Chapter 9. Hydrocarbons

Class 11 Chemistry Chapter 9 Hydrocarbons Internal Questions and Answers :

Problem 9.1 :  Write structures of different chain isomers of alkanes corresponding to the molecular formula  . Also write their IUPAC names.

Answer :  (i) N- Hexane :

 

(ii) 2-Methylpentane :

(iii) 3-Methylpentane

 

(iv) 2,3-Dimethylbutane :

(v) 2,2-Dimethylbutane

Problem 9.2 :  Write structures of different isomeric alkyl groups corresponding to the molecular formula  . Write IUPAC names of alcohols obtained by attachment of  groups at different carbons of the chain.

Answer :

Problem 9.3 :  Write IUPAC names of the following compounds :

(i) 

(ii) 

(iii) tetra – tert-butylmethane .

Answer : (i)   

                 2, 2, 4, 4-Tetramethylpentane .

(ii)   

     

3, 3-Dimethylpentane

(iii) tetra – tert-butylmethane 

   

3,3-Di-tert-butyl -2, 2, 4, 4 –tetramethylpentane .

Problem 9.4 :  Write structural formulas of the following compounds :

(i) 3, 4, 4, 5 -Tetramethylheptane

(ii) 2,5-Dimethyhexane

Answer :  (i) 3, 4, 4, 5 -Tetramethylheptane

    

(ii) 2,5-Dimethyhexane

Problem 9.5 :  Write structures for each of the following compounds. Why are the given names incorrect? Write correct IUPAC names.

(i) 2-Ethylpentane

(ii) 5-Ethyl – 3-methylheptane

Answer :  (i) 2-Ethylpentane : Longest chain is of six carbon atoms and not that of five.

      

Hence, correct name is 3-Methylhexane.

(ii) 5-Ethyl – 3-methylheptane : Numbering is to be started from the end which gives lower number to ethyl group.

    

Hence, correct name is 3-ethyl-5-methylheptane.

Problem 9.6 : Sodium salt of which acid will be needed for the preparation of propane ? Write chemical equation for the reaction.

Answer : The chemical reaction is :

Problem 9.7 :  Write IUPAC names of the following compounds:

Answer : The IUPAC name of the compounds are :

(i)  2,8-Dimethyl-3, 6-decadiene

  

(ii) 1,3,5,7 Octatetraene

     

(iii) 2-n-Propylpent-1-ene

(iv) 4-Ethyl-2,6-dimethyl-dec-4-ene

  

Problem 9.8 : Calculate number of sigma (σ) and pi (π) bonds in the given Compounds .

Answer : (i)  2,8-Dimethyl-3, 6-decadiene

  

σ bonds : 33, π bonds : 2

(ii) 1,3,5,7 Octatetraene

     

σ bonds : 17, π bonds : 4

(iii) 2-n-Propylpent-1-ene

σ bonds : 23, π bond : 1

(iv) 4-Ethyl-2,6-dimethyl-dec-4-ene

  

σ bonds : 41, π bond : 1

Problem 9.9 :  Write structures and IUPAC names of different structural isomers of alkenes corresponding to .

Answer : The different structural isomers of alkanes () and IUPAC name :

Problem 9.10 : Draw  and  isomers of the following compounds. Also write their IUPAC names :

(i) 

(ii) 

Answer: (i)

cis-1,2-Dichloroethene

trans-1,2-Dichloroethene

(ii)

  

cis-3,4-Dimethylhex-3-ene

    

trans-3,4-Dimethylhex-3-ene

Problem 9.11 : Which of the following compounds will show -  isomerism?

(i) 

(ii)

(iii)

(iv)

Answer : (iii)

(iv)

In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom.

(i) 

(ii)

Problem 9.12 :  Write IUPAC names of the products obtained by addition reactions of  to hex-1-ene

(i) in the absence of peroxide and

(ii) in the presence of peroxide.

Answer :

(i) 

 

2-bromohexane

(ii)

 

1-Bromohexane

Problem 9.13 : Write structures of different isomers corresponding to the  member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers?

Answer : The  member of alkyne series has the molecular formula .

The IUPAC name and the isomers of  are : 

The chain and position isomerism shown by different pairs .

Problem 9.14 : How will you convert ethanoic acid into benzene?

Answer :

Class 11 Chemistry Chapter 9 Hydrocarbons Exercises Questions and Answers :

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

Answer: The formation of ethane () during the chlorination of methane () can be explained through a series of free radical substitution reactions. Chlorination of methane typically proceeds in the presence of ultraviolet light or heat, which provides the energy required to initiate the reaction.

Initiation Step: The chlorination process begins with the homolytic cleavage of the chlorine molecule () into two chlorine radicals (  ). This occurs under the influence of heat or UV light.

This produces two chlorine radicals (), which are highly reactive.

Propagation Steps : The chlorine radicals () can then react with methane (). The chlorine radical abstracts a hydrogen atom from methane, forming methyl radical () and hydrogen chloride ():

Now, a methyl radical () is formed. This methyl radical is highly reactive and can react with another molecule of chlorine () to form chloromethane () and another chlorine radical ().

So, at this point, we have chloromethane () and a chlorine radical () again. This chlorine radical can continue the reaction, leading to more chloromethane production.

Formation of Ethane (): The key step in the formation of ethane comes when the methyl radical () generated in the first propagation step reacts with another methyl radical () to form ethane ():

This leads to the formation of ethane ().

Termination Steps : The reaction comes to an end when two radicals combine to form stable molecules. The termination reactions are :

9.2 Write IUPAC names of the following compounds :

Answer: (a) 2-Methylbut-2-ene

(b) Pen-1-ene-3-yne

(c) 1,3 butadiene or But-1,3-diene

(d) 4-Phenyl but-1-ene

(e) 2-Methyl phenol

(f)  5-(2-Methylpropyl)-decane

(g) 4-ethyldeca-1,5,8-triene

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

(a)  (one double bond) (b)  (one triple bond)

Answer:  (a) C₄H₈ (one double bond)

IUPAC name: But-1-ene
Structural formula:

IUPAC name: But-2-ene
Structural formula:

IUPAC Name: 2-Methylprop-1-ene

The structural formula :

(b) C₅H₈ (one triple bond)
Structural formula:
IUPAC name: Pent-1-yne
Structural formula: CH₃-C≡C-CH₂-CH₃
IUPAC name: Pent-2-yne

The structural formula :

IUPAC Name: 3-Methylbut-1-ene

9.4 Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene

Answer:

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

Answer: The alkene A that gives a mixture of ethanal (CH₃CHO) and pentan-3-one (CH₃CH₂COCH₂CH₃) on ozonolysis must have the structure:

CH₃-CH=CH-CH₂-CH₃

IUPAC Name:

The IUPAC name of A is 3-hexene.

9.6 An alkene ‘A’ contains three  , eight  bonds and one  bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Answer:

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

Answer:

9.8 Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene

Answer:

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

Answer:

9.10 Why is benzene extra ordinarily stable though it contains three double bonds?

Answer:

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

Answer:

9.12 Explain why the following systems are not aromatic?

Answer:

9.13 How will you convert benzene into

(i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii) p - nitrotoluene (iv) acetophenone?

Answer:

9.14 In the alkane  , identify 1,2,3 carbon atoms and give the number of H atoms bonded to each one of these.

Answer:

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

Answer:

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:

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

Answer:

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

Answer: The decreasing order of acidic behavior is:

Ethyne (HC≡CH) > Benzene (C₆H₆) > n-Hexane (CH₃(CH₂)₄CH₃)

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

Answer: Benzene undergoes electrophilic substitution reactions easily because its electron-rich π-system attracts electrophiles. The delocalized π-electrons stabilize the intermediate carbocation (arenium ion) formed during the reaction, making the process favorable.

In contrast, nucleophilic substitutions are difficult because benzene's π-electron cloud repels nucleophiles. Additionally, the formation of a high-energy, unstable intermediate (Meisenheimer complex) makes nucleophilic substitution unfavorable unless strong electron-withdrawing groups are present to stabilize the intermediate.

9.20 How would you convert the following compounds into benzene?

(i) Ethyne (ii) Ethene (iii) Hexane

Answer: (i) Ethyne (Acetylene) to Benzene:

Ethyne can be converted into benzene via cyclotrimerization. This involves heating ethyne in the presence of a catalyst.

Reaction: 

Catalyst: Nickel or iron at 400–500°C.

(ii) Ethene (Ethylene) to Benzene:

Ethene can be converted into benzene through aromatization. This involves dehydrogenation and cyclization.

Reaction:

Catalyst: Chromium oxide (Cr₂O₃) or platinum (Pt) at high temperatures.

(iii) Hexane to Benzene:

Hexane can be converted into benzene via catalytic reforming or aromatization. This involves dehydrogenation and cyclization.

Reaction:

Catalyst: Platinum (Pt) or rhenium (Re) on an alumina (Al₂O₃) support.

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

Answer:  Structure of 2-methylbutane:

CH₃-CH(CH₃)-CH₂-CH₃CH₃-CH(CH₃)-CH₂-CH₃

Possible alkenes:

1. 2-Methyl-1-butene:

   CH₂=C(CH₃)-CH₂-CH₃CH₂=C(CH₃)-CH₂-CH₃

2. 2-Methyl-2-butene:

   CH₃-C(CH₃)=CH-CH₃CH₃-C(CH₃)=CH-CH₃

3. 3-Methyl-1-butene:

   CH₂=CH-CH(CH₃)-CH₃CH₂=CH-CH(CH₃)-CH₃

[ Explanation: Hydrogenation of any of these alkenes will add hydrogen across the double bond, resulting in 2-methylbutane.

These are the only alkenes with the same carbon skeleton as 2-methylbutane.]

9.22 Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile,

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

(b) Toluene, ,

Answer:

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

Answer: Out of benzene, m-dinitrobenzene, and toluene, toluene will undergo nitration most easily. This is due to the electron-donating nature of the methyl group (-CH₃) attached to the benzene ring.

[  Explanation:

1. Toluene (C₆H₅CH₃):

   • The methyl group is an electron-donating group (EDG) through hyperconjugation and inductive effects.

   • It activates the benzene ring by increasing the electron density, making it more reactive toward electrophilic substitution reactions like nitration.

   • Nitration occurs predominantly at the ortho and para positions relative to the methyl group.

2. Benzene (C₆H₆):

   • Benzene has no substituents, so it is less reactive than toluene but more reactive than deactivated benzene derivatives like m-dinitrobenzene.

3. m-Dinitrobenzene (C₆H₄(NO₂)₂):

   • The nitro groups (-NO₂) are electron-withdrawing groups (EWGs).

   • They deactivate the benzene ring by reducing the electron density, making it less reactive toward electrophilic substitution.

   • Nitration of m-dinitrobenzene is much slower and requires harsher conditions compared to toluene or benzene.]

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

Answer: A suitable Lewis acid alternative to anhydrous aluminium chloride (AlCl₃) for the ethylation of benzene is ferric chloride (FeCl₃).

FeCl₃ acts as a catalyst by generating the electrophile (ethyl carbocation, CH₃CH₂⁺) from ethyl chloride (CH₃CH₂Cl) in Friedel-Crafts alkylation. It is less reactive than AlCl₃ but is still effective for this reaction.

Reaction:

FeCl₃ is a commonly used Lewis acid in organic synthesis and is particularly useful when milder conditions are required.

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

Answer: The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium to form alkanes. However, it is not preferred for preparing alkanes with an odd number of carbon atoms because it typically produces a mixture of alkanes due to the random combination of alkyl radicals.

Example: If we attempt to prepare propane (C₃H₈) using the Wurtz reaction:

1. React ethyl bromide (CH₃CH₂Br) with sodium.

2. The reaction produces a mixture of alkanes:

   • Butane (C₄H₁₀): Formed by the coupling of two ethyl radicals (CH₃CH₂-CH₂CH₃).

   • Ethane (C₂H₆): Formed by the coupling of two methyl radicals (CH₃-CH₃).

   • Propane (C₃H₈): Formed by the coupling of a methyl radical (CH₃) and an ethyl radical (CH₃CH₂).

The Wurtz reaction does not selectively produce propane. Instead, it yields a mixture of ethane, propane, and butane, making it unsuitable for preparing alkanes with an odd number of carbon atoms. This limitation arises because the reaction cannot control the combination of radicals.