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Class XI Chapter 13 – Hydrocarbons Chemistry
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Question 13.1:
How do you account for the formation of ethane during chlorination of methane?
Answer
Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction
takes place in the given three steps.
Step 1: Initiation:
The reaction begins with the homolytic cleavage of Cl – Cl bond as:
Step 2: Propagation:
In the second step, chlorine free radicals attack methane molecules and break down the
C–H bond to generate methyl radicals as:
These methyl radicals react with other chlorine free radicals to form methyl chloride
along with the liberation of a chlorine free radical.
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:
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:
Class XI Chapter 13 – Hydrocarbons Chemistry
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Hence, by this process, ethane is obtained as a by-product of chlorination of methane.
Question 13.2:
Write IUPAC names of the following compounds:
a.
b.
c.
d.
e.
f.
g.
Answer
(a)
IUPAC name: 2-Methylbut-2-ene
Class XI Chapter 13 – Hydrocarbons Chemistry
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(b)
IUPAC name: Pen-1-ene-3-yne
(c) can be written as:
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
Class XI Chapter 13 – Hydrocarbons Chemistry
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Question 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)
Answer
(a) The following structural isomers are possible for C4H8 with one double bond:
The IUPAC name of
Compound (I) is But-1-ene,
Compound (II) is But-2-ene, and
Compound (III) is 2-Methylprop-1-ene.
(b) The following structural isomers are possible for C5C8 with one triple bond:
Class XI Chapter 13 – Hydrocarbons Chemistry
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The IUPAC name of
Compound (I) is Pent-1-yne,
Compound (II) is Pent-2-yne, and
Compound (III) is 3-Methylbut-1-ene.
Question 13.4:
Write IUPAC names of the products obtained by the ozonolysis of the following
compounds:
(i) Pent-2-ene (ii) 3,4-Dimethyl-hept-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 (I) is ethanal and Product (II)is propanal.
(ii) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:
Class XI Chapter 13 – Hydrocarbons Chemistry
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The IUPAC name 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-1-ene undergoes ozonolysis as:
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The IUPAC name of Product (I)is benzaldehyde and Product (II)is propanal.
Question 13.5:
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 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:
Class XI Chapter 13 – Hydrocarbons Chemistry
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The IUPAC name of ‘A’ is 3-Ethylpent-2-ene.
Question 13.6:
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 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.
Ozonolysis of ‘A’ takes place as:
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The final product is ethanal with molecular mass
Question 13.7:
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 ‘A’. 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:
Question 13.8:
Write chemical equations for combustion reaction of the following hydrocarbons:
Class XI Chapter 13 – Hydrocarbons Chemistry
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(i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene
Answer
Combustion can be defined as a reaction of a compound with oxygen.
(iv)
Toluene
Question 13.9:
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:
Geometrical isomers of hex-2-ene are:
The dipole moment of cis-compound is a sum of the dipole moments of C–CH3 and C–
CH2CH3 bonds acting in the same direction.
The dipole moment of trans-compound is the resultant of the dipole moments of C–CH3
and C–CH2CH3 bonds acting in opposite directions.
Class XI Chapter 13 – Hydrocarbons Chemistry
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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.
Question 13.10:
Why is benzene extra ordinarily stable though it contains three double bonds?
Answer
Benzene is a 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 π
bonds by the lateral overlap of .
The six π’s are delocalized and can move freely about the six carbon nuclei. Even after
the presence of three double bonds, these delocalized π-electrons stabilize benzene.
Question 13.11:
What are the necessary conditions for any system to be aromatic?
Answer
A compound is said to be aromatic if it satisfies the following three conditions:
(i) It should have a planar structure.
(ii) The π–electrons of the compound are completely delocalized in the ring.
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(iii) The total number of π–electrons present in the ring should be equal to (4n + 2),
where n = 0, 1, 2 … etc. This is known as Huckel’s rule.
Question 13.12:
Explain why the following systems are not aromatic?
(i)
(ii)
(iii)
Answer
(i)
For the given compound, the number of π-electrons is 6.
By Huckel’s rule,
4n + 2 = 6
4n = 4
n = 1
For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…). Since
the value of n is an integer, the given compound is aromatic in nature.
(ii)
For the given compound, the number of π-electrons is 4.
By Huckel’s rule,
4n + 2 = 4
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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.
(iii)
For the given compound, the number of π-electrons is 8.
By 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…). Since
the value of n is not an integer, the given compound is not aromatic in nature.
Question 13.13:
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:
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(ii) Benzene can be converted into m-nitrochlorobenzene as:
(iii) Benzene can be converted into p-nitrotoulene as:
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(iv) Benzene can be converted into acetophenone as:
Question 13.14:
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.
Answer
1° carbon atoms are those which are bonded to only one carbon atom i.e., they have
only one carbon atom as their neighbour. The given structure has five 1° carbon atoms
and fifteen hydrogen atoms attached to it.
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2° carbon atoms are those which are bonded to two carbon atoms i.e., they have two
carbon atoms as their neighbours. The given structure has two 2° carbon atoms and four
hydrogen atoms attached to it.
3° carbon atoms are those which are bonded to three carbon atoms i.e., they have three
carbon atoms as their neighbours. The given structure has one 3° carbon atom and only
one hydrogen atom is attached to it.
Question 13.15:
What effect does branching of an alkane chain has on its boiling point?
Answer
Alkanes experience inter-molecular Van der Waals forces. 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.
Question 13.16:
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 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:
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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 as the major product.
This 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:
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.
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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.
Question 13.17:
Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the
result support Kekulé structure for benzene?
Answer
o-xylene has two resonance structures:
All three products, i.e., methyl glyoxal, 1, 2-demethylglyoxal, and glyoxal are obtained
from two Kekule structures. Since all three products cannot be obtained from any one of
the two structures, this proves that o-xylene is a resonance hybrid of two Kekule
structures (I and II).
Question 13.18:
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give
reason for this behaviour.
Answer
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Acidic character of a species is defined 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 the order:
sp3 < sp2 < sp
Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.
Question 13.19:
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic
substitutions with difficulty?
Answer
Benzene is a planar molecule having delocalized 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 electrophilic substitution reactions very easily. Nucleophiles are
electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes
nucleophilic substitutions with difficulty.
Question 13.20:
How would you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene (iii) Hexane
Answer
(i) Benzene from Ethyne:
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(ii) Benzene from Ethene:
(iii) Hexane to Benzene
Question 13.21:
Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Answer
The basic skeleton of 2-methylbutane is shown below:
On the basis of this structure, various alkenes that will give 2-methylbutane on
hydrogenation are:
(a)
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(b)
(c)
Question 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, p-O2N–C6H4–NO2.
Answer
Electrophiles are reagents that participate in a reaction by accepting an 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, E+ (Electrophilic reaction).
(a) The presence of an electron withdrawing group (i.e., NO2–and Cl–) deactivates the
aromatic ring by decreasing the electron density.
Since NO2–group is more electron withdrawing (due to resonance effect) than the Cl–
group (due to inductive effect), the decreasing order of reactivity is as follows:
Chlorobenzene > p – nitrochlorobenzene > 2, 4 – dinitrochlorobenzene
(b) While CH3– is an electron donating group, NO2– group is electron withdrawing.
Hence, toluene will have the maximum electron density and is most easily attacked by
E+.
NO2– is an electron withdrawing group. Hence, when the number of NO2– substituents is
greater, the order is as follows:
Toluene > p–CH3–C6H4–NO2, p –O2 N–C6H4–NO2
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Question 13.23:
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 electron density 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 (NO2–).
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:
Question 13.24:
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
takes place in the presence of a Lewis acid.
Any Lewis acid like anhydrous FeCl3, SnCl4, BF3 etc. can be used during the ethylation of
benzene.
Question 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.
Answer
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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 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.
The boiling points of alkanes (obtained in the mixture) are very close. Hence, it becomes
difficult to separate them
109 [XI – Chemistry]
UNIT-13
HYDROCARBONS
Preparations
• Hydrogenation : Addition of H2 to alkenes and alkynes in presence of
catalyst to form alkanes.
• Wurtz reaction : Treatment of alkyl halides (X = Cl, Br, I) with sodium in
dry ethereal solution to form symmetrical alkanes having double the number
of carbon atoms as present in parent alkyl halide.
• Decarboxylation is the process of removal of CO2 from the carboxylic
acid to form an alkane having one carbon atom less than the parent carboxylic
acid.
• Kolbe’s electrolysis of sodium or potassium salt of carboxylic acid produces
alkane having one carbon atom less than the parent acid.
• Halogenation of alkanes proceeds through free radical chain mechanism.
Monochlorination of 2-methyl butane gives four products.
• Mechanism of Halogenation
(i) Reaction is initiated by homolysis of chlorine molecule in presence of
light or heat
hCl – Cl Cl Cl
(ii) 3 3H C – H Cl CH H – Cl
CH3 CH CH2 CH3
CH3
Cl2 / hCH2CH CH2 CH3
CH3
+ CH3 CH CH2 CH2
CH3
Cl + CH3 CH
Cl
CH CH3
+ CH3 C CH2 CH3
CH3
Cl
Cl
[XI – Chemistry] 110
3 3H C Cl – Cl H C – Cl Cl
(iii) Cl Cl Cl – Cl
3 3H C Cl H C – Cl
3 3 3 3H C CH H C – CH
• Lindlar’s catalyst is the partially deactivated palladised charcoal which
converts unsymmetrical alkynes having no acidic terminal hydrogen to form
cis-alkene.
• Unsymmetrical alkynes having no terminal hydrogen is reduced by Na/liquid
N H3 to form trans-alkene.
• Saytzeff Rule is used in dehydrohalogenation of alkyl halide which is a
-elimination reaction since hydrogen is eliminated from -carbon having
lesser number of hydrogen atoms. (-carbon is the carbon atom which is
next to carbon to which halogen is attached) to form more stable alkene
having more substituted double bond.
An exception to Saytzeff Rule : The most sterically hindered base
+
3 3CH CO K
in tert-butyl alcohol cannot attack at the -carbon atom having
lesser number of hydrogen atoms because of greater crowding at the site of attack.
Therefore, the base, tert-butoxide remove more exposed 1º hydrogen atom from
-carbon.
CH3 CH2 C
CH3
Br
CH3
C2H5 OH C2
H5 ONa
Sterically leasthindered base
CH3CH C
CH3
CH3
(more stable)
+ CH3 CH2 C
CH3
CH2
(less stable)
CH3 CH2 C
CH3
Br
CH3
(CH )3 3 COKCH3CH C
more substituted alkene(Minor)
+ CH3 CH2 C =
CH3
CH2
less substituted alkene(major)
+
(CH )3 3 COH
–
( CH )3 2
111 [XI – Chemistry]
An Exception to Saytzeff Rule
(i) If leaving group is poor like F, then also least substituted albene is formed
as major product. KOH alc
3 2 3 3 2 2H C – CH – CH – CH H C – CH – CH CH
|F
Acid Catalysed Dehydration of Alcohols : Dehydration of 1º alcohols is
an Elimination reaction.
Mechanism :
(i) CH3 CH OH + H O SO H3
+
CH3
CH3 CH OH + O SO H2 3
CH3
OH2H2O + CH3 CH+
OH2
CH3
CH3 CH+
(ii)CH3 CH CH2 + H
+CH2
H
CH3 CH+
Rearrangement of carbocations obtained from 2º or 3º alcohols may occur
by hydride (H) or methanide CH3 shift (both are 1, 2-shifts) to form more
stable carbocations.
CH3 CH2 C
H
CH3
CH2 OHH
+
CH3 CH2C
H
CH3
CH2 – OH2
+
–H O2
CH3 CH2 C
H
CH3
CH2
+CH3 CH2 C
H
CH3
CH2
+
(1º Carbocation, less stable) H shift–
– H+
CH3
C = CH3 CH2 CH2
(Less substituted alkene, less stable) (minor product)
H
CH3 CH–C–CH3
CH3
+
(3º carbocation, more stable)
– H+
[XI – Chemistry] 112
CH3
CH3
– H+
CH = C–CH3
(More substituted alkene, more stable) (major product)
Similarly acid catalysed dehydration of 2-methylbutan-2-ol forms
2-methylbut-2-ene as major product and 2-methylbut-1-ene as a major product.
CH3
CH3
CH3
| | |
CH3–CH
2–C–CH
3 H
CH3–CH = C –CH
3 + CH
3CH
2–C = CH
2
|
OH (minor product) (major product)
Markownikov Rule
Addition of hydrogen bromide to unsymmetrical alkanes
3 2 3 3 3 2
major minor
CH CH CH HBr CH CH Br CH CH CH Br
Mechanism : Markownikov rule proceed via carbocation formation :
1. Larger alkyl group polarise the -bond.
δ+ δ δ+ δ
3 3 3 3 3
Br|
CH C H C H H Br CH C H CH Br CH C HCH
2. H
attacks first and generate more stable carbocation.
3. Brattacks to carbocation to form product.
Another example is :
δ+ δ δ+ δ
3 2 3 3 2 2 3CH CH C H C HCH H Br CH CH C HCH CH Br
3 2 2 3CH –CH –CH–CH –CH
|
Br
Addition of HBr to unsymmetrical alkanes in presence of peroxide
HBr3 3 3 2 2 3 3Benzoyl Peroxide
major minor
CH CH CH CH CH – CH Br CH CH Br CH
113 [XI – Chemistry]
Mechanism of Peroxide effect :
Peroxide effect (Antimarkownikov rule) proceeds via free radical
mechanism.
O O O
|| || ||
C6H
5–C–O–O–C–C
6H
5h
2C6H
5–C–O2C
6H
5+CO
2
One free radical always generate another free radical.
6 5 6 6
HBr
3 2 3 2 3 2 2(More stable radical)
C H HBr C H Br
CH CH CH Br CH CHCH Br CH CH CH Br Br
Friedel-Crafts Reaction :
Introduction of alkyl (R) or acyl group (COR) to benzene nucleus in presence
of anhydrous AlCl3 (Lewis acid) is called Friedel-Crafts rection.
Mechanism :
Step 1. Generation of electrophile, AlCl3 is a Lewis acid and generate
electrophile.
3 3 4 3CH Cl AlCl AlCl C H
Step 2. Attack of electrophile on the ring
Step 3. Abstraction of H+ by the nucleophile
Some special cases of friedel-crafts reaction
+ CH3
+
H CH3 H CH3 H CH3
+
+
+
H CH3
+
H CH3
+ + AlCl4
–
CH3
+ AlCl3 + HCl
[XI – Chemistry] 114
(1) More stable carbocation will form the major product, e.g.,
Directive influence of a functional group in monosubstituted benzene :
Activating groups like NH
2,
NHR,
NR
2,
NHCOCH
3,
OCH
3,
CH
3,
CH
2CH
3 are o-/p-directors.
Halogen atoms deactivates the ring by electron withdrawing inductive effect.
But the halogen atom increase the electron availability at o-and/or/p-positions
by electron donating resonance effect. Hence halogens act also as o-and/or/p-
directors.
Deactivating groups such as NO
2,
CN,
CHO,
COOH,
COR,
CONH2,
COOR and
SO
3H are m-directing groups.
HYDROCARBONS
1. Write IUPAC names of the following organic compounds :
(a) (b) (CH3)4C
CH2
+ CH3CH2CH2Cl AlCl3
CH3 CHCH3
Cumene
+ HCl
CH2CH2Cl + AlCl3 AlCl4
–+ CH3CHCH2
– Hshift
–
CH3–CH–
–CH3
1º carbocation(Less stable)
2º carbocation(more stable)
+ (CH3)2CH2CH2Cl AlCl3
H C–3
+ HCl
CH3
C–CH3
(CH3)2 CHCH2Cl AlCl3
AlCl4 + CH3 –C – CH2
CH3
+
H
Hshift
(CH3)3 C–––
H
O2NOH
O2N
O2N
115 [XI – Chemistry]
(c) (d) 3 32 3CH CH CCHC CH
||O
(e) 3 2 2 2
OH|
CH CH C H CH –CH OH (f)
COOH
|
COOH
(g) (g) CH2 CH CH
2 C
CH
(h) CH3 C(CH
3)2 CH
CH
2(i) CH
3 CH
CH
CH
2COOH
2. Complete the following equations :
(a) 3CH BrNaHC CH A B
(b) 2+ +Hg / H
2333 K
HC CH H O
(c) red hot iron tube
873 KHC CH
(d) 2+ +Hg / H
3333 K
CH C CH HOH
(e) HBrHC CH H Br A B
(f) 2HC CH Br
(g) 2HC CH O
(h) 3CH C CH H Br
(i) 2 2CaC H O
(j) 2NaNHalcohol
2 2KOH
BrCH CH Br A B
3. Explain the following reactions with suitable examples :
(a) Wurtz reaction
(b) Kolbe’s electrolysis
(c) Friedal Craft reaction
OH
COOH
CH3
[XI – Chemistry] 116
4. Define conformations. Write the conformations of ethane. Draw the
structures of various conformers of ethane according to Newman projection
formula. Name the most stable conformer of ethane.
5. State Markovnikov rule. Explain the mechanism of Markovnikov rule. Write
the product when the same reaction is carried with HBr in presence of
Benzoyl peroxide. Explain the mechanism of this reaction also.
6. Complete the following reactions :
(a) C H2CH
2CH
CHCH
3 + HBr
(b) Benzyl
3 2Peroxide
CH CH CH HBr
(c) 2Zn/H O
2 2 3H C CH O A B C
(d) CH3 C = CH
2 + O
3 A 2Zn/H O
B + C CH
3
(e) 2Zn/H O
3 2 3CH CH CH O A B C
(f) CH3 – C = CH
2
+4KMnO /H
|
CH3
(g) 4MnO /OH
3 3CH CH CHCH
(h) 4dil KMnO
2 2 2273 K
CH CH H O
(i) UV
2 500 K3Cl
(j) 2 2 2CH CH Br ( )aq
7. How will you carry out following conversions :
(a) Ethyne into propyne
(b) Ethyne into benzene
(c) Benzene into 4-nitrobromobenzene
(d) Benzene into n-nitrochlorobenzene
(e) But-2-ene into ethanol
(f) Benzene into acetophenone
(g) Benzene into 4-nitrotoluene
(h) Ethane into ethyne
117 [XI – Chemistry]
(i) Ethanoic acid into methane
(j) Methane into ethane
8. Illustrate by drawing resonance structures that OH attached to benzene
ring, is ortho-and para-directing.
9. Illustrate by drawing resonance structure that NO2 attached to benzene
ring is meta-directing.
10. Benzene undergo electrophilic substitution reaction. Explain the mechanism
of electrophilic substitution in the following steps :
(a) Generation of electrophile
(b) Formation of carbocation intermediate
(c) Removal of proton from carbocation intermediate
11. Give reasons :
(a) AlCl3 is used as catalyst in Friedal-Craft reaction.
(b) Wurtz reaction is carried in dry ether.
(c) Wurtz reaction is not preferred for the preparation of alkanes containing
odd number of carbon atoms.
(d) C-C bond length in benzene ring is 139 pm which is in between C-C
single bond 154 pm and C=C double bond 133 pm.
(e) Benzene is extra ordinarly stable although it constants three double
bonds.
(f) trans-2-butene has higher m.p. than cis-isomer.
(g) Chlorobenene is less reactive towards electrophilic aromatic substitu-
tion than benzene.
(h) Ethyne is more acidic than ethene and ethane.
(i) Kharasch effect is spontaneous with HBr only.
(j) OH is activating group while halogens are deactivating group,
however, both are o-/p-director.
12. Give one chemical test to distinguish between compounds of the following
pairs:
(a) Ethane and ethene
(b) Ethene and ethyne
[XI – Chemistry] 118
(c) Benzene and cyclohexene
(d) But-1-yne and But-2-yne
(e) But-1-yne and Buta-1, 3-diene
13. An alkyl dihalide A, C6H
12Br
2 on dehydrobromination with NaNH
2 gave
compound B.C6H
10. ‘B’ on hydration gave 3-methylpentan-2-one. Suggest
suitable structural formulae for compound A and B.
[Ans. (a) CH – CH – C(CH ) – CH – CH3 2 3 2
Br Br
(b) CH3 – CH
2 – CH(CH
3) – C CH]
14. An alkene C8H
16 on ozonolysis form ozonide which on hydrolysis with Zn
dust form an aldehyde and pentan-2-one as products. Draw the structure of
alkene and write the chemical reaction involved.
[Ans.
CH CH CH3 2 2
C = CH – CH – CH ]2 3
H2
15. According to Huckel’s rule what are the necessary conditions for any system
to be aromatic? Explain why the following systems are aromatic or non-
aromatic.
(i) (ii) = CH2 (iii)
16. A hydrocarbon ‘A’ adds on mole of hydrogen in presence of Pt catalyst to
form n-hexane. When ‘A’ is oxidised with KMnO4, a single carboxylic acid
‘B’ containing three carbon atoms is obtained. Identity ‘A’ and ‘B’.
[Ans. A : CH3CH
2CH = CHCH
2CH
3
B : CH3 CH
2COOH]
17. Write the structures of all possible position and chain isomers of hex-1-yne.
18. Show the formation of sigma and pi bonds in ethyne.
19. A hydrocarbon C5H
10 does not react with chorine in dark but gives a single
monochloro compound C5H
9Cl in bright sunlight. Identify the hydrocarbon.
20. Draw the structures of geometrical isomers of but-2-ene. Which of them is
more polar and why ?