NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

NCERT Exercise

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

Ans:

Chlorination of methane proceeds by free radical chain mechanism which involves three steps as follows:

• Initiation:

The reaction begins with the homolytic bond cleavage within Cl – Cl bond that results in the   formation of chlorine free radicals as;

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• Propagation:

In this step, chlorine free radicals formed in the prior step abstracts a hydrogen atom from methane to generate methyl radicals as;

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The above methyl radicals then react with chlorine molecules to form methyl chloride along with the  liberation of a chlorine free radical.

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Thus, methyl and chlorine free radicals set up a chain reaction. When HCl and  $C{{H}_{3}}Cl$  are formed as major products, other higher halogenated compounds are also formed as follows;

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• Termination:

When all the reactants are consumed, the reaction stops and the chain gets to the point of termination. This happens by the  combination of different free radicals. 

Chlorine free radicals combine to form a chlorine molecule. 

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Methyl free radicals  combine to form ethane.

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

2. Write IUPAC names of the following compounds:

a.

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Ans:

The IUPAC name of the above compound is 2-Methylbut-2-ene.

b.

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Ans:

The IUPAC name of the above compound is Pen-1-en-3-yne.

c.

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Ans:

The IUPAC name of the above compound is Buta-1,3-diene or 1,3-Butadiene.

d.

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Ans:

The IUPAC name of the above compound is 4-Phenylbut-1-ene.

e. 

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Ans:

The IUPAC name of the above compound is 2-Methylphenol.

f. 

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Ans:

The above compound can be easily formulated as;

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The IUPAC name of the above compound is 5-(2-Methylpropyl) decane.

g. 

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Ans:

The IUPAC name of the above compound is 4-Ethyldeca-1,5,8-triene.

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) ${{C}_{4}}{{H}_{8}}$  (one double bond)

Ans: 

The structures with their IUPAC names are given as;

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IUPAC name: But-1-ene

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IUPAC name: But-2-ene

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IUPAC name: 2-Methylprop-1-ene

b) ${{C}_{5}}{{H}_{8}}$  (one triple bond)

Ans:

The structures with their IUPAC names are given as;

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IUPAC name: Pent-1-yne

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IUPAC name: Pent-2-yne

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IUPAC name: 3-Methylbut-1-yne

4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

i) Pent-2-ene

Ans: 

The above compound undergo ozonolysis as follows;

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The products of the reaction are ethanal and propanal.

ii) 3,4-Dimethyl-hept-3-ene

Ans:

The above compound undergo ozonolysis as follows;

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The products of reaction are Butan-2-one and Pentan-2-one.

iii) 2-Ethylbut-1-ene

Ans:

The above compound undergo ozonolysis as follows;

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The products of the reaction are Pentan-3-one and methanal.

iv) 1-Phenylbut-1-ene

Ans:

The above compound undergo ozonolysis as follows;

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The products of the reaction are benzaldehyde and propanal.

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

Ans: 

According to the given data the following reaction format can be formed as:

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

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This ozonide is formed as an addition of ozone to reactant ‘A’. Thus, the desired structure of ‘A’ can be obtained by the removal of ozone from the above ozonide. 

The structural formula of ‘A’ is as follows;

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The IUPAC name of the above compound is 3-Ethylpent-2-ene.

6. An alkene ‘A’ contains three C – C, eight C – H $\sigma $  bonds and one C – C $\pi $  bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Ans:

According to 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 8 C–H $\sigma $  bonds which state that there are 8 hydrogen atoms in ‘A’. Also, there are 3 C–C bonds which state that there are 4 carbon atoms present in the same.

Combining the information given, the structure of ‘A’ can be represented as:

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‘A’ has 3 C-C bonds, 8 C-H $\sigma $bonds along with a C-C$\pi $ bond which states that the IUPAC name of the same is But-2-ene.

Now,

Ozonolysis of ‘A’ takes place as;

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Here, we can say that the final product is being proved as ethanal with molar mass 44u.

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

Ans:

According to the given information, propanal and pentan-3-one are the ozonolysis products of an  alkene. Consider the given alkene as ‘A’. 

Now, writing the reverse of the ozonolysis reaction, we get;

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

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We know that, ‘X’ is an addition product of alkene ‘A’ with ozone. Thus, the possible structure of alkene ‘A’ is:

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

i) Butane

Ans: 

The combustion reaction is given as;

\[2{{C}_{4}}{{H}_{10\left( g \right)}}+13{{O}_{2\left( g \right)}}\to 8C{{O}_{2\left( g \right)}}+10{{H}_{2}}{{O}_{\left( g \right)}}+Heat\] 

ii) Pentene

Ans:

The combustion reaction is given as;

\[2{{C}_{5}}{{H}_{10\left( g \right)}}+15{{O}_{2\left( g \right)}}\to 10C{{O}_{2\left( g \right)}}+10{{H}_{2}}{{O}_{\left( g \right)}}+Heat\] 

iii) Hexyne

Ans:

The combustion reaction is given as;

\[2{{C}_{6}}{{H}_{10\left( g \right)}}+17{{O}_{2\left( g \right)}}\to 12C{{O}_{2\left( g \right)}}+10{{H}_{2}}{{O}_{\left( g \right)}}+Heat\] 

iv) Toluene

Ans:

The combustion reaction is given as;

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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 represented as $C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}$. Geometrical isomers of hex-2-ene are:

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The dipole moment of cis-compound is a sum of the dipole moments of$C-C{{H}_{3}}$ and$C-C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}$ bonds acting in the same direction. The dipole moment of a trans-compound is the resultant of the dipole moments of the same bonds acting in opposite directions.

Thus, 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.

Therefore, cis-isomer will have a higher boiling point than trans-isomer.

10. Why is benzene extraordinarily stable though it contains three double bonds?

Ans:

Benzene has resonating structures which define its stability perfectly. They can be represented as;

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All 6 carbon atoms in benzene are $s{{p}^{2}}$  hybridized. The 2 $s{{p}^{2}}$ hybrid orbitals of each carbon  atom overlaps with the $s{{p}^{2}}$ hybrid orbitals of adjacent carbon atoms to form 6 $\sigma $  bonds in  the hexagonal plane. The remaining $s{{p}^{2}}$ hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form 6 sigma C–H bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming 3 $\pi $  bonds by the lateral overlap of adjacent C atoms.

The 6 $\pi $ electrons are delocalized and can move freely about the 6 carbon nuclei. Even after  the presence of 3 double bonds, these delocalized $\pi $-electrons stabilize benzene.

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

Ans:

A compound is only said to be aromatic if it completely satisfies the following conditions;

• It should have a planar structure and should be cyclic.

• The $\pi $-electrons of the compound must be completely delocalized in the ring.

• The total number of $\pi $-electrons present in the ring should be equal to (4n+ 2), where n = 0, 1, 2 … etc. {Huckel’s rule}.

12. Explain why the following systems are not aromatic?

i) 

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Ans: 

In the given compound, one carbon atom is $s{{p}^{3}}$  hybridized which signifies that it is tetrahedral (not planar). As for the compound to be aromatic, it should be planar. Thus, the given compound is not aromatic in nature.

ii)

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Ans:

In the given compound, one carbon atom is $s{{p}^{3}}$ hybridized which signifies that it is tetrahedral (not planar).  As for the compound to be aromatic, it should be planar. 

Also, for the given compound, the number of $\pi -$ electrons is 4 so, by Huckel’s rule;

\[4n+2=4\]

\[n=\frac{1}{2}\]

For a compound to be aromatic, the value of n must be an integer i.e. 0, 1, 2… etc. which is not  satisfied for the given compound. Therefore, it is not aromatic in nature.

iii) 

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Ans:

For the given compound, the number of $\pi -$electrons is 8 so, by Huckel’s rule;

\[4n+2=8\] 

\[n=\frac{2}{3}\] 

For a compound to be aromatic, the value of n must be an integer i.e. 0, 1, 2… etc. which is not  satisfied for the given compound. Therefore, it is not aromatic in nature.

13. How will you convert benzene into

1. p-nitrobromobenzene

Ans:

The reactions are given as;

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2. m-nitrochlorobenzene

Ans:

The reactions are given as;

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3. p -nitrotoluene

Ans:

The reactions are given as;

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4. acetophenone

Ans:

The reactions are given as;

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14. In the alkane $C{{H}_{3}}-C{{H}_{2}}-C{{(C{{H}_{3}})}_{2}}-C{{H}_{2}}-CH{{(C{{H}_{3}})}_{2}}$,  identify $1{}^\circ ,2{}^\circ ,3{}^\circ $  carbon atoms and give  the number of H atoms bonded to each one of these.

Ans:

The given alkane can be represented as;

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• Primary carbon atoms are those which are bonded to only one carbon atom or none. i.e., they have only 1 carbon atom as their neighbor or none (in case of methane). The given structure has 5 primary carbon atoms and 15 hydrogen atoms attached to it.

• Secondary carbon atoms are those which are bonded to 2 carbon atoms i.e., they have 2 carbon atoms as their neighbors. The given structure has 2 secondary carbon atoms and 4 hydrogen atoms attached to it.

• Tertiary carbon atoms are those which are bonded to 3 carbon atoms i.e., they have 3 carbon atoms as their neighbors. The given structure has 1 tertiary carbon atom and only 1 hydrogen atom is attached to it.

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

Ans:

Alkanes mainly experience intermolecular Van der Waals forces. The stronger the force, the greater the boiling point of the alkane.

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

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 a mechanism.

Ans:

Addition of HBr to propene is an example of an electrophilic addition reaction. Hydrogen bromide provides an electrophile, ${{H}^{+}}$ . This electrophile attacks the double bond to form primary and secondary carbocations as shown:

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Secondary carbocations are comparatively more stable than primary carbocations. Thus, the former predominates since it will form at a faster rate. Thus, now $B{{r}^{-}}$ attacks the carbocation to form 2 – bromopropane as the major product.

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This reaction follows Markovnikov’s rule.

Now,

In the presence of benzoyl peroxide, an additional reaction takes place by anti-Markovnikov’s  rule. The reaction follows a free radical chain mechanism as;

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Here, 1 – bromopropane is obtained as the major product.

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

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

Ans:

o-xylene has two resonating structures showing different reactions as follows;

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The three products are formed i.e., methyl glyoxal, 1,2-demethylglyoxal, and glyoxal 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.

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

Ans:

Acidic character of any species is defined on the basis of its ease with which it can lose the H– atoms. The hybridization state of carbon in the given compound is given as;

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According to the hybridization criterion, as the s–character increases the electronegativity of carbon increases and the electrons of C–H  bond pair lie closer to the C atom.

The s–character increases in the order:

\[s{{p}^{3}}<s{{p}^{2}}<sp\] 

Thus, the decreasing order of acidic behavior is Ethyne > Benzene > Hexane.

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

Ans:

Benzene is a planar molecule having delocalized electrons above and below the plane of the ring.  Thus, this makes it electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles. This is the reason; benzene undergoes electrophilic substitution reactions very easily.

Whereas, nucleophiles are electron-rich. Hence, they are repelled by benzene. Therefore, benzene undergoes nucleophilic substitutions with much difficulty.

20. How would you convert the following compounds into benzene?

i) Ethyne

Ans:

The reactions are given as;

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ii) Ethene

Ans:

The reactions are given as;

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iii) Hexane

Ans:

The reactions are given as;

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

Ans:

The structure of 2-methylbutane can be stated as (skeleton);

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On the basis of the above structure, various alkenes that will give 2-methylbutane on hydrogenation  are given as;

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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

Ans:

Electrophiles are reagents that participate in a reaction by accepting an electron pair in order to  bond to the corresponding nucleophiles.

The higher the electron density on a benzene ring, the more reactive is the compound towards an electrophile.

Here,

The presence of an EWG deactivates the aromatic ring by decreasing the electron density. Now, as the $N{{O}_{2}}^{-}$ group is more EWG that $C{{l}^{-}}$ group.

Thus, the decreasing order of EWG is given as;

Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene

b) Toluene, $p-C{{H}_{3}}-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$ , $p-{{O}_{2}}N-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$ .

Ans:

Here, $C{{H}_{3}}^{-}$ is an EDG and $N{{O}_{2}}^{-}$is an EWG. 

Thus, toluene will have the maximum electron density and is most easily attacked by ${{E}^{+}}$ .

The number of $N{{O}_{2}}^{-}$substituents define the order as;

Toluene > $p-C{{H}_{3}}-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$ > $p-{{O}_{2}}N-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$

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

Ans:

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 ($N{{O}_{2}}^{+}$ ).

Now as we know, $C{{H}_{3}}^{-}$group is electron donating and $N{{O}_{2}}^{-}$ is electron withdrawing. Therefore, toluene will have the maximum electron density among the 3 compounds followed by benzene. On the other hand, m– Dinitrobenzene will have the least electron density. Thus, it will undergo nitration with difficulty. Hence, the increasing order of nitration is as;

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

Ans:

The ethylation reaction of benzene involves the addition of an ethyl group on the benzene ring. This reaction is called Friedel-Craft alkylation reaction and takes place in the presence of a Lewis acid.

Any Lewis acid like anhydrous $FeC{{l}_{3}},SnC{{l}_{4}},B{{F}_{3}}$ etc. can be used during the ethylation of  benzene.

25. Why is the Wurtz reaction not preferred for the preparation of alkanes containing an 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:

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This 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.

Example:

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The boiling points of the above alkanes are very close. Hence, it becomes difficult to separate them.

NCERT Solutions for Class 11 Chemistry Chapter 9 - Hydrocarbons

NCERT Solutions for Class 11 Chemistry – Free PDF Download

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Chapter 9 – The Hydrocarbons 

NCERT Solutions for Class 11 Chemistry Chapter 9

• Hydrocarbon

A compound of hydrogen and carbon is called a hydrocarbon.

• Saturated Hydrocarbon

If the hydrocarbon contains C—C single bonds only.

Example: 

Ethane CH3—CH3 

• Unsaturated Hydrocarbon

• Aromatic Hydrocarbon

The Benzene and its derivatives are known as aromatic compounds.

Example:

• Alicyclic Compounds

Cyclic compounds consisting of carbon atoms only are known as carbocyclic or alicyclic compounds.

• Heterocyclic Compounds

Cyclic compounds where the ring atoms are of some element like N, S or O and carbon. They are known as heterocyclic compounds.

• Alkanes

The simplest organic compounds created from only carbon and hydrogen only are called Alkanes.

The general formula of Alkanes is CnHC2n+2 (where n donates = 1, 2, 3, etc.)

Single covalent bonds bond the C atoms in their molecules. As the carbon skeleton of alkanes is completely saturated with hydrogens, they are also known as saturated hydrocarbons. Alkanes have strong C —H and C —C bonds. Hence, this class of hydrocarbons are comparatively chemically inert. So, they are often called paraffin. 

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• Structure

The carbon of Methane forms single bonds with four hydrogen atoms, so it is called Methane. It has a tetrahedral structure.

• Nomenclature Guidelines

Use the following step-by-step process to write the IUPAC names from the structural formulas of the hydrocarbons. Consider the following structural formula:

• Step 1. The first step is to identify the longest chain in the hydrocarbon: longest chain here has seven carbons. The seven carbon chain is heptane.

• Step 2. Number the chain: The chain has to be numbered from left to right, which will give the lowest numbers to the alkyl group.

• Step 3. Identify the alkyl group in the hydrocarbon. So, there are two methyl groups at C-2 and C-3, there is one ethyl group of C-4.

• Step 4. So, the IUPAC name, in this case, is 4-Ethyl-2,3-dimethyl heptane. Points to note: 

1. Always use commas to separate numbers. 

2. Hyphens are used to separate names from numbers.

3. Prefixes di, tri is not taken into consideration in naming substituent names.

• New Man Projections

The molecule is looked at the C—C bond head-on in this projection

• Relative Stability of Conformations

When Ethane is in the staggered form, then there are maximum repulsive forces and stability of the molecule and minimum energy. On the other hand, the eclipsed way the electron clouds of the C-H bond come closer to each other, which will result in increasing the electron cloud repulsion. Hence, the molecule will have more energy and less stability.

• Alkenes

Alkenes are hydrocarbons that have a double bond (C=C) in their C atoms 

The general formula:

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• Structure

Let us consider (H2C=CH2) for visualizing the orbital make up of alkenes.

In ethylene, the C atoms are sp2 hybridized in the ethylene- The bond attaches them. 

The bond derives from the overlap of two sp2 hybrid orbitals. 

The π bond results from the overlap of the unhybridized p-orbitals. Ethylene is a planar molecule.

Points To Be Noted

1. The carbon-carbon double bond in alkenes consists of one σ and one π-bond.

2. Alkanes are less reactive, and alkenes are more reactive due to the availability of n electrons.

• Nomenclature

In the IUPAC system

1. The longest ‘ C chain of which double bond is a part has to be identified.

2. This chain is seen from the end closer to the double bond, and its position is defined by the number of the C atom not which the double bond originates. 

3. First, mention the name of the parent alkene along with the position number of the double bond and then the names of the other substitutes prefixed to it. 

4. The last one of the corresponding alkane is replaced by-a diene to get the name if there is more than one double bond. 

• Isomerism

Structural Isomerism: Structural isomers are those isomers in which the atoms are organized in a different order but having the same molecular formula. Ethene and propene do not show any structural isomers; however, butene has three structural isomers. 

Among them, two are straight-chain structures but having different positions of double bonds in the molecules. 

Geometrical Isomerism: They are molecules that are locked into their spatial positions concerning one another because of the double bond or a ring structure. For example, consider the following two molecules.

This rotation about the carbon-carbon double bond results in geometrical isomerism. An alkene with a formula RCH=CHR can have two stereoisomers, based on if the two alkyl groups are on the opposite or same sides of the double bond. 

But they are called cis-isomer if they are on the same side and they are called trans-isomer if they are on opposite sides.

• Alkynes

Alkynes are defined by the occurrence of a triple bond in the molecule.

CnH2n-2 is the general formula.

The primary member of this is acetylene, HC=CH, and so they are known as the Acetylenes.

Structure: Consider ethyne (HC=CH) for visualizing the orbital make up of ethyne. The c atoms are sp hybridized in the ethyne. They are bonded by a σ-bond and two π-bonds.

The σ -bond comes from the overlap of two sp hybrid orbitals. 

The π bonds are made from the different overlap of the two p-orbitals from the two adjacent C atoms.

The other sp hybrid orbital makes a σ bond with another carbon or hydrogen atom. 

Ethyne is a linear molecule.

(Image to be added soon)

Points To Be Noted

1. The carbon-carbon triple bond consists of one σ and two π bonds in alkynes.

2. Like alkenes, more exposure of π electrons makes the alkylene undergo different reactions.

Nomenclature

IUPAC System

It is obtained by omitting the suffix-ane of the parent alkane and an addition of a new suffix-yne. 

Carbon chain having the triple bond is – numbered from the end closest to this bond. The position of the triple bond is defined by prefixing the number of carbon before the name of the alkyne.

Preparation

From calcium carbide: Ethyne is prepared when calcium carbide is mixed with water. 

The preparation of calcium carbide is as follows:

From Vicinal Dihalides: The alcoholic potassium hydroxide undergoes dehydrohalogenation when reacted with vicinal dihalides. One molecule of hydrogen halide is dropped to make alkyl halide which, when treated with sodamide results in the formation of alkyne.

• Aromatic Hydrocarbons

These hydrocarbons are also called ‘arenes’. Mostly such compounds contain a benzene ring.

Benzenoids are the category of aromatic compounds having benzene rings. On the other hand, aromatic compounds not having the benzene rings are called non- benzenoids.

Structure of Benzene: C6H6 is the molecular formula for Benzene. This shows that Benzene is too unsaturated. 

The Kekule structure given by Kekule shows the chances of two isomeric 1,2-dibromobenzene. From one of them is When the bromine atoms are attached to the double bond C atoms. However, in the other one, they are attached to the single bond C.

Benzene Was An Amalgamation Of These Two Forms

However, this structure was not successful in explaining the unique stability and its inclination to substitution reaction rather than addition reactions.

Resonance Structure of Benzene: Resonance is a phenomenon where two or more structures can be used for a substance involving the same positions of atoms. 

Benzene's Kekule’s structures (1) and (2) show the resonance structures. The original structure – of the molecule is represented by the mixture of these two structures.

Orbital structure of Benzene: All C atoms in Benzene are sp2 hybridized. The hybrid sp2 orbitals overlap with each other and also with s orbitals of the six hydrogen atoms results in the formation C—C and C—H σ-bonds.

X-ray diffraction data show that Benzene is a planar molecule. The data indicate that all the six Carbon bonds length is of the same order (139 pm) which is intermediate between carbon-carbon single bond (154 pm) and a carbon-carbon double bond (133 pm). Hence, the presence of the double bond in Benzene proves the idea of the reluctance of Benzene to show additional reactions under average condition. Therefore, it justifies the peculiar behaviour of Benzene.

Aromaticity: The property of the sp2 hybridized planar rings where the p orbitals permit the cyclic delocalization of π electrons.

Conditions For Aromaticity:

1. An aromatic compound is planar and cyclic. 

2. Every atom has a p orbital in an aromatic ring. These p orbitals should be parallel so that a consistent overlapping is made around the ring.

3. The cyclic π molecular orbital (electron cloud) which is formed by the overlap of p orbitals should contain (4n + 2) π electrons. 

Where n donates any integer (0, 1, 2, 3, etc.). 

This rule is called the Huckel rule.

Preparation of Benzene: There are some laboratory methods for preparing Benzene.

Physical Properties Of Benzene

1. It is a colourless liquid.

2. It is not soluble in water. However, it is soluble in alcohol, ether, chloroform etc.

3. It is a suitable solvent for many inorganic and organic substances, e.g., resins, fats, sulphur and iodine.

4. It produces a luminous flame while burning in contrast to alkanes and alkenes which usually have a bluish flame.

Chemical Properties

Benzene goes through the following chemical reactions.

1. Electrophilic Substitution Reaction

2. Addition Reaction

Electrophilic Substitution Reactions

When too much of chlorine is treated with Benzene in the presence of ALCL3, then Benzene can be chlorinated to hexachlorobenzene (C6Cl6)

Carcinogenicity and Toxicity: Some polynuclear hydrocarbons having more than two Benzene rings mix together and become more toxic, and they have cancer-producing properties. The incomplete combustion of some organic materials like tobacco, coal and petroleum, etc. leads to the formation of this.

Hydrocarbons: They are compounds of carbon and hydrogen only.

Alkane is an open chain saturated compound. 

Unsaturated Compound - Alkynes Aromatic Compound—Benzene and its derivatives and Alkenes are unsaturated compounds, and they are weak acids.

Conformation: The Spatial arrangements made by rotation around sigma bonds.

Eclipsed Conformation: The more repulsion between bond pairs of electrons makes it less stable.

Staggered: The lesser repulsion between bond pairs of electrons makes it more stable.

Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

Check NCERT Chemistry solutions for Class 11 Chapter 9 offered by Vedantu to know more about hydrocarbons The proper explanation of all types of hydrocarbons and their nomenclature in the notes will improve your knowledge about hydrocarbons. Some of the important features of downloading NCERT solutions Class 11 chemistry hydrocarbons notes are: 

• These notes are brought to you by the experienced faculty.  

• The notes will build a proper understanding of the concepts.

• This will help the students to perform better.

Important Study Material Links for Chapter 9: Hydrocarbons

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