# 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.

Intext Questions

Exercise Questions

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 $\dpi{100} SP^{3}$ hybridised carbon atom of an alkyl group.
• in haloarenes, halogen atoms attached to the $\dpi{100} SP^{2}$ 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

## Solutions to In-Text Questions Ex 10.1  to 10.9

(i) 2-Chloro-3-methylpentane

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

Question 10.1      Write structures of the following compounds:

(ii)    1-Chloro-4-ethylcyclohexane

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

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

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

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

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

We obtain four dihalogen derivatives of propane  :-

(i) 1,1 Dibromopropane

(ii) 2, 2 Dibromopropane

(iii) 1, 2 Dibromopropane

(iv) 1, 3 Dibromopropane

(i)       A single monochloride

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.

(ii)      Three isomeric monochlorides.

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.

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

Hence the required isomer is :-

The final products are:-

(ii)

(iii)

(iv)

The obtained product is:-

$CH_3CH_2Br+\ NaI\rightarrow$

(vi)

The obtained product is :-

(i)       Bromomethane, Bromoform, Chloromethane, Dibromomethane.

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

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

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

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

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.

(ii)

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.

(iii)

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.

(i)

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.

(ii)

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.

1st reaction  :-

2nd reaction :-

3rd reaction  :-

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

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

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

(ii)       $CH_{3}CH_{2}CH(CH_{3})CH(C_{2}H_{5})Cl$

3-Chloro-4-methylhexane.  And it is primary alkyl halide.

(iii)     $CH_{3}CH_{2}C(CH_{3})_{2}CH_{2}I$

(iii) 1-Iodo-2, 2-dimethylbutane.  And it is primary alkyl halide.

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

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

(v)      $CH_{3}CH(CH_{3})CH(Br)CH_{3}$

(v) 2-Bromo-3-methylbutane.  And it is secondary alkyl halide.

(vi)      $CH_{3}C(C_{2}H_{5})_{2}CH_{2}Br$

(vi) 1-Bromo-2-ethyl-2-methylbutane.    And it is a primary alkyl halide.

(vii)       $CH_{3}C(Cl)(C_{2}H_{5})CH_{2}CH_{3}$

(vii) 3-Chloro-3-methylpentane. And it is tertiary alkyl halide.

(viii)    $CH_{3}CH = C(Cl)CH_{2}CH(CH_{3})_{2}$

(viii) 3-Chloro-5-methylhex-2-ene.  And it is vinyl halide

(ix)      $CH_{3}CH = CH C (Br)( CH_{3})_{2}$

(ix) 4-Bromo-4-methylpent-2-ene.  And it is allyl halide.

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

(x)  1-Chloro-4-(2-methylpropyl) benzene.  And it is aryl halide.

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

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

(xii)       $o- Br C_{6}H_{4}CH (CH_{3})CH_{2}CH_{3}$

(xii) 1-Bromo-2-(1-methylpropyl) benzene. And it is aryl halide.

Question 10.2     Give the IUPAC names of the following compounds:

(iv)     $(CCl_{3})_{3}CCl$

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

(i)     2-Chloro-3-methylpentane

(i)

(ii)

(iii)     1-Chloro-4-ethylcyclohexane

(iii)

(v)         2-Bromobutane

(v)

(vi)

(vii)     1-Bromo-4-sec-butyl-2-methylbenzene

(vii)

(viii)         1,4-Dibromobut-2-ene

(viii)

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

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.

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.

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

(i)        1-butanol

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

(ii)         1-chlorobutane

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

(iii)     but-1-ene

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

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.

(i)       CH3Br or CH3 I

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.

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

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 $(CH_{3})_{3}CCl$.

(i)     1-Bromo-1-methylcyclohexane

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

(ii)     2-Chloro-2-methylbutane

(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.

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

(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

(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

(ii)

Question 10.11     How will you bring about the following conversions?

(iii)     Propene to 1-nitropropane

(iii)

Question 10.11     How will you bring about the following conversions?

(iv)    Toluene to benzyl alcohol

(iv)

Question 10.11         How will you bring about the following conversions?

(v)      Propene to propyne

(v)

Question 10.11     How will you bring about the following conversions?

(vi)   Ethanol to ethyl fluoride

(vi)

Question 10.11     How will you bring about the following conversions?

(vii)     Bromomethane to propanone

(vii)

Question 10.11     How will you bring about the following conversions?

(viii)    But-1-ene to but-2-ene

(viii)

Question 10.11     How will you bring about the following conversions?

(ix)   1-Chlorobutane to n-octane

(ix)

Question 10.11     How will you bring about the following conversions?

(x)    Benzene to biphenyl.

(x)

Question 10.12     Explain why

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

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

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?

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

(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.

(i)  The obtained product is :-

$\dpi{100} CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone}$

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

(ii)  The obtained product is 2-Methylpropene

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

(iv)  The obtained product is Cyanoethane.

(v) The obtained product is Phenetole.

(vi)  The obtained product is 1-Chloropropane.

(vii)  The obtained product is 1-Bromobutane.

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

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

Question 10.15     Write the mechanism of the following reaction:

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

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

(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

(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

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

(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

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 $C_{6}H_{5}CH_{2}Cl$.

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

The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(ii)      Ethanol to but-1-yne

The reaction mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(iii)       1-Bromopropane to 2-bromopropane

(iii)  The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(iv)      Toluene to benzyl alcohol

(iv) The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

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

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

Question 10.19     How the following conversions can be carried out?

(vii)      Ethanol to propanenitrile

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

Question 10.19     How the following conversions can be carried out?

(viii)      Aniline to chlorobenzene

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

(ix)   The required mechanism is as follows :-

Question 10.19     How the following conversions can be carried out?

(x) The required mechanism is given below :-

Question 10.19     How the following conversions can be carried out ?

(xi)      Ethyl chloride to propanoic acid

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

The mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(xiii)      2-Chloropropane to 1-propanol

The mechanism is :-

Question 10.19     How the following conversions can be carried out?

(xiv)      Isopropyl alcohol to iodoform

(xiv) The proposed mechanism is :-

Question 10.19     How the following conversions can be carried out?

(xv)     Chlorobenzene to p-nitrophenol

(xv) The required mechanism is given below :-

Question 10.19     How the following conversions can be carried out?

(xvi)     2-Bromopropane to 1-bromopropane

The mechanism of the reaction is given below :-

Question 10.19     How the following conversions can be carried out?

(xvii)      Chloroethane to butane

(xvii) The mechanism of the reaction is :-

Question 10.19     How the following conversions can be carried out?

(xviii)      Benzene to diphenyl

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

The mechanism of the given reaction is :-

Question 10.19     How the following conversions can be carried out?

(xx)     Aniline to phenylisocyanide

The mechanism for the given reaction is as follows :-

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.

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,

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

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

The reaction is given below :-

Question 10.22     What happens when

(iv)    ethyl chloride is treated with aqueous KOH

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

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?

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

## NCERT Solutions Class 12 Chemistry

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