NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry chapter 10 Haloalkanes and Haloarenes - We all know about the periodic table and its elements. So, ever wondered that from these elements how Haloalkanes and Halaorenes are actually formed? 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. The same goes for Haloarenes, when a hydrogen atom is replaced by a halogen in an aromatic hydrocarbon then it forms Haloarene or aryl halide. These NCERT Class 12 Chemistry solutions chapter 10 will help in understanding concepts of Haloalkanes and Haloarenes. These NCERT Solutions for Class 12 Chemistry chapter 10 are explained by experts in the easiest way. This will help to resolve all NCERT problems of this chapter.

Also Read| Haloalkanes and Haloarenes Class 12 Notes

Topics and Sub-topics of NCERT Solutions for 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 :-

1650393268093

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 :-

1650393304836

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 :-

1650393334343

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 :-

1650393521544

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 :-

1650393588711

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 + H ₂ SO ₄ → 2KHSO ₄ + 2HI

2HI + H ₂ SO ₄ → I ₂ + SO ₂ + H ₂ O

Question 10.3 Write structures of different dihalogen derivatives of propane.

Answer :

We obtain four dihalogen derivatives of propane :-

(i) 1,1 Dibromopropane

1650393649596

(ii) 2, 2 Dibromopropane

1650393673880

(iii) 1, 2 Dibromopropane

1650393696576

(iv) 1, 3 Dibromopropane

1650393717664

Question 10.4 Among the isomeric alkanes of molecular formula C ₅ H ₁₂ , 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.

1650393974824

Question 10.4 Among the isomeric alkanes of molecular formula C ₅ H ₁₂ , 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.

1650394033275

Question 10.4 Among the isomeric alkanes of molecular formula C ₅ H ₁₂ , 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) 1650394128047

Answer:

The final products are:-

1650394146630

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

1650394177371

Answer :

1650394198247

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

(iii) 1650394239508

Answer :

1650394258543

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

(iv) 1650394292762

Answer :

The obtained product is:-

1650394311632

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

$CH_3CH_2Br+\ NaI\rightarrow$

Answer : 1653564082205

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

(vi) 1650394413285

Answer :

The obtained product is :-

1650394434422

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.

$CH_{3}CH_{2}CH_{2}CH_{2}Br$

Answer :

In this case, the rate of 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.

1650394520742

Answer :

The rate of 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.

1650394551180

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 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) 1650394593342

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?

1650394629230

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:

1650394661923

Answer :

1st reaction :-

1650394679729

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:

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

Answer :

3-Chloro-4-methylhexane. And it is primary 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:

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

Answer :

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

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

Answer :

(v) 2-Bromo-3-methylbutane. And it is 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:

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

Answer :

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

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

Answer :

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

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

Answer :

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

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

Answer :

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

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

Answer :

(x) 1-Chloro-4-(2-methylpropyl) benzene. And it is aryl 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 10.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

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

Answer :

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

Question 10.2 Give the IUPAC names of the following compounds:

(i) $CH_{3}CH(Cl)CH (Br)CH_{3}$

Answer :

(i) 2-Bromo-3-chlorobutane

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)

1650394770185

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

(ii) p-Bromochlorobenzene

Answer :

(ii)

1650394830725

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

(iii) 1-Chloro-4-ethylcyclohexane

Answer :

(iii)

1650394865688

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

(iv) 2-(2-Chlorophenyl)-1-iodooctane

Answer :

(iv)

1650394905647

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

(v) 2-Bromobutane

Answer :

(v)

1650394945214

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

(vi) 4-tert-Butyl-3-iodoheptane

Answer :

(vi)

1650394997618

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

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

Answer :

(vii)

1650395035475

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

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

Answer :

(viii)

1650395068985

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

(i) CH ₂ Cl ₂ (ii) CHCl ₃ (iii) CCl ₄

Answer :

The order of dipole moment will be :- CH ₂ Cl ₂ > CHCl ₃ > CCl _4.

The reason for the above order is given as- CCl ₄ is a symmetrical compound so its dipole moment will be zero. In case of CHCl ₃ , 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 CH ₂ Cl ₂ , both Cl groups contribute to the dipole moment so it has the highest dipole moment among all.

Question 10.5 A hydrocarbon C ₅ H ₁₀ does not react with chlorine in dark but gives a single monochloro compound C ₅ H ₉ Cl in bright sunlight. Identify the hydrocarbon.

Answer :

We are given the formula C ₅ H ₁₀ 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.

1650395110987

Question 10.6 Write the isomers of the compound having formula .

Answer :

The isomers of the compound are :-

(i) 1-Bromobutane

1650395155565

(ii) 2-Bromobutane

1650395182967

(iii) 1-Bromo-2-methylpropane

1650395202259

(iv) 2-Bromo-2-methylpropane

1650395222249

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 :-

1650395262688

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 :-

1650395301827

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 :-

1650395383932

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 S _N 2 reaction with –OH?

(i) CH ₃ Br or CH ₃ I

Answer :

In this case, we have the same alkyl group but different halide ions. For this rate of S _N 2 reaction increases with increase in atomic mass. So, CH ₃ I will react faster than CH ₃ Br.

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

(ii) or

Answer :

In this case, the hindrance will be deciding factor for the rate of S _N 2 reaction because hindrance will directly affect the attack of the nucleophile. So will react faster as compared to _.

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 :

1650395452626

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

1650395489418

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)

1650395539877

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

1594913117082 1594913112687

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)

1650395582853

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

1650395602563 and 1650395617100

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 :-

1650395649100

Now,

1650395667267

1650395692814

Question 10.11 How will you bring about the following conversions?

(ii) Ethane to bromoethene

Answer :

(ii)

1650395734449

Question 10.11 How will you bring about the following conversions?

(iii) Propene to 1-nitropropane

Answer :

(iii)

1650395766265

Question 10.11 How will you bring about the following conversions?

(iv) Toluene to benzyl alcohol

Answer :

(iv)

1650395803620

Question 10.11 How will you bring about the following conversions?

(v) Propene to propyne

Answer :

(v)

1650395841286

Question 10.11 How will you bring about the following conversions?

(vi) Ethanol to ethyl fluoride

Answer :

(vi)

1650395904233

Question 10.11 How will you bring about the following conversions?

(vii) Bromomethane to propanone

Answer :

(vii)

1650395947461

Question 10.11 How will you bring about the following conversions?

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

Answer :

(viii)

1650396011008

Question 10.11 How will you bring about the following conversions?

(ix) 1-Chlorobutane to n-octane

Answer :

(ix)

1650396049768

Question 10.11 How will you bring about the following conversions?

(x) Benzene to biphenyl.

Answer :

(x)

1650396088832

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 sp ² hybridized carbon atom whereas, in cyclohexyl chloride, the Cl-atom is attached to a sp ³ hybridized carbon atom. It is known that sp ² hybridized carbon has more s-character than sp ³ 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.

1650396174518

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 ₄ :- 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 :-

$CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone}$ 18566

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

1594913141655 1594913139667

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.

1594913157886 1594913156062

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.

1650396358369

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 S _N 2 mechanism. The mechanism for the given reaction is shown below :-

1650396425250

Question 10.16 Arrange the compounds of each set in order of reactivity towards S _N 2 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.

1650396424113

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 S _N 2 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.

1650396475386

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 S _N 2 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:-

1650396510971

It is clear that carbocation of is more stable.

Hence will hydrolyse faster than _.

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.

1650396563342 1650396578894 1650396597407

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 :-

1650396628364

Question 10.19 How the following conversions can be carried out?

(ii) Ethanol to but-1-yne

Answer :

The reaction mechanism is given below :-

1650396659481

Question 10.19 How the following conversions can be carried out?

(iii) 1-Bromopropane to 2-bromopropane

Answer :

(iii) The mechanism is given below :-

1650396693799

Question 10.19 How the following conversions can be carried out?

(iv) Toluene to benzyl alcohol

Answer :

(iv) The mechanism is given below :-

1650396725677

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 :-

1650396774093

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 :-

1650396805595

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 :-

1650396845862

Question 10.19 How the following conversions can be carried out?

(viii) Aniline to chlorobenzene

Answer :

The mechanism of the reaction is given below :-

1650396881053

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 :-

1650396915499

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 :-

1650396949386

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 :-

1650396978762

Question 10.19 How the following conversions can be carried out?

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

Answer :

The mechanism is given below :-

1650397010184

Question 10.19 How the following conversions can be carried out?

(xiii) 2-Chloropropane to 1-propanol

Answer :

The mechanism is :-

1650397041445

Question 10.19 How the following conversions can be carried out?

(xiv) Isopropyl alcohol to iodoform

Answer :

(xiv) The proposed mechanism is :-

1650397073455

Question 10.19 How the following conversions can be carried out?

(xv) Chlorobenzene to p-nitrophenol

Answer :

(xv) The required mechanism is given below :-

1650397106405 `

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 :-

1650397139391

Question 10.19 How the following conversions can be carried out?

(xvii) Chloroethane to butane

Answer :

(xvii) The mechanism of the reaction is :-

1650397178117

Question 10.19 How the following conversions can be carried out?

(xviii) Benzene to diphenyl

Answer :

The mechanism for the given reaction is :-

1650397215170

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 :-

1650397251052

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 :-

1650397282194

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.

1650397317223

1650397335438

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:-

1650397373001

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

1650397392860

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:-

1650397448020

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 :-

1650397481230

Question 10.22 What happens when

(iii) chlorobenzene is subjected to hydrolysis

Answer :

The reaction is given below :-

1650397519695

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 :-

1650397551997

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:-

1650397586595

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:-

1650397621876

Some Important Basic Points of the NCERT Class 12 Chemistry Solutions Chapter 10

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.
For better understanding use NCERT Solutions for Class 12 Chemistry chapter 10 PDF, students can download it by using an online webpage to PDF converter

NCERT Solutions Class 12 Chemistry

Chapter 1	NCERT solutions for class 12 chapter 1 The Solid State
Chapter 2	NCERT solutions for class 12 chemistry chapter 2 Solutions
Chapter 3	NCERT Solutions for class 12 chemistry chapter 3 Electrochemistry
Chapter 4	NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics
Chapter 5	NCERT Solutions for 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	NCERT solutions for class 12 chemistry chapter 7 The P-block elements
Chapter 8	NCERT Solutions for 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	NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers
Chapter 12	NCERT Solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids
Chapter 13	NCERT solutions for class 12 chemistry chapter 13 Amines
Chapter 14	NCERT solutions for class 12 chemistry chapter 14 Biomolecules
Chapter 15	NCERT Solutions for 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

NCERT Solutions for class 12 biology
NCERT solutions for class 12 maths
NCERT solutions for class 12 chemistry
NCERT Solutions for class 12 physics

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

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Revision will be quite easier because the detailed solutions will help you to remember the concepts and get good marks in your class.
Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for Class 12 Chemistry and you are good to go.

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NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

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

More About NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes And Haloarenes

Some Important Basic Points of the NCERT Class 12 Chemistry Solutions Chapter 10

NCERT Solutions Class 12 Chemistry

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