NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

 

NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes- When a hydrogen atom is replaced by a halogen atom (F, Cl, Br, I) in an aliphatic hydrocarbon it results in formation of Haloalkane or alkane halide. Similarly, when a hydrogen atom is replaced by a halogen in an aromatic hydrocarbon then it forms Haloarene or aryl halide. Such common facts are exaplained in detail in NCERT solutions for class 12 chemistry chapter 10 haloalkanes and haloarenes. Haloalkanes and Haloarenes have many uses in our day to day life, for example, chloramphenicol​​​​​​, a haloarene is used for the treatment of typhoid fever. Our body produces thyroxine hormone which contains iodine and deficiency of iodine can cause a disease called goiter. In this chapter, you will study methods of preparation, physical and chemical properties and uses of haloalkanes and haloarenes. You will find all the CBSE NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes here for free. This chapter carries 4 marks in the CBSE board exams hence it becomes necessary for you to find all the solutions of NCERT class 12 chemistry chapter 10 haloalkanes and haloarenes to score good marks. In this chapter, there are 9 intext questions and  22 questions in the exercise.

The CBSE NCERT Solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes are explained by chemistry experts to leave you with no doubts about the question. These NCERT solutions will help you in your preparation for CBSE Board exams as well as in competitive exams like JEE, NEET, BITSAT, and KVPY,  etc. The seven sub-topics of NCERT solutions for Class 12 Chemistry chapter 10 Haloalkanes and Haloarenes covers important concepts like IUPAC nomenclature and preparation of haloalkanes and haloarenes and discuss how to use stereochemistry as a tool in the understanding the reaction mechanism and applications of organo-metallic compounds. At the end of the chapter, the environmental effects of polyhalogen compounds also highlighted. Halogenated compounds remain in the environment due to their resistance to breakdown by the bacteria in the soil.

 

Some important basic points of the Chapter 10 Haloalkanes and Haloarenes-

  • The replacement of one or more than one H-atom in an aliphatic hydrocarbon by halogen atoms results in the formation haloalkane or alkyl halide and the replacement of H-atoms in an aromatic hydrocarbon by halogen atoms results in the formation of haloarene or aryl halide.
  • In haloalkanes, halogen atoms attached to the SP^{3} hybridised carbon atom of an alkyl group.
  • in haloarenes, halogen atoms attached to the SP^{2} hybridised carbon atoms of an aryl group.

Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes-

10.1 Classification

10.2 Nomenclature

10.3 Nature of C–X Bond

10.4 Methods of Preparation of Haloalkanes

10.5 Preparation of Haloarenes

10.6 Physical Properties

10.7 Chemical Reactions

Find NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes below:

Solutions to In-Text Questions Ex 10.1  to 10.9

Question 1 Write structures of the following compounds:

(i) 2-Chloro-3-methylpentane

Answer:

The structure of 2-Chloro-3-methylpentane is given below :-

Question 10.1      Write structures of the following compounds:

        (ii)    1-Chloro-4-ethylcyclohexane

Answer:

The structure of  1-Chloro-4-ethylcyclohexane is given below :- 

                                                                                  

Question 10.1      Write structures of the following compounds:

 (iii)       4-tert. Butyl-3-iodoheptane

Answer:

The structure of 4-tert. Butyl-3-iodoheptane is given below :-

                                                      

 

Question 10.1       Write structures of the following compounds:

       (iv)    1,4-Dibromobut-2-ene

Answer:

The structure of 1,4-Dibromobut-2-ene is given below :- 

                                      

Question 10.1       Write structures of the following compounds: 

        (v)      1-Bromo-4-sec. butyl-2-methylbenzene

Answer:

The structure of  1-Bromo-4-sec. butyl-2-methylbenzene is shown below :- 

                                                      

Question 10.2  Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer:

We don't use sulphuric acid because it acts as an oxidising agent and the required alkyl iodide is not produced. The reactions are given below :-

                      2KI  +  H2SO4     →    2KHSO4  + 2HI

                      2HI  + H2SO4    →  I2  +  SO2  +  H2O

Question 10.3      Write structures of different dihalogen derivatives of propane.

Answer:

We obtain four dihalogen derivatives of propane  :-

(i) 1,1 Dibromopropane           

                                                

(ii) 2, 2 Dibromopropane  

                                                   

(iii) 1, 2 Dibromopropane 

                                                 

(iv) 1, 3 Dibromopropane

                                              

Question 10.4      Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

       (i)       A single monochloride

Answer:

In this we have to find an isomer in which replacement of any hydrogen atom gives the singel compound for all replacements.

So the isomer is Neopentane.

                                                  

Question 10.4      Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

        (ii)      Three isomeric monochlorides.

Answer:

For the given condition we must have three different hydrogens so that we can get three different monochlorides on the replacement.

Thus the isomer is n-pentane.

                                

Question 10.4      Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields

      (iii)      Four isomeric monochlorides.

Answer:

For four monochlorides we need four different hydrogens which can be replaced by chlorine.

Hence the required isomer is :-

                                             

Question 10.5      Draw the structures of major monohalo products in each of the following reactions:

     (i)     

Answer:

The final products are:-

                                            

Question 10.5      Draw the structures of major monohalo products in each of the following reactions:

     (ii)         

Answer:

                            

 

Question 10.5      Draw the structures of major monohalo products in each of the following reactions:

     (iii)        

Answer:

         

Question 10.5      Draw the structures of major monohalo products in each of the following reactions:

   (iv)      

 

Answer:

The obtained product is:-

     

Question 10.5      Draw the structures of major monohalo products in each of the following reactions:

        (vi)     

Answer:

The obtained product is :-

                

 

Question 10.6      Arrange each set of compounds in order of increasing boiling points.

        (i)       Bromomethane, Bromoform, Chloromethane, Dibromomethane.

Answer:

It is known that boiling point increases with increase in molecular mass when the alkyl group is the same.

So the order of increasing boiling point is Chloromethane <  Bromomethane  <  Dibromomethane  <  Bromoform

 

Question 10.6      Arrange each set of compounds in order of increasing boiling points.

        (ii)      1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Answer:

In the given compounds the halide groups are same. In these cases, the boiling point depends on the bulkiness of the alkyl group. The boiling point increases with an increase in the chain length. Also, the boiling point decreases with an increase in branching.

So the order is :-    1- Chlorobutane  >  1- Chloropropane  >  Isopropyl Chloride

 

Question 10.7      Which alkyl halide from the following pairs would you expect to react more rapidly by an S_{N}2 mechanism? Explain your answer.

         (i)      CH_{3}CH_{2}CH_{2}CH_{2}Br  or 

Answer:

In this case, the rate of S_{N}2   reaction will depend on the hindrance of the substrate.

Since 1- Bromobutane is a 1^{\circ} alkyl halide and 2- Bromobutane is a 2^{\circ} alkyl halide hence 2- Bromobutane gives more hindrance to the nucleophile.

Hence  1- Bromobutane reacts faster.

 

Question 10.7   Which alkyl halide from the following pairs would you expect to react more rapidly by an S_{N}2 mechanism? Explain your answer.

            (ii)    

Answer:

The rate of S_{N}2 reaction decreases with increase in hindrance to the attack of the nucleophile.

So 2-bromobutane will react faster than 2-bromo-2-methylpropane in the nucleophilic attack.

 

Question 10.7      Which alkyl halide from the following pairs would you expect to react more rapidly by an S_{N}2 mechanism? Explain your answer.

      (iii)      

Answer:

In these kinds of cases, we see where is the substituent is attached i.e., how far from the halide group. It can be clearly seen that the methyl group attached in 1-bromo-2-methylbutane is near than that attached in 1-bromo-3-methyl butane.

Hence the rate of S_{N}2 reaction will be faster in case of 1-bromo-3-methylbutane.

 

Question 10.8      In the following pairs of halogen compounds, which compound undergoes faster S_{N}1 reaction?

      (i)      

                        

Answer:

In S_{N}1 reactions, we see the formation of carbocation and this is the rate determining step for this kind of reactions. So the compound having more stable carbocation will react faster. In the given case 2- Chloro, 2- Methylpropane we have 3^{\circ} carbon whereas in 3- Chloropentane we have 2^{\circ} carbon.

 

Question 10.8      In the following pairs of halogen compounds, which compound undergoes faster S_{N}1 reaction?

       (ii)     

Answer:

In S_{N}1 reactions, we see the formation of carbocation and this is the rate determining step for these kinds of reactions.So the compound having more stable carbocation will react faster. Hence 2-Chloroheptane will react faster than 1- Chlorohexane.

 

Question 10.9     Identify A, B, C, D, E, R and R1 in the following:

            

Answer:

1st reaction  :-         

                                   

2nd reaction :-             

                                      

3rd reaction  :-   

                                       

NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes- Exercise Questions

Question 10.1      Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

       (i)      \left ( CH_{3} \right )_{2}CHCH (Cl)CH_{3}

Answer:

(i) 2-Chloro-3-methylbutane.   And it is a secondary alkyl halide.

 

Question 10.1    Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

       (iv)     (CH_{3})_{3}CCH_{2}CH(Br)C_{6}H_{5}

Answer:

(iv)  1-Bromo-3, 3-dimethyl-1-phenylbutane.   And it is secondary benzyl halide.

 

Question 10.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

        (xi)     m- Cl C_{6}H_{4}CH_{2}CH (CH_{3})_{2}

Answer:

(xi) 1-Chloromethyl-3-(2, 2-dimethylpropyl) benzene.  And it is primary benzyl halide.

 

Question 10.2     Give the IUPAC names of the following compounds:     

          (ii)     CHF_{2}CBrCIF

Answer:

1-Bromo-1-chloro-1, 2, 2-trifluroethane

 

Question 10.2     Give the IUPAC names of the following compounds:

           (iii)    Cl CH_{2}C \equiv C CH_{2}Br

 

Answer:

 1-Bromo-4-chlorobut-2-yne

 

Question 10.2     Give the IUPAC names of the following compounds:

           (iv)     (CCl_{3})_{3}CCl

Answer:

2-(Trichloromethyl)-1, 1, 1, 2, 3, 3, 3-heptachloropropane

 

Question 10.2     Give the IUPAC names of the following compounds:

           (v)     CH_{3}(p- Cl C_{6}H_{4})_{2}CH(Br)CH_{3}

Answer:

2-Bromo-3, 3-bis(4-Chlorophenyl) butane

 

Question 10.2     Give the IUPAC names of the following compounds:

        (vi)     (CH_{3})_{3}CCH = CCl C_{6}H_{4}I - p

Answer:

 1-Chloro-1-(4-iodophenyl)-3, 3-dimethylbut-1-ene

 

Question 10.3     Write the structures of the following organic halogen compounds.

         (i)     2-Chloro-3-methylpentane

Answer:

(i)   

                       

 

Question 10.3     Write the structures of the following organic halogen compounds.

     (viii)         1,4-Dibromobut-2-ene

Answer:

(viii)  

                    

 

Question 10.4 Which one of the following has the highest dipole moment?

             (i) CH2Cl2         (ii) CHCl3         (iii) CCl4

Answer:

The order of dipole moment will be :-                CH2Cl2    >  CHCl3   >   CCl4.

The reason for the above order is given as- CCl4 is a symmetrical compound so its dipole moment will be zero. In case of CHCl, one of the Cl cancels dipole moment of the opposite Cl atom, so net dipole moment is just due to one Cl. In the case of CH2Cl2 , both Cl groups contribute to the dipole moment so it has the highest dipole moment among all.

 

Question 10.5    A hydrocarbon C5H10  does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Answer:

We are given the formula C5H10  which can be either of an alkene or of cycloalkane. Since the hydrocarbon doesn't react with chlorine in dark thus it cannot be alkene. So the only option left out is cyclopentane.

                                                             

 

Question 10.6     Write the isomers of the compound having formula C_{4}H_{9}Br.

Answer:

The isomers of the compound C_{4}H_{9}Br  are :-

  (i)  1-Bromobutane

                    

(ii)   2-Bromobutane

                      

(iii)  1-Bromo-2-methylpropane

                        

(iv)  2-Bromo-2-methylpropane

                          

 

Question 10.7     Write the equations for the preparation of 1-iodobutane from:

        (i)        1-butanol 

Answer:

(i)  The procedure given below can be used :-

                   

 

Question 10.7     Write the equations for the preparation of 1-iodobutane from

         (ii)         1-chlorobutane

Answer:

(ii) The required product can be obtained as shown below :-

                                   

 

Question 10.7     Write the equations for the preparation of 1-iodobutane from

          (iii)     but-1-ene

Answer:

(iii) The required product is obtained by following procedure :-                   

                                       

 

Question 10.8   What are ambident nucleophiles? Explain with an example.

Answer:

The ambident nucleophiles are those nucleophiles which have two nucleophilic sites through which they can attack. For e.g Nitrile ion can attack through both nitrogen atom (forms nitroalkanes) and an oxygen atom (forms alkyl nitrites), thus it is an ambident nucleophile.

 

Question 10.9     Which compound in each of the following pairs will react faster in SN2 reaction with –OH?

         (i)       CH3Br or CH3 I 

Answer:

In this case, we have the same alkyl group but different halide ions. For this rate of SN2 reaction increases with increase in atomic mass. So,  CH3I will react faster than CH3Br.

 

Question 10.9     Which compound in each of the following pairs will react faster in SN 2 reaction with –OH?

        (ii)     (CH_{3})_{3}CCl or CH_{3}Cl

Answer:

In this case, the hindrance will be deciding factor for the rate of  SN 2 reaction because hindrance will directly affect the attack of the nucleophile. So  CH_{3}Cl  will react faster as compared to (CH_{3})_{3}CCl.

 

Question 10.10     Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

             (i)     1-Bromo-1-methylcyclohexane 

Answer:

                      

 In this compound, it is clear that we have identical \beta hydrogen, therefore, dehalogenation of the given compound gives the same alkene.

                            

Question 10.10     Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: 

          (ii)     2-Chloro-2-methylbutane

Answer:

(ii)   

                                   

In this compound we have two kind of \beta hydrogen. So dehalogenation will give two kind of alkenes, namely 2-Methylbut-2-ene and 2-Methylbut-1-ene.

                    

The major product of this reaction will be 2-Methylbut-2-ene as the number of \alpha - hydrogens attached to double bonded carbon are more in case of this compound.

 

Question 10.10     Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the  major alkene: 

          (iii)     2,2,3-Trimethyl-3-bromopentane

Answer:

(iii)     

                                         

 In this compound we have two type of \beta - hydrogen thus dehalogenation we get two types of products namely 3, 4, 4-Trimethylpent-2-ene and 2-Ethyl-3,3-dimethylbut-2-ene.

                    and               

Here 3, 4, 4-Trimethylpent-2-ene will be major product, since the \alpha - hydrogen attached to the double bond are greater.

 

Question 10.11     How will you bring about the following conversions?

        (i)      Ethanol to but-1-yne 

Answer:

(i)  The conversion will take place by following procedure :-                           

                                  

     Now,                  

                                 

                                

 

Question 10.11     How will you bring about the following conversions?

          (ii)     Ethane to bromoethene 

Answer:

(ii)    

                 

 

Question 10.11     How will you bring about the following conversions?

         (iii)     Propene to 1-nitropropane

Answer:

(iii)   

                                           

 

Question 10.11     How will you bring about the following conversions?

          (iv)    Toluene to benzyl alcohol 

Answer:

(iv)

                          

 

Question 10.11         How will you bring about the following conversions?

           (v)      Propene to propyne

Answer:

(v)   

           

 

Question 10.11     How will you bring about the following conversions?

          (vi)   Ethanol to ethyl fluoride

Answer:

(vi)  

                                   

 

Question 10.11     How will you bring about the following conversions?

        (vii)     Bromomethane to propanone 

Answer:

(vii)

                                

 

Question 10.11     How will you bring about the following conversions?

       (viii)    But-1-ene to but-2-ene 

Answer:

(viii)   

                                  

 

 

Question 10.11     How will you bring about the following conversions?

          (ix)   1-Chlorobutane to n-octane 

Answer:

(ix) 

                            

 

Question 10.11     How will you bring about the following conversions?

          (x)    Benzene to biphenyl.

Answer:

(x)  

                           

Question 10.12     Explain why

         (i)     the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

Answer:

We know that the Cl-atom in chlorobenzene is attached to a sp2 hybridized carbon atom whereas, in cyclohexyl chloride, the Cl-atom is attached to a sp3 hybridized carbon atom. It is known that sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Thus, chlorobenzene is more electronegative than cyclohexyl chloride. 

Apart from this, the - R effect of the benzene ring of chlorobenzene results in decreasing the electron density of the C - Cl bond near the Cl-atom. And the C - Cl bond in chlorobenzene becomes less polar. Due to these reasons, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

 

Question 10.12     Explain why

         (ii)     alkyl halides, though polar, are immiscible with water? 

Answer:

For being soluble in the water we have a condition that the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are held together by dipole-dipole interactions and there are polar molecules. Similarly, the intermolecular force of attraction present between the water molecules is hydrogen bonding. The new force of attraction after we dissolve solute in water i.e., between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. That is why alkyl halides (though polar) are immiscible with water.

 

Question 10.12     Explain why

        (iii)     Grignard reagents should be prepared under anhydrous conditions?

Answer:

This is done because in presence of moisture, they react to give alkane.

                          

 

Question 10.13      Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Answer:

(i)   Freon-12 or dichlorodifluoromethane is generally known as CFC. It is used in refrigerators and air conditioners as a refrigerant. It is also used in body sprays, hair sprays, etc. But it has environmental impacts as it damages the ozone layer.

(ii)  DDT or p, p-dichlorodiphenyltrichloroethane is one of the best-known insecticides which was used very widely all over the world. It is very effective against mosquitoes, insects and lice. But it has harmful effects.

(iii)  CCl :-   It is mostly used for manufacturing refrigerants for refrigerators and air conditioners.  It is also used as a solvent in the manufacture of pharmaceutical products. In the early years, carbon tetrachloride was widely used as a cleaning fluid and a fire extinguisher.

(iv) Iodoform was used earlier as an antiseptic. And this antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

          (i)       CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone} 

Answer:

(i)  The obtained product is :-

                                            CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone}  

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

          (ii)     (CH_{3})_{3}CBr+KOH \xrightarrow[heat]{ethanol}

Answer:

(ii)  The obtained product is 2-Methylpropene           

                      

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

         (iii)     CH_{3}CH(Br)CH_{2}CH_{3}+ NaOH \overset{water}{\rightarrow}

Answer:

(iii)  The obtained product is Butan-2-ol.

           

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

          (iv)     CH_{3}CH_{2}Br+KCN \overset{aq. ethanol}{\rightarrow}

Answer:

(iv)  The obtained product is Cyanoethane.

                              Bromobutane to cyanoethane

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

           (v)     C_{6}H_{5}ONa + C_{2}H_{5}Cl \rightarrow 

Answer:

(v) The obtained product is Phenetole.

                         Sodium phenoxide to phenetole

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

          (vi)    CH_{3}CH_{2}CH_{2}OH+SOCl_{2}\rightarrow 

Answer:

(vi)  The obtained product is 1-Chloropropane.

           1- propanol to 1-Chloropropane

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

        (vii)       CH_{3}CH_{2}CH = CH _{2}+HBr \overset{peroxide}{\rightarrow} 

Answer:

(vii)  The obtained product is 1-Bromobutane.

                            But-1-ene to 1-bromobutane

 

Question 10.14     Write the structure of the major organic product in each of the following reactions:

      (viii)       CH_{3}CH = C(CH_{3})_{2}+ HBr \rightarrow

Answer:

The obtained product is 2-Bromo-2-methylbutane.

                              

 

Question 10.15     Write the mechanism of the following reaction:

                   nBuBr + KCN \overset{EtOH -H_{2}O}{\rightarrow} nB_{4}CN  

Answer:

The reaction will proceed through SN2 mechanism. The mechanism for the given reaction is shown below :-

                       

Question 10.16     Arrange the compounds of each set in order of reactivity towards SN2 displacement:

         (i)     2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

Answer:

(i)  Here the deciding factor the rate of reaction will be a steric hindrance.

                    

It is clear from the above that the order of hindrance is:-   

1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane

So the order of rate of reaction will be:-        

 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

 

Question 10.16     Arrange the compounds of each set in order of reactivity towards SN2 displacement:

         (ii)     1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane 

Answer:

(ii) In this case, also, the order of the rate of reaction will be decided from the steric hindrance factor.

                 

It is clear from the above that the steric hindrance in 2-Bromo-2-methylbutane highest (note that hindrance of the carbon attached to halide ion is seen). So the order of the rate of reaction is:-  

                    2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

 

Question 10.16     Arrange the compounds of each set in order of reactivity towards SN2 displacement:

        (iii)     1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Answer:

(iii)  The steric hindrance is the deciding factor here.

The order of steric hindrance is :-     

1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane< 1-Bromo-2, 2-dimethylpropane

Thus the order of the rate of reaction will be : - 

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane

 

Question 10.17     Out of C_{6}H_{5}CH_{2}Cl  and C_{6}H_{5}CHCl C_{6}H_{5} , which is more easily hydrolysed by aqueous KOH.

Answer:

Hydrolysis by KOH will take place by the formation of a carbocation in its rate-determining step. So the compound having stable carbocation will hydrolyse faster.

The carbocations of both the compounds are given below:-

            

It is clear that carbocation of C_{6}H_{5}CHCl C_{6}H_{5} is more stable. 

Hence  C_{6}H_{5}CHCl C_{6}H_{5} will hydrolyse faster than C_{6}H_{5}CH_{2}Cl.

 

Question 10.18     p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.

Answer:

The structures of  o-Dichlorobenzene,  m-Dichlorobenzene and  p-Dichlorobenzene are given below.

                                                                   

We can see that p-Dichlorobenzene is a very symmetric structure thus packing of it will be maximum. As a result more and more energy will be required to break bonds (during boiling). Thus boiling point is high for p-Dichlorobenzene.                     

 

Question 10.19     How the following conversions can be carried out?

           (i)    Propene to propan-1-ol 

 

Answer:

   The mechanism is given below :-

                               

Question 10.19     How the following conversions can be carried out?

         (ii)      Ethanol to but-1-yne

Answer:

   The reaction mechanism is given below :-

                        

 

 

Question 10.19     How the following conversions can be carried out?

      (iii)       1-Bromopropane to 2-bromopropane

Answer:

(iii)  The mechanism is given below :- 

                     

 

Question 10.19     How the following conversions can be carried out? 

         (iv)      Toluene to benzyl alcohol

Answer:

(iv) The mechanism is given below :-  

                       

Question 10.19     How the following conversions can be carried out?

           (v)     Benzene to 4-bromonitrobenzene 

Answer:

The mechanism for the given reaction is as follows  :- 

                    

Question 10.19     How the following conversions can be carried out?

         (vi)    Benzyl alcohol to 2-phenylethanoic acid 

Answer:

(vi) The mechanism for the reaction is given below :- 

                            

 

Question 10.19     How the following conversions can be carried out?

      (vii)      Ethanol to propanenitrile 

Answer:

(vii) The mechanism of the given reaction reaction is :-

                      

 

Question 10.19     How the following conversions can be carried out?

        (viii)      Aniline to chlorobenzene 

Answer:

The mechanism of the reaction is given below :- 

                     

 

Question 10.19     How the following conversions can be carried out ?

        (ix)      2-Chlorobutane to 3, 4-dimethylhexane

Answer:

(ix)   The required mechanism is as follows :-

             

Question 10.19     How the following conversions can be carried out?

         (x)      2-Methyl-1-propene to 2-chloro-2-methylpropane 

Answer:

(x) The required mechanism is given below :- 

              

 

Question 10.19     How the following conversions can be carried out ?

        (xi)      Ethyl chloride to propanoic acid 

Answer:

 The mechanism of the given reaction is :- 

                               

 

Question 10.19     How the following conversions can be carried out?

        (xii)     But-1-ene to n-butyliodide

Answer:

The mechanism is given below :-               

                      

 

Question 10.19     How the following conversions can be carried out?

        (xiii)      2-Chloropropane to 1-propanol

Answer:

The mechanism is :-

                       

 

Question 10.19     How the following conversions can be carried out?

        (xiv)      Isopropyl alcohol to iodoform

Answer:

(xiv) The proposed mechanism is :-

                                  

Question 10.19     How the following conversions can be carried out?

        (xv)     Chlorobenzene to p-nitrophenol

Answer:

(xv) The required mechanism is given below :-

                           

 

Question 10.19     How the following conversions can be carried out?

        (xvi)     2-Bromopropane to 1-bromopropane

Answer:

 The mechanism of the reaction is given below :- 

                              

 

Question 10.19     How the following conversions can be carried out?

      (xvii)      Chloroethane to butane

Answer:

(xvii) The mechanism of the reaction is :-

                                       

 

Question 10.19     How the following conversions can be carried out?

     (xviii)      Benzene to diphenyl

Answer:

 The mechanism for the given reaction is :-

                            

 

Question 10.19     How the following conversions can be carried out?

      (xix)      tert-Butyl bromide to isobutyl bromide

Answer:

The mechanism of the given reaction is :-

                                          

 

Question 10.19     How the following conversions can be carried out?

        (xx)     Aniline to phenylisocyanide

Answer:

The mechanism for the given reaction is as follows :- 

                       

Question 10.20    The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

Answer:

In an aqueous solution, KOH almost completely dissociates into OH - ions. We know that OH - ions are strong nucleophile, which leads the alkyl chloride to undergo a reaction to form alcohol. But an alcoholic solution of KOH contains alkoxide (RO ) ion, which is a strong base. Thus, it can remove hydrogen from the β-carbon of the alkyl chloride and form an alkene.  The OH - ion is a weaker base than RO - ion. The basic character of OH - ion decreases in aqueous solution. Therefore, it cannot remove hydrogen from the β-carbon.

 

Question 10.21     Primary alkyl halide C_{4}H_{9}Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C_{8}H_{18} which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Answer:

With the given formula we have two alkyl halides They are n - butyl bromide and isobutyl bromide.

For the first set of reaction we get two possibilities:- 

                                          

Therefore compound (d) is 2, 5-dimethylhexane.

                                

                                   

 

Question 10.22     What happens when

         (i)       n-butyl chloride is treated with alcoholic KOH,

Answer:

When n-butyl chloride is treated with alcoholic KOH the following reaction occurs:-

                               

 

Question 10.22     What happens when

        (ii)     bromobenzene is treated with Mg in the presence of dry ether

Answer:

 When bromobenzene is treated with Mg in the presence of dry ether the following reaction occurs :-

                            

 

Question 10.22     What happens when

         (iii)    chlorobenzene is subjected to hydrolysis

Answer:

The reaction is given below :- 

                                              

 

Question 10.22     What happens when

        (iv)    ethyl chloride is treated with aqueous KOH

Answer:

When ethyl chloride is treated with aqueous KOH the following reaction occurs :-

                                       

 

Question 10.22     What happens when

         (v)    methyl bromide is treated with sodium in the presence of dry ether

Answer:

When methyl bromide is treated with sodium in the presence of dry ether then following reaction occurs:- 

                                      

 

Question 10.22     What happens when

        (vi)      methyl chloride is treated with KCN?

Answer:

When methyl chloride is treated with KCN the following reaction occurs:-    

                            

 

NCERT Solutions Class 12 Chemistry

Chapter 1

CBSE NCERT solutions for class 12 chapter 1 The Solid State

Chapter 2

NCERT solutions for class 12 chemistry chapter 2 Solutions

Chapter 3

Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry

Chapter 4

CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Chapter 5

Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry

Chapter 6

NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements

Chapter 7

CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements

Chapter 8

Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements

Chapter 9

NCERT solutions for class 12 chemistry chapter 9 Coordination compounds

Chapter 10

NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes

Chapter 11

CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers

Chapter 12

Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids

Chapter 13

NCERT solutions for class 12 chemistry chapter 13 Amines

Chapter 14

CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules

Chapter 15

Solutions of NCERT class 12 chemistry chapter 15 Polymers

Chapter 16

NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

NCERT Solutions for Class 12 Subject wise

Solutions of NCERT class 12 biology

NCERT solutions for class 12 maths

CBSE NCERT solutions for class 12 chemistry

Solutions of NCERT class 12 physics

Benefits of NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes

  • The comprehensive answers given in the NCERT solutions for class 12 chemistry chapter 10 Haloalkanes and Haloarenes will be helpful in understanding the chapter easily.

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