Hydrocarbons Note

Subject Topic Student Manual

Chemistry Hydrocarbons - Notes PU – Notes

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Hydro carbon – 2

Points and formulae to remember

Cycloalkanes are saturated hydro carbons containing ring structures. They have general formula

2n nC H

Example: 3 6C H cyclo propane

4 6C H cyclo butane

Lower member of cyclo alkane are unstable so they undergo ring opening reactions.

Example: Hydrogenation in presence of Ni : catalyst

As the ring size increases temperature required to break the ring increases, due to increase in

stability of cyclo alkane.

Bayer strain theory

Postulate:

1. All cyclo alkane have a planar structure

2. Any deviation of bond angles from the normal tetrahedral angle incurred during the

formation of cycloalkanes would impose internal strain on the ring, which makes the

molecule unstable.

3. Higher the strain, lesser is the stability of cycloalkane

4. Higher the stability of cycloalkane, easier is its formation

The internal strain in cycloalkane can be calculated in terms of angle strain.

Angle strain o 11109 28 bond angle in cycloalkane

2

Cyclo alkane Bond angle Angle stain

Cyclo propane 60 124 44

Cyclo butane 90 19 44

Cyclo pentane 108 10 44

Condition of Baeyer strain theory

Higher the angle strain less is the stability and vice versa

Baeyer strain theory explains the stability of cycloalkanes upto cyclo pentane. However, it fails to

explain the stability of cyclo hexane and higher member

2 H

2 H

2 H

Ni 80 C

Ni 200 C

Ni 300 C

3 2 3CH CH CH

3 2 2 3CH CH CH CH

3 2 2 2 3CH CH CH CH CH

(1)

(2)

(2)



2

According to sachse-mohr theory in cycloalkanes having six or more carbon atoms, the carbon

atoms lie in different planes. Hence the rings become free from strain. These rings are called

strainless rings by attaining nonplanar structure.

The two non planar conformation are

(a) chair form (b) boat form

Here chair form is more stable since all the H and C stay away from each other and no

interaction.

Benzene is quite stable. It does not decolorize cold aqueous potassium permanganate and also

Bromine in carbon tetrachloride.

Kekule in 1865, proposed the cyclic planar structure for benzene

Bond length in benzene is found to be same for all c – c bonds which lies in below that of c – c

single bond (0.154nm) and that of c – c double bond (0.134 nm)

In benzene all the six carbon atoms are 2sp hybridised Benzene contain 12 bond and 3 bond. It

is a planar molecule with a bond angle of 120

Electrophile - H , Cl , Br , 2NO , 3CH , 3SO , 3BF , 3AlCl , etc

Electrophiles in different reactions are

Chlorination - Cl

Nitration - 2NO

Sulphonation - 3SO

Friedell – Crafts reaction - 3CH

Catalyst used in different reaction

Chlorination - 3FeCl

Nitration - 2 4H SO

Friedel – Crafts reaction – Anhydrous 3AlCl

General mechanism of electrophilic substitution reaction:

Step 1: Generation of an eletrophile

Spontaneous dissociation: E Nu E Nu

Acid catalysed dissociation: Catalyst

E Nu C E NuC

Step 2: Attack by the electrophile on the electrons of benzene

E

E H



3

Stabilisation by resonance

Step 3: Elimination of a proton

Hydrocarbon – 2

1. What are cycloalkane?

Ans: Cycloalkanes are saturated hydrocarbons containing ring structure

2. Give the general formula of cycloalkane

Ans: 2n nC H

3. Name the least stable cycloalkane

Ans: Cyclo propane

4. Name the cycloalkane which undergoes hydrogenation at lowest temperature why?

Ans: Cyclo propane, due to high angle strain or least stability

5. What is normal tetrahedral bond angle?

Ans: 1109 28

6. Define angle strain

E H

A

E

HA

E H

E H

E H

E H

(Resonance Hybrid)



4

Ans: It is half the difference between normal tetrahedral angle and bond angle in

cycloalkanes or Angle strain 11109 28 bond angle in cycloalkanes

2

7. Give an expression to calculate the angle strain in cycloalkane

Ans: Angle strain 11109 28 bond angle in cycloalkanes

2

8. Which is the most stable conformation for cyclohexane?

Ans: chair form

9. What are aromatic hydrocarbon?

Ans: Aromatic hydrocarbon are the compounds having at least one benzene ring.

10. What is the hybridisation of carbon atoms in benzene?

Ans. 2sp

11. Why benzene undergoes electrophilic substitution reaction?

Ans: Delocalized electrons of benzene attracts the electrophile and hence it undergoes

electrophilic substitution reaction.

12. State Huckel rule.

Ans: It states that, the compound which has 4 2n e or delocalized electron then it is

aromatic.

Where 1,2,3 no of ringn n

13. Name the electrophile used in the following reaction

(a) chlorination (b) Nitration

(c) Sulphonation (d) Friedel –craft’s reaction

(a) Ans: CI (b) Ans: 2NO

(c)

Ans: 3SO

(d) Ans 3CH

14. What are electrophile?

Ans: The substance or species which is electron deficient.



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15. What is the role of sulphuric acid in nitration?

Ans: Act as catalyst.

16. What is the catalyst used in friedel-crafts reaction?

Ans: Anhydrous 3AlCl

17. How do you prepare propane from cyclopropane?

Ans: Cyclopropane when heated with hydrogen in presence of Nickel: catalyst at 80 c

80 /

2 3 2 3.c NiH CH CH CH

18. Why cyclopentane require high temperature of 300C than cyclopropane which is at 80

C.

Ans: Due to less angle strain and high stability of cyclopentane than cyclopropane.

19. Name the compound obtained when ozone reacts with benzene?

Ans: Benzene triozonide

20. According to Baeyer’s strain theory which cyclo alkane has least angle strain?

Ans: Cyclopentane.

21. Give one limitation of Baeyer’s strain theory?

Ans: Applicable only for lower cycloalkane till cyclopentane having planar structure. But not

for higher cycloalkane, having nonplanar structure.

Two Mark Questions:-

1. Calculate the angle strain in the following cycloalkane.

(a) Cyclopropane (b) Cyclobutane (c) Cyclopentane (d) Cyclohexane

(a) Cyclopropane:- Angle strain11

109 28 bond angle2 . [½ Mark (Formula)]

Angle strain 1 11

109 28 60 24 442

[½ mark Answer]

Calculation can be done for b, c & d. [½ mark substitution]



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2. What are the limitation of Baeyer’s strain theory? [½ mark each]

(a) The theory proposes a planar model to all cycloalkanes.

(b) The theory is invalid for cyclohexane & higher cycloalkane

(c) Cyclopropane can be more easily formed than butane, which contradict to the theory.

(d) Presence of some substituents stabilise the cycloalkane which cannot be explained by the

theory.

3. Explain Sache-Mohr theory of strainless rings?

Ans: According to this concept a cycloalkane ring becomes free from strain when all the

carbon atoms of the ring are not forced into one plane.

When the carbon atoms are present in different planes the normal tetrahedral angle can be

retained giving a strainless ring.

This strainless ring formation can be seen in higher cycloalkane from cyclohexane. Two non

planar structure of cyclohexane. [1mark]

In chair form of cyclohexane the four carbon 2 3 5 6, , ,c c c c are in the same plane. Carbon atom

1C is above the plane and 4C below the plane. Hence hydrogen atoms are in the opposite

corner. So there is no interaction between them. Hence it is highly stable.

In boat form of cyclohexane the four carbon 2 3 5 6, , ,c c c c are in the same plane. Carbon atoms

1 4&C C are above the plane. Hence hydrogen atoms attached to 1 4&C C have interaction

which reduces stability. Hence less stable than chair form. [1mark]

H H 1

H

4

2

H

6

H H

5

H

H

H 3

H 4

H

H H

4

H

H

H

5

H 3

H

2 H

H

1 H

H 6 H

Chain form Boat form

H



7

In both cases the bond angle of 1109 28 is retained.

4. Write the chair and boat form of cyclolecane

Ans: refer for structure in question (3) 1Mark

5. What are the requirements of a molecule to be aromatic?

1) The compound should have ring structure with alternate single and double bond so that

there is delocalization of electron

2) They should have planar structure

3) It should obey Huckel rule, i.e. if should have 4 2n e , where 0,1,2.....n

6. Give a reaction to show that a molecule of benzene contains only three double bonds. One

mole of benzene on catalytic hydrogenation reacts with 3 mole of hydrogen to form

cyclohexane.

7. Give a reaction to show that all the hydrogen atoms in benzene are similar.

Ans: Benzene react with chlorine to give one meno substituted product, chloro benzene in

presence of 3Fecl .

8. Explain with an example each:

(a) Nitration:

When benzene is heated with a mixture of concentrated nitric and concentrated sulphuric

acid to 050 55to c to give nitro benzene 1 Mark

1 Mark

(b) Chlorination: Benzene reacts with chlorine in presence of 3FeCl at 0180 C to from chloro

benzene 1 Mark

1 Mark

03180 /

2Fecl

Cl

Cl

HCl

02

1803

NiH

32

Feclcl

+ HClCl

2 43

concH SOHNO

2NO

2H O



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(c) Sulphonation:

Benzene reacts with sulphuric acid at 080 c to from benzene sulphonic acid 1 Mark

9. How do you convert benzene to toluene:

Benzene react with methyl chloride in the presence of anhydrous 3AlCl , which act as catalyst

to from toluene. 1 Mark

1 Mark

Note: For method of preparation or chemical property, 1 Mark is allotted for explanation and

1 Mark for balanced chemical equation with all conditions.

0802 4

cH SO

3SO H

2H O

33

anhy AlclCl CH

3CH

HCl



9

Three or Four question: 12

1. Explain Baeyers strain theory.

T, P and Catalyst

(1) The carbon atoms constituting the ring lie in the same plane and thus, all the cyuoalkanes

are planar. 1 Mark

(2) The deviation of bond angle from the normal tetrahedral angle 0 1109 28 causes a strain

in the ring and this is called angle strain.

(3) Greater the angle strain, greater is the instability of the ring.

(4) Higher the stability of the ring, greater would be its ease of formation.

2. Give the structural elucidation of benzene. 1 2 Mark

(1) The elemental analysis and molecular analysis shows the formula of benzene in 6 6C H

(2) One mole of benzene on catalytic hydrogenation reacts with 3mole of hydrogen,

chlorination with 3mole of chlorine and ozanalysis with 3mole of ozone to form

cyclohexane, benzenehexachloride and benzene respectively, show presence of 3 double

bonds.

(3) It does not decolourise alkaline 4KMnO , 2Br in 4CCl shows that the 3 double bonds are

not free to react

(4) In presence of 3FeCl catalyst benzene reacts with 2Cl to form one type of

monosubstituted chloro benzene. This shows that all the hydrogen are identical in

benzene.

(5) Kekule proposed a structure of benzene with alternate single and double bond

(6) From the above structure the disubstituted product obtained at ortho position should

have two different product

But both are found to be identical.

or

Cl

Cl Cl

Cl

I II



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(7) Kekule suggested that double band are in rapid oscillation and two structures are in

dynamic equilibrium

(8) Dynamic equilibrium in both structure is possible then the 3 single bond should be of

1.54 and 3 double bond should be of 1.34 nm. But in benzene it was found to 1.38 nm.

Hence the concept of equilibrium was ruled out. The new concept of resonance was

consider to be as resonance hybride due to delocalization of e .

3. Describe the molecular orbital theory for the structure of benzene

The electronic configuration of carbon in the ground state is

In the excited state of carbon, one of the 2s electrons gets promoted 2p orbital.

So the electronic configuration of carbon in the excited state is

In benzene all the six carbon atoms undergo 2sp hybridization giving

three 2sp hybrid orbitals. These three hybrid orbitals are present in a

plane with angle of separation of 120 . Two of such orbitals of each

carbon atoms overlap with the hybrid orbitals of two adjacent carbon

atoms forming two carbon-carbon sigma bonds. The third 2sp hybrid

orbital of each carbon atom overlaps with the 1s orbita l of hydrogen

forming carbon-hydrogen sigma bond. Hence each carbon atom

forms three sigma bonds.

Now each carbon atom contains one unhybridised p orbital containing an unpaired electron and lying

perpendicular to the plane of hybrid orbitals. These p orbitals overlap sideways forming pi bonds. The

p orbitals on each of the six carbon atoms can overlap sideways with the p orbital present on either side in



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two different ways as shown below.

This causes resonance to take place and the electrons get completely

delocalised. The electrons can freely move about the six carbon nuclei

instead not only two carbon nuclei. Hence benzene is a planar molecule,

with the electron cloud above and below the plane of the ring as

shown in the adjacent structure

The orbital in each of carbon which is not hybridised is used for bond in benzene.

The e in benzene is in continuous delocalization in the module forming electron cloud.

4. Explain the mechanism of

(a) Chlorination:

Benzene reacts with chlorine in presence of 3Fecl to form chloro benzene.

Mechanism:

Step1: Generation of the electrophile cl

34

Cl Cl FeCl Cl FeCl

1 Mark

1S

1S

1S

1S

1S

1S

H H

H

H

H

H

2SP

2SP2SP

2SP

2SP

2SP

2SP

2SP2SP

2SP

2SP

2SP



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Step 2: Attack of the electrophile Cl to benzene

1 Mark

Resonance stabilization

Step 3: Elimination of proton

1 Mark

(b) Nitration

Mechanism:

Step1: Generation of electrophile 2NO 1 Mark

3 2 4 2 2 4HNO H SO NO H O HSO

Step2: Attack of electrophile 2NO to benzene 1 Mark

Resonance stabilization:

+

H

+

H

+

H

+

HNO2 NO2 NO2 NO2

Cl

Cl

Cl

Cl

Cl

Cl

ClH

4 FeCl

Cl

3 FeCl HCl

H

2NO

22NO



13

Step3:

Elimination of proton

1 Mark

(c) Sulphonation

Mechanism:

Step1: Generation of electrophile 3SO

2 4 3 3 42H SO SO H O HSO 1 Mark

Step 2: Attack of electrophile 3SO on benzene

Resonance hybrid

1 Mark

Step 3:

1 Mark

+

H

+ +

NO2NO2

4HSO H2SO4

3SO +

3SO

+

3SO

H

+

3SO

H

+

H3SO

3SO

+H

4HSO 2 4H SO

3SO3SO



14

Step 4: Benzene sulphonate anion is protorated by 3H o to form Benzene sulphonic acid

3H O

3SO H3SO

2H O

(d) Friedel-Graft Alkylation reaction:-

Benzene reacts with alkyl halide in presence of anhydrous 3AlCl to form alkyl benzene

Mechanism:

Step 1: Generation of electrophile

Methyl chloride reacts with anhydrous 3AlCl to form methyl carbocation

3 3 3 4CH Cl AlCl CH AlCl

Step -2: Attack of electrophile on benzene

The electrophile methyl carbocation attacks electrons of benzene to form arenium ion

Stabilized by reactions:

Step 3: Elevation of proton

The arenium ion formed loses a proton to 4AlCl to regain the aromatic character and to form

tolerance as the final product and regenerating the catalyst.

3 CH

3CH

3CH

3CH

3CH

3CH

3CH H

4 AlCl

3CH

3 AlCl HCl



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