NCERT Exemplar Class 12 Chemistry Solutions chapter 10 Haloalkanes and Haloarenes

NCERT Exemplar Class 12 Chemistry Solutions chapter 10 Haloalkanes and Haloarenes

Edited By Sumit Saini | Updated on Sep 16, 2022 05:49 PM IST | #CBSE Class 12th
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NCERT exemplar Class 12 Chemistry solutions chapter 10 is helpful for board exam students and competitive exam preparation as well. Experts in Chemistry field have prepared the NCERT exemplar Class 12 Chemistry chapter 10 solutions as per the latest NCERT pattern. Class 12 Chemistry NCERT exemplar solutions chapter 10 are fairly detailed and accurately explained. It aids students to grasp the concepts very well. NCERT exemplar Class 12 Chemistry solutions chapter 10 pdf download can also be used by students by downloading the page as pdf using online tools.

This Story also Contains
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: MCQ (Type 1)
  2. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: MCQ (Type 2)
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Short Answer Type
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Matching Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Assertion and Reason Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Long Answer Type
  7. Major Topics and Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes
  8. NCERT Exemplar Class 12 Chemistry Solutions Chapter 10 - Learning Outcome
  9. NCERT Exemplar Class 12 Chemistry Solutions
  10. Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes.

Also, check - NCERT Solutions for Class 12 for other subjects.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: MCQ (Type 1)

Question:1

The order of reactivity of following alcohols with halogen acids is
(A) CH_{3}CH_{2}-CH_{2}-OH
(B)
(C)

(i) (A) > (B) > (C)
(ii) (C) > (B) > (A)
(iii) (B) > (A) > (C)
(iv) (A) > (C) > (B)
Answer:

The alcohol reacts with the halogen acids in an order of 3^{\circ}> 2^{\circ} > 1^{\circ}.
Hence correct option is (ii) (C) > (B) > (A)

Question:2

Which of the following alcohol will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

Answer:

The tertiary alcohol reacts the most with concentrated HCl, as the tertiary carbocation is most stable. Therefore, for tertiary alcohol, the room temperature is enough for the reaction. However, for primary and secondary alcohols, there is a need for the presence of a catalyst (ZnCl_2).
Hence option (iv) is correct.

Question:3

Identify the compound Y in the following reaction:

Answer:

(a) Dissolved or suspended in cold aqueous mineral acid, a primary aromatic amine turns into a diazonium salt, after treating with sodium nitrite. The diazonium group is replaced by -Cl as soon as the cuprous chloride is added with the previously prepared diazonium salt. Hence, the ‘Y’ here in the equation is chlorobenzene.

Question:4

Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is
(a) electrophilic elimination reaction
(b) electrophilic substitution reaction
(c) free radical addition reaction
(d) nucleophilic substitution reaction
Answer:

(b) This type of reaction is called an electrophilic substitution reaction.

Question:5

Which of the following is a halogen exchange reaction?
(i)RX+NaI\rightarrow RI+NaX
(ii)

(iii) R-OH+HX\overset{ZnCl_{2}}{\rightarrow} R-X+H_{2}O
(iv)

Answer:

(a) Halogen exchange reaction means one halide is replaced by the other. This reaction is also called the Finkelstein reaction.

Question:6

Which reagent will you use for the following reaction?
CH_3CH_2CH_2CH_3 \rightarrow CH_3CH_2CH_2CH_2Cl + CH_3CH_2CHClCH_3

(i) Cl_{2}/UV light
(ii) NaCl + H_2SO_4
(iii) Cl_{2} gas in dark
(iv) Cl_{2} gas in the presence of iron in dark
Answer:

(a) Here, it is a free radical substitution reaction. That means, this reaction happens in the presence of UV light/peroxides/ high temperature. These are all free radical generators.

Question:7

Arrange the following compounds in the increasing order of their densities.

(i) (a) < (b) < (c) < (d)
(ii) (a) < (c) < (d) < (b)
(iii) (d) < (c) < (b) < (a)
(iv) (b) < (d) < (c) < (a)
Answer:

(a) Density increases with increase in molecular mass. The order of molecular masses is (a) < (b) < (c) < (d)
Hence option (i) (a) < (b) < (c) < (d) is correct.

Question:8

Arrange the following compounds in increasing order of their boiling points
(i)
(ii) CH_{3}CH_{2}CH_{2}CH_{2}Br
(iii)

(i) (b) < (a) < (c)
(ii) (a) < (b) < (c)
(iii) (c) < (a) < (b)
(iv) (c) < (b) < (a)

Answer:

Option (iii) (c) < (a) < (b) is correct. Here the given alkanes are isomeric. Therefore isomeric alkane’s boiling point decreases with increase in branching. When the branching of an alkane increases, the surface area decreases. This causes a decrease in intermolecular force.

Question:9

In which of the following molecules carbon atom marked with an asterisk (*) is asymmetric?

(i) (a), (b), (c), (d)
(ii) (a), (b), (c)
(iii) (b), (c), (d)
(iv) (a), (c), (d)

Answer:

(ii) An Asymmetric carbon atom has linked with four different groups or atoms with all of its four valencies. Here, the carbon atoms of (i), (ii), and (iii), all have linked with four different groups of atoms. Hence, those are asymmetric. Whereas, in (iv), the carbon atom linked with two of its valencies with two similar hydrogen atoms. Therefore, it is not an asymmetric carbon atom.

Question:10

Which of the following structure is enantiomeric with the molecule (A) given below:


Answer:

(i) Here the structure (i) is the enantiomer of compound (A). Here the position two groups, i.e., CH3 and C2H5 in (i) is exactly reversed of the given sample (A) at the chiral carbon.

Question:11

Which of the following is an example of vic-dihalide?
(a) Dichloromethane
(b) 1, 2-dichloroethane
(c) Ethylidene chloride
(d) Allyl chloride
Answer:

(b) 1, 2-Dichloroethane is the example of a vic-dihalide. This is because the presence of two Cl atoms on vicinal carbon atoms (adjacent).

Question:12

The position of -Br in the compound CH_3CH = CHC(Br)(CH_3)_2, can be classified as.
(a) allyl
(b) aryl
(c) vinyl
(d) secondary
Answer:

(a) The given sample is an allylic compound. Here the Br atom is attached next to double-bonded carbon.

Question:14

Ethylidene chloride is a/an ________.
(a) vic-dihalide
(b) gem-dihalide
(c) allylic halide
(d) vinylic halide
Answer:

(ii) Ethylidene chloride is a gem-dihalide. This is because the halogen atoms of CH_3- CHCl_2 are connected to the same carbon atom.

Question:15

What is A in the following reaction?

Answer:

(c) Here, according to Markownikow's rule, HCl will be attached to the doubly bonded carbons. Therefore, for the carbon atom that has a lesser number of hydrogens, the addition of a negative addendum will take place.

Question:16

A primary alkyl halide would prefer to undergo __________ .
(a) S_N1 reaction
(b) S_N2 reaction
(c) \alpha-Elimination
(d) Racemisation
Answer:

(b) S_N2 reaction happens through the formation of the transition state. Therefore a primary alkyl halide would prefer to undergo S_N2 reaction due to less steric hindrance.

Question:17

Which of the following alkyl halides will undergo S_N1 reaction most rapidly?
(a) (CH_3)_3C-F
(b) (CH_3)_3C-Cl
(c) (CH_3)_3C-Br
(d)(CH_3)_3C-I
Answer:

(d) (CH_3)_3C-I will undergo S_N1 reaction most readily. Here, the C-l bond is the weakest because the size difference between carbon and iodine is huge.

Question:18

Which is the correct IUPAC name of ?
(i) 1-Bromo-2-ethylpropane
(ii) 1-Bromo-2-ethyl-2-methylethane
(iii) 1-Bromo-2-methylbutane
(iv) 2-Methyl-1-bromobutane
Answer:

(c) The correct IUPAC name of the given compound is 1-Bromo-2-methyl butane.

Question:19

What should be the correct IUPAC name for diethyl bromomethane?
(a) 1-Bromo-1, 1-diethylmethane
(b) 3-Bromopentane
(c) 1-Bromo-1-ethylpropane
(d) 1-Bromopentane
Answer:

(b) IUPAC name of Diethyl Bromomethane is 3-Bromopentane.

Question:21

Chloromethane on treatment with excess of ammonia yields mainly
(i) N,N- Dimethylmethanamine
(ii) N-methylmethanamine (CH_{3}-NH-CH_{3})
(iii) Methanamine (CH_{3}-NH_{2})
(iv) Mixture containing all of these mixtures in equal proportion
Answer:

(iii) Chloromethane on treatment with an excess of ammonia yields mainly methenamine.
CH_3Cl + NH_3 \rightarrow CH_3NH_2 + HCl
However, if the two reactants are present in the same amount, the mixture of the primary, secondary, and tertiary amine is obtained.

Question:22

Molecules whose mirror image is non-superimposable over them are known as chiral. Which of the following molecules is chiral in nature?
(a) 2-Bromobutane
(b) 1-Bromobutane
(c) 2-Bromopropane
(d) 2-Bromopropan-2-ol
Answer:

Question:23

Reaction of C_6H_5CH_2Br with aqueous sodium hydroxide follows _________.
(a)S_N1 mechanism
(b)S_N2 mechanism
(c) Any of the above two depending upon the temperature of the reaction
(d) Saytzeff rule
Answer:

(a) Reaction of C_6H_5CH_2Br with aqueous sodium hydroxide follows the S_N1 mechanism since the carbocation formed C_6H_5CH_2^{\bigoplus } is resonance stabilized cation.
Benzylic halides are highly reactive in nature towards the S_N1 reaction.

Question:24

Which of the carbon atoms present in the molecule given below are asymmetric?

(i) a, b, c, d
(ii) b, c
(iii) a, d
(iv) a, b, c
Answer:

(b) Carbon has four valencies. When all those four valencies are attached with four different atoms or groups then the carbon is called asymmetric/ chiral carbon. Here, ‘b’ and ‘c’ are attached to four different atoms/ groups. Hence, these are asymmetric.

Question:25

Which of the following compounds will give a racemic mixture on nucleophilic substitution by OH^{-} ion?

(i) (a)
(ii) (a), (b), (c)
(iii) (b), (c)
(iv) (a), (c)
Answer:

(a) As the alkyl halide in it contains asymmetric carbon, therefore, a mixture of an enantiomer is formed during the S_N1 reaction.

Question:26

In the question arrange the compounds in increasing order of the rate of reaction towards nucleophilic substitution.

(i) (a) < (b) < (c)
(ii) (c) < (b) < (a)
(iii) (a) < (c) < (b)
(iv) (c) < (a) < (b)
Answer:

(iii) (-NO_2) is an electron-withdrawing group and when it is at ortho and para position it causes nucleophilic substitution. Where the effect of this same electron-withdrawing group is very less when it is at the meta position.

Question:27

In the questions, arrange the compounds in increasing order of the rate of reaction towards nucleophilic substitution.

(i) (a) < (b) < (c)
(ii) (a) < (c) < (b)
(iii) (c) < (b) < (a)
(iv) (b) < (c) < (a)
Answer:

(iv) The rate of nucleophilic substitution decreases when there is an electron releasing group at ortho or para positions.

Question:28

In the questions, arrange the compounds in increasing order of the rate of reaction towards nucleophilic substitution.

(i) (c) < (b) < (a)
(ii) (b) < (c) < (a)
(iii) (a) < (c) < (b)
(iv) (a) < (b) < (c)

Answer:

(iv) The reactivity of aryl halides increases with the increase in electron-withdrawing group. Therefore, the more the number of electron-withdrawing groups, the more the rate of nucleophilic substitution.

Question:29

In the questions, arrange the compounds in increasing order of the rate of reaction towards nucleophilic substitution.

(i) (a) < (b) < (c)
(ii) (b) < (a) < (c)
(iii) (c) < (b) < (a)
(iv) (a) < (c) < (b)
Answer:

(c) Electron releasing groups increase the reactivity of aryl halides, less is the number of electron releasing groups,the less is the rate towards nucleophilic substitution.

Question:30

Which is the correct increasing order of boiling points of the following compounds?
1-Iodobutane, 1-Bromobutane, 1-Chlorobutane, Butane

(a) Butane < 1-Chlorobutane < 1-Bromobutane < 1 -Iodobutane
(b) 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane < Butane
(c) Butane < 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane
(d) Butane < 1-Chlorobutane < 1-Iodobutane < 1-Bromobutane
Answer:

(a) The intermolecular force of attraction will increase with the increase in surface area. As a result, the boiling point will also increase. As the boiling point increases with the molecular mass for the similar alkyl halide. The Iodine has the highest atomic mass. Therefore, the boiling point of 1-iodobutane is the highest.

Question:31

Which is the correct increasing order of boiling points of the following compounds?
1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene

(a) Bromobenzene < 1-Bromobutane < 1-Bromopropane < 1-Bromoethane
(b) Bromobenzene < 1-Bromoethane < 1 -Bromopropane < 1-Bromobutane
(c) 1-Bromopropane < 1-Bromobutane < 1-Bromoethane < Bromobenzene
(d) 1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene
Answer:

d) Boiling point increases with an increase in molecular mass for the same types of alkyl halide.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: MCQ (Type 2)

Question:32

Which of the statements are correct about above reaction?

(i) (a) and (e) both are nucleophiles.
(ii) In (c) carbon atom is sp3 hybridised.
(iii) In (c) carbon atom is sp2 hybridised.
(iv) (a) and (e) both are electrophiles.

Answer:

(i), (iii)
OH^{-} and Cl^{-}are nucleophiles. In (c), the C atom is sp2 hybridized due to the formation of C – OH bond and breaking of C – Cl bond simultaneously.

Question:33

Which of the following statements are correct about this reaction?

(i) The given reaction follows S_{N}2 mechanism.
(ii) (b) and (d) have opposite configuration.
(iii) (b) and (d) have same configuration.
(iv) The given reaction follows S_{N}1 mechanism.

Answer:

(i), (ii) In the given reaction, alkyl halide is primary in nature. Here, a transition state is observed in which one bond is broken and one bond is formed simultaneously i.e., in one step. So, it follows the S_{N}2 mechanism.
In this mechanism, the nucleophile attacks the carbon at 180^{\circ} to the leaving group. So the reactant and product have opposite configurations.

Question:34


Which of the following statements are correct about the reaction intermediate?

(i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms.
(ii) Intermediate (c) is unstable because carbon atom is sp^{2} hybridised.
(iii) Intermediate (c) is stable because carbon atom is sp^{2} hybridised.
(iv) Intermediate (c) is less stable than the reactant (b).
Answer:

(i), (iv) are correct options.

Question:35

Which of the following statements are correct about the mechanism of this reaction?


(i) A carbocation will be formed as an intermediate in the reaction.
(ii) OH^{-} will attach the substrate (b) from one side and Cl^{-} will leave it simultaneously from other side.
(iii) An unstable intermediate will be formed in which OH^{-} and Cl^{-} will be attached by weak bonds.
(iv) Reaction proceeds through S_N1 mechanism.
Answer:

(i), (iv) are correct. As the given reaction is a tertiary halide, therefore S_N1 mechanism will be present and as an intermediate, carbocation is formed.

Question:36


Which of the following statements are correct about the kinetics of this reaction?

(i) The rate of reaction depends on the concentration of only (b).
(ii) The rate of reaction depends on concentration of both (a) and (b).
(iii) Molecularity of reaction is one.
(iv) Molecularity of reaction is two.
Answer:

(i), (iii) is the correct statements.

Question:37

Haloalkanes contain halogen atom(s) attached to sp3 hybridized carbon atoms of an alkyl group. Identify haloalkane from the following compounds.
(i) 2-Bromopentane
(ii) Vinyl chloride (chloroethene)
(iii) 2-chloroacetophenone
(iv) Trichloromethane

Answer:

(i), (iv) are the haloalkanes from the above list.

Question:38

Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements.
(i) Both the compounds form same product on treatment with alcoholic KOH.
(ii) Both the compounds form same product on treatment with aq.NaOH.
(iii) Both the compounds form same product on reduction.
(iv) Both the compounds are optically active.

Answer:

(i), (iii) are correct.

Question:39

Which of the following compounds are gem-dihalides?
(i) Ethylidene chloride
(ii) Ethylene dichloride
(iii) Methylene chloride
(iv) Benzyl chloride

Answer:

(i), (iii) are Gem-dihalides.

Question:40

Which of the following are secondary bromides?
(a) (CH_3 )_{2}CHBr
(b) (CH_3 )_3C CH_2Br
(c) CH_3 CH(Br)CH_2CH_3
(d) (CH_3 )_2CBrCH_2CH_3
Answer:

(i), (iii) are secondary bromides.

Question:41

Which of the following compounds can be classified as aryl halides?
(i) p-ClC_6H_4CH_2CH(CH_3)_2
(ii) p-CH_3CHCl(C_6H_4)CH_2CH_3
(iii) o-BrH_2C-C_6H_4CH(CH_3)CH_2CH_3
(iv) C_6H_5Cl
Answer:

(i), (iv) are classified as aryl halides as there is a bond between a halogen atom and an atomic ring.

Question:42

Alkyl halides are prepared from alcohol by treating with
(i) HCl + ZnCl_2
(ii) RedP + Br_2
(iii) H_2SO_4 + KI
(iv) all the above
Answer:

(i), (ii) are correct options

NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Short Answer Type

Question:44

Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenas with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does the preparation of aryl iodides require the presence of an oxidizing agent?

Answer:

Iodination reactions are reversible in nature as because HI is formed in the process. To keep the reaction in forward direction, we need to remove the HI by oxidation process by adding some oxidizing agents like HIO_{4}.

Question:45

Out of o- and p-bromobenzene, which one has a higher melting point, and why?
Answer:

p-Dibromobenzene has a higher melting point. Because, in a crystal lattice, the symmetry of p-bromobenzene makes it fit better. As a result, a higher temperature is required to break the bond between the molecules and hence the melting point is higher.

Question:46

Which of the compounds will react faster in S_N1 reaction with the OH^{-} ion?
CH_3 - CH_2 - Cl\; or\; C_6H_5 - CH_2 - Cl

Answer:

C_6H_5- CH_2 - Cl will react faster in S_N1 reaction with the OH^{-} ion. This is because there is enough stability of the carbocation in this compound. Here, due to resonance, the C_6H_5 group is already stable and it is attached with CH_{2} and makes the whole structure stable.

Question:47

Why does iodoform have appreciable antiseptic properties?
Answer:

As because there are liberal free iodines, iodoform is an applicable antiseptic.

Question:48

Haloarenes are less reactive than haloalkanes and haloalkenes. Explain.
Answer:

This is because there is a resonance stabilization formed in the aryl ring. Where, C-Cl bond acts as a partial double bond because of resonance. Therefore, they are less reactive in nucleophilic substitution.

Question:49

Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.
Answer:

Aryl bromides and chlorides can be prepared in the presence of Lewis acid catalysts (iron or iron chloride) from the electrophilic substitution of arenas with bromine and chlorine respectively.

Question:50

Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H_{2}SO_4. Explain why?
(a) CH_{3}CH_{2}CH_{2}OH
(b)

Answer:

If we mix NaBr and H_2SO_4, we will get Br_{2} gas. Here, the product (b) will not react with Br_{2} gas. This is because it has already formed a stable structure due to resonance.

Question:51

Which of the products will be the major product in the reaction given below? Explain.

Answer:

As per Markovnikov's rule B will be the major product

Question:52

Why is the solubility of haloalkanes in water very low?
Answer:

The solubility of the haloalkanes in water is low. This is because, to make a haloalkane soluble in water, we need energy. Which will help to overcome the attractions between the haloalkane molecules and also help to break the hydrogen bonds between water molecules.

Question:53

Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing, or meta directing.

Answer:


The functional group present in this ring is ortho-para directing. This is because the density of electrons at that position is more.

Question:54

Classify the following compounds as primary, secondary and tertiary halides.
(i) 1-Bromobut-2-ene
(ii) 4-Bromopent-2-ene
(iii) 2-Bromo-2-methylpropane

Answer:

(i) 1-Bromobut-2-ene: It is a primary halide.
(ii) 4-Bromopent-2-ene: Here Bromine is attached to the secondary carbon. Hence, it is a secondary halide.
(iii) 2-Bromo-2-methylpropane: Here Bromine is attached to the tertiary carbon. Hence, it is a tertiary halide.

Question:55

Compound ‘A’ with molecular formula C_{4}H_{9}Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
(ii) Out of these two compounds, which one will be converted to the product with an inverted configuration.

Answer:


(i) The rate of reaction with aqueous KOH of compound ‘A’ only depends upon the concentration of ‘A’, therefore, the reaction mechanism is S_N1 and ‘A’ is 2-Bromo-2-methylpropane(tertiary bromide).
Whereas, ‘B’ is optically active and is an isomer of ‘A’. Therefore, ‘B’ must be 2-Bromobutane. As the concentration of ‘B’ is responsible for the rate of reaction of ‘B’ with aqueous KOH, therefore, the reaction mechanism is S_N2
(ii) Because of the S_N2 reaction, compound ‘B’ will have an inversion of configuration and turn out to be an inverted product.

Question:56

Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, C_{7}H_8 is treated with Cl_2 in the presence of FeCl_3.
Answer:

Toluene(C_6H_5CH_3) is the compound that has a molecular formula C_{7}H_8. When toluene is treated with Cl_2 in the presence of FeCl_3 , we get o-chlorotoluene and p-chlorotoluene where the p-isomer predominates. This is because of the -CH_3 group is o-, p-directing.

Question:58

Which of the following compounds will have the highest melting point and why?

Answer:

The structure (II) has symmetry of para-positions. Therefore, it fits into crystal lattice better than other isomers and has highest melting point.

Question:59

Write down the structure and IUPAC name for neo-pentylbromide.
Answer:


1-Bromo-2,2-dimethylpropane is IUPAC name for neo-pentylbromide.

Question:60

A hydrocarbon of molecular mass 72 g/ mol gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.
Answer:

We can say pentane (C_5H_{12}) has a molecular mass of 72 g/ mol. Therefore, the isomer of pentane with single monochloro derivative should conisists of 12 hydrogens equivalent.

Question:61

Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.
Answer:

Here, two different alkenes such as methylene cyclohexane and 1-Methylcyclohex-1-ene can be found.

Question:62

Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason.
(i) 1-Bromobutane
(ii) 2-Bromobutane
(iii) 2-Bromo-2-methylpropane
(iv) 2-Chlorobutane
Answer:

(iii) 2-Bromo-2-methylpropane
This comlpex will easily react with the aqueous KOH. Here, tertiary carbocation will be formed and this will be most stable.

Question:63

Why can aryl halides not be prepared by reaction of phenol with HCl in the presence of ZnCl_{2} ?
Answer:

In the presence of ZnCl_{2}, phenol will not react with HCl. This is because there is a partial double bond characteristics present between the benzene ring and O, which will produce between benzene ring and OH group due to resonance. As a result aryl halide will not be prepared.

Question:64

Which of the following compounds would undergo S_{N}1 reaction faster and why?

Answer:

B will undergo S_{N}1 reaction faster. This is because the when Cl is removed, carbocation formed and stabilized by resonance.

On the other hand , the carbonation formed during the reaction (A) is not resonance stituation.

Question:65

Allyl chloride is hydrolysed more readily than n-propyl chloride. Why?
Answer:

Allyl chloride becomes high reactive in nature due to the formation of carbocation by hydrolysis, which is very much stable due to resonance. In case of n-propyl chloride , there will be no such stability exists due to carbocation.

Question:66

Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?
Answer:

It is neessary to avoid even traces of moisture because the Grignard reagents are highly reactive in nature. Those reagents can react with even traces of water and produce corresponding hydrocarbons.
RMgX + H_2O \rightarrow RH + Mg(OH)X

Question:67

How do polar solvents help in the first step in S_{N}1 mechanism?
Answer:

Through formation of carbocation, the mechanism of the S_{N}1 moves forward. In this process, the breaking of C-halogen bond takes place. As a reuslt, the solvetion of the halide ions is done with the protons of the of the protic solvent. In this way, polar solvents help in ionisations step by stabilizing the ions by solvation.

Question:68

Write a test to detect the presence of double bond in a molecule.
Answer:

If a molecule contains a double bond, then it can be detected easily by bromine water test and Baeyer’s test. The bromine water or aqueous KMnO_{4} becomes coloerless.

Question:69

Diphenyls are a potential threat to the environment. How are these produced from aryl halides?
Answer:

Diphenyls(ex: p,p ‘-dichlorodiphenyltrichloroethane., i.e DDT) are a serious threat to the environment. This is because those are chemicaly stable and soluble in fat. For the long term presence of it in the atmosphere is extremely dangerous.
Di-phenyls can be prepared from aryl halides by the following two methods:

Question:70

What are the IUPAC names of the insecticide DDT and benzenehexachloride? Why is their use banned in India and other countries?
Answer:

IUPAC name of DDT is 2, 2-bis (4-chlorophenyl)-1, 1, 1-trichloroethane that of benzene hexachloride is 1, 2, 3, 4, 5, 6-hexachlorocyclohexane.
They are non-biodigredable and toxic at the same time. They are soluble in fat and their concentration keeps increasing in the food chain. This is why their use is banned in India and other countries.

Question:71

Elimination reactions (especially \beta-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. Specify the reagents used in both cases.

Answer:

Alkyl halides responds in nucleophilic substitution as well as elimination (\beta-elimination) reaction.
Although, by setting the reaction condition and right choice of the reagent, a particular product can be obtained. Normally, elemination reaction suits the strong and larger bases along with high tempareture. Whereas substitution reaction is suitable for the weaker and smaller bases in lower temperature.
Ex:
CH_{3}CH_{2}Br\xrightarrow[alc. KOH]{473-573K}CH_{2}=CH_{2} (Elimination)
CH_{3}CH_{2}Br\xrightarrow[aq. KOH]{373K}CH_{3}-CH_{2}OH (Substitution)
Nucleophilic Subsitution : Reagents used are nucleophilies like ^{-}{O}H, NH_{3}, ^{-}{C}\equiv N, O=N-O, ^{-}OR, etc at lower temperature under goes Substitution reaction.

Question:72

How will you obtain mono bromobenzene from aniline?
Answer:

Mono bromobenzene from aniline:

Question:73

Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution.

Answer:

When the nucleophile is attached at the carbon carrying -Cl, due to resonance a stabilised intermediate compound is found. As the -NO_{2} is electron withdrawing in nature, the nucleophile will get attached to the benzene ring easily. When the molecule contains more number of -NO_{2} groups, the nucleophile will be attached more easily. Hence, the order of reactivity is III > II > I.

Question:74

tert-Butylbromide reacts with aq. NaOH by S_N1 mechanism while n-butylbromide reacts by S_N2 mechanism. Why?
Answer:

In general, the S_N1 reaction goes forward by the formation of carbocation. The tert-butyl bromide will form 3^{\circ} carbocation by losing the Br^{-} ion, which is actually stable. Therefore, it reacts with aqueous KOH by S_N1 mechanism as:

On the other hand, due to unstable in nature, n-Butyl bromide does not form 1° n-Butyl carbocation by ionization. Therefore, it folows the mechanismof S_N2 mechanism. This happens through a transition state where OH^{-} ion form (nucleophilic attack) in the remote side with simultaneous expulsion of Br^{-} ion from the front side.

Question:76

Discuss the nature of C – X bond in the haloarenes.
Answer:

Resonance effect: C – X bond holds the nature of a partial double bond character and as a result, C – X bond is very difficult to break.
(i) C – X bond in haloarenes is extremely less reactive towards nucleophilic
(ii) In C – X bond, C atom attached to halogen is sp2 hybridised. A carbon that is sp2 hybridised with a greater s-character is more electronegative in nature and it can hold the electron pair of C – X bond more tightly than sp3 hybridised carbon in haloalkanes with less s-character.

Question:77

How can you obtain iodoethane from ethanol when no other iodine containing reagent except Nal is available in the laboratory?
Answer:

Ethanol can be converted into chloroethane with the help of HCl + ZnCl_2 and Cl in the chloroethane can be replaced by I with the help of NaI.
C_{2}H_{5}OH+HCl \overset{ZnCl_2}{\rightarrow}C_{2}H_{5}Cl\overset{NaI}{\rightarrow}C_{2}H_{5}I

Question:78

Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger nucleophile in aqueous medium? Give reason for your answer.
Answer:

A cyanide ion acts as a stronger nucleophile from the carbon end. This is because it will form a C – C bond which is more stable (bond between two similar atoms) than C – N bond.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Matching Type

Question:79

Match the compounds given in Column I with the effects given in Column II.

Column I

Column II

i. Chloramphenicol

a. Malaria

ii. Thyroxine

b. Anaesthetic

iii. Chloroquine

c. Thypoid fever

iv. Chloroform

d. Goiter


e.Blood substituent

Answer:

(i \rightarrow c); (ii \rightarrow d); (iii \rightarrow a); (iv \rightarrow b)
(i) Chloramphenicol is a broad spectrum antibiotic which is used to treat the typhoid fever.
(ii) Thyroxine is a hormone secreted by thyroid gland. When the thyroxine is secreted more, then the patient will notice an enlarged thyroid gland which is called goiter.
(iii) Chloroquine is used to prevent malaria parasite plasmodium vivax in the blood.
(iv) Chloroform is trichloromethane (CHCl_3). It is used as an anaesthetic.

Question:80

Match the items of Column I and Column II.

column I

Column II

i. S_N1 reaction

a. vic-dibromides

ii. Chemicals in fire extingusiher

b. gem - dihalides

iii. Bromination of alkene

c. Racemination

iv. Alkylidene halides

d. Saytzeff rule

v. Elimination of HX from alkylhalide

e. Chlorobromocarbons

Answer:

(i \rightarrow c), (ii \rightarrow e), (iii \rightarrow a), (iv \rightarrow b), (v \rightarrow d)
(i) When any mixture has two enantiomers in equal proportions, then there will be zero optical rotation observed. These kinds of mixtures are called racemic mixtures and this process of conversion of enantiomers into a racemic mixture is known as racemisation. When alkyl halide follows SN1 mechanism then racemisation is observed.
(ii) Chlorofluorocarbons are used in fire extinguishers.
(iii) Halogen atoms are present on the adjacent carbon atom in vic-dihalides. vic-dihalides are observed from the Bromination of alkanes.
(iv) Alkylidene halides are also known as gem-dihalides. Halogen atoms are present on the same carbon atom in case of gem-dihalides.
(v) Elimination of HX from alkylhalide follows Saytzeff rule. This rule states that “in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms”.

Question:81

Match the reactions given in column I with the types of reactions given in column II.

Column I

Column II

i.

a. Aryl halide

ii. CH_{2}=CH-CH_{2}-X

b. Alkyl halide

iii.

c. Vinyl halide

iv. CH_{2}=CH-X

d. Allyl halide

Answer:

(i \rightarrow b), (ii \rightarrow d), (iii \rightarrow a), (iv \rightarrow c)
(i) Halogen atom in alkyl halide is bonded to a carbon atom that is sp3 hybridised, where further addition of alkyl group may happen. CH_3-CH (X) - CH_3 is alkyl halide.
(ii) Halogen atom in allyl halide is bonded to a carbon atom next to C=C double bond that is sp^3 hybridised. CH_2= CH - CH_2- X is allyl halide.
(iii) Halogen atoms in aryl halide are bonded to a carbon atom that is sp^2 hybridised aromatic ring. C_6H_5X is aryl halide.
(iv) Halogen atoms in vinyl halide is bonded to a sp^2 hybridized carbon atom of a C=C double bond. CH_2=CH-X is vinyl halide.

Question:82

Match the reactions given in column I with the types of reactions given in column II.

Answer:

(i \rightarrow b), (ii \rightarrow d), (iii \rightarrow e), (iv \rightarrow a), (v \rightarrow c)
(i) Here, substitution takes place when an electrophile Cl^+ attacks on the benzene ring.
(ii) Here, according to Markownifoff’s rule, on the double bonded carbons the addition of HBr takes place followed by electrophilic addition.
(iii) Here, the reactant is secondary halide in nature and the hydroxyl ion will substitute the halogen. Here the reaction mechanism is S_N1 due to the presence of secondary halide.
(iv) Here the aromatic ring and the halogen atom are attached directly. The halogen of the given compound is substituted by the OH^- group here, due to nucleophilic substitution.
(v) It follows Saytzeff rule, i.,e elimination reaction.

Question:83

Match the structures given in Column I with the names given in Column II.

Answer:

(i \rightarrow a), (ii \rightarrow c), (iii \rightarrow b), (iv \rightarrow d)
(i) A: 4-bromopent-2-ene.
(ii) B: 4-bromo-3-methylpent-2-ene.
(iii) C: 1-bromo-2-methylbut-2-ene.
(iv) D: 1 -bromo-2-methylpent-2-ene.

Question:84

Match the reactions given in Column I with the names given in Column II.

Answer:

(i\rightarrow b), (ii \rightarrow a), (iii \rightarrow d), (iv \rightarrow c)
(i) Alkyl arene is produced by the mixture of an alkyl halide and aryl halides when treated with sodium in dry ether. This reaction is called the Wurtz-Fittig reaction.
(ii) Analogous compound is found from the aryl halides when it is treated with sodium in dry ether. Here, two aryl groups are attracted together. This reaction is called Fittig reaction.
(iii) Chlorobenzene or bromobenzene is formed from the diazonium salt when it is treated with cuprous chloride or cuprous bromide. This reaction is called Sandmeyer’s reaction.
(iv) In dry acetone, alkyl chlorides and sodium iodide react with each other and form Alkyl iodides. The reaction is called the Finkelstein reaction.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Assertion and Reason Type

Question:85

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): Phosphorus chlorides (tri and penta) are preferred over thionyl chloride for the preparation of alkyl chlorides from alcohols.
Reason (R): Phosphorus chlorides give pure alkyl halides.

Answer:

(ii) To convert alcohol to alkyl halide, thionyl chloride is the best halogen carrier.Because all the byproducts are in gaseous state. As a result, pure form of alkyl halide can be prepared.

Question:86

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): The boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF
Reason (R): The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.
Answer:

(v) The order of boiling point :
RI > RBr > RCl > RF
As the halides are polar molecules, therefore, their boiling temperature is always more than the boiling temperature of the hydrocarbons.

Question:87

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): KCN reacts with methyl chloride to give methyl isocyanide
Reason (R):CN^{-} is an ambident nucleophile.
Answer:

(iv) R-Cl + KCN \rightarrow R-CN + KCl
Alkyl Cyanide
Methyl cyanide is also obtained as CN^{-} group present here is an ambident nucleophile.

Question:88

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane.
Reason (R): In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide.
Answer:

Option (i) is correct.
In the Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon contain double the number of carbon atoms present in the halide.
tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane.

Question:89

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): Presence of a nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution.
Reason (R): Nitro group, being an electron withdrawing group decreases the electron density over the benzene ring.

Answer:

(i) Nitro group used to decrease the electron density in the ring as it is an electron withdrawing group. As a result, this will cause nucleophilic substitution due to an increase in reactivity of haloarenes.

Question:90

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): In mono haloarenes, further electrophilic substitution occurs at ortho and para positions.
Reason (R): Halogen atom is a ring deactivator.
Answer:

(v) Due to (+M) or (+R) effect, the halogens are usually ortho-para directing in nature. However, due to high electronegativity, they are also deactivating in nature.

Question:91

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): Aryl iodides can be prepared by reaction of arenes with iodine in the presence of an oxidising agent.
Reason (R): Oxidising agent oxidises I_{2} into HI.
Answer:

(iii) HI converts into I_{2} by the oxidising agent such as HIO_3. Although, HI can convert aryl halides into arenes.
5HI + HIO_3 \rightarrow 3H_2O + 3I_2

Question:92

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): It is difficult to replace chlorine by -OH in chlorobenzene in comparison to that in chloroethane.
Reason (R): Chlorine-carbon (C – Cl) bond in chlorobenzene has a partial double bond character due to resonance.
Answer:

(i) There is a partial double bond formed due to resonance in between benzene ring and halogen. As a result, it is difficult for the aryl halide to take part in nucleophilic substitution as compared to the alkyl halide.

Question:93

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): Hydrolysis of (-)-2-bromooctane proceeds with inversion of configuration.
Reason (R): This reaction proceeds through the formation of a carbocation.

Answer:

(iii) Here, S_N2 mechanism will be seen due to the hydrolysis of 2-bromooctane. As a result, the configuration will be inverted. Here no carbocation will be formed.

Question:94

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question:
(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
Assertion (A): Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene.
Reason (R): -NO_2 group is a m-directing group.
Answer:

(iv) Nitration of chlorobenzene results in ortho and para nitro chlorobenzene. Although, the nitro groups are meta directing in nature.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 10: Long Answer Type

Question:95

Some alkyl halides undergo substitution whereas some undergo elimination reaction on treatment with bases. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer:

Primary alkyl halides prefer SN2 mechanism in which a nucleophile attacks at 180? to the halogen atom. A transition state is formed in which carbon is bonded to two nucleophiles and finally halogen atom is pushed out. Below is the mechanism-

Hence, in SN2 mechanism, substitution takes place. Tertiary alkyl halides follow SN1 mechanism. In this case, tert alkyl halides form 3? carbocations. If the reagent used is a weak base then substitution occur while if it is a strong base than instead of substitution elimination occur.

The reagent used is aq. KOH. It is a weak base so, substitution takes place.

As alc. KOH is a strong base, so elimination competes over substitution and alkene is formed

Question:96

Some halogen containing compounds are useful in daily life. Some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to a great extent. Name the class of these halocompounds. In your opinion, what should be done to minimize harmful effects of these compounds
Answer:

Uses of halogen containing compounds are as follows:
Dichloromethane is mainly used as a solvent. Although it can be used as a paint remover, finishing solvent, metal cleaner, propellant in aerosols, solvent in case of drug manufacturing, etc.
Trichloromethane is used as a fat solvent. However, it can be used as solvent for the iodine, alkaloids, and other substances.
Triiodomethane is used as an antiseptic.
But some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to great extent. These are as follows
(i) Tetrachloromethane
• CCl3 is released into the air and starts depleting the ozone layer in the atmosphere. As a result UV rays will come to the earth easily and humans will get affected. This will cause skin cancer, functional disorder, eye diseases, and also the damage in the immune system. These UV rays not only affect the human but also cause damage to plants, and other animals too.
(ii) Freons
• Freon-113, releases and will move to the top of the atmosphere. Here, it generates Cl atoms in order to damage the ozone layer. As a result of this depletion UV rays enter our atmosphere and become responsible for damage to a great extent.
(iii) p – p’ – Dichlorodiphenyltrichloroethane (DDT)
• DDT can not be destroyed and removed from the atmosphere because it is not completely biodegradable. It is soluble in fats and creates a chain. When DDT goes to the body of a human, then it will affect the reproductive system.
In order to get rid of the harmful effects of the following substances, i.e., freons, hydrofluorocarbons, fluorocarbons, we should lower the use of these substances.

Question:97

Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides? How can we enhance the reactivity of aryl halides?
Answer:

Aryl halides are less reactive towards nucleophilic substitution reactions for these reasons mentioned as follows:
(i) In case of haloarenes, the benzene ring is attached with the lone pair electron of the halogen in resonance. As a result, C-Cl bond acts as a partial double bond and increases the strength of the bond. This bond is difficult to substitute by nucleophilic substitution method. Hence, they usually react less in the above mentioned reaction.
(ii) In case of haloarenes, the sp^2 hybridized carbon atom is attached to a halogen. An sp^2 hybridized carbon atom is more electronegative and it holds the C-Cl pair strongly and forces the C-Cl bond shorter than the haloalkanes.
Method to increase the reactivity:
The reactivity of the aryl halides can be increased when there is an electron withdrawing group(NO_2) at the ortho and para position. This presence of this electron withdrawing group at the above mentioned position withdraws electron density in the benzene ring. As a result, it will be easier for the nucleophile to attack. As a result carbocation is formed through resonance.

So, in case of o- and p-chlorobenzenes, the carbon atom bearing (NO_2) group has a negative charge due to the resonating structure. Therefore, the (NO_2) group along with the π-electrons of the benzene ring stabilizes the carbon ions. However, none of the resonating structures carries the negative charge along with the carbon atom bearing -NO_2 group, in case of m-nitrochlorobenzene. Therefore, the negative charge does not stabilize by the nitro group at meta position. However, the p-electrons of the benzene ring stabilize the carbanion. In other words, the carbanions formed from o-nitrochlorobenzene and p-nitrochlorobenzene are more stable than those formed from m-nitrochlorobenzene.
Thus, the presence of electron withdrawing groups at o- and p-positions (but not at m-positions) w.r.t. The halogen atom activates the aryl halides towards nucleophilic substitution reaction.

Major Topics and Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

  • Haloalkanes and Haloarenes
  • Classification
  • Nomenclature
  • Nature of C–X Bond
  • Methods of Preparation of Haloalkanes
  • Preparation of Haloarenes
  • Physical Properties
  • Chemical Reactions
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NCERT Exemplar Class 12 Chemistry Solutions Chapter 10 - Learning Outcome

  • In Class 12 Chemistry NCERT exemplar solutions chapter 10 the students will learn that both haloalkanes and Haloarenes are hydrocarbons with one or more hydrogen atoms replaced. Haloalkanes are more reactive than Haloarenes.

  • Every topic in this chapter helps students learn about their uses, such as, Haloalkanes are used as refrigerants and as propellants.

  • They are also used for generating plastic aerosols. Haloalkanes are used for dry-cleaning and so on.

  • Haloarenes are used as flame retardants, they are important components of fire extinguishers, used widely in pharmaceuticals.

  • They react with different metals such as sodium and magnesium. Overall, NCERT exemplar Class 12 Chemistry solutions chapter 10 apart from being informative, it is also fun and interesting for all those chemistry lovers who love to find different reactions and compounds.

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NCERT Exemplar Class 12 Chemistry Solutions


Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes.

  • These are a few important topics of Haloalkane and Haloarenes include Nucleophilic substitution, Insolubility in water, Mono Haloalkanes, Mono Haloarenes, Haloalkane- Benzylic Halide and Haloarenes- Vinyl Halide.

  • Students must also study the concept of hybridization.

NCERT Exemplar Class 12 Solutions

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Frequently Asked Questions (FAQs)

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The direct PDF download link to be available. Till then, students can use save as a web page function which is available in their browser. 

2. What is the name of Class 12 Chemistry Chapter 10?

The name of the 10 chapter of Class 12 Chemistry is Haloalkanes and Haloarenes.

3. Can I use these solutions for entrance exam preparation?

Yes, these solutions are quite helpful in board exam preparation.

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hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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