NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

Question 9.1 How do you account for the formation of ethane during the chlorination of methane?

Answer:

Chlorination of methane occurs by a free radical mechanism and it takes place in three steps-
(i) Initiation-
First, homolytic cleavage of the $Cl-Cl$ bond as

(ii) Propagation-
Chlorine-free radicals attack methane molecules and generate methyl-free radicals as

This methyl radical reacts with other molecules of chlorine ( $Cl-Cl$ ) to form methyl chloride and liberate chlorine-free radicals.

Then the chlorine free radical reacts with methyl chloride and this way propagation occurs.

(iii) Termination-
Ethane will be produced as a final product in this step. When two methyl free radicals react with each other, ethane will be formed.

Question 9.2(a) Write IUPAC names of the following compounds

$C{H}_{3}CH=C(C{H}_3{)}_2$

Answer:

The parent chain has 4 carbons and the double bond is at C-2, so it but-2-ene and also methyl group is also present C-2.

$C{H}_{3}CH=C(C{H}_3{)}_2$
The IUPAC name of the given compound is 2-methylbut-2-ene

Question 9.2(b) Write IUPAC names of the following compounds

$C{H}_2=CH-C\equiv C-C{H}_3$

Answer:

Both alkene and alkyne are present but the priority will go to the alkene group.

$C{H}_2=CH-C\equiv C-C{H}_3$
The IUPAC name of the compound is Pent-1-ene-3-yne

Question 9.2(c) Write IUPAC names of the following compounds

Answer:

The IUPAC name of the compound is 1, 3-Butadiene or Buta-1, 3-diene. It is a symmetrical compound so the numbering cam be done from either end.

Question 9.2(d) Write IUPAC names of the following compounds :

Answer:

The parent chain will be the aliphatic carbon and benzene will be the side group. The IUPAC name of the compound is 4-phenylbut-1-ene.

Question 9.2(e) Write IUPAC names of the following compounds :

Answer:

Here there are two functional groups hydroxy and methyl. The priority will go to the hydroxy group.

The IUPAC name of the compound is 2-methylphenol

Question 9.2 (f) Write IUPAC names of the following compounds :

Answer:

The parent chain has ten carbons so decane (longest chain). The IUPAC name of the compound is 5-(2-methylpropyl)-decane.

Question 9.2(g) Write IUPAC names of the following compounds :

Answer:

The parent chain has ten carbons and the numbering starts where the alkene give the least number. The IUPAC name of the compound is 4-ethyl-1,4,8-deca triene.

Question 9.3 For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of the double or triple bond as indicated :

$C_4{H}_8$ ( one double bond)

Answer:

$C_4{H}_8$ the following isomers are possible with one double bond ;

	But-1-ene
	But-2-ene
	2-Methylprop-1-ene

Question 9.4(i) Write IUPAC names of the products obtained by the ozonolysis of the following compounds :

Pent-2-ene

Answer:

Ozonolysis of Pent-2-ene gives two product,s both are aldehyde compounds. In this process, the ozone molecules attach to the double bond of the molecule and break it into two products.

The IUPAC name of the compounds-(i) ethanal (ii) propanal

Question 9.4 (ii) Write IUPAC names of the products obtained by the ozonolysis of the following compounds :

3,4-Dimethylhept-3-ene

Answer:

Ozonolysis of 3,4-dimethylhept-3-ene gives two products of keto-compounds. The ozone molecules attach to the double bond.

The IUPAC name of the compound is -(i) Butan-2-one (ii) Pentan-2-one

Question 9.4(iii) Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

2-Ethylbut-1-ene

Answer:

Ozonolysis of 2-ethylbut-1-ene gives two products, one is a keto compound and another is an aldehyde.

.
The IUPAC name of one compound is Pentan-3-one and the other compound is methanal.

Question 9.4 (iv) Write IUPAC names of the products obtained by the ozonolysis of the following compounds :

1-Phenylbut-1-ene

Answer:

Ozonolysis of 1-Phenylbut-1-ene gives two aldehydes. One is aromatic in nature and the other is an aliphatic aldehyde.

The IUPAC name of the compound is-(i) Benzaldehyde (ii) propanal

Question 9.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3- one. Write structure and IUPAC name of ‘A’.

Answer:

In the process of ozonolysis, an ozonide, a cyclic ring structure intermediate, is formed, which undergoes cleavage to give the product. The compound A produces pentan-3-one and ethanal. So, the possible structure of A should be

Thus, by removing the ozone from ozonide, we can get the parent alkene structure.

Question 9.6 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’.

Answer:

As per the given data, compound A on ozonolysis gives two moles of aldehyde, having a molar mass of 44u. It indicates that the compound is symmetrical. So, the possible general structure of
A = $YC=CY$
There are eight $C-H$ sigma bonds, which means eight hydrogen atoms in the structure A. Also, three $C-C$ bonds that indicates the presence of four carbon atoms in A.
Now, by combining all the observations, the structure of the A would be

The ozonolysis reaction is shown below-

The atomic mass of ethanal is 44 u.

Question 9.7 Propanal and pentane-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Answer:

In the process of ozonolysis, an ozonide (a cyclic ring structure intermediate) is formed, which undergoes cleavage to give the product. The parent compound produces pentan-3-one and propanal. So, the possible structure should be

Here in the above structure, the right side will give Pent-3-one and the left-hand side will give propanal.

Thus, by removing the ozone from ozonide, we can get the parent alkene structure.
Therefore, the structure of the parent alkene is

(3-ethyl-3-hexene)

Question 9.8(i) Write chemical equations for the combustion reaction of the following hydrocarbons:

Butane

Answer:

Combustion is a chemical reaction in which a substance reacts with oxygen releasing energy in the form of heat and light. It is commonly seen in burning fuels like wood, petrol or gas.

Combustion means the reaction of a compound with dioxygen( $O_2$)
$C{H}_3-(C{H}_2{)}_2-C{H}_3(g)+13O_2(g)\to 8C{O}_2(g)+10H_{2}O(g)+Heat$

Question 9.8(ii) Write chemical equations for combustion reaction of the following hydrocarbons:

Pentene

Answer:

Combustion means the reaction of a compound with dioxygen( $O_2$ ). Combustion reaction of the Pentene-
$C{H}_3-(C{H}_2{)}_3-C{H}_3(g)+15O_2(g)\to 10C{O}_2(g)+10H_{2}O(g)+Heat$

Question 9.8 (iii) Write chemical equations for combustion reaction of the following hydrocarbons:

Hexyne

Answer:

Combustion means the reaction of the given compound with dioxygen($O_2$) that gives carbon dioxide, water molecules and produces some amount of heat.

Structure of hexyne

The reaction

$2C_6{H}_{10}+17O_2\to 12C{O}_2+10H_{2}O+heat$

Question 9.8(iv) Write chemical equations for combustion reaction of the following hydrocarbons:

Toluene

Answer:

Combustion is the reaction of the given compound with dioxygen( $O_2$ ) to give carbon dioxide, water molecules and produces some amount of heat.

Structure of toluene

$C_6{H}_5(C{H}_3)+9O_2\to 7C{O}_2+4H_{2}O+heat$

Question 9.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Answer:

The structure of the hex-2-ene is shown here-

Now, the Geometrical, cis and trans isomers -


Cis-isomer has a higher boiling point than trans-isomers due to more dipole-dipole interactions between the molecules. The cis-form is more polar than the trans-form because it has a higher dipole moment. The trans-form is almost non-polar.

Question 9.10 Why is benzene extraordinarily stable though it contains three double bonds?

Answer:

Benzene is a hybrid of various resonating structures.

Each carbon of benzene is in $s{p}^2$ hybridization. Two $s{p}^2$ hybrid orbitals of each carbon overlap with the adjacent carbon atoms' orbital, resulting in six C-C sigma bonds that all are in a hexagonal plane and the remaining $s{p}^2$ orbital overlaps with the s-orbitals of hydrogen and forms C-H sigma bonds.

In benzene, six C-H sigma bonds present. And the remaining unhybridised $p$-orbital (which are perpendicular to the plane) formed $\pi$-bond by lateral overlapping. The possibility of forming $\pi$ -bond is six ($C_1-C_2,C_3-C_4,C_6-C_1,/C_2-C_3,C_4-C_6,C_6-C_1$).
There are 6 $\pi$ electrons, whichare delocalised and move freely about the six carbon nuclei, and the presence of these delocalised $\pi$ electrons in benzene makes it more stable.

Question 9.11 What are the necessary conditions for any system to be aromatic?

Answer:

The necessary conditions for any system to be aromatic are

1. The cyclic compound should be a planner
2. The complete(continuous) delocalisation of $\pi$ -electrons in the ring
3. Follow the Huckel rule- it states that the ring has (4n+2) $\pi$ -electrons, where n =integer(n = 0, 1, 2, 3....)

Question 9.12 (i) Explain why the following systems are not aromatic?

Answer:

Not an aromatic compound because the $pi$-electrons in the ring are not in complete conjugation. And it is a non-planner structure.

Question 9.12(ii) Explain why the following systems are not aromatic?

Answer:

There is no complete conjugation of a $\pi$ electron in the ring as there is sp3 hybridized carbon between them. And also it does not obey huckle rule [(4n+2) $\pi$ ]electron.

Question 9.12(iii) Explain why the following systems are not aromatic?

Answer:

It disobeys the Hückel rule of (4n+2) $\pi$ electrons. According to this rule, it should have 2, 6, 10 .... number of $\pi$ electrons but it has 8 $\pi$ electrons. Also, this structure becomes non-planar.

Question 9.13(i) How will you convert benzene into

p-nitrobromobenzene

Answer:

Bromination of a benzene ring in the presence of anhydrous $FeC{l}_3$ and $B{r}_2$ gives bromobenzene. Then, treating bromobenzene with conc. nitric acid in the presence of sulphuric acid gives p-nitrobenzene. Halogens are ortho and para-directing groups.

Question 9.13 (ii) How will you convert benzene into

m- nitrochlorobenzene

Answer:

Benzene on treatment with conc. nitric acid and sulphuric acid give nitrobenzene, which on further treatment with chlorine in the presence of anhydrous aluminium chloride ($AlC{l}_3$) gives m-nitrochlorobenzene. Here, Cl will attach at meta position as NO₂ is a meta directing group.

Question 9.13(iii) How will you convert benzene into

p - nitrotoluene

Answer:

Alkylation of benzene in the presence of anhydrous aluminium chloride gives methylbenzene and $HCl$. When methylbenzene reacts with conc. nitric acid and sulphuric acid it gives a mixture of para and ortho products of nitrotoluene, which on distillation gives $p$-nitrotoluene.

Question 9.13 (iv) How will you convert benzene into

acetophenone

Answer:

Benzene on reacting with an acyl chloride in the presence of anhydrous aluminium chloride, gives acetophenone with hydrochloric acid as a by-product.

Question 9.14 In the alkane H₃C- CH₂ -C(CH₃ )₂-CH₂ -CH(CH₃)₂ , identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.

Answer:

$1^0$ carbon are those that are directly bonded with only one carbon atom; the given structure has five $1^0$ carbon atoms and 15 hydrogens attached to it.
$2^0$ carbon are those that are connected with only two carbon atoms. In the above structure, there are two $2^0$ carbons present, and four hydrogens attached to them.
$3^0$ are those that connect with three carbon atoms. In the above structure, 1 carbon atom is $3^0$ carbon atom and only one hydrogen is attached to it.

Question 9.15 What effect does branching of an alkane chain has on its boiling point?

Answer:

With an increase in the branching of the alkane, the boiling point of the alkane decreases. Alkanes experience intermolecular van der Waals forces. The strong is the force, strong will be the boiling point. When we increase the branching, the surface area of the molecule decreases; as a result, the van der Waals force also decreases.

Question 9.16 Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Answer:

Addition of $HBr$ to propene-
In this addition, an electrophile $H^{+}$ attacks the double bond of the alkene to form $1^0$ and $2^0$ carbocations-

A secondary carbocation is more stable than a primary carbocation. Thus bromide ion attacks the carbocation to form 2-bromopropane as a major product. (This mechanism is followed by Markovnikov's rule)

Question 9.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Answer:

Ortho-xylene has two resonant structures, so,

Since all three products, methylglyoxal, 1,2-methylglyoxal and glyoxal cannot be obtained from either of the two structures (i and ii). Hence, we can say that o-xylene is a resonant hybrid of two Kekule structures (I and II)

Question 9.18 Arrange benzene, $n$-hexane and ethyne in decreasing order of acidic behaviour. Also, give a reason for this behaviour.

Answer:

The acidic character of a species is defined on the basis of the ease with which it can lose its H+ ions.

The acidic character decreases in the order: Ethyne > Benzene > Hexane.

As the s-character decreases, carbon electronegativity decreases. Higher the s-character, more will be the negativity so the negative charge developing on carbon will be more stabilized. So the H that is attached to the alkyne will leave more readily.

Question 9.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

Answer:

In benzene, the $pi$-electrons are delocalised above and below the ring. Thus, it is an electron-rich species. In nucleophilic substitution, the attacking species is electron-rich in nature, so it becomes challenging to attack benzene because there will be repulsion. On the other hand, when the attacking agent is an electrophile, which is electron-deficient, they are easily attracted by the benzene.

Question 9.20 (i) How would you convert the following compounds into benzene?

Ethyne

Answer:

Benzene can be formed by the cyclic polymerisation of ethyne. Ethyne on passing through a red-hot tube(made of iron) at 873K forms benzene. Three molecules of ethyne polymerise to form benzene.

Question 9.20(ii) How would you convert the following compounds into benzene?

Ethene

Answer:

By converting ethene to ethyne by reacting with bromine in the presence of carbon tetrachloride. And then heating in the presence of alc. KOH followed by $NaN{H}_2/liq.N{H}_3$. Cyclic polymerisation of ethyne gives us benzene.

Question 9.20(iii) How would you convert the following compounds into benzene?

Hexane

Answer:

The cyclisation of hexane in the presence of $C{r}_2{O}_3$ produces cyclohexane, which on aromatisation gives benzene.

Question 9.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Answer:

Structures of all the alkenes that on hydrogenation, give 2-methylbutane.
The general structure of the 2-methylbutane is shown here;

As per the above structure, the following alkene compounds produce 2-methylbutane by hydrogenation are

Question 9.22 (a) Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E +

Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene

Answer:

Electrophiles are electron-deficient species, so they want a nucleophile that donates electrons to them. The higher the electron density on a benzene ring, the higher the reactivity towards electrophiles.
$N{O}_2$ is an electron-withdrawing group; it deactivates the benzene ring towards electrophiles by decreasing the electron density of the ring.
Decreasing order of their reactivity with an electrophile($E^{+}$)
Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene

Question 9.22 (b) Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E +

Toluene, $p-{\mathrm{H}}_3\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{NO}}_2,p-{\mathrm{O}}_2\mathrm{N}-{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{NO}}_2$.

Answer:

Electrophiles are electron-deficient species, so they want a nucleophile that donates electrons to them. The higher the electron density on a benzene ring, the higher the reactivity towards electrophiles.
Since the methyl group is an electron-donating group, it increases the electron density on the benzene ring. The more the number of electron withdrawing groups, the less reactive towards electrophiles.

Therefore, the decreasing order of reactivity towards electrophile -
toluene $>p-{\mathrm{CH}}_3{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{NHO}}_2>p-{\mathrm{O}}_2{\mathrm{NC}}_6{\mathrm{H}}_4-{\mathrm{NO}}_2$

Question 9.23 Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?

Answer:

Nitration occurs by an electrophilic substitution reaction, in which an electron-rich species is attacked by an electron-deficient molecule known as an electrophile. In nitration, $N{O}_2^{+}$ is used as an electrophile. Here methyl group is electron electron-donating group, and the nitro group is electron electron-withdrawing group. So, the benzene ring attached with the $-C{H}_3$ group has high electron density, and the ring that is attached with the nitro group has the least electron density. Hence, toluene undergoes nitration most easily.

Question 9.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

Answer:

Lewis acid like anhydrous ferric chloride $(FeC{l}_3)$, stannic chloride $(SnC{l}_4)$ , $B{F}_3$ etc. can be used instead of aluminium chloride.

Question 9.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing an odd number of carbon atoms? Illustrate your answer by taking one example.

Answer:

Wurtz reaction-

Wurtz reaction is not preferred for the preparation of alkanes containing an odd number of carbon atoms because if we take two dissimilar alkyl halides as reactants, the product will be a mixture of alkanes and alkene may also form due to free radical mechanism. example- bromomethane and iodoethane.

All the products in the mixture have nearly the same boiling point. So, the separation will be difficult.