{"id":866665,"date":"2025-08-16T15:18:29","date_gmt":"2025-08-16T09:48:29","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866665"},"modified":"2025-08-16T15:18:29","modified_gmt":"2025-08-16T09:48:29","slug":"ncert-solutions-class-11-chemistry-part-ii-chapter-9-hydrocarbons-free-pdf","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-9-hydrocarbons-free-pdf\/","title":{"rendered":"NCERT Solutions Class 11 Chemistry (Part-II) Chapter 9: Hydrocarbons (Free PDF)"},"content":{"rendered":"\n

NCERT Solutions for Class 11 Chemistry Chapter 9: Hydrocarbons (Free PDF) are a valuable resource for students preparing for school exams and competitive exams. This chapter covers different types of hydrocarbons, such as alkanes, alkenes, alkynes, and aromatic hydrocarbons, along with their nomenclature, properties, reactions, and practical applications. The step-by-step NCERT solutions make revising key concepts easier, strengthen problem-solving skills, and ensure better exam preparation.<\/p>\n\n\n\n\n\n\n

Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n

Chapter 1<\/a><\/strong><\/td>Chapter 2<\/a><\/strong><\/td>Chapter 3<\/a><\/strong><\/td>Chapter 4<\/strong><\/a><\/td>Chapter 5<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

NCERT Solutions Class 11 Chemistry (Part-2) Chapter 9: Hydrocarbons<\/h2>\n\n\n\n

Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-2) Chapter 9: Hydrocarbons.<\/p>\n\n\n\n

Exercises<\/h2>\n\n\n\n
    \n
  1. How do you account for the formation of ethane during the chlorination of methane?<\/li>\n\n\n\n
  2. Write IUPAC names of the following compounds:
    (a) CH\u2083CH=C(CH\u2083)\u2082
    (b) CH\u2082=CH-C\u2261C-CH\u2083
    (c) CH\u2082-CH(CH\u2083)\u2082
    (d) \u2013CH\u2082\u2013CH\u2082\u2013CH=CH\u2082
    (e) CH\u2083(CH\u2082)\u2084CH(CH\u2082)\u2083CH\u2083
    (f) CH\u2082\u2013CH(CH\u2083)\u2082
    (g) CH\u2083\u2013CH=CH\u2013CH\u2082\u2013CH=CH\u2013CH\u2013CH\u2082\u2013CH=CH\u2082
    \u2003\u2003|
    \u2003\u2003C\u2082H\u2085<\/li>\n\n\n\n
  3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bonds as indicated:
    (a) C\u2084H\u2088 (one double bond)
    (b) C\u2085H\u2088 (one triple bond)<\/li>\n\n\n\n
  4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
    (i) Pent-2-ene
    (ii) 3,4-Dimethylhept-3-ene
    (iii) 2-Ethylbut-1-ene
    (iv) 1-Phenylbut-1-ene<\/li>\n\n\n\n
  5. An alkene \u2018A\u2019 on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and the IUPAC name of \u2018A\u2019.<\/li>\n\n\n\n
  6. An alkene \u2018A\u2019 contains three C\u2013C, eight C\u2013H \u03c3 bonds, and one C\u2013C \u03c0 bond. \u2018A\u2019 on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of \u2018A\u2019.<\/li>\n\n\n\n
  7. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?<\/li>\n\n\n\n
  8. Write chemical equations for the combustion reaction of the following hydrocarbons:
    (i) Butane
    (ii) Pentene
    (iii) Hexyne
    (iv) Toluene<\/li>\n\n\n\n
  9. Draw the cis and trans structures of hex-2-ene. Which isomer will have a higher b.p. and why?<\/li>\n\n\n\n
  10. Why is benzene extraordinarily stable, though it contains three double bonds?<\/li>\n\n\n\n
  11. What are the necessary conditions for any system to be aromatic?<\/li>\n\n\n\n
  12. How will you convert benzene into:
    (i) p-nitrobromobenzene
    (ii) m-nitrochlorobenzene
    (iii) p-nitrotoluene
    (iv) acetophenone?<\/li>\n\n\n\n
  13. In the alkane H\u2083C\u2013CH\u2082\u2013C(CH\u2083)\u2082\u2013CH\u2082\u2013CH(CH\u2083)\u2082, identify 1\u00b0, 2\u00b0, 3\u00b0 carbon atoms and give the number of H atoms bonded to each one of these.<\/li>\n\n\n\n
  14. What effect does branching of an alkane chain have on its boiling point?<\/li>\n\n\n\n
  15. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give a mechanism.<\/li>\n\n\n\n
  16. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support the Kekul\u00e9 structure for benzene?<\/li>\n\n\n\n
  17. Arrange benzene, n-hexane, and ethyne in decreasing order of acidic behaviour. Also, give a reason for this behaviour.<\/li>\n\n\n\n
  18. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?<\/li>\n\n\n\n
  19. How would you convert the following compounds into benzene?
    (i) Ethyne
    (ii) Ethene
    (iii) Hexane<\/li>\n\n\n\n
  20. Write structures of all the alkenes that, on hydrogenation, give 2-methylbutane.<\/li>\n\n\n\n
  21. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E\u207a:
    (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
    (b) Toluene, p-CH\u2083\u2013C\u2086H\u2084\u2013NO\u2082, p-O\u2082N\u2013C\u2086H\u2084\u2013NO\u2082
    <\/li>\n\n\n\n
  22. Out of benzene, m-dinitrobenzene, and toluene, which will undergo nitration most easily, and why?<\/li>\n\n\n\n
  23. Suggest the name of a Lewis acid other than anhydrous aluminium chloride that can be used during the ethylation of benzene.<\/li>\n\n\n\n
  24. Why is the Wurtz reaction not preferred for the preparation of alkanes containing an odd number of carbon atoms? Illustrate your answer by taking one example.<\/li>\n<\/ol>\n\n\n\n

    Also Read: <\/strong>NCERT Class 11 Sociology Chapter 2: Terms, Concepts, and Their Use in Sociology Notes (Free PDF)<\/strong><\/a><\/p>\n\n\n\n

    Solutions<\/h2>\n\n\n\n
      \n
    1. Ethane is formed during the chlorination of methane because two methyl free radicals, produced in the chain reaction, combine:
      CH\u2083\u2022 + CH\u2083\u2022 \u2192 C\u2082H\u2086<\/li>\n\n\n\n
    2. IUPAC names:<\/strong>
      <\/strong> (a) 2-Methylbut-2-ene
      (b) Pent-1-en-3-yne
      (c) 2-Methylprop-1-ene
      (d) But-1-ene
      (e) 2-Methyloctane
      (f) 2-Methylpropane
      (g) 5-Ethyl-deca-1,3,5,8-tetraene<\/li>\n\n\n\n
    3. Isomers:<\/strong>
      <\/strong> (a) C\u2084H\u2088 (one double bond) \u2192 But-1-ene, But-2-ene, 2-Methylprop-1-ene
      (b) C\u2085H\u2088 (one triple bond) \u2192 Pent-1-yne, Pent-2-yne, 3-Methylbut-1-yne<\/li>\n\n\n\n
    4. Ozonolysis products:<\/strong>
      <\/strong> (i) Pent-2-ene \u2192 2 moles of ethanal
      (ii) 3,4-Dimethylhept-3-ene \u2192 2,3-Dimethylbutan-2-one + Ethanal
      (iii) 2-Ethylbut-1-ene \u2192 Propanal + Ethanal
      (iv) 1-Phenylbut-1-ene \u2192 Benzaldehyde + Propanal<\/li>\n\n\n\n
    5. Ethanal + Pentan-3-one \u2192 Alkene is Hex-3-ene.
      Structure: CH\u2083\u2013CH\u2082\u2013CH=CH\u2013CH\u2082\u2013CH\u2083<\/li>\n\n\n\n
    6. Two molecules of ethanal (M = 44 u) form \u2192 Alkene is But-2-ene.<\/li>\n\n\n\n
    7. Propanal + Pentan-3-one \u2192 Alkene is Hex-3-ene.<\/li>\n\n\n\n
    8. Combustion reactions:<\/strong>
      <\/strong> (i) 2C\u2084H\u2081\u2080 + 13O\u2082 \u2192 8CO\u2082 + 10H\u2082O
      (ii) C\u2085H\u2081\u2080 + 7.5O\u2082 \u2192 5CO\u2082 + 5H\u2082O
      (iii) 2C\u2086H\u2081\u2080 + 17O\u2082 \u2192 12CO\u2082 + 10H\u2082O
      (iv) C\u2087H\u2088 + 9O\u2082 \u2192 7CO\u2082 + 4H\u2082O<\/li>\n\n\n\n
    9. Cis-hex-2-ene \u2192 bulky groups on the same side<\/li>\n<\/ol>\n\n\n\n

      Trans-hex-2-ene \u2192 bulky groups on opposite sides
      Trans-isomer has a higher boiling point<\/strong> due to better packing and higher intermolecular forces.<\/p>\n\n\n\n

        \n
      1. Benzene is extraordinarily stable due to resonance and delocalisation of \u03c0-electrons<\/strong> (aromatic stabilization).<\/li>\n\n\n\n
      2. Conditions for aromaticity:<\/li>\n<\/ol>\n\n\n\n