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Solutions

Overview

Step-by-step NCERT solutions for Hydrocarbons (Chapter 9, NCERT Class 11 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the Hydrocarbons textbook chapter.

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What these solutions cover

All 25 questions in Hydrocarbons are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

  1. How do you account for the formation of ethane during chlorination of methane?
  2. Write IUPAC names of the following compounds:
    • (a) CH₃CH=C(CH₃)₂
    • (b) CH₂=CH-C≡C-CH₃
    • (c) CH₂=CH-CH=CH₂ (buta-1,3-diene bond-line structure)
    • (d) benzene ring bearing a -CH₂-CH₂-CH=CH₂ substituent
    • (e) benzene ring with -CH₃ and -OH on adjacent (ortho) carbons
    • (f) CH₃(CH₂)₄CH[CH₂CH(CH₃)₂](CH₂)₃CH₃
    • (g) CH₃-CH=CH-CH₂-CH=CH-CH(C₂H₅)-CH₂-CH=CH₂
  3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated:
    • (a) C₄H₈ (one double bond)
    • (b) C₅H₈ (one triple bond)
  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
  5. An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.
  6. An alkene A contains three C-C sigma bonds, eight C-H bonds and one C=C pi bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of 'A'.
  7. Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene?
  8. Write chemical equations for combustion reaction of the following hydrocarbons:
    • (i) Butane
    • (ii) Pentene
    • (iii) Hexyne
    • (iv) Toluene
  9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
  10. Why is benzene extraordinarily stable though it contains three double bonds?
  11. What are the necessary conditions for any system to be aromatic?
  12. Explain why the following systems are not aromatic:
    • (i) Cyclobutadiene
    • (ii) A ring compound with an sp³ carbon (CH₂ group in the ring)
    • (iii) A bicyclic non-planar system
  13. How will you convert benzene into
    • (i) p-nitrobromobenzene
    • (ii) m-nitrochlorobenzene
    • (iii) p-nitrotoluene
    • (iv) acetophenone?
  14. In the alkane H₃C-CH₂-C(CH₃)₂-CH₂-CH(CH₃)₂, identify primary, secondary, tertiary and quaternary carbon atoms and give the number of H atoms bonded to each one.
  15. What effect does branching of an alkane chain have on its boiling point?
  16. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
  17. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
  18. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
  19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
  20. How would you convert the following compounds into benzene?
    • (i) Ethyne
    • (ii) Ethene
    • (iii) Hexane
  21. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
  22. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile E⁺:
    • (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
    • (b) Toluene, p-nitrotoluene, p-dinitrobenzene
  23. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
  24. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
  25. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Sample

Worked examples

3 questions from Hydrocarbons, solved step by step.

Exercises

How do you account for the formation of ethane during chlorination of methane?

Chlorination of methane proceeds by a free radical chain mechanism with three stages: initiation, propagation, and termination.

Initiation: Cl-Cl bond undergoes homolysis in the presence of UV light to form two chlorine free radicals.

Cl₂ --(hv)-→ 2 Cl·

Propagation: Cl· abstracts a hydrogen from CH₄ to form a methyl free radical (CH₃·).

CH₄ + Cl· → CH₃· + HCl

CH₃· + Cl₂ → CH₃Cl + Cl·

Termination: Two methyl free radicals combine with each other in one of the chain-terminating steps.

CH₃· + CH₃· → CH₃-CH₃ (ethane)

This coupling of two methyl radicals is a termination step that consumes the radicals and produces ethane as a by-product. Although ethane is formed in small amounts, it is an inevitable product of the termination stage of the free radical mechanism.

Answer: Ethane forms in the termination step of the free radical chain mechanism: two methyl radicals (CH₃·) combine to give CH₃-CH₃.

Exercises

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (a) C₄H₈ (one double bond) (b) C₅H₈ (one triple bond)

(a) C₄H₈ with one double bond (alkene isomers):

1. But-1-ene: CH₂=CH-CH₂-CH₃

2. But-2-ene: CH₃-CH=CH-CH₃ (exists as cis and trans geometrical isomers)

3. 2-methylpropene: CH₂=C(CH₃)-CH₃

These are all the possible open-chain structural isomers of C₄H₈ with one C=C double bond.

(b) C₅H₈ with one triple bond (alkyne isomers):

Degree of unsaturation of C₅H₈ = (2×5+2-8)/2 = 2; one triple bond accounts for both units, so no rings or double bonds.

1. Pent-1-yne: CH≡C-CH₂-CH₂-CH₃

2. Pent-2-yne: CH₃-C≡C-CH₂-CH₃

3. 3-methylbut-1-yne: CH≡C-CH(CH₃)-CH₃

Answer: C₄H₈ (one double bond): but-1-ene, but-2-ene, 2-methylpropene. C₅H₈ (one triple bond): pent-1-yne, pent-2-yne, 3-methylbut-1-yne.

Exercises

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

Ozonolysis (O₃/Zn-H₂O) cleaves C=C double bond; each carbon of the double bond becomes a carbonyl group (aldehyde if it bore one H, ketone if it bore no H).

(i) Pent-2-ene: CH₃-CH=CH-CH₂-CH₃

Cleavage at C₂=C₃: CH₃CHO + CH₃CH₂CHO

Products: ethanal (CH₃CHO) + propanal (CH₃CH₂CHO)

(ii) 3,4-Dimethylhept-3-ene: CH₃CH₂-C(CH₃)=C(CH₃)-CH₂CH₂CH₃

Cleavage at C₃=C₄: CH₃CH₂-CO-CH₃ + CH₃-CO-CH₂CH₂CH₃

Products: butan-2-one (CH₃CH₂COCH₃) + pentan-2-one (CH₃COCH₂CH₂CH₃)

(iii) 2-Ethylbut-1-ene: CH₂=C(CH₂CH₃)-CH₂CH₃

Cleavage at C₁=C₂: HCHO + (CH₃CH₂)CO(CH₂CH₃)

Products: methanal (HCHO, formaldehyde) + pentan-3-one (CH₃CH₂-CO-CH₂CH₃)

(iv) 1-Phenylbut-1-ene: C₆H₅-CH=CH-CH₂CH₃

Cleavage at C₁=C₂: C₆H₅-CHO + CH₃CH₂-CHO

Products: benzaldehyde (C₆H₅CHO) + propanal (CH₃CH₂CHO)

Answer: (i) ethanal + propanal (ii) butan-2-one + pentan-2-one (iii) methanal + pentan-3-one (iv) benzaldehyde + propanal

All 25 questions in Hydrocarbons are solved step by step in the Solutions PDF above.

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