Answer

To account for the formation of ethane during the chlorination of methane, we can break down the process into several key steps: initiation, propagation, and termination. Here’s a step-by-step explanation:

Step 1: Initiation

The chlorination process begins with the initiation step, where chlorine gas $\left(\mathrm{Cl}_2\right)$ is exposed to energy (usually heat or light, represented as Hv ). This energy causes the $ \mathrm{Cl}-\mathrm{Cl}$ bond to break through a process called homolytic cleavage, resulting in the formation of two chlorine free radicals $(\mathrm{Cl} \bullet)$.

Step 2: Propagation

In the propagation step, one of the chlorine free radicals $(\mathrm{Cl} \cdot)$ reacts with methane $\left(\mathrm{CH}_4\right)$. The chlorine radical abstracts a hydrogen atom from methane, resulting in the formation of a methyl radical $\left(\mathrm{CH}_3 \bullet\right)$ and hydrochloric acid (HCl).

Reaction:

$ \mathrm{Cl} \bullet+\mathrm{CH}_4 \rightarrow \mathrm{CH}_3 \bullet+\mathrm{HCl} $

The newly formed methyl radical $\left(\mathrm{CH}_3 \bullet\right)$ can now react with another chlorine molecule. The methyl radical attacks another $ \mathrm{Cl}_2$ molecule, leading to the formation of chloromethane $\left(\mathrm{CH}_3 \mathrm{Cl}\right)$ and regenerating another chlorine radical.

Reaction:

$ \mathrm{CH}_3 \bullet+\mathrm{Cl}_2 \rightarrow \mathrm{CH}_3 \mathrm{Cl}+\mathrm{Cl} \bullet $

Step 3: Formation of Ethane

The key to understanding the formation of ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$ lies in the termination step. If two methyl radicals $\left(\mathrm{CH}_3 \bullet\right)$ collide, they can combine to form ethane.

Reaction:

$ \mathrm{CH}_3 \bullet+\mathrm{CH}_3 \bullet \rightarrow \mathrm{C}_2 \mathrm{H}_6 $

Step 4: Termination

In the termination step, free radicals combine to form stable products. Various combinations can occur, including: Two methyl radicals forming ethane $\left(\mathrm{C}_2 \mathrm{H}_6\right)$. A chlorine radical combining with a methyl radical to form chloromethane $\left(\mathrm{CH}_3 \mathrm{Cl}\right)$. Two chlorine radicals recombining to form chlorine gas $\left(\mathrm{Cl}_2\right)$.

Answer

(a) $CH3CH=C(CH3)2$

The IUPAC name of the compound is : 2 -Methylbut-2-ene

(b) ${C} H_2={C} H-{C} \equiv {C}-{C} H_3$

The IUPAC name of the compound is : Pen-1-ene-3-yne

(c)

The IUPAC name of the compound is : 1,3-Buitadiene or Buta-1,3-diene

(d)

The IUPAC name of the compound is :1-(but-3-enyl)benzene

(e)

The IUPAC name of the compound is : 2-Methyl phenol

(f)

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

(g)

$CH_3-CH=CH-CH_2-CH=CH-\underset{\large{C_2H_5}}{\underset{|\quad}{CH}}-CH_2-CH=CH_2$

The IUPAC name of the compound is : 4-Ethyldeca-1,5,8-trieyl

Answer

(a) The following structural isomers are possible for $C_4 H_8$ with one double bond :

(I) $ H_2 {C}= {C} H -{C} H_2-{C} H_3 $

(II) ${C} H_3-{C} H={C} H-{C} H_3$

(III) ${C} H_2=\underset{CH_3}{\underset{\text{|}}C}-{C} H_3$

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 $C_5 C_8$ with one triple bond :

(I) $ H {C} \equiv {C} H_2-{C} H_2-{C} H $

(II) $ H_3 {C}-{C} \equiv {C}-{C} H_2-{C} H_3 $

(III) $ H_3 {C-} \underset{CH_3}{\underset{\text{|}}{{C}}}H-{C} \equiv {C}H $

The IUPAC name of

Compound (I) is Pent-1-yne,

Compound (II) is Pent-2-yne, and

Compound (III) is 3-Methylbut-1-yne.

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 :

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 :

The IUPAC name of Product (I) is benzaldehyde and Product (II) is propanal.

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:

The IUPAC name of ‘A’ is 3-Ethylpent-2-ene.

Answer

As per the given information, ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass $44 \hspace{0.5mm} u$. 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 $\sigma$ 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 $\sigma$ bonds, and one C-C $\pi$ bond.

Hence, the IUPAC name of ’ $A$ ’ is But-2-ene.

Ozonolysis of ‘A’ takes place as :

The final product is ethanal with molecular mass $=[(2 \times 12)+(4 \times 1)+(1 \times 16)]$

$ \hspace{8.3cm} =44 \hspace{0.5mm} u$

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 :

Answer

Combustion can be defined as a reaction of a compound with oxygen.

(i) Butane

$ \underset{Butane} {2 C_4 H _{10}}(g) +13 O _{2}(g) \longrightarrow 8 CO _{2}(g)+10 H_2 O _{}(g)+\text{ Heat } $

(ii) Pentene

$ \underset{Pentene} {2 C_5 H _{10}}(g) +15 O _{2}(g) \longrightarrow 10 CO _{2}(g)+10 H_2 O _{}(g)+\text{ Heat } $

(iii) Hexyne

$ \underset{Hexyne}{2 C_6 H _{10}}(g)+17 O _{2}(g) \longrightarrow 12 CO _{2}(g)+10 H_2 O _{}(g)+\text{ Heat } $

(iv) Toluene

$ \underset{Toluene} {C _ 7 H _ 8}(g)+9 O _{2}(g) \longrightarrow 7 CO _{2}(g)+4 H_2 O _{}(g)+$ Heat

Answer

Hex-2-ene is represented as :

$ CH_3-CH=CH-CH_2-CH_2-CH_3 $

Geometrical isomers of hex-2-ene are :

The dipole moment of cis-compound is a sum of the dipole moments of C- $CH_3$ and C- $CH_2 CH_2 CH_3$ bonds acting in the same direction.

The dipole moment of trans-compound is the resultant of the dipole moments of C- $CH_3$ and $C- CH_2 CH_2 CH_3$ bonds acting in opposite directions.

Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipoledipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than transisomer.

Answer

Benzene is a hybrid of resonating structures given as :

All six carbon atoms in benzene are $s p^{2}$ hybridized. The two $s p^{2}$ hybrid orbitals of each carbon atom overlap with the $s p^{2}$ hybrid orbitals of adjacent carbon atoms to form six 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 six sigma $ C- H $ bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three $\pi$ bonds by the lateral overlap of

$C_1-C_2, C_3-C_4, C_5-C_6$ or $C_2-C_3, C_4-C_5, C_6-C_1$.

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

Answer

A compound is said to be aromatic if it satisfies the following three conditions :

(i) It should have a planar structure.

(ii) The $ \pi $ -electrons of the compound are completely delocalized in the ring.

(iii) The total number of $\pi$ -electrons present in the ring should be equal to $(4 n+2)$, where $n=0,1,2 \ldots$ etc. This is known as Huckel’s rule.

Answer

(i)

For the given compound, the number of $\pi$-electrons is six. But only four $\pi$-electrons are present within the ring. Also there is no conjugation of $\pi$-electrons within the ring and the compound is not planar in shape. Hence, the given compound is not aromatic in nature.

(ii)

For the given compound, the number of $\pi$-electrons is four.

By Huckel’s rule,

$4 n+2=4$

$4 n=2$

$n=\dfrac{1}{2}$

For a compound to be aromatic, the value of $n$ must be an integer $(n=0,1,2 \ldots)$, which is not true for the given compound. Hence, it is not aromatic in nature.

(iii)

For the given compound, the number of m-electrons is eight.

By Huckel’s rule,

$4 n+2=8$

$4 n=6$

$n=\dfrac{3}{2}$

For a compound to be aromatic, the value of $n$ must be an integer $(n=0,1,2 \ldots)$. Since the value of $n$ is not an integer, the given compound is not aromatic in nature.

Answer

(i) Benzene can be converted into $p$-nitrobromobenzene as :

(ii) Benzene can be converted into $m$-nitrochlorobenzene as :

(iii) Benzene can be converted into $p$-nitrotoulene as :

(iv) Benzene can be converted into acetophenone as :

Answer

$1^{\circ}$ 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 $1^{\circ}$ carbon atoms and fifteen hydrogen atoms are attached to it.

$2^{\circ}$ 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 $2^{\circ}$ carbon atoms and four hydrogen atoms are attached to it.

$3^{\circ}$ 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 $3^{\circ}$ carbon atom and only one hydrogen atom is attached to it.

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.

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 $1^{\circ}$ and $2^{\circ}$ carbocations as shown :

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.

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.

Answer

o-Xylene may be regarded as a resonance hybrid of the following two kekule structures. Ozonolysis of each one of these gives two products as shown below :

Thus, in all, three products are formed . Since, all the three products cannot be obtained from any one of the two Kekule structures, this shows that o-xylene is a resonance hybrid of the two Kekule structures (I and II)

Answer

The hybridization state of carbon in these three compounds is

Since, s-electrons are closer to the nucleus, therefore, as the s-character of the orbital making the C - H bond increases, the electrons of C-H bond lie closer and closer to the carbon atom.

In other words, the partial +ve charge on the H -atom increases and hence the acidic character increases as the s-character of the orbital increases.

Thus, the acidic character decreases in the order: Ethyne > Benzene > Hexane

Answer

Benzene $( C_6 H_6 )$ is a planar, cyclic molecule with a ring structure. Each carbon atom in benzene is $ \mathrm{sp}^2$ hybridized, leading to a planar arrangement.

The benzene ring contains delocalized $\pi$ electrons, which are spread over the entire ring structure. This delocalization creates a region of high electron density above and below the plane of the ring.

Electrophilic Substitution Reactions:

Electrophiles are species that are electron-deficient and seek out electrons. Since benzene is electron-rich due to its delocalized electrons, it readily attracts electrophiles.

The presence of electron-rich $\pi$ electrons makes benzene highly reactive towards electrophiles, facilitating electrophilic substitution reactions. This is why benzene undergoes electrophilic substitution reactions easily.

Nucleophilic Substitution Reactions:

Nucleophiles are species that are electron-rich and have a tendency to donate electrons. Since benzene itself is electron-rich, it repels other electron-rich species (nucleophiles).

This repulsion makes it difficult for nucleophiles to attack the benzene ring, resulting in nucleophilic substitution reactions being less favorable or difficult to carry out.

So, benzene undergoes electrophilic substitution reactions easily due to its electron-rich nature, which attracts electron-deficient electrophiles. Conversely, nucleophilic substitution reactions are difficult because the electron-rich nucleophiles are repelled by the electron-rich benzene.

Answer

(i) Benzene from Ethyne :

(ii) Benzene from Ethene :

(iii) Hexane to Benzene

Answer

The basic skeleton of 2-methylbutane is shown below :

$\underset{\text{2-Methylbutane}}{CH_3 - \stackrel{\large{CH_3}}{\stackrel{|\quad }{CH}}-CH_2-CH_3}$

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

Answer

(a)

Chlorobenzene $ (C_6H_5Cl) $

2,4-Dinitrochlorobenzene $ (C_6H_3Cl(NO_2)_2) $

p-Nitrochlorobenzene $ (C_6H_4Cl(NO_2)) $

Nature of Substituents:

Chlorine $ (Cl) $ is an electron-withdrawing group $ (EWG) $ due to its electronegativity.

Nitro groups $ (NO_2) $ are strong electron-withdrawing groups.

Analyze the Electron Density on the Benzene Ring:

Chlorobenzene has one EWG $ (Cl) $ , which decreases the electron density but not significantly.

p-Nitrochlorobenzene has two $ EWGs $ $ (Cl ~and ~ NO_2)$ , which will further decrease the electron density.

2,4-Dinitrochlorobenzene has three EWGs $ (Cl ~and ~ two ~ NO_2) $ , leading to the lowest electron density.

Order of Reactivity:

More electron density on the benzene ring leads to higher reactivity with electrophiles.

Therefore, the order of decreasing reactivity is:

$ Chlorobenzene > p-Nitrochlorobenzene > 2,4-Dinitrochlorobenzene $

(b)

Compounds: Toluene, $ p-H_3C-C_6H_4-NO_2 ~, ~ p-O_2N-C_6H_4-NO_2 $

Toluene $ (C_6H_5CH_3) $

$ p-H_3C-C_6H_4-NO_2 $ (p-Nitrotoluene)

$ p-O_2N-C_6H_4-NO_2 $ (p-Dinitrobenzene)

Nature of Substituents:

Methyl $ (CH_3) $ in toluene is an electron-donating group $ (EDG) $ .

Nitro groups $ (NO_2) $ are strong electron-withdrawing groups.

Analyze the Electron Density on the Benzene Ring:

Toluene has an $ EDG $ $ (CH_3) $ , increasing the electron density and making it very reactive.

p-Nitrotoluene has one $ EDG $ $ (CH_3) $ and one $ EWG $ $ (NO_2) $ , leading to moderate electron density.

p-Dinitrobenzene has two $ EWGs $ $ (NO_2) $ , significantly decreasing the electron density.

Order of Reactivity:

The order of decreasing reactivity with electrophiles is:

$ Toluene > p-Nitrotoluene > p-Dinitrobenzene $

Answer

To determine which compound among benzene, m-dinitrobenzene, and toluene undergoes nitration most easily, we need to analyze the electronic effects of the substituents on the benzene ring. Nitration is an electrophilic substitution reaction where an electrophile (the nitronium ion, $ NO_2^+ $ ) attacks the benzene ring.

Nitration is an electrophilic substitution reaction where the nitronium ion $ (NO_2^+) $ acts as the electrophile. The reaction requires a benzene ring that is electron-rich to facilitate the attack by the electrophile.

Benzene:

Benzene $ (C_6H_6) $ has no substituents. It is a neutral compound and has a moderate electron density. Therefore, it can undergo nitration, but not as easily as compounds with electron-donating groups.

Toluene:

Toluene $ (C_6H_5CH_3) $ has a methyl group $ (-CH_3) $ as a substituent. The methyl group is an electron-donating group through the $ +I $ (inductive) effect. This increases the electron density on the benzene ring, making it more reactive towards electrophiles like the nitronium ion. Thus, toluene will undergo nitration more easily than benzene.

m-Dinitrobenzene:

m-Dinitrobenzene $ (C_6H_3(NO_2)_2) $ has two nitro groups $ (-NO_2) $ in the meta position. Nitro groups are strong electron-withdrawing groups due to their $ -M $ (mesomeric) and $ -I $(inductive) effects. They decrease the electron density on the benzene ring, making it less reactive towards electrophilic substitution. Therefore, m-dinitrobenzene will undergo nitration the least easily among the three compounds.

Comparing Reactivity:

Toluene (most reactive due to electron-donating methyl group)

Benzene (moderately reactive)

m-Dinitrobenzene (least reactive due to electron-withdrawing nitro groups)

So , toluene will undergo nitration most easily, followed by benzene, and then m-dinitrobenzene.

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 $FeCl_3, SnCl_4, BF_3 $ etc. can be used during the ethylation of benzene.

Answer

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 :

$ \underset {Bromomethane}{CH_3-Br} +2 Na+Br-CH_3 \xrightarrow{\text{ Dry ether }} \underset {Ethane}{CH_3-CH_3} +2 NaBr $

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