Class 11 - Chemistry - Hydrocarbons

Question 13.1

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

Answer.

Chlorination of methane occurs by the following mechanism. It is a free radical reaction mechanism.

 Class_11_Chemistry_HydroCarbons_Equation_1

Question 13.2

Write IUPAC names of the following compounds:

Class_11_Chemistry_HydroCarbons_Equation_2

Class_11_Chemistry_HydroCarbons_Equation_3

 

 

Question 13.3

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)C4H8 (one double bond)

(b)C5H8 (one triple bond)

Answer.

 

 Class_11_HydroCarbons_Structure_3

Question 13.4

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.

 

 Class_11_HydroCarbons_Structure_4

Question 13.5

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

Answer.

An alkene (A) when subjected to ozonloysis cleaves at the double bond and give rise to carbonyl compounds.

Class_11_HydroCarbons_Ozonolysis

Since, given compound is ethanal and pentan-3-one, alkene can be

Class_11_HydroCarbons_Ozonolysis_1

 

Question 13.6

An alkene ‘A’ contains three C-C, eight C-H sigma bond 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.

  • An aldehyde with molar mass of 44 u is ethanal.
  • Two moles of ethanal side to side with their oxygen atoms pointing towards each other.

           CH3CH=O   O=CHCH3

         Ethanal        Ethanal

 

  • Remove the oxygen atoms and join them by double bond. Structure of alkene A is as follows.

Hence, there are 3 C-C, 8 C-H a-bonds and one C-C pi bond.

 Class_11_HydroCarbons_Structure_Of_But-2-ene

 

Question 13.7

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

Answer.

  Class_11_HydroCarbons_Ozonolysis_2

Question 13.8

Write chemical equations for combustion reaction of the following hydrocarbons:

  • Butane (ii) Pentene (iii) Hexyne (iv) Toulene

Answer.

 

 Class_11_HydroCarbons_Ozonolysis_3

Question 13.9

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

Answer.

Structures of cis- and trans-isomer of hex-2-ene are as below.

Class_11_HydroCarbons_Cis_Trans_Structure_of_Hex-2-ene

Cis-Hex-2-ene have higher boiling point and higher dipole moment.

Trans-Hex-2-ene has lower dipole moment and lower boiling point.

Boiling point depends on dipole-dipole interactions.

 

 

Question 13.10

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

Answer.

Extra stability of benzene:

  1. In benzene, each of the 6-Catoms is sp2 hybridized with one p-orbital in each C-atom left unhybridized.
  2. While two of the sp2 orbitals form bonds with 2C atoms, third is involved in bonding with H-atom.
  3. Hence, three valencies of C are satisfied.
  4. Unhybridized p-orbital containing one electron is left for bonding.
  5. Each p-orbital can overlap with adjacent C atom and result in bonding.
  6. Pie electrons are more localized between just 2 C-atoms.
  7. These electrons are shared and attracted by all the six C-atoms.
  8. Three pi bonds are not localized but are spread over the entire molecule.

 Class_11_HydroCarbons_Cis_Trans_Structure_4

 

Question 13.11

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

Answer.

  1. It should have a single cyclic cloud of delocalized n-electrons above and below the plane of the molecule.
  2. It should be planar because complete delocalization of n-electrons is possible only if the ring is planar to allow cyclic overlap of p-orbitals.
  3. It should contain Huckel number of electrons, that is, (4n +2)

Where, n = number of electrons (1, 2, 3, etc.)

  

Question 13.12

Explain why the following systems are not aromatic?

Class_11_HydroCarbons_Aromatic_Structures

Answer.

  • Due to the presence of sp3 hybridized carbon, system is not planar. It have six n-electrons
  • but system is not fully conjugated since all these do not form single cyclic electron cloud which
  • surrounds all the atoms of the ring. Hence, it is not an aromatic compound.
  • Due to the presence of sp3-C, system is not planar. It contains only four n-electrons.
  • Hence, the system is not aromatic because it does not contain planar cyclic cloud having (4n+2) n-electrons.
  • It is not planar but it tub-shaped, therefore, it is non-planar.have 8 n-electrons.

 

 

Question 13.13

How will you convert benzene into

  • p-nitrobromobenzene
  • m-nitrochlorobenezene
  • p-nitrotoulene
  • acetophenone

Answer.

(i)

  • Major product is p-Bromonitrobenzene
  • Minor product is o-Bromonitrobenzene

       Class_11_HydroCarbons_BromonitroBenzene

(ii) 

          Class_11_HydroCarbons_NitrochloroBenzene

  • Major product is p-Nitrotoulene
  • Minor product is o-Nitrotoulene

       Class_11_HydroCarbons_Nitrotoulene

(iv)

     Class_11_HydroCarbons_Acetophenone

 

Question 13.14

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

Answer.

Expanded formula is given below:

 Class_11_HydroCarbons_Structure_5

 

Question 13.15

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

Answer.

Branching of carbon atom chain, decreases boiling point of alkane. Reduced surface area results in

decreased Van der Waal’s interaction and finally leads to lower boiling point.

  

Question 13.16

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 is an ionic electrophilic addition reaction in which H+ is an electrophile which adds to gives a more stable 2o carbocation.

During the next step, carbocation is rapidly attached by the nucleophile Br-ion to give 2-bromopropane.

Class_11_HydroCarbons_Structure_of_2-Bromopropane

In the presence of benzoyl peroxide, reaction is electropilic. Br is free radical which is obtained by the action of benzoyl peroxide on HBr.

Class_11_HydroCarbons_Structure_of_Benzene

Br radical adds to propene in such a way to generate the more stable 2o free radical.

In the next step, free radical obtained rapidly abstract a hydrogen atom from HBr to give 1-Bromopropane.

Class_11_HydroCarbons_Structure_of_1-Bromopropane

Both these reactions are electrophilic addition reactions, but gives different products due to different order of addition of H and Br atoms.

 

 

Question 13.17

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

Answer.

o-xylene is regarded as a resonance hybrid of the following two Kekule structures.

Ozonolysis of these products is given below:

Class_11_HydroCarbons_Structure_of_O-Xylene

All the three products formed cannot be obtained from anyone of two Kekule structures.

This shows that o-xylene is the resonance hybrid of two kekule structures.

 

 

Question 13.18

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

Answer.

s-electrons are closer to the nucleus. As the s-character of the orbital making C-H bond increases,

the electrons of C-H bond lie closer to the C-atom.

So the partial positive charge on H-atom and acidic character increases as s-character of the orbital increases.

Hybridization state of C in three carbon compounds are

Class_11_HydroCarbons_Structure_6

Decreasing order of acidic character is

Ethyne > Benzene > Hexane

 

 

Question 13.19

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

Answer.

Due to the presence of electron cloud containing 6-electrons below and above the plane of benzene ring,

it is rich a source of electrons.

It attracts electrophiles reagents towards it and repels nucleophiles.

Benzene does not undergoes electrophilic substitution reactions and nucleophilic substitutions easily.

 

 

Question 13.20

How would you convert the following compounds into benzene?

  • Ethyne
  • Ethene
  • Hexane

Answer.

 

 Class_11_HydroCarbons_Structure_7

Question 13.21

Write structures of all alkenes which in hydrogenation give 2-methylbutane.

Answer.

Structures which produce 2-methyl butane on hydrogenation are

 Class_11_HydroCarbons_Structure_8

 

Question 13.22

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

  • Chlorobenzene, 2,4-dinitrobenzene, p-nitrochlorobenzene
  • Toulene, p-H3C-C6H4-NO2, p-O2N-C6H4-NO2.

Answer.

  • Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene

Higher electron density in benzene ring, more reactive is the compound towards these reactions.

NO2 is more powerful electron-withdrawing group than Cl. Hence, more the number of nitro groups, less reactive is the compound.

  • Toulene > p-H3C-C6H4-NO2 > p-O2N-C6H4-NO2

Methyl group is electron donating but NO2 group is electron withdrawing.

 

 

Question 13.23

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

Answer.

Methyl group is electron donating and nitro group is electron withdrawing.

Maximum electron density will be in toluene followed by benzene and least in m-dinitrobenzene.

Toulene > Benzene > m-dinitrobenzene

 

Question 13.24

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

Answer.

Anhydrous FeCl3 (Ferric Chloride) can be used.

  

Question 13.25

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.

Wurtz reaction is used for preparation of alkanes.

For preparing alkanes having odd number of C-atoms, a mixture of alkyl halides have to be used.

Class_11_HydroCarbons_Wurtz_Reaction

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