Chapter 9: Hydrocarbons

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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.

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

b.

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

c.

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

d.

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

e. 

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

f. 

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

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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

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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

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The above compound undergo ozonolysis as follows;

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The products of the reaction are ethanal and propanal.

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The above compound undergo ozonolysis as follows;

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

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The above compound undergo ozonolysis as follows;

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

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The above compound undergo ozonolysis as follows;

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The products of the reaction are benzaldehyde and propanal.

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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.

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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.

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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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The combustion reaction is given as;

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

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The combustion reaction is given as;

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

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The combustion reaction is given as;

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

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The combustion reaction is given as;

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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.

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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.