Important Questions Class 12 Chemistry Chapter 11

Important Questions Class 12 Chemistry Chapter 11

Important Questions For CBSE Class 12 Chemistry Chapter 11 – Alcohols, Phenols And Ethers

Alcohols, phenols and ethers are some of the most important compounds studied in organic chemistry. Students learn the IUPAC system of nomenclature, various reactions involved in the preparation of the aforementioned compounds, their structures and properties, and some chemical reactions on the basis of their functional groups in Class 12 Chemistry Chapter 11. Important Questions Class 12 Chemistry Chapter 11 will help students practise questions that are important from the perspective of the board exam. These questions are selected by subject matter experts after consulting the question papers of the past years. The latest CBSE Syllabus has been followed in this question set. So, students can rely on them to enhance their preparation for the exam.

CBSE Class 12 Chemistry Chapter-11 Important Questions

Study Important Questions For Class 12 Chemistry Chapter 11 – Alcohols, Phenols And Ethers

Q.1. Write the IUPAC names of the following compounds.


Ans. C6H5OC3H7 → Propoxybenzene

CH3CH2OCH2CH2CH2Cl → 3-Ethoxy Chlorophen

CH3OCH3 → Methoxymethane

C6H5O(CH2)6-CH3 →1-Phenoxyheptane

CH3OCH2CH2CH3 →1-Methoxypropen

Q.2. Describe the following reactions.

Hydroboration oxidation of alkanes

Williamson Synthesis

Reimer-Tiemann reaktion

Kolbe’s reaction

Friedel-Crafts acylation of anisole

Ans. (i) Hydroboration Oxidation Of Alkanes: This is a two-step process to produce alcohol from alkenes. This method follows the Anti-Markovnikov rule as the hydrogen atom dissociated from BH3 or BHR2 attaches itself to the most substituted carbon present in the double bond of alkene and the boron atom attaches itself to the least substituted carbon.

(ii) Williamson Synthesis: This is a method of producing ether by permitting the alkyl group halides to react with sodium alkoxides. This is an SN2 reaction where the alkoxide ion displaces the halide ion. Generally, primary halides are used in this reaction to obtain a better result. Tertiary alkyls do not produce any ether.

(iii) Reimer-Tiemann Reaction: This is a common substitution reaction to produce salicylaldehyde from phenol. The reaction is carried out in the presence of a biphasic solvent because hydroxides do not dissolve in chloroform easily. The reaction needs heat to initiate but once it gets started it generates heat that increases the rate of reaction.

(iv) Kolbe’s Reaction: This is an additional reaction where sodium phenoxide and carbon dioxide are heated together at 180⁰C – 200⁰C temperature under pressure to produce salicylic acid.

(v) Friedel-Crafts Acylation Of Anisole: The aromatic ring of anisole is converted into an aryl ketone through Friedel-Crafts Acylation process. This reaction requires acid chloride (R−COCl) and a Lewis acid (for example, AgCl3) to act as catalysts.

Q.3. Alcohols have higher boiling points than ethers of comparable molecular masses. What is the reason for this?

Ans. The presence of the intermolecular hydrogen bond makes the boiling points of alcohols high whereas this type of bond is not present in ethers. Therefore, alcohols have higher boiling points than ethers of comparable molecular masses.

Q.4. Phenols are more acidic than alcohols. Explain.

Ans. The acidic nature of phenol and alcohol depends on their reactivity towards the ionisation reaction of the O−H bond. The ionisation reaction of alcohols can be expressed by the following equation:

H2O + R−OH (alcohol) ⇋ H3O+ + R−O−(alkoxide)

The tendency of alkyl to release electrons destabilises the alkoxide ion of alcohol. Therefore, alcohols never ionise in water.

However, the phenoxide ion produced by the ionisation of phenol in water is stabilised through delocalisation of the negative charge produced in the aqueous solution of phenol. Hence, phenol shows more acidic nature by completing the ionisation process.

Q.5. What is the chemical name of picric acid? How can it be prepared?

Ans. The chemical name of picric acid is: 2, 4, 6 −trinitro phenol.

Generally, picric acid is prepared from phenol in laboratories. Phenol (C6H5OH) is added to concentrated sulphuric acid (H2SO4) to produce disulphonic acid. When this newly formed acid is added to concentrated nitric acid (HNO3) it produces picric acid. Phenol cannot be added to concentrated nitric acid directly as the picric acid produced in this way becomes a weak acid.

Q.6. The phenyl methyl ether reacts with HI to produce phenol and iodomethane, not methanol and iodobenzene. Why?

Ans. HI is a strong acid so it easily produces hydrogen ions. The oxygen atoms present in phenyl methyl ether contain lone pair electrons. So, the hydrogen ions attack the oxygen atoms in phenyl methyl ether.

When oxygen forms a triple bond and has a positive charge, it becomes stable. As a result, nucleophilic addition occurs.

The charge on the methyl group is still unstable and there is resonance between oxygen and the benzene ring; it undergoes the SN2 process. The carbon atoms of the methyl group form 5 bonds when attacked by the nucleophiles.

As both iodine ions and benzene rings are big in size there will be a steric repulsion between them. Thus, phenol and methyl iodide are produced as a result.

Q.7. How can the mixture of o-nitrophenol and p-nitrophenol be separated?

Ans. The diluted nitric acid in reaction with phenol at 298 K temperature produces a mixture of ortho- and para-nitrophenol. These isomers can be separated by steam distillation process. O-Nitrophenol is a steam volatile because of intramolecular hydrogen bonding; whereas p-Nitrophenol is less volatile due to intermolecular hydrogen bonding that causes the association of molecules.

Q.8. Name the chemical test commonly used to distinguish between the following pairs of compounds.

  1. n-Propyl alcohol and isopropyl alcohol.
  2. Methanol and ethanol
  3. Cyclohexanol and phenol
  4. Phenol and anisole
  5. Ethanol and diethyl ether

Ans. a. n-Propyl alcohol and isopropyl alcohol can be distinguished by the Lucas Test.

  1. Methanol and ethanol can be distinguished by the Iodoform Test.
  2. Cyclohexanol and phenol can be distinguished by FeCl3 test.
  3. Phenol and anisole can be distinguished by FeCl3 test.
  4. Ethanol and diethyl ether can be distinguished by the iodoform test.

Q.9. Identify the compounds.

(i) An alcohol A, in reaction with acidified potassium dichromate, produces B.

(ii) When the alcohol is dehydrated with concentrated H2SO4, it produces C.

(iii) Compound D is produced in the reaction of C and aqueous solution of H2SO4. D is an isomer of A.

(iv) D is an oxidation resistant compound while A gets oxidised easily.

Ans. (i) Given that, the alcohol A gets oxidised easily and produces compound B in the presence of acidified potassium dichromate. The compound A, is a primary alcohol 2-methyl propan-1-ol.

(ii)The byproduct B, is carboxylic acid, is 2-methyl propan-1-oic acid.

(iii) When dehydrated with concentrated H2SO4, B further produces C, that is, 2-methyl prop-1-ene. This is an alkene.

(iv) When the alkene reacts with the aqueous solution of H2SO4 it produces an alcohol,2-methyl propan-2-ol, which is an isomer of 2-methyl propan-1-ol.

Q.10. An organic compound A turns the colour of aqueous solution of FeCl3 violet. When A is treated with NaOH in presence of CO2 under pressure at 400 K temperature, it produces compound B which is a non-steroidal anti-inflammatory drug. If acidified, B produces C which in reaction with acetyl chloride gives compound D which is prescribed by doctors to alleviate pain. Can you recognise the compounds?

Ans. (i) Only phenol produces different colours when it reacts with an aqueous solution of FeCl3. So, compound A must be phenol.

(ii) Compound B corresponds to sodium salicylate.

(iii) Compound C is salicylic acid, which is also used in skin products.

(iv) Compound D is 2-acetoxy benzoic acid, commonly known as aspirin.

Q.11. What is Lucas Reagent? How does anhydrous ZnCl2 distinguish between 1⁰, 2⁰, 3⁰ alcohols?

Ans. Lucas reagent is a mixture of ZnCl2 and HCl. ZnCl2 acts as a Lewis acid due to its vacant d-orbital. The oxygen from the -OH group forms a coordinate bond with Zn. Thus, oxygen acquires the positive charge while Zn ion acquires the negative. Thus, the oxygen of the -OH group acts as a good leaving agent and the reaction progresses. However, every alcohol does not react with Lucas reagent immediately. By observing the time taken to complete the reaction, the primary, secondary and tertiary alcohols can be distinguished. For example, tertiary alcohols are more reactive than primary alcohols at room temperature.

Q.12. What are the determining factors of the solubility of alcohols in water?

Ans. The solubility of alcohol in water depends on the following factors:

  • Propensity to create hydrogen bond: Alcohols show an inclination to form hydrogen bond with the water molecules which makes them soluble in water.
  • Size of the alkyl and aryl groups: The solubility of alcohols in water decreases with the increase in the size of alkyl and aryl groups. The intermolecular hydrogen bonding between alcohol and water molecules makes alcohols of low molecular mass liquid in water. This is another reason for the solubility of alcohol in water.
  • Molecular mass of alcohols: The greater the molecular mass is, the less is the polar -OH group in alcohols. The solubility of alcohols decreases with the decrease in polar effect. So, alcohols with less molecular mass are more soluble in water than alcohols with large molecular masses.

Q.13. In a reaction with active metals like Na and k, alcohols produce alkoxides. Write down the increasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.

Ans. Na is basic in nature whereas alcohols are acidic in nature. Therefore, the alcohol with highest acidity will react with Na the fastest. In tertiary alcohols, the alkyl groups increase electron density on the oxygen which reduces the acidic nature of the alcohol by decreasing the polarity of O-H bond. So, if the alcohols are arranged in increasing order based on their reactivity, the right sequence will be: tertiary > secondary > primary.

Q.14. What is meant by optical activity in organic chemistry? Which compounds show this property?

Ans. Optical activity is the ability of an organic compound to rotate the polarised plane clockwise or anticlockwise when its solution is placed on the path of a plane-polarised light ray. The compounds that show optical activity are: butan 1-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol, butan-2-ol.

FAQs (Frequently Asked Questions)

1. Why should students practise Important Questions?

Alcohols, Phenols, Ethers” is an important chapter in organic chemistry. All of these compounds have carbon and hydrogen common in them. Students must conceptually understand these compounds and differentiate them. As seen in the case of IUPAC nomenclature, though compounds are made up of the same elements, their positions in the carbon chain change the properties of the compounds. Students must practise the equations several times, especially those related to the conversion from one compound to another, to grasp the chapter very well. These concepts are important from the exam point as well. Therefore, students should practise important questions to understand these topics and revise them effectively.


2. Why is ortho nitrophenol called steam volatile?

Ortho nitrophenol is called steam volatile because of its low boiling point.

3. Which process is useful to prepare absolute alcohol from a rectified spirit?

Azeotropic distillation is used to prepare absolute alcohol from a rectified spirit.

4. What are the byproducts of the reaction between phenyl methyl ether and hydroiodic acid?

When phenyl methyl ether and hydroiodic acid react with each other they produce phenol and iodomethane.

5. How can students benefit from the Important Questions provided by Extramarks?

Students should practise more questions to avoid common mistakes in the exam. They can practise the set of Important Questions provided by Extramarks. Here are some of the benefits:-

  • This question set has been prepared by subject matter experts.
  • The CBSE Syllabus and questions of past years have been consulted before preparing the set of questions.
  • These questions also include solutions. Students can refer to them and improve the quality of their answers.