Important Questions Class 11 Chemistry Chapter 13

Important Questions Class 11 Chemistry Chapter 13 – Hydrocarbons

Chemistry helps you to comprehend changes in the natural and physical world. The field of Chemistry is   interesting and students are intrigued to learn more. Understanding the characteristics of materials and converting them into new, and useful chemicals is made possible by chemistry but certain concepts are challenging to comprehend. Therefore, students need to make extra effort to understand its reactions and chemical formulas. Chapter 13 of Class 11 Chemistry is about ‘Hydrocarbons’.

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We at Extramarks take our role seriously and try to aid students with valuable guidance and support to help them excel in the board exams. The in-house subject experts highlight the key concepts and questions from each chapter to help students with their studies. The students will become more familiar with final exam questions by solving questions from our question bank of Important Questions Class 11 Chemistry Chapter 13. Chemistry requires deep conceptual understanding so cramming answers won’t help students in the long run. The only way is to solve regular questions and get their doubts cleared. The old adage says the more you practice, the better you’ll get. Extramarks experts believe that students must study regularly with concentration and focus to get 100% marks in Chemistry.

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Important Questions Class 11 Chemistry Chapter 13 – With Solutions

Students can refer to Extramarks NCERT solutions which comprise chapter notes, revision notes, and solved exam-related questions. To regularly practise questions, students can rely on our question bank of Important Questions  Chapter 13 Class 11 Chemistry. The questions cover all aspects of the chapter and will help students to revise all concepts from the hydrocarbons, important reactions of alkanes and alkenes and their characteristics. 

 Here is a list of questionnaires and their answers from our Chapter 13 Class 11 Chemistry Important Questions.

Question 1: Arrange the following hydrogen halides in order of their decreasing reactivity with propene.

(i) HCl > HBr > HI

(ii) HBr > HI > HCl

(iii) HI > HBr > HCl

(iv) HCl > HI > HBr

Answer 1: (iii) HI > HBr > HCl

Explanation: Bond strength and bond dissociation energy are two factors that impact the reactivity of hydrogen halides. We are aware that the size of the halogen atom increases while the bond dissociation energy and bond strength decreases in halogen halides. Hence, the reactivity increases.

Question 2: Arrange the following carbanions in order of their decreasing stability.

(i) A > B > C

(ii) B > A > C

(iii) C > B > A

(iv) C > A > B

Answer 2: (ii) B > A > C

Explanation: In C, the +I effect of the CH3 group directly affects the negatively charged carbon, reducing the stability of the carbon anion. Additionally, the +I effect is present in A, but it is further away from the negatively charged carbon, while it is absent in B.

Question 3: Arrange the following alkyl halides in decreasing order of the rate of – elimination reaction with alcoholic KOH.

(B) CH3-CH2-Br      

(C) CH3-CH2-CH2-Br

(i) A > B > C

(ii) C > B > A

(iii) B > C > A

(iv) A > C > B

Answer 3: (iv) A > C > B

Explanation: KOH, an alkene, is created when alkyl halides are heated with alcohol by getting rid of one molecule of halogen acid. The beta carbon atom   30  > 20  >  10 loses its hydrogen atom. The nature of alkyl groups determines the rate of reaction.

Question 4: Which of the following reactions of methane is incomplete combustion:

(iii) CH4  +  O2  →  C(s)    + 2H2O(l)

(iv) CH4    +   2O2  →   CO2(g)   +  2H2O(l)

Answer 4: (iii) CH4  +  O2  →  C(s)    + 2H2O(l)

Explanation:  Carbon black is produced when there is not enough oxygen or air to complete the combustion process.

Hence,  CH4  +  O2  →  C(s)    +   2H2O(l)

Question 5: Arrange the following in decreasing order of their boiling points.

(A) n-butane

(B) 2–methylbutane

(C) n-pentane

(D) 2,2–dimethylpropane

(i) A > B > C > D

(ii) B > C > D > A

(iii) D > C > B > A

(iv) C > B > D > A

Answer 5: (iv) C > B > D > A

Explanation: We are aware that the boiling point depends on both the molar mass and surface area.

It implies that branching will cause the boiling point to drop (surface area decreases on branching). As a result, n-butane has the lowest boiling point while n-pentane has the highest. Since the other two alternatives have branches, 2-methyl butane boils at a higher temperature than 2,2-dimethylpropane.

Question 6: The increasing order of reduction of alkyl halides with zinc and dilute HCl is

(i) R–Cl < R–I < R–Br

(ii) R–Cl < R–Br < R–I

(iii) R–I < R–Br < R–Cl

(iv) R–Br < R–I < R–Cl

Answer 6:  (ii) R-Cl < R-Br < R-I

Explanation: Since we are aware that halogen reactivity diminishes as the group size increases, the reduction of alkyl halides with ZnCl/HCl occurs in the opposite order.

Reactivity of reduction 1bond strength of C-X

Reactivity of reduction 1bond strength  C-X

It also depends on the size of the halogen. As a result, we can conclude that reducing binding strength is necessary to increase reactivity.

Question 7:

The correct IUPAC name of the following alkane is

(i) 3,6 – Diethyl – 2 – methyloctane

(ii) 5 – Isopropyl – 3 – ethyloctane

(iii) 3 – Ethyl – 5 – isopropyloctane

(iv) 3 – Isopropyl – 6 – ethyloctane

Answer 7:  (i) 3,6-Diethyl-2-methyloctane

Explanation: The alkane is octane because it has 8 carbon atoms, making it the longest chain. Carbon 2 has a methyl group, and carbon 3 and carbon 6 have ethyl groups. Since there are two ethyl groups, it will be diethyl, which is alphabetically before methyl. The side chains on carbon atoms 2, 3, and 6 adhere to the lowest sum rule.

Question 8:

The addition of HBr to 1-butene gives a mixture of products A, B and C

(C) CH3 – CH2 -CH2– CH2-Br

The mixture consists of

(i) A and B as major and C as minor products

(ii) B as major, A and C as minor products

(iii) B as minor, A and C as major products

(iv) A and B as minor and C as major products

Answer 8: (i) A and B as major and C as minor products.

Explanation: The major product is 2-Bromobutane, and the minor product is I-Bromobutane, according to Markovnikov’s rule. It is asymmetrical, butane-1. Due to the chiral nature of the carbon in 2-bromobutane, there are two enantiomers.

Question 10: What are cycloalkanes?

Answer 10: When carbon atoms form closed chains or ring compounds, which are similar to alkanes, all carbon is connected with single bonds in such a way that it gives a closed structure, and cycloalkanes are formed.

Some common examples of cycloalkanes are cyclopentane, Cyclobutane, cyclohexane, cycloheptane, cyclooctane, etc

Question 11: What is hydrogenation?

Answer 11: Dihydrogen gas is added to gas and alkenes in the presence of finely divided catalysts like Pt, Pd, or Ni to create alkanes. This process is known as hydrogenation.

Question 12: Methane does not react with chlorine in the dark. Why?

Answer 12: Chlorination of Methane is a substitution process that uses free radicals. The reaction does not occur because chlorine cannot be converted into free radicals in the dark.

Question 13: State Le Chatelier’s principle.

Answer 13: It states that when one of the factors defining a system’s equilibrium conditions changes, the system will adjust in a way that lessens or cancels out the effect of the change.

Question 14: If Qc < Kc,  when we continuously remove the product, what would be the direction of the reaction?

Answer 14: When a product is continuously removed, Qc remains smaller than Kc, and the reaction moves in the direction of the reactant.

Question 15: What is -elimination reaction?

Answer 15: In the -elimination reaction, the hydrogen atom is removed from the -carbon atom next to the carbon atom to which the halogen is attached.

Question 16: Why do alkynes not show geometrical isomerism?

Answer 16: The structure of alkynes is linear. Therefore, they are unable to show geometric isomerism.

Question 17: Although benzene is highly saturated, it does not undergo addition reactions. Give reasons.

Answer 17: In contrast to olefins, benzene -electrons are moved via resonance, making them less reactive for subsequent reactions.

Question 18:Unsaturated compounds undergo addition reactions. Justify.?

Answer 18: Two- or three-fold carbon bonds can be found in unsaturated hydrocarbon molecules. The -bond is a multiple bond that adds across several bonds when it becomes unstable.

Question 19: Cyclobutane is less reactive than cyclopropane. Justify.

Answer 19: Cyclobutane has a C-C-C bond angle of 900, while cyclopropane has a C-C-C bond angle of 600. It proves that there is a lesser variation in bond angles between cyclobutane (109028’) and cyclopropane. In other words, cyclopropane is more reactive than cyclobutane because it is under far greater stress.

Question 20:When alkanes are heated, the C-C bonds rather than the C-H bonds break. Give reasons.

Answer 20: When the alkanes are heated, the C-C bond instead of the C-H bond breaks because the C-C bond (H=83 KCal/mole) has a lower energy bond than the C-H bond (H=90 KCal/mole).

Question 21: State Markovnikov’s Rule.

Answer 21: According to Markovnikov’s Rule, when a polar compound is added to an unsymmetrical alkene or alkyne, the positive part goes to the carbon atom with the most substitutes, and the negative part goes to the carbon atom with the least substitutes.

Question 22: How will you distinguish between butene-1 and butene-2?

Answer 22: By ozonolysis or oxidation with an acidic KMnO4 solution, which yields unique carbonyl compounds, butenes 1 and 2 can be distinguished from one another.

The chemical representation is as follows:

CH3-CH2-CH=CH2  →   CH3CH2CHO    +    HCHO

CH3-CH=CH-CH3    → CH3CHO    +  CH3CHO

Question 23: State Kharasch effect.

Answer 23: It asserts that the addition of HBr (but not HCl or HI) to unsymmetrical alkenes breaks Markownikov’s rule when peroxides, such as benzoyl peroxide, are present.

The reaction is as follows:

CH3CH=CH2      +  HBr (peroxide)  →    CH3-CH2-CH3

Question 24: What do you understand by resonance energy?

Answer 24: The difference in energy between the most stable contributing structure and the resonance hybrid is known as resonance energy. Benzene has resonance energy of 147 KJ/mole.

Question 25: How is the aromaticity of a compound judged?

Answer 25: The following characteristics determine a compound’s aromaticity:

• Planarity
• Delocalisation of the  -electrons in the ring completely.
• 4n+2 electrons are present in the ring, where n is an integer (0, 1, 2, etc.).

This is generally known as  the Huckel Rule.

Question 26: Explain the term polymerisation with two examples.

Answer 26: In the right conditions, polymerisation is the process of joining two or more molecules of unsaturated chemicals to produce a larger complex. The process is called polymerisation, and the end result is referred to as a polymer.

Addition polymerisation: Nothing is lost during addition polymerisation since the larger molecule (polymer) is an exact multiple of the smaller molecule.

Condensation polymerisation: Water, hydrochloric acid, and other molecules are frequently lost during the condensation polymerisation process. During this type of polymerisation, the polymer is not a precise multiple of the smaller molecule.

Question 27: Ethyne is acidic in nature in comparison to ethene and ethane. Why is it so?

Answer 27: Hydrogen atoms are joined to sp hybridised carbon atoms in ethyne, whereas they are joined to sp2 hybridised carbon atoms in ethene and sp3 hybridised carbons in ethane.

Due to the largest percentage of s-character (50%) in ethyne molecules, the carbon atoms’ sp hybridised orbitals have the highest electronegativity (50%) of any orbital.

In comparison to the sp2 hybridised orbitals of carbon in ethene and the sp3 hybridised orbitals of carbon in ethane, this attracts the shared pair of the C-H bond of ethyne more strongly. Hydrogen atoms can be released as protons more readily in the ethyne molecule as compared to ethene and ethane.

Question 28: Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why?

Answer 28:  The maximum distance between the hydrogen atoms and the electron bonding pairs, which results in the least amount of repulsion, makes the staggered conformation the most stable. The other intermediate conformation (Gauche or skew form), which is between them, is less stable than the eclipsed shape.

Question 29: Despite their – I effect, halogens are o- and p-directing in haloarenes. Explain.

Answer 29: The effects of halogens on benzene rings are -I and +R. The +R effect raises the electron density on ortho and para locations, while the -I effect deactivates the ring. Halogens are hence ortho- and para-directing.

Question 30: Why does the presence of a nitro group make the benzene ring less reactive than the unsubstituted benzene ring? Explain.

Answer 30: An electron-withdrawing group is the Nitro group (-R and -I effects). By reducing the nucleophilicity for subsequent substitution, it deactivates the ring.

Question 31: What products are formed when zinc reacts with

(i) vicinal C2H4Br2 and

Answer 31:  CH2Br – CH2Br + Zn →  CH2 = CH2 + ZnBr2.

(ii) CH3CHBr – CH2Br.

Answer:  CH3 – CHBr – CH2Br + Zn → CH3 — CH = CH2 + ZnBr2.

Question 32: How will you prepare Alkanes by

(i) Wurtz reaction

Answer 32: 

Methods of preparation of Alkanes

Wurtz reaction. After being treated with sodium in dry ether, alkyl halides produce higher alkanes, ideally with an even number of carbon atoms.

(ii) Decarboxylation of sodium salts of fatty acids

Answer: 

When heated with soda lime (a solution of NaOH and CaO), sodium salts of fatty acids produce alkanes with one fewer carbon atom than the carboxylic acid. The procedure of decarboxylation involves eliminating carbon dioxide from a carbonyl group.

iii) Kolbe’s electrolytic method.

Answer: 

Kolbe’s electrolytic method: Alkanes with an even number of carbon atoms are produced at the anode by electrolysing an aqueous solution of sodium or potassium salt of a carboxylic acid.

The probable mechanism for the reaction is

Question 34: What effect does branching of an alkane chain have  on its boiling point?

Answer 34: Intermolecular Van der Waals forces are present in alkanes. The boiling point of the alkane will increase with increasing force. A smaller area of contact occurs from the molecule’s surface area, decreasing as branching increases. This results in a decreased Van der Waals force that can be resisted at a substantially lower temperature. As a result, the boiling point of an alkane chain drops as branching increases.

Question 35: Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? 

Answer 35: Planar molecules like benzene have electrons that are delocalised above and below the plane of the ring. It is so rich in electrons. It is therefore extremely alluring to organisms lacking in electrons or electrophiles. It is hence particularly susceptible to electrophilic substitution reactions. Electrophiles have a lot of electrons. Benzene repels them as a result. As a result, nucleophilic replacements of benzene are challenging.

Question 36: Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. 

Answer 36: An ethyl group is added to the benzene ring as part of the methylation reaction. A Friedel-Craft alkylation reaction is what is known as such a reaction. A Lewis acid is necessary for this reaction to occur. Any Lewis acid, including anhydrous FeCl3, SnCl4, BF3, and others, may be used to methylate benzene.

Question 37: Why is Wurtz reaction not preferred for the preparation of alkanes containing an odd number of carbon atoms? Illustrate your answer with  one example.

Answer 37: The synthesis of symmetrical alkanes (alkanes with an even number of carbon atoms) is only possible by the Wurtz reaction, which uses two identical alkyl halides as reactants to create an alkane with twice as many carbon atoms. 

Example: 

Because the Ethane Wurtz reaction produces a mixture of alkanes when two different alkyl halides are utilised as the reactants, it cannot be used to create unsymmetrical alkanes. Free radical species are involved in the reaction. Therefore a side reaction that results in an alkene also takes place. For instance, a mixture of alkanes is produced when bromoethane and iodoethane combine.

Question 38: Arrange the following set of compounds in order of their decreasing

relative reactivity with an electrophile, E+

(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene

(b) Toluene, p-H3C – C6H4 – NO2, p-O2N – C6H4 – NO2

Answer 38:

Reagents called electrophiles to take part in reactions by attaching themselves to nucleophiles by accepting two electrons.

The more electrons there are on a benzene ring, the more electrophilic (E+) the molecule is (Electrophilic reaction).

(a) Due to the presence of an electron-withdrawing group (NO2 – and Cl –), which deactivates the aromatic ring, the aromatic ring’s electron density lowers.

The increasing sequence of reactivity is as follows because the Cl – group is less electron-withdrawing (due to the inductive effect) than the NO2 – group (due to the resonance effect):

2, 4 – dinitrochlorobenzene < p – nitrochlorobenzene < Chlorobenzene

(b) CH3– is an electron-donating group, whereas NO2- is an electron-withdrawing group.

As a result, toluene has the highest electron density and is most susceptible to E+ attack. NO2- is an electron-removing group, hence. As a result, when there are more NO2 alternatives, the order is as follows:

p-O2N – C6H4 – NO2 < p – H3C – C6H4 – NO2  < Toluene.

Question 39: List the names of some Lewis acids which can be used during the ethylation of benzene in a Friedel-Craft alkylation reaction.

Answer 39:  Acyl benzene is produced when benzene reacts with an acyl halide or acid anhydride in the presence of Lewis acids (AlCl3) (or benzene ring). Such a process is referred to as a Friedel-Craft alkylation reaction. In the presence of a Lewis acid, the reaction takes place.

Any Lewis acid, including anhydrous AlCl3, FeCl3, SnCl4, BF3, etc., may be used to ethylate benzene in the Friedel-Craft alkylation reaction.

Question 40: Why is benzene extraordinarily stable though it contains three double bonds?

Answer 40: The resonating structure of benzene can be represented as:

The benzene molecule contains six sp2 hybridised carbon atoms. Six sigma bonds are created in the hexagonal plane by the two sp2 hybrid orbitals of each carbon atom overlapping with the sp2 hybrid orbitals of nearby carbon atoms. Six sigma bonds are created when the remaining sp2 hybrid orbitals on each carbon atom overlap with the hydrogen s-orbital. The option exists for the remaining unhybridised carbon atoms to create three pi bonds via the lateral overlap of C1-C2, C3-C4, C5-C6, or C2-C3, C4-C5, C6-C1.

Due to their delocalisation, the six pi can freely migrate among the six carbon nuclei. The benzene is stabilised by these delocalised pi- electrons even when there are three double bonds present.

Question 41: Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Answer 41: According to the information provided, an alkene’s ozonolysis produces propanal and pentan-3-one. Please use the alkene supplied, “A.” When we write the ozonolysis reaction backwards, we get:

On the cleavage of ozonide “X,” the products are produced. As a result, both products are present in ‘X’ in cyclic form. The following is a possible representation of ozonide’s structure::

Alkene “A” and ozone now produce “X,” a byproduct. The following is an example of an alkene’s potential structure:

Question 42: Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Answer 42: 

Hex-2-ene is represented as

CH3– CH=CH-CH2-CH2-CH3

Geometrical isomers of hex-2-ene are:                                     

The total of the dipole moments of the C-CH3 and C-CH2 CH2 CH3 bonds acting in the same direction makes up the dipole moment of the cis-compound.

The dipole moments of the C-CH3 and C-CH2CH2CH3 bonds operating in opposition to one another create the trans-dipole compound’s moment.

The cis-isomer is, therefore, more polar than the trans-isomer. The intermolecular dipole-dipole interaction is stronger, and the boiling point rises with increasing polarity. As a result, the boiling point of the cis-isomer will be greater than the trans-isomer.

Question 43: How will you convert benzene into

(i) p-nitrobromobenzene

(ii) m-nitrochlorobenzene

(iii) p -nitrotoluene

(iv) acetophenone

Answer 43: 

(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:

Question 44: In the alkane H3C– CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°, and 3° carbon atoms and give the number of H atoms bonded to each one of these.

Answer 44: 

One-carbon-atom bonds, or having just one carbon atom as a neighbour, are referred to as 1° carbon atoms. The provided structure is made up of fifteen hydrogen atoms and five carbon atoms at 1°.

Carbon atoms with a 2° link to another carbon atom have another two carbon atoms as their neighbours. Four hydrogen atoms are 

joined to the two 2° carbon atoms in the provided structure.

Three carbon atoms are connected to three carbon atoms, making them the neighbours of three carbon atoms. Only one hydrogen atom and one 3° carbon atom make up the given structure.

Question 45: Addition of HBr to propene yields 2-bromopropane, while in benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give a mechanism.

Answer  45: An example of an electrophilic substitution reaction is the addition of HBr to propene.

H+, an electrophile, is produced by hydrogen bromide. As demonstrated, this electrophile attacks the double bond to produce the 1° and 2° carbocations.

Compared to primary carbocations, secondary carbocations are more stable. As a result, the former prevails since it will form more quickly. As a result, in the following step, Br- attacks the carbocation to produce the main product, 2 – bromopropane.

This reaction adheres to Markovnikov’s rule, according to which the carbon atom with the fewest hydrogen atoms receives the addendum’s negative portion.

Benzoyl peroxide causes an additional reaction that goes against Markovnikov’s law. The process of the reaction is a free radical chain as follows:

Compared to primary radicals, secondary free radicals are more stable. As a result, the former prevails since it forms more quickly. As a result, the main product is 1 – bromopropane.

Br free radical behaves like an electrophile in the presence of peroxide. As a result, when HBr is added to propene, two distinct compounds are produced in both the presence and absence of peroxide.

Question 46: Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Answer 46: 

O-xylene has two resonance structures:

Two Kekule structures are used to produce all three products, methyl glyoxal, 1, 2-dimethylglyoxal, and glyoxal. This demonstrates that o-xylene is a resonance hybrid of two Kekule structures because none of the three products can be produced by either of the two structures alone (I and II).

Question 47: Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also, give the reason for this behaviour.

Answer 47: 

A species’ acidic nature is determined by how readily it may shed its H-atoms.

The hybridisation state of carbon in the given compound is:

The electronegativity of carbon rises with increasing s-character as the C-H bond pair’s electrons move in closer to the carbon atom. As a result, the H-partial atom’s positive charge rises and H+ ions are liberated.

The s–character increases in the order:

sp3 < sp2 < sp

Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

Question 48: Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

Answer 48:  Planar molecules like benzene have electrons that are delocalised above and below the plane of the ring. It is so rich in electrons. It is therefore extremely alluring to organisms lacking in electrons or electrophiles.

It is hence particularly susceptible to electrophilic substitution reactions. Electrophiles have a lot of electrons. Benzene repels them as a result. As a result, nucleophilic replacements of benzene are challenging.

Question 49: How would you convert the following compounds into benzene?

(i) Ethyne

(ii) Ethene

(iii) Hexane

Answer 49: 

(i)Benzene from Ethyne:

(ii)Benzene from Ethene:

(iii)Hexane to Benzene

Question 50: Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Answer 50: 

The skeleton of 2-methylbutane is shown below:

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

(a)

(b)

(c)

Question 51: Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+

(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene

(b) Toluene, p-H3 C – C6H4 – NO2, p-O2N – C6H4– NO2.

Answer 5511: Reagents called electrophiles take part in reactions by bonding with nucleophiles by accepting an electron pair.

The benzene ring’s electron density determines how reactive the molecule is to an electrophile, E+ (Electrophilic reaction).

(a) By lowering the electron density, electron-withdrawing groups like NO2- and Cl– deactivate the aromatic ring.

The decreasing sequence of reactivity is as follows because the NO2 – group is more electron withdrawing (due to resonance effect) than the Cl- group (due to inductive action):

Chlorobenzene > p – nitrochlorobenzene > 2, 4 – dinitrochlorobenzene

(b) The electron-withdrawing NO2- group contrasts with the electron-donating CH3– group. Toluene will have the highest electron density and be susceptible to E+ attack.

An electron-withdrawing group is NO2. Consequently, the sequence is as follows when there are more NO2- substituents:

Toluene > p– CH3–C6H4 – NO2, p –O2 N– C6H4 – NO2

Question 52: Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?

Answer 52: The compound’s ability to generate nitrates depends on the amount of electron density there is. Examples of nitration reactions include electrophilic substitution reactions in which a nitronium ion attacks an electron-rich species (NO2–).

Now, the CH3– group donates electrons while the NO2- group withdraws them. Thus, of the three chemicals, toluene will have the highest electron density, followed by benzene. m- Dinitrobenzene, on the other hand, will have the lowest electron density. As a result, nitration will be challenging. Consequently, the nitration is in increasing sequence as follows:

Question 53: Why is Wurtz reaction not preferred for the preparation of alkanes containing an odd number of carbon atoms? Illustrate your answer with one example.

Answer 53: 

Wurtz reaction can only be used to create symmetrical alkanes (alkanes with an even number of carbon atoms). Two identical alkyl halides are used as reactants in the reaction, and an alkane with twice as many carbon atoms is produced.

Example:

Because the Wurtz reaction produces a mixture of alkanes when two different alkyl halides are utilised as the reactants, it cannot be used to create unsymmetrical alkanes. Free radical species are involved in the reaction, therefore a side reaction that results in an alkene also takes place. For instance, a mixture of alkanes is produced when bromoethane and iodoethane combine.

Alkanes produced from the mixture have extremely similar boiling temperatures. Consequently, it becomes challenging to distinguish them.

Question 54: Which of the following alkenes on ozonolysis give a mixture of ketones only?

Answer 54:  (iii) and (iv)

Explanation: Alkenes produce two molecules of carbonyl compounds when subjected to ozonolysis, depending on the groups or atoms attached. If the alkyl groups on the double-bonded carbon atoms are present, ketones are produced.

Question 55: Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?

Answer 55: 

On the reduction of 2-Butyne, butene-2 is produced, with the methyl groups either on the same side or the opposite side to display geometric isomers.

Question 56: Rotation around the carbon-carbon single bond of ethane is not completely free. Justify the statement.

Answer 56: 

(i)Alkanes feature a carbon-carbon sigma bond in which the distribution of the sigma molecular orbit’s electrons is symmetrical around the bond’s internuclear axis, i.e., not distributed as a result of rotation about its axis.

(ii) As a result, the C-C single bond is allowed to rotate freely, resulting in various spatial configurations of atoms that are interchangeable.

These atomic configurations are known as conformations, rotamers, or conformers.

(iii) Alkenes have an endless number of conformations because of rotation around C-C bonds. However, due to a mild repulsive contact between nearby bonds, rotation around a C-C single bond is not entirely free and is hampered by a tiny energy barrier of 1–20 kJ mol-1. The strain is referred to as a torsion.

Question 57: The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same, and the bond energy of HCl, HBr and HI is 430.5 kJ mol-1, 363.7 kJ mol-1 and 296.8 kJ mol-1 respectively. What will be the order of reactivity of these halogen acids?

Answer 57: 

The dissociation enthalpy of H-X determines how reactive hydrogen halides are. Hydrogen halides are more reactive in the following order: HI > HBr > HCl. To create alkyl halides, they combine with alkanes.

Due to the bond enthalpy of HI, HBr, and HCl, the order of their reactivity is 

HCl < HBr < HI.

Question 58: Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give a reason.

Answer 58: 

The methoxy group makes anisole more reactive than benzene because it is an electron releasing group that increases the benzene ring’s electron density due to the +R effect.

Since the Cl group produces the +R and -I effects, it is less reactive than methoxybenzene. Additionally, because it has the -I and -R effect, chlorobenzene is more reactive than the group.

Question 59: Despite their – I effect, halogens are o- and p-directing in haloarenes. Explain.

Answer 59: 

Halogens are ortho and para directing because they have a +R and -I effect. Now, halogens present on the benzene ring have these effects: the -I effect deactivates the ring, and the +R effect increases the electron density on ortho and para positions.

Question 60: Why does the presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.

Answer 60: 

Due to the presence of the nitro group, which possesses -I and -R effects, the benzene ring deactivated and the electron density decreased in the ortho and para positions relative to the meta locations.

Question 61: Suggest a route for the preparation of nitrobenzene starting from acetylene.

Answer 61: 

When ethanol is pushed through an extremely hot iron tube at 873 K, it undergoes cyclic polymerization and produces benzene, from which nitrobenzene can be made through nitration.

Question 62: Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.

Answer 62: 

Tertiary free radicals are more stable and sit stabilised due to hyperconjugation and nine hydrogen, whereas free radicals with one hydrogen and one hyper conjugative structure are less stable.

Question 63: Suggest a route to prepare ethyl hydrogen sulphate (CH3 – CH2 – OSO2-OH) starting from ethanol (C2H5OH).

Answer 63: 

Preparing ethyl hydrogen sulphate starting from ethanol.

Step I-

Protonation of alcohol and formation of a carbocation.

H2SO4 →  H+   +   –OSO2OH

CH3  – CH2 – O – H  +  H+    →  CH3  –  CH2 – +OH2

CH3 – CH2 – + OH2  →  CH3  – + CH2   + H2O

Step II-

Attack of nucleophile

HO  – SO2 – O–     +    +CH2   – CH3     →   CH3  –   CH2  – O – SO2 – OH

Question 64: An alkyl halide  C5H11Br (A) reacts with ethanolic KOH to give an alkene ‘B’, which reacts with Br2 to give a compound ‘C’, which on dehydrobromination gives an alkyne ‘D’. On treatment with sodium metal in liquid ammonia one mole of ‘D’ gives one mole of the sodium salt of ‘D’ and half a mole of hydrogen gas. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify A, B, C and D. Give the reactions involved.

Answer 64: 

To identify A, B, C and D, the reactions involved are-

C5H11Br(A)    +  alc. KOH  →  C5H10 (B)

C5H10 (B)  + Br /CS2    →  C5H10Br2(C)

C5H10Br2(C)   +   alc. KOH   →  C5H8 (D) Alkyne

2C5H8   + 2 Na   →    2 C5H7Na  + H2

All the compounds above must be straight-chain as hydrogenation of alkyne (D) gives straight-chain alkane. It is clear that D is terminal alkyne as alkyne gives sodium alkenyde.

Question 65: An unsaturated hydrocarbon ‘A’ adds two molecules of H2 and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of ‘A’, write its IUPAC name and explain the reactions involved.

Answer 65: 

Two molecules of hydrogen add on ‘A’; this shows that ‘A’ is either an alkadiene or an alkyne.

On ozonolysis, compound A gives

CH3CHO  +  O = CH – CH2 – CH2 – CHO   +   O = C ( CH3)2

Hence, the structure of A (Its IUPAC name will be 2-methyl octa 2,6 diene) is,

Question 66: In the presence of peroxide, addition of HBr to propene takes place according to anti-Markovnikov’s rule, but peroxide effect is not seen in the case of HCl and HI. Explain.

Answer 66: 

(i)

(ii)

(iii)

(iv)

The peroxide effect only occurs in the case of HBr because the H-Cl link is stronger than the H-Br bond. Additionally, in the case of HI, it is not seen.

While the H-I bond is weaker and iodine free radicals unite to produce dimer iodine molecules because the H-Br link has less bond energy than the H-Cl bond, the H-Cl bond is not cleaved by the free radical.

Benefits of Solving Important Questions Class 11 Chemistry Chapter 13

Students can face difficulty understanding the complex concepts in Chemistry. One way to tackle this is to practice questions given in our Chemistry Class 11 Chapter 13 important questions. The questions given in Class 11 Chemistry Chapter 13 important questions cover all the main topics, and these questions are created from an exam perspective and are likely to come in the exam. Solving important questions gives students a competitive edge. 

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• Exam questions are created exclusively by the subject experts for the board and entrance exams, and students will revise everything and cover the entire syllabus. 
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