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Hydrocarbons

CBSE · Class 11 · Chemistry

NCERT Solutions for Hydrocarbons — CBSE Class 11 Chemistry.

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EXERCISES

9.1How do you account for the formation of ethane during chlorination of methane?Show solution
During chlorination of methane, free radicals are formed. In the termination step, two methyl radicals can combine:

CH˙3+CH˙3→CH3−CH3\dot{\mathrm{CH}}_3 + \dot{\mathrm{CH}}_3 \rightarrow \mathrm{CH}_3-\mathrm{CH}_3

So ethane is formed as a by-product during the radical chain reaction.

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9.4(i)Pent-2-eneShow solution
Pent-2-ene has structure CH3−CH=CH−CH2−CH3\mathrm{CH_3-CH=CH-CH_2-CH_3}. The longest chain has 5 carbon atoms and the double bond gets the lowest possible number, so it is at carbon 2.

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9.4(ii)3,4-Dimethylhept-3-eneShow solution
The given compound is named by selecting the longest chain containing the double bond. The parent chain has 7 carbons and the double bond is at carbon 3. There are methyl groups at carbons 3 and 4, so the name is 3,4-dimethylhept-3-ene.

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9.4(iii)2-Ethylbut-1-eneShow solution
For this structure, the longest chain containing the double bond has 4 carbon atoms, so the parent is but-1-ene. There is an ethyl substituent at carbon 2. Hence the name is 2-ethylbut-1-ene.

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9.4(iv)1-Phenylbut-1-eneShow solution
In the structure, the longest chain containing the double bond is but-1-ene. A phenyl group is attached to carbon 1. Therefore the IUPAC name is 1-phenylbut-1-ene.

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9.5An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.Show solution
On ozonolysis, the alkene gives ethanal and pentan-3-one.

- Ethanal (CH3CHO\mathrm{CH_3CHO}) means one side of the double bond had **CH3_3 and H.
- Pentan-3-one** (CH3CH2COCH2CH3\mathrm{CH_3CH_2COCH_2CH_3}) means the other carbon of the double bond had two ethyl groups.

So the alkene is:

CH3CH=C(CH2CH3)2\mathrm{CH_3CH=C(CH_2CH_3)_2}

Its IUPAC name is 3-ethylhex-2-ene.

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9.6An 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 IUPAC name of 'A'.Show solution
An aldehyde of molar mass 44 u is ethanal (CH3CHO\mathrm{CH_3CHO}). If ozonolysis gives two moles of ethanal, the alkene must be but-2-ene:

CH3−CH=CH−CH3\mathrm{CH_3-CH=CH-CH_3}

This fits the bond count too: three C-C bonds, eight C-H sigma bonds and one C=C pi bond.

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9.7Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?Show solution
The ozonolysis products are propanal (CH3CH2CHO\mathrm{CH_3CH_2CHO}) and pentan-3-one (CH3CH2COCH2CH3\mathrm{CH_3CH_2COCH_2CH_3}).

So the alkene must be:

CH3CH2CH=C(CH2CH3)2\mathrm{CH_3CH_2CH=C(CH_2CH_3)_2}

The longest chain containing the double bond has 6 carbon atoms, and the substituent arrangement gives the IUPAC name 3-ethylhex-3-ene.

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9.8Write chemical equations for combustion reaction of the following hydrocarbons:Show solution
For combustion, hydrocarbons react with oxygen to give carbon dioxide and water.

(i) Butane

C4H10+132O2→4CO2+5H2O\mathrm{C_4H_{10} + \tfrac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O}

(ii) Pentene

C5H10+152O2→5CO2+5H2O\mathrm{C_5H_{10} + \tfrac{15}{2}O_2 \rightarrow 5CO_2 + 5H_2O}

(iii) Hexyne

C6H10+172O2→6CO2+5H2O\mathrm{C_6H_{10} + \tfrac{17}{2}O_2 \rightarrow 6CO_2 + 5H_2O}

(iv) Toluene

C7H8+9O2→7CO2+4H2O\mathrm{C_7H_8 + 9O_2 \rightarrow 7CO_2 + 4H_2O}

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9.9Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?Show solution
The two geometrical isomers of hex-2-ene are:

- cis-hex-2-ene: the two larger groups are on the same side of the double bond.
- trans-hex-2-ene: the two larger groups are on opposite sides of the double bond.

The cis isomer has the higher boiling point because it is more polar than the trans form, so intermolecular attraction is greater.

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9.10Why is benzene extra ordinarily stable though it contains three double bonds?Show solution
Benzene is extraordinarily stable because its **six π\pi electrons are delocalised over the entire ring. All six carbon atoms are sp2sp^2 hybridised**, and the pp orbitals overlap to form a single continuous **c0c0-electron cloud above and below the ring.

This resonance stabilisation** makes benzene much more stable than a normal compound with three isolated double bonds.

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9.11What are the necessary conditions for any system to be aromatic?Show solution
For a system to be aromatic, it must have:

1. Planarity
2. **Complete delocalisation of c0c0 electrons in the ring
3. 4n+24n+2 c0c0 electrons** in the ring, where n=0,1,2,…n=0,1,2,\dots

This is called Hückel's rule.

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9.12Explain why the following systems are not aromatic?Show solution
The given systems are not aromatic because they do not satisfy one or more of the conditions for aromaticity. In general, a system must be planar, have **complete delocalisation of c0c0 electrons, and obey the 4n+24n+2 rule. The shown systems fail to meet these requirements, so they are not aromatic**.

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9.13How will you convert benzene into
9.15What effect does branching of an alkane chain has on its boiling point?
9.16Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
9.17Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
9.18Arrange benzene, nn-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
9.19Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
9.20How would you convert the following compounds into benzene?
9.21Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
9.22(a)Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+\mathrm{E}^+
9.22(b)Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+\mathrm{E}^+
9.23Out of benzene, mm-dinitrobenzene and toluene which will undergo nitration most easily and why?
9.24Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
9.25Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

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Frequently Asked Questions

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Understand the core concepts first, then work through the 76 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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