• Home
  • Courses
  • Online Test
  • Contact
    Have any question?
    +91-8287971571
    contact@dronstudy.com
    Login
    DronStudy
    • Home
    • Courses
    • Online Test
    • Contact

      Class 12 CHEMISTRY – JEE

      • Home
      • All courses
      • Chemistry
      • Class 12 CHEMISTRY – JEE
      CoursesClass 12ChemistryClass 12 CHEMISTRY – JEE
      • 1. Solid State
        11
        • Lecture1.1
          Crystalline & Amorphous Solid 50 min
        • Lecture1.2
          Law of Crystallography 01 hour
        • Lecture1.3
          Bravius lattice & Important Terms of solid state 48 min
        • Lecture1.4
          Type of Cubic crystal & Closest packed St. 01 hour
        • Lecture1.5
          Tetrahedral & Octahedral Void 38 min
        • Lecture1.6
          Type of Voids & Radius Ratio 44 min
        • Lecture1.7
          Type of ionic solid 59 min
        • Lecture1.8
          Defect in Solid 48 min
        • Lecture1.9
          Metallic Bonding 52 min
        • Lecture1.10
          Chapter Notes – Solid State
        • Lecture1.11
          NCERT Solutions – Solid State
      • 2. Solution and its C.P
        9
        • Lecture2.1
          Condition of solution formation, TD of Solution, Factors affecting solubility-Henary’s Law 55 min
        • Lecture2.2
          Colligative Properties, Raoult’s Law 49 min
        • Lecture2.3
          Relative lowering of V.P. & Problems 45 min
        • Lecture2.4
          Non ideal solution, Azeotropic Solution 46 min
        • Lecture2.5
          Elevation in B.P., Depression in F.P. 47 min
        • Lecture2.6
          Osmotic Pressure, Abnormal C.P. & Van’t Hoff Factor 59 min
        • Lecture2.7
          Solution – Ostwald Walker Exp. 13 min
        • Lecture2.8
          Chapter Notes – Solution and its C.P
        • Lecture2.9
          NCERT Solutions – Solution and its C.P
      • 3. Chemical Kinetics
        10
        • Lecture3.1
          Rate of reaction 37 min
        • Lecture3.2
          Differential Rate Law 38 min
        • Lecture3.3
          Integrated Rate Law 56 min
        • Lecture3.4
          Integrated Rate problems 53 min
        • Lecture3.5
          Pseudo order Reaction 40 min
        • Lecture3.6
          Reaction Mechanism 47 min
        • Lecture3.7
          Collision Model 34 min
        • Lecture3.8
          Arhenius Equation 34 min
        • Lecture3.9
          Chapter Notes – Chemical Kinetics
        • Lecture3.10
          NCERT Solutions – Chemical Kinetics
      • 4. Electrochemistry
        13
        • Lecture4.1
          Introduction & Galvanic cell 32 min
        • Lecture4.2
          Cell Notation & Cell Reaction 35 min
        • Lecture4.3
          Electrode & Cell Potential 38 min
        • Lecture4.4
          Electrochemical series 39 min
        • Lecture4.5
          The Nernst Equation 39 min
        • Lecture4.6
          Concentration cell, Battery, Corrosion 52 min
        • Lecture4.7
          Electrolysis 20 min
        • Lecture4.8
          Faraday Law 45 min
        • Lecture4.9
          Resistance & Conductance 40 min
        • Lecture4.10
          Molar & Eq. Conductance, Kohlraush’s Law 29 min
        • Lecture4.11
          Problems on Resistance & Conductance 23 min
        • Lecture4.12
          Chapter Notes – Electrochemistry
        • Lecture4.13
          NCERT Solutions – Electrochemistry
      • 5. Surface Chemistry
        11
        • Lecture5.1
          Introduction & Surface tension & surface energy 33 min
        • Lecture5.2
          Adsorption 47 min
        • Lecture5.3
          Factors affecting Adsorption 39 min
        • Lecture5.4
          Catalysis 34 min
        • Lecture5.5
          Type of Catalysis & Enzyme Catalysis 41 min
        • Lecture5.6
          Colloidal Solution 57 min
        • Lecture5.7
          Type of Colloidal Solution 43 min
        • Lecture5.8
          Properties of Colloidal Solution 50 min
        • Lecture5.9
          Protective Colloids 58 min
        • Lecture5.10
          Chapter Notes – Surface Chemistry
        • Lecture5.11
          NCERT Solutions – Surface Chemistry
      • 6. Alcohol & Ether
        8
        • Lecture6.1
          Preparation 35 min
        • Lecture6.2
          Physical Properties & Oxidation Of Alcohol 29 min
        • Lecture6.3
          Hydrates, Acetal, Ketal 38 min
        • Lecture6.4
          Tests Of Alcohol 47 min
        • Lecture6.5
          Ether Preparation & Its Properties 33 min
        • Lecture6.6
          Thiol & Thioether 16 min
        • Lecture6.7
          Chapter Notes – Alcohol & Ether
        • Lecture6.8
          NCERT Solutions – Alcohol & Ether
      • 7. Aldehyde & Ketone
        10
        • Lecture7.1
          Preparation 33 min
        • Lecture7.2
          Physical Properties, Beckmann Rearrangement, Witting Reaction 46 min
        • Lecture7.3
          Schmidt Reaction, Bayer Villegar Oxidation 22 min
        • Lecture7.4
          Aldol Condensation Reaction 40 min
        • Lecture7.5
          Cannizzaro Reaction 32 min
        • Lecture7.6
          Acyloin, Benzoin, Clasien, Perkin Condensation 28 min
        • Lecture7.7
          Reformasky Reaction, Tischenko Reaction 20 min
        • Lecture7.8
          Tests-8 40 min
        • Lecture7.9
          Chapter Notes – Aldehyde & Ketone
        • Lecture7.10
          NCERT Solutions – Aldehyde & Ketone
      • 8. Acid & derivatives
        4
        • Lecture8.1
          Preparation 31 min
        • Lecture8.2
          Chemical Reactions Of Acids 31 min
        • Lecture8.3
          Arndt Eistert, Curtius, Hvz, Hoffmann Reaction 19 min
        • Lecture8.4
          Acid Derivatives 38 min
      • 9. Nitrogen containing compounds
        4
        • Lecture9.1
          Alkyl Nitrites, Nitro Alkane 27 min
        • Lecture9.2
          Alkane Nitrile & Isonitrile 20 min
        • Lecture9.3
          Amine Preparation 24 min
        • Lecture9.4
          Properties Of Amines 13 min
      • 10. Aromatic Compounds
        7
        • Lecture10.1
          Benzene 41 min
        • Lecture10.2
          Aromatic Hydrocarbon 29 min
        • Lecture10.3
          Aryl Halides 18 min
        • Lecture10.4
          Phenol 40 min
        • Lecture10.5
          Aromatic Aldehyde 39 min
        • Lecture10.6
          Aniline 32 min
        • Lecture10.7
          Phenyl Diazonium Salts 37 min
      • 11. Biomolecules
        14
        • Lecture11.1
          Introduction & Types Of Carbohydrates 47 min
        • Lecture11.2
          D-glucose & D-fructose 50 min
        • Lecture11.3
          Reactions Of D-glucose & D-fructose 32 min
        • Lecture11.4
          Reactions Of D-glucose & D-fructose 23 min
        • Lecture11.5
          Sucrose, Maltose, Lactose 31 min
        • Lecture11.6
          Starch, Cellulose, Glycogen 27 min
        • Lecture11.7
          Reducing Sugar, Mutarotation, Osazone Formation 40 min
        • Lecture11.8
          Problems On Carbohydrates 41 min
        • Lecture11.9
          Amino Acids 48 min
        • Lecture11.10
          Peptides 47 min
        • Lecture11.11
          Proteins 18 min
        • Lecture11.12
          Enzyme & Vitamins 30 min
        • Lecture11.13
          Nucleic Acid 36 min
        • Lecture11.14
          Chapter Notes – Biomolecules
      • 12. Polymer Chemistry
        6
        • Lecture12.1
          Polymerisation Addition Reaction 32 min
        • Lecture12.2
          Coordination Addition, Condensation Reaction 24 min
        • Lecture12.3
          Division Of Polymer 41 min
        • Lecture12.4
          Examples Of Polymer 31 min
        • Lecture12.5
          Examples Of Polymer 31 min
        • Lecture12.6
          Chapter Notes – Polymer Chemistry
      • 13. Practical Organic Chemistry
        4
        • Lecture13.1
          Poc Qualitative Analysis 23 min
        • Lecture13.2
          Poc Qualitative Analysis 20 min
        • Lecture13.3
          Poc Quantitative Analysis 29 min
        • Lecture13.4
          Poc Quantitative Analysis 20 min
      • 14. P block elements II
        13
        • Lecture14.1
          VA – Elemental Properties of N family 51 min
        • Lecture14.2
          VA – Compounds of N family 43 min
        • Lecture14.3
          VA – N & Its compounds 45 min
        • Lecture14.4
          VA – Oxides & Oxyacids of Nitrogen 55 min
        • Lecture14.5
          VA – P & its compounds 31 min
        • Lecture14.6
          VA – Oxides & Oxyacids of P 31 min
        • Lecture14.7
          VIA 1 – Elemental Properties of O-Family 36 min
        • Lecture14.8
          VIA 2 – compounds of VIA elements 41 min
        • Lecture14.9
          VIA 3 – Oxygen & Ozone 47 min
        • Lecture14.10
          VIA 4 – Sulphur & oxides of Sulphur 37 min
        • Lecture14.11
          VIA 5 – Sulphuric Acid 25 min
        • Lecture14.12
          Chapter Notes – P block elements
        • Lecture14.13
          NCERT Solutions – P block elements
      • 15. P block elements III
        5
        • Lecture15.1
          VIIA 1 – elemental properties of Halogen 40 min
        • Lecture15.2
          VIIA 2 – Compounds of Halogen 49 min
        • Lecture15.3
          VIIA 3 – Chlorine & its Compounds 41 min
        • Lecture15.4
          VIIIA 1 – Properties of Noble Gas 34 min
        • Lecture15.5
          VIIIA 2 – Compounds of Noble Gas 34 min
      • 16. D block metals
        8
        • Lecture16.1
          D block – Elemental Properties 55 min
        • Lecture16.2
          Elemental Properties 01 hour
        • Lecture16.3
          Elemental Properties 53 min
        • Lecture16.4
          KMnO4 & K2Cr2O7 47 min
        • Lecture16.5
          Problems 40 min
        • Lecture16.6
          Problems 20 min
        • Lecture16.7
          Chapter Notes – The d-and f-Block Elements
        • Lecture16.8
          NCERT Solutions – The d-and f-Block Elements
      • 17. F block metals
        3
        • Lecture17.1
          Lanthanoids 52 min
        • Lecture17.2
          Actinoids 48 min
        • Lecture17.3
          Problems 42 min
      • 18. Co-ordination compounds
        17
        • Lecture18.1
          Introduction of Complex Compound, Ligands 42 min
        • Lecture18.2
          Classification of Ligands, Denticity 35 min
        • Lecture18.3
          Nomenclature of Complex Compounds 46 min
        • Lecture18.4
          Nomenclature of Complex Compounds 2 40 min
        • Lecture18.5
          Bonding in Complex Compound, Primary & Secondary Valency 44 min
        • Lecture18.6
          Concept of EAN 29 min
        • Lecture18.7
          VBT in Complex Compounds 58 min
        • Lecture18.8
          Examples on VBT in complex compounds 31 min
        • Lecture18.9
          CFT in Complex Compounds 43 min
        • Lecture18.10
          CFT for Octahedral & Tetrahedral Complex 35 min
        • Lecture18.11
          Colour & Stability of Complex Compounds 28 min
        • Lecture18.12
          Structural Isomerism in Complex Compounds 49 min
        • Lecture18.13
          Geometrical Isomerism in Complex Compounds 43 min
        • Lecture18.14
          Optical Isomerism in Complex Compounds, use of Complex 01 hour
        • Lecture18.15
          Organometallic Compounds 29 min
        • Lecture18.16
          Chapter Notes – Co-ordination compounds
        • Lecture18.17
          NCERT Solutions – Co-ordination compounds
      • 19. Environmental Chemistry
        4
        • Lecture19.1
          Introduction & Air Pollution 35 min
        • Lecture19.2
          Air Pollution 20 min
        • Lecture19.3
          Water Pollution 23 min
        • Lecture19.4
          Soil Pollution, Prevention of Pollution 16 min

        NCERT Solutions – Alcohol & Ether

        11.1. Write IUPAC names of the following compounds:

        (i)

        (ii)

        (iii)

        (iv)

        (v)

        (vi)

        (vii)

        (viii)

        (ix)

        (x) 

        (xi) 

        (xii)

        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

        11.2. 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

        (i)

        (ii)

        (iii)

        (iv)

        (v)

        (vi)

        (vii)

        (viii)

        (ix)

        (x)

        11.3. (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 11.3 (i) as primary, secondary and tertiary alcohols.

        Answer

        (i) The structures of all isomeric alcohols of molecular formula, C5H12O are shown below:

        (a) 

        Pentan-1-ol (1°)

        (b)

        2-Methylbutan-1-ol (1°)

        (c)

        3-Methylbutan-1-ol (1°)

        (d)

        2, 2-Dimethylpropan-1-ol (1°)

        (e)

        Pentan-2-ol (2°)

        (f)

        3-Methylbutan-2-ol (2°)

        (g)

        Pentan-3-ol (2°)

        (h)

        2-Methylbutan-2-ol (3°)

        (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

        11.4. 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

        Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane.

        11.5. 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.

        11.6. 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.

        11.7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.

        Answer

        11.8. 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.

        11.9. 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.

        11.10. 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.

        11.11.Write the mechanism of hydration of ethene to yield ethanol.

        Answer

        The mechanism of hydration of ethene to form ethanol involves three steps.

        Step 1:

        Protonation of ethene to form carbocation by electrophilic attack of H3O+:

        Step 2:

        Nucleophilic attack of water on carbocation:

        Step 3:

        Deprotonation to form ethanol:

        11.12. You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.

        Answer

         

        11.13.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

        (i) By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenylethanol can be synthesized.

        (ii) When chloromethylcyclohexane is treated with sodium hydroxide, cyclohexylmethanol is obtained.

        (iii) When 1-chloropentane is treated with NaOH, pentan-1-ol is produced.

        11.14. 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:

        (i) Phenol reacts with sodium to give sodium phenoxide, liberating H2.

        (ii) Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products.

        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.

        11.15.Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

        Answer

        The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O−H bond. As a result, it is easier to lose a proton. Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.

        On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the O−H bond and hence, the proton cannot be given out easily.

        For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

        11.16. 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. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.

        As a result, the benzene ring is activated towards electrophilic substitution.

        11.17. 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) 

        (ii)

        (iii)

        (iv)

         

        11.18. Explain the following with an example.

        (i) Kolbe’s reaction.

        (ii) Reimer-Tiemann reaction.

        (iii) Williamson ether synthesis.

        (iv) Unsymmetrical ether.

        Answer

        (i) 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.

        (ii) 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.

        (iii) Williamson ether synthesis:

        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.

        (iv) Unsymmetrical ether:

        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).

        11.19.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:

        Protonation of ethanol to form ethyl oxonium ion:

        Step 2:

        Formation of carbocation (rate determining step):

        Step 3:

        Elimination of a proton to form ethene:

        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.

        11.20. 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.

        (ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

        (iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.

        (iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.

        11.21. 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

        (ii) Pyridinium chlorochromate (PCC)

        (iii) Bromine water

        (iv) Acidified potassium permanganate

        (v) 85% phosphoric acid

        (vi) NaBH4 or LiAlH4

        11.22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

        Answer

        Ethanol undergoes intermolecular H-bonding due to the presence of −OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxymethane.

         

        11.23. Give IUPAC names of the following ethers:

        (i)

        (ii)

        (iii)

        (iv)

        (v)

        (vi)

        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

        11.24. 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) 

        (ii)

        (iii)

        (iv)

        11.25. 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.

        11.26. 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

        11.27. 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 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.

        11.28. Write the equation of the reaction of hydrogen iodide with:

        (i) 1-propoxypropane

        (ii) Methoxybenzene and

        (iii) Benzyl ethyl ether

        Answer

        (i)

        (ii)

        (iii)

        11.29. 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.

        (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.

        11.30. 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−:

        Step3:

        When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide

        11.31. 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)

        (ii)

        (iii)

        (iv)

        11.32. Show how would you synthesise the following alcohols from appropriate alkenes?

        (i)

        (ii)

        (iii)

        (iv)

        Answer

        The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed hydration of appropriate alkenes.

        (i)

        (ii)

        (iii)

        Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan-3-ol.

        Thus, the first reaction is preferred over the second one to get pentan-2-ol.

        (iv)

        11.33. 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

        The mechanism of the given reaction involves the following steps:

        Step 1: Protonation

        Step 2: Formation of 2° carbocation by the elimination of a water molecule

        Step 3: Re-arrangement by the hydride-ion shift

        Step 4: Nucleophilic attack

        Prev Chapter Notes – Alcohol & Ether
        Next Preparation

        Leave A Reply Cancel reply

        Your email address will not be published. Required fields are marked *

        All Courses

        • Backend
        • Chemistry
        • Chemistry
        • Chemistry
        • Class 08
          • Maths
          • Science
        • Class 09
          • Maths
          • Science
          • Social Studies
        • Class 10
          • Maths
          • Science
          • Social Studies
        • Class 11
          • Chemistry
          • English
          • Maths
          • Physics
        • Class 12
          • Chemistry
          • English
          • Maths
          • Physics
        • CSS
        • English
        • English
        • Frontend
        • General
        • IT & Software
        • JEE Foundation (Class 9 & 10)
          • Chemistry
          • Physics
        • Maths
        • Maths
        • Maths
        • Maths
        • Maths
        • Photography
        • Physics
        • Physics
        • Physics
        • Programming Language
        • Science
        • Science
        • Science
        • Social Studies
        • Social Studies
        • Technology

        Latest Courses

        Class 8 Science

        Class 8 Science

        ₹8,000.00
        Class 8 Maths

        Class 8 Maths

        ₹8,000.00
        Class 9 Science

        Class 9 Science

        ₹10,000.00

        Contact Us

        +91-8287971571

        contact@dronstudy.com

        Company

        • About Us
        • Contact
        • Privacy Policy

        Links

        • Courses
        • Test Series

        Copyright © 2021 DronStudy Pvt. Ltd.

        Login with your site account

        Lost your password?

        Modal title

        Message modal