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Class XI Chapter 13 – Hydrocarbons Chemistry

Page 1 of 23

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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:


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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


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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


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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:


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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:


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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:


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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:


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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.


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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)


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)


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


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+


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


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


(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


(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


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


(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


(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 ?

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