NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Chemistry
Chapter : 13
Chapter Name : Hydrocarbons

Q13.1 How do you account for the formation of ethane during chlorination of methane ?

Answer. The reaction begins with the homolytic cleavage of Cl — Cl bond as:
Step 1: Initiation:
hv
˙ ˙
Cl − Cl ⟶ Cl + C

Step 2: Propagation:
In the second step, chlorine free radicals attack methane molecules and break down the bond
to generate methyl radicals as:
hv
˙ ˙
CH4 + C1 ⟶ CH3 + H − Cl

These methyl radicals react with other chlorine free radicals to form methyl chloride along
with the liberation of a chlorine free radical.
˙ ˙
CH3 + Cl − Cl ⟶ CH3 − Cl + C1

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCI and
CH C are the major products formed, other higher halogenated compounds are a so formed
3

as:
˙ ˙
CH3 Cl + C1 ⟶ CH2 Cl + HCl
˙ ˙
CH2 Cl + Cl − Cl ⟶ CH2 Cl2 + Cl

Step 3: Termination:
Formation of ethane is a result of the termination of chain reactions taking place as a result of
the consumption of reactants as:
˙ ˙
Cl + Cl ⟶ Cl − Cl
˙ ˙
H3 C + CH3 ⟶ H3 C − CH3

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

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Q13.2 Write IUPAC names of the following compounds :

Answer. (a)

IUPAC name: 2-Methylbut-2-ene

(b)
IUPAC name: Pen-1 -ene-3-yne

(c)
IUPAC name: 1, 3-Butadiene or Buta-1,3-diene

(d)
IUPAC name: 4-Phenyl but-1-ene

(e)
IUPAC name: 2-Methyl phenol

(f)
IUPAC name: 5-(2-Methylpropyl)-decane

(g)
IUPAC name: 4-Ethyldeca-1, 5, 8 triene

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Q13.3 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) C H (one double bond)
4 8

(b) C H (one triple bond)
5 8

Answer. (a) The following structural isomers are possible for C H with one double bond:
4 8

The IUPAC name of
Compound (I) is But-1 -ene,
Compound (II) is But-2-ene, and
Compound (Ill) is 2-Methylprop-1 -ene.

(b) The following structural isomers are possible for C C with one triple bond .
5 8

The IUPAC name of
Compound (I) is Pent-1 -yne,

Compound (II) is Pent-2-yne, and
Compound (Ill) is 3-Methylbut-1-ene.

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Q13.4 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

Answer. (i ) Pent-2-ene undergoes ozonolysis as:

The IUPAC name of Product (1) is ethanol and Product (II)is propanel.

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

The IUPAC næ•ne of Product (I)is butan-2-one and Product (II) is Pentan-2-one.

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

The IUPAC name of Product (I)is pentan-3-one and Product (II)is methanal.

(iv) 1 -Phenylbut-l -ene undergoes ozonolysis as:

The IUPAC name of Product (I)is benzaldehyde and Product (II)is propenel,

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Q13.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3- one. Write
structure and IUPAC name of ‘A’.

Answer.

During ozonolysis, an ozonide having a cyclic structure is formed as an intermediate which
undergoes cleavage to give the nal 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 addition of ozone to The desired structure of 'A' can be obtained by
the removal of ozone from the ozonide. Hence. the structural formula of 'A'

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

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Q13.6 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 IUPAC name of ‘A’.

Answer. As per the given information, 'A' on ozonolysis gives two moles of an aldehyde of
molar mass 44 u, 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. Hence,
the structure Of 'A' can be represented as: XC = CX There are eight C-H bonds. Hence, there are
8 hydrogen atoms in 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, B C—H c bonds, and one C—C n bond Hence, the IUPAC name of 'A' is
But-2-ene. Ozonolysis of 'A' takes place as:

The nal product is ethanal with molecular mass
= [(2 × 12) + (4 × 1) + (1 × 16)]

=44u

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Q13.7 Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the
structural formula of the alkene?

Answer. As per the given information, propanal and pentan-3-one are the ozonolysis products
of an alkene, Let the given alkene be "Airiting the reverse Of the ozonolysis reaction, we

The products are obtained on the cleavage of ozonide 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:

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Q13.8 Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene

Answer. Combustion can be de ned as a reaction of a compound with oxygen.

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Q13.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and
why?

Answer. Hex-2-ene is represented as:
H3 C − HC = CH − CH2 − CH3

Geometrical isomers of hex-2-ene ere:

The dipole moment of cis-compound is sum of the dipole moments of C − CH and 3

C − CH CH , bonds acting in the same direction. The dipole moment of trans-compound is
2 3

the resultant Of the dipole moments Of C − CH and C − CH CH bonds acting in opposite
3 2 3

directions. Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the
greater is the dipole-dipole interaction and the higher Will be the boiling point, Hence, cis-
isomer will have a higher boiling point than trans-isomer.

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Q13.10 Why is benzene extra ordinarily stable though it contains three double bonds?

Answer. Benzene is hybrid of resonating structures given as:

All six carbon atoms in benzene are sp2 hybridized. 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. The remaining unhybridized p-orbital of

carbon atoms has the possibility of forming three n Bonds by the overlap of
C − C , C − C , C − C or C − C , C − C , C − C
1 2 3 4 5 6 2 3 4 3 6 1

The six Π s are delocalized and can move freely ±out the six carbon nuclei. Even after the
′

presence of the three double bonds, these delocalized n-electrons stabilize benzene

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Q13.11 What are the necessary conditions for any system to be aromatic?

Answer. A compound is said to be aromatic if it satis es the following three conditions:
(i) It should have a planar structure.
(ii) The n-electrons of the compound are completely delocalized in the ring.
(iii) The total number of n—electrons present in the ring should be equal to (4n + 2),
Where n= 0, 1, 2 …. etc. This is known as Huckel's rule.

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Q13.12 Explain why the following systems are not aromatic?

Answer. (i)

For the given compound, the number Of electron is 6.
By Huckel rule,
4n+2=6
4n=4
n=1
For a compound to be aromatic. the value of n must be an integ« (n = 0,1,2,3….) since the
value Of n is integer. the given compound is aromatic in nature

(ii)
For the given compound, the number of n-electrons is 4,
By Huckel's rule,
4r,+2=4
4n=2
n=0.5
For a compound to be aromatic, the value of n must be an integer (n = 0,1,2…..) whichis not
true for the given compound. Hence, it is not aromatic in nature.

(iii)

For the given compound, the number of (I-electrons is 8.
By huckel's rule,
4n+2 =8
4n=6
3
n =
2

For a compound to be aromatic, the value of n must be an integer (n= 0,1,2…….) Since the
value of n Is not an integer, the given compound is not aromatic in nature.

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Q13.13 How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m- nitrochlorobenzene
(iii) p - nitrotoluene
(iv) acetophenone?

Answer. (i) Benzene can be converted into p-nitrobromobenzene as:

(ii) Benzene can be converted into m-nitrochlorobenzene as:

(iii) Benzene can be converted into p-nitrotoluene as:
2 carbon atoms ye those which are bonded to two carbon atoms i.e., they have two cæbon
∘

atoms as their neighbours. The given structure has two 2 carbon atoms fou hydrogen atoms
∘

attached to it. 3 carbon atoms are those which ye bonded to three carbon atoms i.e., hey have
∘

three atoms as their neighbours. The given structure has 30 carbon atcvn only one hydrogen
atom is attached to it.

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Q13.14 In the alkane H C − CH − C(CH ) – CH – CH (CH ) , identify 1°,2°,3° carbon
3 2 3 2 2 3 2

atoms and give the number of H atoms bonded to each one of these.

Answer.
1 carbon atoms those which are bonded to only one carbon atoms i.e., they have only me
∘

cubm atom as their neighbour. The given structure has ve 1 carbon atoms and fteen
∘

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

This reaction follows Markovnikov's rule Khere 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:

Secondary free radicals are more stable than primary radicals. Hence, the former
predominates since it forms at a faster rate. Thus, 1 - bromopropane is obtained as the major
product.

In the presence of peroxide, 3r 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.

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Q13.15 What effect does branching of an alkane chain has on its boiling point?

Answer. Alkanes experience inter-molecular Waals The stronger the forces, the greater Will be
boiling point Of the As branching increases, the surface area of the molecule decreases which
results in a small of contact. As a result, the der Waals force also decreases which car' be
overcome at a relatively temperature. Hence, the boiling point Of chain decreases with an
increase in brmching.

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Q13.16 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.

Answer. Addition of HBr to propene is an ionic electrophilic addition reaction in which the
electrophile. i,e. H rst adds to give a more stable 2 carbocation. In the 2nd step, the
+ ∘

carbocation is rapidly attacked by the nucleophile Br~ ion to give 2-bromopropane.

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Q13.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does
the result support Kekulé structure for benzene?

Answer. 0-xylene has two resonance structures:

All three products, i.e., methyl glyoxal, l, 2—methylgLyoxal, glyoxal are obtained from two
Kekule structures. Since all three products cannot be obtained from one of
the two structures, this proves that o-xylene is a hybrid of two Kekule structures (I and II).

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Q13.18 Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also
give reason for this behaviour.

Answer. Acidic character of a species is de ned on the basis of ease with which it can lose its
H- atoms. The hybridization state of carbon in the given compound is:

As the s—character increases, the electronegativity Of carbon increases and the electrons of
C-H bond pair lie closer to the carbon atom. As a result, partial positive charge of H-atom
increases and H ions are set free. The s—character increases in order : sp < sp < sp
+ 3 2

Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

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Q13.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic
substitutions with dif culty?

Answer. Benzene is a planar molecule having delocalized electrons above below the pine of
ring. Hence. it is electron-rich. As a result. it is hVily attractive to electron de cient species
i.e., electrophiles. it undergoes electrophilic substitution reactions easily. nucleophiles are
electron-rich. Hence, they repelled by benzene. Hence, benzene undergoes nucleophilic
substitutions with dif culty.

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Q13.20 How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene
(iii) Hexane

Answer. (i) Benzene from Ethene:

(ii)Benzene from Ethyne:

(iii)Hexane to Benzene :

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Q13.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Answer. The basic skeleton of 2- methylbutane is shown below:

(a)
On the basis of this structure, various alkenes that will give 2-methylbutane on hydrogenation
are:

(a)

(b)

(c)

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Q13.22 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 − H C − C H − N O , p − O N − C H − N O .
3 6 4 2 2 6 4 2

Answer. Electrophiles ye reagents that participate in a reaction by accepting electron pair in
order to bond to nucleophiles. The higher the electron density on a benzene ring, the more
reactive is the compound Towards an electrophile. (Electrophilic reaction).

(a) The presence of electron (i.e.. NO and Cl ) deactivates the aromatic ring by decreasing
−

2
−

the electron density. SinceNO -group is more electron withdrawing (due to resonance effect)
−

2

than the Cl ) (due to inductive effect), the decreasing cyder of reactivity is as follows:
−

Chlorobenzene > p — nitrochlorobenzene > 2, 4 — dinitrochlorobenzene

(b) While CH − is an electron donating group,NO - group is electron withdrawing. Hence,
−
3 2

toluene will have the maximum electron density and is most easily attacked by E NO is
+ −

2

electron withdrawing voup_ Hence, when the of NO substituents is greater, the order is as
−

2

follows:
toluene > p − CH3 − C6 H4 − NO2t , ρ − O2 N − C6 H4 − NO2

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Q13.23 Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most
easily and why?

Answer. The ease of nitration depends on the presence of On the compound to form nitrates.
Nitration reactions are examples of electrophilic substitution reactions Where an electron-
rich species is attacked by a nitronium ion NO . Now, CH.— group is electron donating and
−

2

is electron withdrawing. toluene will have the maximum electron density the three
−
NO
2

compounds followed by benzene. On the other hand, m- Dinitrobenzene Will have the least
electron density. Hence, it will nitration with dif culty. Hence, the increasing order Of
nitration is as follows:

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Q13.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can
be used during ethylation of benzene.

Answer. The ethylation reaction of benzene involves the addition of an ethyl group on the
benzene ring. Such a reaction is called a Friedel-Craft alkylation reaction. This reaction t*es
place in the presence of a Lewis acid. Any Lewis acid like anhydrous FeCl , SnCl , BF etc.
3 4 3

be used drawing the ethylation of benzene.

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Q13.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing odd
number of carbon atoms? Illustrate your answer by taking one example.

Answer. Wurtz reaction is limited for the synthesis of symmetrical alkanes (alkanes Kith 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 cannot be used for the preparation Of unsymmetrical alkanes because if two
dissimilar alkyl halides are taken as the reactants, then a mixture of alkanes is obtained as the
products. Since the reaction involves free radical species. a side reaction also Occurs to
produce an alkene. For example, the reaction Of bromomethane and iodoethane gives a
mixture of alkanes.

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