question 7.9 (2) Write the equations involved in the following reactions: Kolbe’s reaction
Answer :
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.
Answer :
1.Reaction of ethanol with hydrogen bromide.
2. Reaction of 3-methylpentan-2-ol with sodium
3. Reaction of product formed in 1st and reaction with the product formed in the 2nd reaction.
Question 7.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?
Answer :
(ii) is appropriate because
Question 7.12 (2) Predict the products of the following reactions:
Answer :
The reaction between ethoxybenzene and HBr is
Question 7.12 (3) Predict the products of the following reactions:
Answer :
Reaction between ethoxybenzene and
Question 7.12 (4) Predict the products of the following reactions:
Answer :
Reaction between ter - butyl ethyl ether and HI
NCERT Solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers - Exercise Questions
Question 7.1 (1) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 2,2,4-Trimethylpentan-3-ol
Question 7.1 (2) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 5-Ethylheptane-2,4-diol
Question 7.1 (3) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is Butane-2,3-diol
Question 7.1 (4) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is Propane-1,2,3-triol
Question 7.1 (5) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 2-Methylphenol
Question 7.1 (6) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 4-Methylphenol
Note : Also called p-cresol
Question 7.1 (7) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 2,5-Dimethylphenol
Question 7.1 (8) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 2,6-Dimethylphenol
Question 7.1 (9) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 1-Methoxy-2-methylpropane
Question 7.1 (10) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is Ethoxybenzene
Question 7.1 (11) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 1-Phenoxyheptane
Question 7.1 (12) Write IUPAC names of the following compounds:
Answer :
IUPAC name of the given compound is 2-Ethoxybutane
Question 7.2 Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
Answer :
Structure of 2-Methylbutan-2-ol
Question 7.2 Write structures of the compounds whose IUPAC names are as follows: (ii) 1-Phenylpropan-2-ol
Answer :
Structure of 1-Phenylpropan-2-ol
Question 7.2(iii) Write structures of the compounds whose IUPAC names are as follows: (iii) 3,5-Dimethylhexane –1, 3, 5-triol
Answer :
structure of 3,5-Dimethylhexane –1, 3, 5-triol
Question 7.2(iv) Write structures of the compounds whose IUPAC names are as follows: (iv) 2,3 – Diethylphenol
Answer :
structure of 2,3 – Diethylphenol
Question 7.2(v) Write structures of the compounds whose IUPAC names are as follows:
(v) 1 – Ethoxypropane
Answer :
structure of 1 – Ethoxypropane
Question 7.2(vi) Write structures of the compounds whose IUPAC names are as follows:
(vi) 2-Ethoxy-3-methylpentane
Answer :
structure of 2-Ethoxy-3-methylpentane
Question 7.2(vii) Write structures of the compounds whose IUPAC names are as follows:
(vii) Cyclohexylmethanol
Answer :
Structure of Cyclohexylmethanol
Question 7.2(viii) Write structures of the compounds whose IUPAC names are as follows:
(viii) 3-Cyclohexylpentan-3-ol
Answer :
The structure of 3-Cyclohexylpentan-3-ol is as follows:
Question 7.2(ix) Write structures of the compounds whose IUPAC names are as follows:
(ix) Cyclopent-3-en-1-ol
Answer :
structure of Cyclopent-3-en-1-ol
Question 7.2(x) Write structures of the compounds whose IUPAC names are as follows:
(x) 4-Chloro-3-ethylbutan-1-ol.
Answer :
structure of 4-Chloro-3-ethylbutan-1-ol
Question 7.3 (i) Draw the structures of all isomeric alcohols of molecular formula
and give their IUPAC names.
Answer :
The structures of all isomeric alcohols of C 5 H 12 O are given below:
![]() | Pentan-1-ol |
![]() | Pentan-2-ol |
![]() | Pentan-3-ol |
![]() | 3-Methylbutan-1-ol |
![]() | 3-Methylbutan-1-ol |
![]() | 2,2-Dimethylpropan-1-ol |
![]() | 3-Methylbutan-2-ol |
![]() | 2-Methylbutan-2-ol |
Question 7.3 (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.
Answer :
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
Question 7.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer :
Propanol forms intermolecular H-bonds because of the presence of -OH group while butane cannot. To break these bonds, extra energy will be required. This causes a higher boiling point for propanol as compared to butane.
Question 7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer :
Alcohols form hydrogen bonds with water due to the presence of –OH group whereas hydrocarbons cannot. Due to this inter molecular hydrogen bonding, alcohols are more soluble in water.
Question 7.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer :
Hydroboration-oxidation reaction also called HBO reaction is the addition of borane followed by oxidation to produce alcohol.
Eg: Hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH 3 ) 2 to form trialkyl borane. This addition product is oxidized by hydrogen peroxide in the presence of aqueous sodium hydroxide to form propan-1-ol.
Question 7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula,
Answer :
The structures and IUPAC names of monohydric phenols of molecular formula,
Phenol:
Answer :
Due to inter-molecular H bonding in para-nitrophenol, it gets tightly bonded with water. But ortho nitrophenol has intra-molecular H bonding and hence is steam volatile.
Question 7.9 Give the equations of reactions for the preparation of phenol from cumene.
Answer :
Cumene (isopropylbenzene) is oxidised in the presence of air to form cumene hydroperoxide. On treating with dilute acid it is converted to phenol and acetone.
Question 7.10 Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer :
Chlorobenzene, when fused with NaOH, produces sodium phenoxide which on acidification produces Phenol.
Question 7.11 Write the mechanism of hydration of ethene to yield ethanol.
Answer :
Ethanol is yielded from ethene by acid catalysed hydration.
The mechanism:
Step 1. Protonation of alkene to form carbocation by electrophilic attack of hydronium ion.
Step 2. Nucleophilic attack of water on carbocation .
Step 3. Deprotonation to form an alcohol.
Question 7.12 You are given benzene,
Answer :
When benzene reacts with
Question 7.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
Answer :
Styrene on acid catalysed hydration gives 1-phenylethanol.
Question 7.13 (2) Show how will you synthesise:
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
Answer :
On adding NaOH to chloromethylcyclohexane, cyclohexy methanol is formed.
Question 7.13 Show how will you synthesise:
(iii) pentan-1-ol using a suitable alkyl halide?
Answer :
when 1-chloropentane reacts with NaOH it gives pantan-1-ol
Question 7.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer :
1. Phenol reacts with sodium to give sodium phenoxide, liberating hydrogen gas.
2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water.
Phenol is more acidic than ethanol. This is because phenol after losing a proton becomes phenoxide ion which is highly stable due to resonance whereas ethoxide ion does not.
Question 7.15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?
Answer :
Ortho-nitrophenol is more acidic than ortho- methoxyphenol . The presence of the nitro group, which is an electron withdrawing group, at the ortho position decreases the electron density in the O-H bond. Also, the o- nitrophenoxide ion formed after the loss of protons is stabile due to resonance. Hence, ortho nitrophenol is a stronger acid. Whereas the methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond.
Question 7.16 Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Answer :
The -OH group is an electron-donating group (EDG). Thus, it increases the electron density in the benzene ring in the resonance structure of phenol. As a result, the benzene ring is activated towards electrophilic substitution.
Question 7.17(i) Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline
Answer :
Oxidation of propane-1-ol with alkaline
Question 7.17(ii) Give equations of the following reactions:
Answer :
Bromine in
Question 7.17(iii) Give equations of the following reactions:
(iii) Dilute HNO3 with phenol.
Answer :
When dilute HNO3 reacts with phenol it gives o-bromo phenol and p-bromo phenol
Question 7.17(iv) Give equations of the following reactions:
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer :
Treating phenol with chloroform in presence of aqueous NaOH.
This reaction is known as the Reimer-Tiemann reaction.
Question 7.18(i) Explain the following with an example.
(i) Kolbe’s reaction.
Answer :
Kolbe’s reaction: Phenol with carbon dioxide under pressure followed by treating the product with sulphuric acid produces Ortho-hydroxybenzoic acid (salicylic acid). Phenoxide ion generated is more reactive than phenol towards electrophilic aromatic substitution. Hence, it undergoes electrophilic substitution with carbon dioxide, a weak electrophile.
Question 7.18(ii) Explain the following with an example.
(ii) Reimer-Tiemann reaction.
Answer :
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
Question 7.18(iii) Explain the following with an example. (iii) Williamson ether synthesis.
Answer :
Williamson ether synthesis is a reaction forming ether from a primary alkyl halide via S N 2 reaction.
Question 7.18(iv) Explain the following with an example.(iv) Unsymmetrical ether.
Answer :
If the alkyl or aryl groups attached to the oxygen atom are different, then it is mixed or unsymmetrical ether.
Eg:
Question 7.19 . Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer :
Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid. Ethanol undergoes dehydration by heating it with concentrated sulphuric acid at 443 K.
Question 7.20(i) How are the following conversions carried out?
(i) Propene
Answer :
Acid catalysed hydration of propene produces propan-2-ol.
Question 7.20(ii) How are the following conversions carried out?
(ii) Benzyl chloride
Answer :
Benzyl chloride treated with NaOH followed by acidification produces benzyl alcohol.
Question 7.20(iii) How are the following conversions carried out?
(iii) Ethyl magnesium chloride
Answer :
Ethyl magnesium chloride treated with formaldehyde followed by hydrolysis produces propan-1-ol.
Question 7.20(iv) How are the following conversions carried out?
(iv) Methyl magnesium bromide
Answer :
Methyl magnesium bromide treated with propane, gives 2-methylpropane-2-ol on hydrolysis.
Question 7.21(i) Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
Answer :
Acidic/neutral/alkaline potassium permanganate
Question 7.21(ii) Name the reagents used in the following reactions:
(ii) Oxidation of a primary alcohol to aldehyde.
Answer :
The reagent used for oxidation of primary alcohol to aldehyde is Pyridinium chlorochromate (PCC) .
Question 7.21(iii) Name the reagents used in the following reactions:
(iii) Bromination of phenol to 2,4,6-tribromophenol.
Answer :
Reagents used in the bromination of phenol to 2,4,6-tribromophenol is Bromine water
Question 7.21(iv) Name the reagents used in the following reactions:
(iv) Benzyl alcohol to benzoic acid.
Answer :
Reagent used in the benzyl alcohol to benzoic acid isAcidified
Question 7.21(v) Name the reagents used in the following reactions:
(v) Dehydration of propan-2-ol to propene.
Answer :
Reagents used in the dehydration of propan-2-ol to propene is Concentrated Phosphoric acid.
Question 7.21(vi) Name the reagents used in the following reactions:
(vi) Butan-2-one to butan-2-ol.
Answer :
Reagent used in the butan-2-one to butan-2-ol is
Question 7.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer :
Ethanol undergoes intermolecular hydrogen bonding due to the presence of -OH group. Therefore, extra energy is required to break those hydrogen bonds. Whereas methoxymethane does not make H-bonds and hence ethanol has a higher boiling point than methoxymethane.
Question 7.23(i) Give IUPAC names of the following ethers:
Answer :
IUPAC names of the given ether is 1-Ethoxy-2-methylpropane
Question 7.23(ii) Give IUPAC names of the following ethers:
Answer :
IUPAC names of the given ether is 2-Chloro-1-methoxyethane
Question 7.23(iii) Give IUPAC names of the following ethers:
Answer :
IUPAC names of the given ether is 4-Nitroanisole
Question 7.23(iv) Give IUPAC names of the following ethers:
Answer :
IUPAC names of the given ether is 1-Methoxypropane
Question 7.23(v) Give IUPAC names of the following ethers:
Question 7.23(vi) Give IUPAC names of the following ethers:
Answer :
IUPAC names of the given ether is Ethoxybenzene
Question 7.24(i) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
Answer :
Names of reagents and equations for the preparation of the 1-Propoxypropane ether by Williamson’s synthesis:-
Question 7.24(ii) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(ii) Ethoxybenzene
Answer :
Names of reagents and equations for the preparation of the Ethoxybenzene ether by Williamson’s synthesis:-
with NaBr as side product.
Question 7.24(iii) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(iii) 2-Methoxy-2-methylpropane
Answer :
Names of reagents and equations for the preparation of the2-Methoxy-2-methylpropane ether by Williamson’s synthesis:-
Question 7.24(iv) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (iv) 1-Methoxyethane
Answer :
Names of reagents and equations for the preparation of the1-Methoxyethane ether by Williamson’s synthesis:-
Question 7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer :
Williamson synthesis involves
Question 7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Answer :
Propan-1-ol on dehydration using protic acids such as sulphuric acid gives 1-propoxypropane.
Mechanism of this reaction:
Formation of protonated alcohol.
Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction.
Formation of ethene by the elimination of a proton.
Question 7.27 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 a primary alcohol is an S N 2 reaction. In case of secondary or tertiary alcohols, the alkyl group is hindered and hence elimination dominates substitution. Therefore alkenes are formed in place of ethers.
Question 7.28(i) Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane
Answer :
1-propoxypropane reacts with HI to give propan-1-ol and 1-iodopropane as the products.
Question 7.28(ii) Write the equation of the reaction of hydrogen iodide with: (ii) methoxybenzene
Answer :
Methoxybenzene reacts with HI to give phenol and iodomethane.
Question 7.28(iii) Write the equation of the reaction of hydrogen iodide with: (iii) benzyl ethyl ether.
Answer :
Benzyl ethyl ether reacts with HI to give benzyl iodide and ethanol.
Question 7.29(i) Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution
Answer :
Due to the +R effect of the alkoxy group, it increases the electron density of the benzene ring pushing electrons into the ring making the benzene ring activated towards electrophilic substitution reactions.
Question 7.29(ii) Explain the fact that in aryl alkyl ethers (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer :
The above resonating structures shows that the electron density increases more at the ortho and para positions as compared to the meta positions. Hence, the alkoxy group directs the incoming substituents to ortho and para positions in the benzene ring.
7.30. Write the mechanism of the reaction of HI with methoxymethane.
Answer :
Following is the mechanism:
1. Protonation of methoxymethane
2. Nucleophilic attack of
3. If HI is in excess, then methanol formed in step 2 reacts with another HI molecule and gets converted to methyl iodide at a high temperature.
Question 7.31(i) Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
Answer :
Fridel Craft reaction(Alkylation):
Question 7.31(ii) Write equations of the following reactions: (ii) Nitration of anisole.
Answer :
Nitration of anisole:
Question 7.31(iii) Write equations of the following reactions: (iii) Bromination of anisole in ethanoic acid medium.
Answer :
Bromination of anisole in ethanoic acid medium:
Question 7.31(iv) Write equations of the following reactions: (iv) Friedel-Craft’s acetylation of anisole.
Answer :
Friedel-Craft’s acetylation of anisole:
Question 7.32 Show how would you synthesise the following alcohols from appropriate alkenes?
Answer :
(i)
(ii)
(iii)
(iv)
Question 7.33 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 3rd carbon atom.
Answer :
Mechanism for the reaction 3-methylbutan-2-ol is treated with HBr
Alcohols Phenols and Ethers Class 12 NCERT solutions have answers to a total of 12 topic-wise questions and 33 questions in the exercise. In the CBSE board exam, the weightage of this chapter is 4 marks hence it is recommended to solve all the exercises of the book to get good marks. You will find all the NCERT solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols, and Ethers here free of cost.
By referring to the NCERT solutions for class 12, students can understand all the important concepts and practice questions well enough before their examination. Alcohols are the basic compound for the formation of detergents, phenols are the basic compound for the formation of antiseptics and ethers are the basic compound for the formation of fragrances.
7.1 Classification |
7.2 Nomenclature |
7.3 Structures of Functional Groups |
7.4 Alcohols and Phenols |
7.5 Some Commercially Important Alcohols |
7.6 Ether |
Chapter 1 | NCERT Solution for chapter 1 Solutions |
Chapter 2 | NCERT Solution for chapter 2 Electrochemistry |
Chapter 3 | NCERT Solution for chapter 3 Chemical Kinetics |
Chapter 4 | NCERT Solution for chapter 4 The d and f block elements |
Chapter 5 | NCERT Solution for chapter 5 Coordination compounds |
Chapter 6 | NCERT Solution for chapter 6 Haloalkanes and Haloarenes |
Chapter 7 | NCERT Solution for chapter 7 Alcohols, Phenols and Ethers |
Chapter 8 | NCERT Solution for chapter 8 Aldehydes, Ketones and Carboxylic Acids |
Chapter 9 | NCERT Solution for chapter 9 Amines |
Chapter 10 | NCERT Solution for chapter 10 Biomolecules |
For complete solutions refer to this website: https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry
Important topics:
This chapter holds weightage of 4 percent in NEET examination. Practice NEET previous year papers, NCERT question and NCERT exemplar questions for a good score.
Questions worth 5-6 marks are asked from this chapter in CBSE Board exam. Follow the NCERT syllabus for good score in board exam.
This chapter holds the weightage of 4-5 Marks in JEE Main
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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
Type C: For candidates scoring between 40% and 50%.
Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
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