CBSE · NCERT · Class 11 Chemistry · Chapter 9

NCERT Solutions: Class 11 Chemistry Chapter 9 - Hydrocarbons

21 textbook Q&A21 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Hydrocarbons, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 21
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1Exercises21 questions
Q.9.1How do you account for the formation of ethane during chlorination of methane ?v
Solution

Chlorination of methane proceeds by a free-radical chain mechanism. Chlorine radicals abstract hydrogen from methane to form methyl radicals: CH4 + Cl· → CH3· + HCl. Two methyl radicals may combine in a termination step: CH3· + CH3· → CH3CH3. This accounts for ethane formation.

Answer:

Ethane forms by coupling of methyl free radicals.

Q.9.3For 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)v
Solution

For C4H8 with one double bond, possible open-chain alkenes are CH2=CHCH2CH3, CH3CH=CHCH3 and CH2=C(CH3)2. For C5H8 with one triple bond, possible open-chain alkynes are HC≡CCH2CH2CH3, CH3C≡CCH2CH3 and HC≡CCH(CH3)2.

Answer:

(a) But-1-ene, but-2-ene and 2-methylpropene. (b) Pent-1-yne, pent-2-yne and 3-methylbut-1-yne.

Q.9.4Write 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-enev
Solution

Ozonolysis cleaves the C=C bond and converts each alkene carbon into a carbonyl carbon. Pent-2-ene gives ethanal and propanal. 3,4-Dimethylhept-3-ene gives CH3COCH2CH3 and CH3COCH2CH2CH3. 2-Ethylbut-1-ene gives methanal from the terminal carbon and pentan-3-one from the substituted carbon. 1-Phenylbut-1-ene gives benzaldehyde and propanal.

Answer:

(i) Ethanal and propanal. (ii) Butan-2-one and pentan-2-one. (iii) Methanal and pentan-3-one. (iv) Benzaldehyde and propanal.

Q.9.5An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.v
Solution

Rejoin the carbonyl carbons of ethanal (CH3CHO) and pentan-3-one (CH3CH2COCH2CH3) by replacing C=O bonds with a C=C bond. This gives CH3CH=C(CH2CH3)2. The longest chain containing the double bond has five carbons with an ethyl substituent at C-3, so the name is 3-ethylpent-2-ene.

Answer:

A is CH3CH=C(CH2CH3)2, named 3-ethylpent-2-ene.

Q.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’.v
Solution

The aldehyde of molar mass 44 u is ethanal, CH3CHO. Since ozonolysis gives two moles of ethanal, the alkene must be symmetrical: CH3CH=CHCH3. This alkene has three C-C sigma bonds, eight C-H sigma bonds and one π bond. Its IUPAC name is but-2-ene.

Answer:

But-2-ene.

Q.9.7Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?v
Solution

Propanal contributes an alkene carbon bearing H and ethyl, while pentan-3-one contributes an alkene carbon bearing two ethyl groups. Rejoining the carbonyl carbons gives CH3CH2CH=C(CH2CH3)2, whose IUPAC name is 3-ethylhex-3-ene.

Answer:

CH3CH2CH=C(CH2CH3)2.

Q.9.8Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluenev
Solution

Complete combustion converts all carbon to CO2 and hydrogen to H2O. Balancing C, H and then O gives the equations listed.

Answer:

(i) 2C4H10 + 13O2 → 8CO2 + 10H2O; (ii) 2C5H10 + 15O2 → 10CO2 + 10H2O; (iii) 2C6H10 + 17O2 → 12CO2 + 10H2O; (iv) C7H8 + 9O2 → 7CO2 + 4H2O.

Q.9.10Why is benzene extra ordinarily stable though it contains three double bonds?v
Solution

Benzene is planar, cyclic and fully conjugated with six π electrons, satisfying Huckel's 4n+2 rule for n = 1. The π electrons are delocalised over the whole ring, giving resonance stabilisation. Therefore benzene is much more stable than a normal triene with three isolated double bonds.

Answer:

Because benzene is aromatic and has delocalised π electrons.

Q.9.11What are the necessary conditions for any system to be aromatic?v
Solution

For aromaticity, the system must form a ring, all ring atoms must have p orbitals allowing continuous overlap, the ring must be planar enough for delocalisation, and the delocalised π electron count must satisfy Huckel's rule, 4n+2 where n is 0, 1, 2 and so on.

Answer:

It must be cyclic, planar, fully conjugated and contain (4n+2) π electrons.

Q.9.13How will you convert benzene into (i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii) p - nitrotoluene (iv) acetophenone?v
Solution

(i) Benzene + Br2/FeBr3 → bromobenzene; nitration with HNO3/H2SO4 gives mainly o- and p-nitrobromobenzene, from which the para isomer is separated. (ii) Benzene → nitrobenzene by nitration; NO2 is meta-directing, so chlorination with Cl2/FeCl3 gives m-nitrochlorobenzene. (iii) Benzene + CH3Cl/AlCl3 → toluene; nitration gives chiefly o- and p-nitrotoluene, from which the para isomer is separated. (iv) Benzene + CH3COCl/AlCl3 → acetophenone.

Answer:

(i) Brominate, then nitrate and isolate para product. (ii) Nitrate, then chlorinate. (iii) Methylate, then nitrate and isolate para product. (iv) Friedel-Crafts acylation with acetyl chloride.

Q.9.14In 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.v
Solution

A primary carbon is attached to one carbon, secondary to two, tertiary to three and quaternary to four. The five methyl carbons are primary and carry 3 H each. The two -CH2- carbons are secondary and carry 2 H each. The -CH(CH3)2 carbon is tertiary and carries 1 H. The central C attached to two methyl groups and two chain carbons is quaternary and carries no hydrogen.

Answer:

There are five 1° carbons, each bonded to 3 H; two 2° carbons, each bonded to 2 H; one 3° carbon, bonded to 1 H. The central C in C(CH3)2 is 4° and has no H.

Q.9.15What effect does branching of an alkane chain has on its boiling point?v
Solution

Branching makes an alkane molecule more compact and decreases its surface area. Weaker London dispersion forces then operate between molecules, so the boiling point decreases as branching increases among isomeric alkanes.

Answer:

Branching lowers the boiling point.

Q.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.v
Solution

Without peroxide, propene is protonated to form the more stable secondary carbocation, CH3-CH+-CH3, which is attacked by Br- to give 2-bromopropane. In the presence of benzoyl peroxide, Br radicals form. Br· adds to propene so that the more stable secondary carbon radical is produced: Br· + CH3CH=CH2 → CH3-CH·-CH2Br. This radical abstracts H from HBr to give CH3CH2CH2Br and regenerate Br·. Hence the product is 1-bromopropane.

Answer:

Without peroxide, Markovnikov electrophilic addition gives 2-bromopropane. With peroxide, free-radical anti-Markovnikov addition gives 1-bromopropane.

Q.9.18Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.v
Solution

Acidity of C-H bonds increases with the s-character of the carbon orbital holding the bonded electron pair: sp > sp2 > sp3. Ethyne has sp C-H bonds, benzene has sp2 C-H bonds and n-hexane has sp3 C-H bonds. Therefore the decreasing acidity order is ethyne, benzene, n-hexane.

Answer:

Ethyne > benzene > n-hexane.

Q.9.19Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?v
Solution

The π electron cloud of benzene attracts electrophiles, so electrophilic attack occurs readily. After substitution, loss of H+ restores aromaticity. Nucleophiles are repelled by the electron-rich ring, and nucleophilic substitution would require difficult disruption of aromatic stabilisation or a suitable leaving group/electron-withdrawing activation.

Answer:

Benzene is electron-rich and aromatic; electrophilic substitution preserves aromaticity, while nucleophilic substitution is difficult.

Q.9.20How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexanev
Solution

(i) 3HC≡CH → C6H6 on passing ethyne through a red-hot iron tube. (ii) CH2=CH2 + Br2 → BrCH2CH2Br; on treatment with alcoholic KOH, the vicinal dibromide gives ethyne, which trimerises to benzene. (iii) n-Hexane undergoes dehydrocyclisation/aromatisation over catalysts such as Cr2O3/V2O5/Mo2O3 at high temperature to give benzene.

Answer:

(i) Trimerise ethyne. (ii) Convert ethene to ethyne, then trimerise. (iii) Aromatise hexane.

Q.9.21Write structures of all the alkenes which on hydrogenation give 2-methylbutane.v
Solution

Hydrogenation keeps the carbon skeleton unchanged and converts C=C to C-C. For the 2-methylbutane skeleton, placing one double bond at non-equivalent adjacent carbon positions gives CH2=C(CH3)CH2CH3, CH2=CHCH(CH3)CH3 and (CH3)2C=CHCH3.

Answer:

2-methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene.

Q.9.22Arrange 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-H3C-C6H4-NO2, p-O2N-C6H4-NO2.v
Solution

Electron-withdrawing nitro groups deactivate the ring toward electrophilic substitution; more nitro groups mean lower reactivity. Alkyl groups activate the ring, so toluene is most reactive in (b). A nitro group reduces reactivity, and two nitro groups reduce it still further.

Answer:

(a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. (b) Toluene > p-nitrotoluene > p-dinitrobenzene.

Q.9.23Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?v
Solution

The methyl group in toluene activates the benzene ring through +I effect and hyperconjugation, increasing electron density and making electrophilic attack easier. Benzene is less reactive than toluene. Nitro groups are strongly deactivating, so m-dinitrobenzene is least reactive.

Answer:

Toluene undergoes nitration most easily.

Q.9.24Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.v
Solution

Friedel-Crafts alkylation needs a Lewis acid to generate the electrophile from an alkyl halide. Besides anhydrous AlCl3, Lewis acids such as FeCl3, BF3 or SnCl4 can be used; FeCl3 is one suitable example.

Answer:

Anhydrous ferric chloride, FeCl3, can be used.

Q.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.v
Solution

Wurtz reaction couples alkyl halides with sodium. A single alkyl halide gives an even-carbon symmetrical alkane. To prepare an odd-carbon alkane such as propane, one would use CH3Br and C2H5Br, but coupling gives ethane (CH3-CH3), propane (CH3-CH2CH3) and butane (C2H5-C2H5). Because the desired product is mixed with other alkanes, Wurtz reaction is not preferred for odd-carbon alkanes.

Answer:

Because preparing an odd-carbon alkane requires two different alkyl halides, which gives a mixture of products.