NCERT Solutions for Class 11 Chemistry Chapter 13: Hydrocarbons

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

    Ans. The chlorination of methane follows free radical mechanism. The whole reaction takes place in three steps as follows:

    Mechanism :
    Step I. Initiation : In the initiation step, the homolytic cleavage of chlorine molecule into chlorine radical takes place.

    $$\text{Cl}—\text{Cl} \xrightarrow{\text{Homolytic fission}} 2\text{Cl}\\ \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space Chlorine free radical}$$

    Step II. Propogation : In this step, chlorine free radical react with methane molecule and break down the C—H bond thereby forming methyl radicals.

    These methyl radicals further react with chlorine molecules giving methyl chloride and chlorine free radicals.

    $$ \text{CH}_3 — \text{Cl} — \text{Cl}   \xrightarrow{} \text{CH}_3\text{Cl} + \text{Cl} $$

    Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and CH3Cl are the major products formed, other higher halogenated compounds are also formed as:

    $$ \text{CH}_3\text{Cl}  + \text{Cl}   \xrightarrow{} \text{CH}_2\text{Cl} + \text{HCl} $$

    $$\text{CH}_2\text{Cl} + \text{Cl} — \text{Cl} \xrightarrow{} \text{CH}_2\text{Cl}_2 + \text{Cl}$$

    Step III. Termination : The chain reaction finally terminates giving ethane and chlorine as a result of the consumption of reactants.

    $$\text{CH}_3 +\text{ CH}_3 \xrightarrow{} \text{CH}_3 —  \text{CH}_3 \\ \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space Ethane}$$

    $$\text{CH}_3  + \text{Cl} \xrightarrow{} \text{CH}_3 — \text{Cl};$$

    $$ \text{Cl} + \text{Cl} \xrightarrow{} \text{Cl} — \text{Cl}$$

    In the chain termination step, the two CH3 free radicals combine together to form an ethane (CH3 – CH3) molecule. Hence, by free radical mechanism, ethane is obtained as the by- product of chlorination of methane.

    13.2. Write IUPAC names of the following compounds:

    (a)  CH3CH = C(CH3)2

    (b) CH2 = CH – C ≡ C – CH3





    IUPAC Name : 2-Methylbut-2-ene

    (b) $$ ^1 \text{CH}_2 = ^2 \text{CH} — ^3 \text{C} ≡ ^4 \text{C} — ^5 \text{CH}_3 \\ \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space Pent-1-ene-3-yne}$$

    IUPAC Name : Pen-1-ene-3-yne



    $$ \text{H}_2 \space ^1\text{C} = ^2\text{CH} — ^3\text{CH} = ^4\text{CH}_2$$

    IUPAC Name : 1, 3-Butadiene or Buta-1,3-diene






    IUPAC Name : 5-(2-Methylpropyl)-decane


    13.3. 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)

    Ans. Isomers of hydrocarbons are compounds having the same molecular formula but different structural formulas.

    (a) C4H8 is Butene (an alkene) containing one double bond. Its isomers are as follows :


    $$\text{CH}_3—\text{CH}_2—\text{CH} = \text{CH}_2$$



    $$\text{CH}_3—\text{CH} = \text{CH} —\text{CH}_3$$





    (b) C5H8 is Pentyne (an alkyne) containing one triple bond. Its isomers are as follows :


    13.4. 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

    Ans. Ozonolysis of alkenes refers to the addition of an ozone molecule to an alkene to create ozonide, followed by the cleavage of the ozonide by Zn-H2O to smaller molecules. The ozonolysis reaction involves two steps :

    1. Reaction with ozone (O3)

    2. Reaction with Zn in the presence of H2O

    (i) Pent-2-ene undergo ozonolysis as:

    (ii) 3,4-Dimethylhept-3-ene undergo ozonolysis as:


    (iii) 2-Ethylbut-1-ene undergo ozonolysis as:


    (iv) 1-Phenylbut-1-ene undergo ozonolysis as:

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


    During ozonolysis an ozonide having a cyclic structure is formed as an intermediate which undergoes cleavage to give the final products. Ethanal and pentan-3-one are obtained from the intermediate ozonide. Hence, the expected structure of the ozonide is:


    This ozonide is formed as an addition of ozone to ‘A’. The desired structure of ‘A’ can be obtained by the removal of ozone from the ozonide. Hence, the structural formula of ‘A’ is:


    The IUPAC name of ‘A’ is 3-Ethylpent-2-ene.

    13.6. An alkene ‘A’ contains three C—C, eight C—H σ bonds and one C—C p bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44u. Write IUPAC name of ‘A’.

    Ans. It is given that an alkene ‘A’ contains three C–C, eight C–H bonds, and one C–C π-bond. This alkene ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 amu The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Thus, the structure of ‘A’ can be represented as: 

                              XC = CX

    There are eight C—H σ bonds. Hence, there are 8 hydrogen atoms in ‘A’. Also, there are three C—C bonds. Hence, there are four carbon atoms present in the structure of ‘A’.
    Combining the inferences, the structure of ‘A’ can be represented as


    ‘A’ has 3 C—C bonds, 8 C—H σ bonds and one C—C π bond. Hence, the IUPAC name of ‘A’ is But-2-ene.

    Verification - Ozonolysis of ‘A’ i.e., But-2-ene takes place as:


    Thus, the final product is ethanal with molecular mass:

    = [(2 × 12)+ (4 × 1) + (1 × 16)]
    = 44 u

    13.7. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

    Ans. Let the given alkene be ‘A’. It is given that alkene ‘A’ on ozonolysis produces propanal and pentan-3-one. By writing the reverse of the ozonolysis reaction, we get:


    The products are obtained on the cleavage of ozonide ‘X’. Hence, ‘X’ contains both products in the cyclic form. The possible structure of ozonide can be represented as:


    Now, ‘X’ is an addition product of alkene ‘A’ with ozone. Therefore, the possible structure of alkene ‘A’ is:


    The IUPAC name of an alkene ‘A’ is 3-Ethylhex-3-ene.

    Verification : Since a mixture of aldehyde and ketone is produced, it means the given alkene is an unsymmetrical alkene containing one branch on the double-bonded carbon atom.

    Step 1: Write the structures of propanal and pentan-3-one with their oxygen atoms facing each other, we have,


    Step 2: Remove oxygen atoms and joins the two fragments by a double bond, the structure of the alkene is,


    13.8. Write chemical equations for combustion reaction of the following hydrocarbons:

    (i) Butane

    (ii) Pentene

    (iii) Hexyne

    (iv) Toluene

    Ans. Combustion reaction : The reaction of one mole of organic compound in the presence of air (especially O2) producing carbon dioxide and water is known as a combustion reaction.

    $$(i)\underset{Butane}{C_4H_10(g)}+13/2O_2(g)\xrightarrow{\Delta}4CO_2(g)\underset{+heat}{+5H_2O(g)}\\(ii) \underset{Pentene}{C_5H_10(g)}+15/2O_2(g)\xrightarrow{\Delta} 5CO_2(g)\underset{+Heat}{+5H_2O(g)}\\(iii)\underset{Hexyne} {C_6H_{10}(g)}+17/2O_2(g)\xrightarrow{\Delta}6CO_2(g)\underset{+Heat}{+5H_2O(g)}$$


    13.9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

    Ans. Hex-2-ene is CH3—CH2—CH2—CH2—CH= CH—CH3

    Due to differences in the spatial arrangement of groups around double bonds, alkene generally shows geometrical isomerism. The structure of cis-and trans-isomer of hex-2-ene are:


    For a molecule, the higher the bond polarity, the greater will be intermolecular dipole-dipole interaction, and hence, the higher will be its boiling point.

    The dipole moment of cis-compound is a sum of the dipole moments of C—CH2 and C— CH2CH2CH3 bonds acting in the same direction. The dipole momentof trans-compound is the resultant of the dipole moments of C—CH3 and C—CH2CH3 bonds acting in opposite directions. Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.

    13.10. Why is benzene extra ordinarily stable though. It contains three double bonds?

    Benzene is a hybrid of resonating structures given as:

    Hybrid structure
    Benzene having molecular formula C6H6 contains sp2 -hybridised six carbon atoms. The two sp2 hybrid orbitals of each carbon atom overlap with the sp2 hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining sp2 hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six sigma C–H bonds.
    hybrid orbitals in planar arrangement.

    The remaining unhybridised p-orbital of carbon atoms has the possibility of forming three π bonds by the lateral overlap of:

    The remaining unhybridised

    The six π’s are delocalised and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalised π-electrons stabilise benzene.

    Thus, in benzene, stability is mainly achieved through electron resonance or delocalisation because all six electrons of the three double bonds are entirely delocalised forming a single lowest energy molecular orbital that surrounds all of the ring’s carbon atoms, therefore, it is exceptionally stable.

    13.11. What are the necessary conditions for any system to be aromatic?

    Ans. Huckel predicted that electrons in a cyclic conjugated having a certain amount of electrons are totally delocalised based on molecular orbital theory, making these compounds exceptionally stable. These compounds are known as aromatic compounds.

    The following are the requirements for a molecule to be aromatic:

    (i) Aromatic compounds are cyclic.
    (ii) These compounds have conjugated π-bonds.
    (iii) Aromatic rings are planar.
    (iv) According to Huckel’s rule, the aromatic compounds must have (4n + 2) π electrons in the ring systems. (where n = 0, 1, 2, 3,...) Examples are :


    13.12. Explain why the following systems are not aromatic?


    Ans. (i) The system is not planar due to the presencem of an sp3 -hybridised carbon. Although it contains six-electrons, the system is not fully conjugated because the six -electrons do not form a single cyclic electron cloud that surrounds all of the ring’s atoms. As a result, it is not an aromatic substance.

    (ii) For a given compound, the number of π- electrons is 4.


    According to Huckel’s rule,
    4n + 2 = 4
    4n = 2

    For a compound to be aromatic, the value of n must be an integer (n= 0, 1, 2,....), which is not true for the given compound. Hence, it is not aromatic in nature. Further, due to the presence of an sp3 - carbon, the system is not planar. Hence, the compound is not aromatic in nature.


    (iii) For a given compound, the number of π-electrons is 8.
    According to Huckel’s rule,

    4n + 2 = 8
    4n = 6

    For a compound to be aromatic, the value of n must be an integer (n= 0, 1, 2,....), which is not true for the given compound. Hence, it is not aromatic in nature.

    Further, the given compound is not planar rather it is tub-shaped. It is, therefore, a non-planar system having 8 p-electrons. Therefore, the molecule is not aromatic
    since it does not contain a planar cyclic cloud having (4n + 2)π-electrons.

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

    Ans. (i) Electrophilic aromatic substitution of benzene with bromine in presence of ferric bromide gives bromobenzene. Nitration with conc. nitric acid and concentrated sulphuric acid give a mixture of ortho-bromonitrobenzene and para-bromo nitrobenzene which are separated by fractional distillation. Thus, the sequence of reaction is:


    (ii) Nitration of benzene with conc nitric acid and conc sulphuric acid gives nitrobenzene. Chlorination with chlorine in presence of anhydrous aluminum chloride gives meta nitro chlorobenzene.Thus, the sequence of reaction is:

    FC, Alkylation

    (iii) Benzene reacts with methyl chloride in presence of anhydrous aluminum chloride to give toluene. After that, toluene reacts with a nitrating mixture to give a mixture of p nitrotoluene and o- nitrotoluene. The mixture of p-nitrotoluene and o-nitrotoluene can be separated by fractional distillation. Thus, the sequence of reaction is :

    FC, Alkylation

    (iv) For the preparation of acetophenone from benzene, Friedel crafts acylation reaction is used. In this reaction, benzene is treated with an acyl halide or acid anhydride in the presence of Lewis acids (AlCl3) to yield acyl benzene or acetophenone.


    13.14. In the alkane H3C CH2–C(CH3)2–CH2CH(CH3)2, identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.

    Ans. The degree of carbon is decided by the number of alkyl groups directly attached to it. 1° carbon is that carbon atom that is bonded to only one carbon atom, i.e., they have only one carbon atom as its neighbor. The given structure has five 1° carbon atoms and fifteen hydrogen atoms attached to it.

    2° carbon is that carbon atom that is bonded to two carbon atoms, i.e., they have two carbon atoms as its neighbour. The given structure has two 2° carbon atoms and four hydrogen atoms attached to it.

    3° carbon is that carbon atom that is bonded to three carbon atoms, i.e., they have three carbon atoms as its neighbour. The given structure has one 3° carbon atoms and only one hydrogen atom attached to it.

    13.15. What effect does branching of an alkane chain has on its boiling point?

    Ans. The boiling point of the substance depends upon the type of intermolecular forces of attraction Due to the non-polar nature of alkanes, the intermolecular forces that bind the two alkane molecules together are Van der Waal’s forces of attraction. The stronger the force, the greater will be the boiling point of the alkane.

    As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.

    13.16. Addition of HBrt opropene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

    Ans. Addition of HBr to propene is an example of an electrophilic substitution reaction. Hydrogen bromide provides an electrophile, H+. This electrophile attacks the double bond to form 1° and 2° carbocations as shown :

    Primary carbocation

    Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br– attacks the carbocation to form 2-bromopropane asthe major product.


    The reaction follows Markovnikov’s rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms.

    In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov’s rule. The reaction follows a free radical chain mechanism as:

    primary free radical

    The reaction follows Markovnikov’s rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms.

    In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov’s rule. The reaction follows a free radical chain mechanism as:

    1-Bromopropane (major product)

    In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide.

    13.17. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?

    Ans. The resonance hybrid of the two Kekule structures of o-Xylene and ozonolysis of each one of these gives two products is given below:

    1, 2-Dimethylglyoxal

    Thus, in all, three products are formed i.e., methyl glyoxal, 1,2-dimethylglyoxal, and glyoxal are obtained from two Kekule structures. Since all the three products cannot be obtained from any one of the two Kekule structures, this shows that o-xylene is a resonance hybrid of the two Kekule structures (I and II).

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

    Ans. The tendency of species to lose H-atoms is known as its acidic character. The hybridisation state of carbon in these three compounds is:

    Type of hybridisation

    In hydrocarbons, acidity of the hydrogen atoms attached to the carbon depends on the state of hybridisation of that carbon. The increasing order of electronegativity is sp3 < sp2 < sp. More electronegative carbon means the electrons of C—H bond are more closure to carbon and hence hydrogen can show more acidic character. In
    ethyne, hydrogen is attached to sp hybridised carbon, in benzene it is sp2 and in n-hexane it is sp3 . So, the order of acidity is :

    Ethyne > Benzene > n-Hexane

    13.19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

    Ans. Benzene is a planar molecule having delocalised electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles.

    Therefore, it undergoes electrophillic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophillic substitutions with difficulty.

    13.20. How would you convert the following compounds into benzene?
    (i) Ethyne
    (ii) Ethene
    (iii) Hexane.


    (ii) Ethene is first converted into ethyne and then to benzene as shown below:

    red hot iron tube

    (iii) When hexane vapours are run over a heated Cr2O3, Mo2O3, and V2O5 catalyst at 773 K and 10–20 atm pressure, cyclisation and aromatisation occur concurrently, yielding benzene.


    13.21. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

    Ans. The basic skeleton of 2-methyl butane is :


    The structures of various alkenes that yield 2-methyl butane on hydrogenation can be obtained by putting double bonds at different positions in the skeleton of 2-methylbutane i.e.,


    13.22. 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–H3C–C6H4–NO2,


    Ans. Electrophiles are the reagents that accept the electron pair while nucleophiles are the reagents that donate the electron pair. The electron density on a benzene ring describes its reactivity towards an electrophile, E+.
    (i) Electrophilic substitution reactions are the most common benzene reactions. The higher the electron density on a benzene ring, the more reactive is the compound towards an electrophile, E+ (Electrophilic reaction). This is because NO2 is a more effective electron-withdrawing group than Cl, the molecule becomes less reactive as the number of nitro groups increases. As a result, the overall reactivity decreases in order:
    Chlorobenzene > p nitrochlorobenzene > 2, 4-dinitrochlorobenzene.
    (ii) The CH3 group is an electron donating group, while the NO2 group is electron withdrawing. As a result, toluene will have the maximum electron density, followed by
    p-nitro toluene and p-dinitrobenzene. As a result, total reactivity diminishes in the following order:

    Toluene > p–H3C–C6H4 –NO2 > p– O2N– C6H4 –NO2.

    13.23. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?

    Ans. Nitration reaction are examples of electrophilic substitution reactions where an electron-rich species is attacked by nitronium ion (NO2+). The ease of nitration depends on the presence of electron density on the compound to form nitrates.

    Now —CH3 group is electron donating and NO2, is electron withdrawing. Therefore, toluene will have the maximum electron density among the three compounds followed by benzene. On the other hand, m-dinitrobenzene will have the least electron density. Hence, it will undergo nitration with difficulty. Hence, the increasing order of nitration is as follows :


    Toluene > Benzene > m-dinitrobenzene.

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

    Ans. The addition of an ethyl group on the benzene ring is termed as ethylation reaction of benzene. Such a reaction is generally called a Friedel-Crafts alkylation reaction. These reactions always take place in the presence of a Lewis acid. Some lewis acids that can be used for ethylation of benzene are Anhydrous FeCl3, SnCl4, BF3, etc.

    13.25. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

    Ans. Wurtz reaction is limited for the synthesis of symmetrical alkanes (alkanes with an even number of carbon atoms). In the reaction, two similar alkyl halides are taken as reactants and an alkane, containing double the number of carbon atoms, are formed.
    Example :


    Wurtz reaction is not preferred for the preparation of alkanes containing odd number of carbon atoms because this reaction involves the union of two alkyl halide molecules. If the two alkyl halides are different then there is a possibility of formation of three products so the required product is obtained in lesser amount. For example,

    In order to prepare alkane with odd number of carbon atoms, two different haloalkanes are needed; one with odd number and the other with even number of carbon atoms. For example, bromoethane and 1-bromopropane will give pentane as a result of the reaction.


    But side products will also be formed when the members participating in the reaction react separately. For example, bromoethane will give butane and 1-bromopropane will give rise to hexane.


    Thus, mixture of butane, pentane and hexane will be formed. It will be quite difficult to separate the individual components from the mixture.

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