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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.
Contents
Explore Notes of Class 11 Chemistry
NCERT Solutions Class 11 Chemistry (Part-2) Chapter 9: Hydrocarbons
Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-2) Chapter 9: Hydrocarbons.
Exercises
- How do you account for the formation of ethane during the chlorination of methane?
- Write IUPAC names of the following compounds:
(a) CH₃CH=C(CH₃)₂
(b) CH₂=CH-C≡C-CH₃
(c) CH₂-CH(CH₃)₂
(d) –CH₂–CH₂–CH=CH₂
(e) CH₃(CH₂)₄CH(CH₂)₃CH₃
(f) CH₂–CH(CH₃)₂
(g) CH₃–CH=CH–CH₂–CH=CH–CH–CH₂–CH=CH₂
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C₂H₅ - 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) C₄H₈ (one double bond)
(b) C₅H₈ (one triple bond) - 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 - An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and the IUPAC name of ‘A’.
- An alkene ‘A’ contains three C–C, eight C–H σ bonds, and one C–C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of ‘A’.
- Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
- Write chemical equations for the combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene - Draw the cis and trans structures of hex-2-ene. Which isomer will have a higher b.p. and why?
- Why is benzene extraordinarily stable, though it contains three double bonds?
- What are the necessary conditions for any system to be aromatic?
- How will you convert benzene into:
(i) p-nitrobromobenzene
(ii) m-nitrochlorobenzene
(iii) p-nitrotoluene
(iv) acetophenone? - In the alkane H₃C–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂, identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.
- What effect does branching of an alkane chain have on its boiling point?
- 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.
- Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support the Kekulé structure for benzene?
- Arrange benzene, n-hexane, and ethyne in decreasing order of acidic behaviour. Also, give a reason for this behaviour.
- Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
- How would you convert the following compounds into benzene?
(i) Ethyne
(ii) Ethene
(iii) Hexane - Write structures of all the alkenes that, on hydrogenation, give 2-methylbutane.
- 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-CH₃–C₆H₄–NO₂, p-O₂N–C₆H₄–NO₂ - Out of benzene, m-dinitrobenzene, and toluene, which will undergo nitration most easily, and why?
- Suggest the name of a Lewis acid other than anhydrous aluminium chloride that can be used during the ethylation of benzene.
- 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.
Also Read: NCERT Class 11 Sociology Chapter 2: Terms, Concepts, and Their Use in Sociology Notes (Free PDF)
Solutions
- Ethane is formed during the chlorination of methane because two methyl free radicals, produced in the chain reaction, combine:
CH₃• + CH₃• → C₂H₆ - IUPAC names:
(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 - Isomers:
(a) C₄H₈ (one double bond) → But-1-ene, But-2-ene, 2-Methylprop-1-ene
(b) C₅H₈ (one triple bond) → Pent-1-yne, Pent-2-yne, 3-Methylbut-1-yne - Ozonolysis products:
(i) Pent-2-ene → 2 moles of ethanal
(ii) 3,4-Dimethylhept-3-ene → 2,3-Dimethylbutan-2-one + Ethanal
(iii) 2-Ethylbut-1-ene → Propanal + Ethanal
(iv) 1-Phenylbut-1-ene → Benzaldehyde + Propanal - Ethanal + Pentan-3-one → Alkene is Hex-3-ene.
Structure: CH₃–CH₂–CH=CH–CH₂–CH₃ - Two molecules of ethanal (M = 44 u) form → Alkene is But-2-ene.
- Propanal + Pentan-3-one → Alkene is Hex-3-ene.
- Combustion reactions:
(i) 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
(ii) C₅H₁₀ + 7.5O₂ → 5CO₂ + 5H₂O
(iii) 2C₆H₁₀ + 17O₂ → 12CO₂ + 10H₂O
(iv) C₇H₈ + 9O₂ → 7CO₂ + 4H₂O - Cis-hex-2-ene → bulky groups on the same side
Trans-hex-2-ene → bulky groups on opposite sides
Trans-isomer has a higher boiling point due to better packing and higher intermolecular forces.
- Benzene is extraordinarily stable due to resonance and delocalisation of π-electrons (aromatic stabilization).
- Conditions for aromaticity:
- Planarity
- Complete delocalisation of π-electrons
- (4n + 2) π electrons (Hückel rule)
- Conversions:
(i) Benzene → Nitrobenzene → Bromination → p-Nitrobromobenzene
(ii) Benzene → Nitrobenzene → Chlorination → m-Nitrochlorobenzene
(iii) Benzene → Toluene → Nitration → p-Nitrotoluene
(iv) Benzene → Acylation with CH₃COCl/AlCl₃ → Acetophenone - H₃C–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂
- 1° carbons: 5 (with 11 H atoms)
- 2° carbons: 2 (with 2 H atoms)
- 3° carbon: 1 (with 1 H atom)
- Branching decreases the boiling point due to a decrease in surface area and van der Waals forces.
- Without peroxide: Markovnikov addition → 2-Bromopropane
With peroxide: Anti-Markovnikov (free radical mechanism) → 1-Bromopropane
- Ozonolysis of o-xylene → Glyoxal + Phthalaldehyde → supports Kekulé structure due to delocalisation.
- Order: C₂H₂ > C₆H₆ > C₆H₁₄
Reason: Greater % of s-character → more electronegativity → more acidic. - Benzene undergoes electrophilic substitution easily due to high electron density (π-cloud). Nucleophilic substitution is difficult due to the stability of delocalised electrons.
- Conversions:
(i) Ethyne → Benzene (trimerisation)
(ii) Ethene → Benzene (cyclic polymerisation)
(iii) Hexane → Benzene (aromatisation at 773 K, 10–20 atm, Pt catalyst) - Alkenes give 2-Methylbutane on hydrogenation:
- 2-Methylbut-1-ene
- 2-Methylbut-2-ene
- 3-Methylbut-1-ene
- Reactivity order:
(a) Chlorobenzene > p-Nitrochlorobenzene > 2,4-Dinitrochlorobenzene
(b) Toluene > p-CH₃–C₆H₄–NO₂ > p-O₂N–C₆H₄–NO₂ - Toluene > Benzene > m-Dinitrobenzene
Reason: –CH₃ is activating, –NO₂ is deactivating. - Ferric chloride (FeCl₃) can also be used as a Lewis acid in ethylation.
- The Wurtz reaction is not preferred for odd carbon alkanes because a mixture of products forms. Example: CH₃Br + C₂H₅Br + 2Na → C₃H₈ + C₂H₆ + C₄H₁₀
Also Read: NCERT Class 11 Sociology Chapter 2 Terms, Concepts, and Their Use in Sociology Solutions (Free PDF)
Download NCERT Solutions Class 11 Chemistry (Part-2) Chapter 9: Hydrocarbons
Download the Solutions of Other Chapters of Class 11 Chemistry
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