Hydrocarbons

Chlorination of methane takes place following a free radical chain mechanism. The whole reaction mechanism can be represented by the following three steps below: -

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 the 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:-

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

The parent chain in this compound is an alkene with a double bond between 1^st and 2^nd carbon and one methyl group attached to the 2^nd carbon atom. Hence, the name of the compound is - 2-Methylbut-2-ene. The numbering is done as follows : -

The given compound contains both double and triple bond, a double bond between 1^st and 2^nd carbon and the triple bond between 3^rd and 4^th carbon. The numbering is done as follows : -

Since we know a double bond is given more preference than triple bond so, the name of the compound is - Penta-1-ene-3-yne.

The parent chain in this compound is an alkene with a double bond between 1^st and 2^nd carbon and between 3^rd and 4^th carbon. Hence, the name of the compound is - Buta-1,3-diene. The numbering is done as follows : -

The given compound consists of a phenyl ring and a butane side chain with a double bond between 1^st and 2^nd carbon. Hence, the name of the compound is - 4-Phenyl but-1-ene.

The given compound consists of a benzene ring with a methyl substituent at 2^nd carbon and hydroxyl substituent at 1^st carbon. Hence, the name of the compound is - 2-Methyl phenol. The numbering is done as follows : -

The given compound : -

This is a simple hydrocarbon alkane chain with Methylpropyl substitution at 5^th carbon. Hence, the name of the compound is- 5-(2-Methylpropyl)-decane.

The parent chain in this compound is an alkene with a double bond between 5^th and 6^th, 8^th and 9^th carbon and an ethyl substitution at 4^th carbon. Hence, the name of the compound is- 4-Ethyldeca-1, 5, 8-triene. The numbering is done as follows : -

(a) All possible isomers of C₄H₈ having one double bond are as follows : -

(i)

(ii)

(iii)

(b) All possible isomers of C₅H₈ having one triple bond are as follows : -

(i)

(ii)

(iii)

The reaction of Pent-2-ene with ozone is as shown below : -

Hence, the IUPAC names of the products obtained by the ozonolysis are ethanol and propanal.

The reaction of 3,4-Dimethylhept-3-ene with ozone is as shown below : -

Hence, the IUPAC names of the products obtained by the ozonolysis of 3,4-Dimethylhept-3-ene are a product (I) is butan-2-one and Product (II) is Pentan-2-one.

The reaction of 2-Ethylbut-1-ene with ozone is as shown below : -

Hence, the IUPAC names of the products obtained by the ozonolysis of 2-Ethylbut-1-ene are product (I) is pentan-3-one and Product (II) is methanal.

The reaction of 1-Phenylbut-1-ene with ozone is as shown below : -

Hence, the IUPAC names of the products obtained by the ozonolysis of 1-Phenylbut-1-ene are a product (I) is benzaldehyde and Product (II) is propanal.

According to the question the alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3- one. This could be represented as follows: -

Now, we know that during the process of 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 could be:-

This ozonide is being 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:-

3-Ethylpent-2-ene

According to the question alkene ‘A’ on ozonolysis gives two moles of an aldehyde having a 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 drawn above, the structure of ‘A’ can be represented as:

But-2-ene

Ozonolysis of ‘A’ takes place in the following manner:-

The final product is ethanal with molecular mass 44u.

- According to the question 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 predicted as:-

Now, ‘X’ is an addition product of alkene ‘A’ with ozone. Hence, the possible structure of alkene ‘A’ is:-

A combustion reaction takes place in presence of oxygen.

(i) Combustion reaction of Butane is as follows : -

(ii) Combustion reaction of Pentene is as follows : -

(iii) Combustion reaction of Hexyne is as follows : -

(iv)

(v) Combustion reaction of Toluene is as follows : -

Toluene

Hex-2-ene is represented by the following structure:-

H₃C — HC = CH — CH₂ — CH₃

The geometrical isomers of hex-2-ene are:

The dipole moment of cis-compound is actually the sum of the dipole moments of the C–CH₃ and C–CH₂CH₃ bonds acting in together 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.

Hence, cis-isomer is more polar than trans-isomer.

Now, the higher the polarity, the greater is the intermolecular dipole-dipole interaction in between and the higher will be the boiling point.

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

Benzene is the hybrid of the resonating structures shown below:-

Now, all the six carbon atoms in benzene are sp₂ hybridized. The two sp₂ hybrid orbitals of each carbon atom overlap effectively 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 orbitals and their electrons are delocalized and hence can move freely about the six carbon nuclei. Even after the presence of the three double bonds, these delocalized π-electrons stabilize benzene.

The necessary conditions for any system to be aromatic are as follows : -

(i) It should have a planar structure.

(ii) The π–electrons of the compound should be completely delocalized in the ring.

(iii) The total number of π–electrons present in the ring should be equal to (4n + 2),

where n = 0, 1, 2 … etc which is known as Huckel’s rule.

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…).

In this case, the value of n is 1 which is an integer, hence the given compound is aromatic in nature.

For the given compound, the number of π-electrons is 4.

By Huckel’s rule,

⇒ 4n + 2 = 4

⇒ 4n = 2

⇒ n = 1/2

For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…).

This is not true for the given compound as it is a fraction. Hence, it is not aromatic in nature.

For the given compound, the number of π-electrons is 8.

By Huckel’s rule,

⇒ 4n + 2 = 8

⇒ 4n = 6

⇒ n = 3/2

For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…).

This is not true for the given compound as it is a fraction. Hence, it is not aromatic in nature.

Benzene can be converted into p-nitrobromobenzene by following the following reaction procedure:-

Benzene can be converted into m-nitrochlorobenzene by following the following reaction procedure:-

Benzene can be converted into p-nitrotoulene by following the following reaction procedure:-

Benzene can be converted into acetophenone by following the following reaction procedure:-

The given alkane is H₃C – CH₂ – C(CH₃)₂ – CH₂ – CH(CH₃)₂ having the following structure: -

Now, 1° carbon atoms are those which are bonded to only one carbon atom i.e., they have only one carbon atom as their neighbour. Our given structure has five 1° carbon atoms and fifteen hydrogen atoms attached to it.

2° carbon atoms are those which are bonded to two carbon atoms i.e. they have two carbon atoms as their immediate neighbours. Our 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 immediate neighbours. Our given structure has one 3° carbon atom and only one hydrogen atom is attached to it.

Alkanes experience inter-molecular Van der Waals forces. The stronger the force will be, the greater would be the boiling point of the alkane.

As branching increases, the surface area of the molecule decreases resulting 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.

Addition of HBr to propene occurs through an electrophilic substitution reaction mechanism.

This occurs via the following steps : -

Step – 1:

Hydrogen bromide provides an electrophile, H⁺. This electrophile attacks the double bond to form 1° and 2° carbocations as shown below:-

Step – 2: 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:

Now, 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.

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.

o-xylene has the following two resonance structures:-

We can see that all the 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 represented as I and II above.

The acidic character of a species is defined as the ease with which it can lose its H^- atoms.

The hybridization state of carbon in the given compounds is as shown below:-

Now, as the s–character increases, the electronegativity of carbon increases and the electrons of C–H bond pair lies closer to the carbon atom. As a result, the 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.

Benzene is a planar molecule having delocalized electrons above and below the plane of its ring. Hence, it is an electron-rich species, due to which it is highly attractive to electron deficient

species i.e., electrophiles.

Therefore, it undergoes electrophilic substitution reactions very easily.

Nucleophiles are electron-rich species. Hence, they are repelled by benzene(an electron-rich species). Hence, benzene undergoes nucleophilic substitutions with difficulty.

Conversion of Ethylene into benzene takes place by the following reaction procedure : -

Conversion of Ethene into benzene takes place by the following reaction procedure : -

Conversion of Hexane into benzene takes place by the following reaction procedure : -

The parent skeleton structure of 2-methylbutane is : -

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

(i)

(ii)

(iii)

We know that electrophiles are the reagents or species that participate in a reaction by accepting an electron pair in order to bond with nucleophiles.

Also, the higher the electron density of a benzene ring, the more reactive is the compound towards an electrophile, E⁺ (Electrophilic reaction).

(a) The presence of an electron withdrawing group like 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 the reactivity is as follows:-

Chlorobenzene > p – nitrochlorobenzene > 2, 4 – dinitrochlorobenzene

(b) We know that CH₃^- is an electron donating group and NO₂^{- the} 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 decreasing order of the reactivity order is as follows:

Toluene > p-H₃C – C₆H₄ – NO₂> p-O₂N – C₆H₄ – NO₂

The ease with which a nitration reaction takes place depends upon 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 an electron withdrawing group. 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 comparatively more difficulty.

Hence, the increasing order of nitration is as follows:-

The ethylation reaction of benzene happens when an ethyl group is added to 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.

Wurtz reaction is limited for the synthesis of symmetrical alkanes (alkanes having an even number of carbon atoms) In the wurtz reaction, two similar alkyl halides are taken as reactants and an alkane, containing double the number of carbon atoms, are formed as the product, as shown in the reaction below : -

Wurtz reaction cannot be used for the preparation of unsymmetrical alkanes because if two dissimilar alkyl halides are taken as the reacting species, 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. This could be easily understood by the the reaction of bromomethane and iodoethane giving a mixture of alkanes as shown below : -

The alkanes that are obtained in the mixture have very close boiling point making their separation rather more difficult.

Hence, Wurtz reaction is not preferred for the preparation of alkanes containing odd number of carbon atoms.