NCERT Exemplar Class 12 Chemistry Solutions chapter 10 Haloalkanes and Haloarenes

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

Question:1

The order of reactivity of the following alcohols with halogen acids is
(A) $C{H}_{3}C{H}_2-C{H}_2-OH$
(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 the correct option is (ii) (C) > (B) > (A)

Question:2

Which of the following alcohols 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 the 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 $(ZnC{l}_2)$.
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 treatment with sodium nitrite. The diazonium group is replaced by -Cl as soon as the cuprous chloride is added to 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)$RX+NaI\to RI+NaX$
(ii)

(iii) $R-OH+HX\stackrel{ZnC{l}_2}{\to }R-X+H_{2}O$
(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?
$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_3\to C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Cl+C{H}_{3}C{H}_{2}CHClC{H}_3$
(i) $C{l}_2$/UV light
(ii) $NaCl+H_{2}S{O}_4$
(iii) $C{l}_2$ gas in dark
(iv) $C{l}_2$ 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 an 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) $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br$
(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 is 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 is 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 structures 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 an example of a vic-dihalide. This is because ofthe presence of two Cl atoms on vicinal carbon atoms (adjacent).

Question:12

The position of -Br in the compound $C{H}_{3}CH=CHC(Br)(C{H}_3{)}_2$, 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:13

Chlorobenzene is formed by the reaction of chlorine with benzene in the presence of $AlC{l}_3$. Which of the following species attacks the benzene ring in this reaction?
(a) $C{l}^{-}$
(b) $C{l}^{+}$
(c) $AlC{l}_3$
(d) $[AlC{l}_4{]}^{-}$
Answer:

(b) Here, the $C{l}^{+}$ attacks the benzene ring to form chlorobenzene.

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 $C{H}_3-CHC{l}_2$ 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) $S_{N}1$ reaction
(b) $S_{N}2$ reaction
(c) $\alpha$-Elimination
(d) Racemisation
Answer:

(b) $S_{N}2$ reaction happens through the formation of the transition state. Therefore a primary alkyl halide would prefer to undergo $S_{N}2$ reaction due to less steric hindrance.

Question:17

Which of the following alkyl halides will undergo $S_{N}1$ reaction most rapidly?
(a) $(C{H}_3{)}_{3}C-F$
(b) $(C{H}_3{)}_{3}C-Cl$
(c) $(C{H}_3{)}_{3}C-Br$
(d)$(C{H}_3{)}_{3}C-I$
Answer:

(d) $(C{H}_3{)}_{3}C-I$ will undergo $S_{N}1$ 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-methylene
(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) The IUPAC name of Diethyl Bromomethane is 3-Bromopentane.

Question:20

The reaction of toluene with chlorine in the presence of iron and in the absence of light yields _________ .

Answer:

Question:21

Chloromethane on treatment with excess of ammonia yields mainly
(i) N,N- Dimethylmethanamine
(ii) N-methylmethanamine $(C{H}_3-NH-C{H}_3)$
(iii) Methanamine $(C{H}_3-N{H}_2)$
(iv) Mixture containing all of these mixtures in equal proportion
Answer:

(iii) Chloromethane on treatment with an excess of ammonia yields mainly methenamine.
$C{H}_{3}Cl+N{H}_3\to C{H}_{3}N{H}_2+HCl$
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

The reaction of $C_6{H}_{5}C{H}_{2}Br$ with aqueous sodium hydroxide follows _________.
(a)$S_{N}1$ mechanism
(b)$S_{N}2$ mechanism
(c) Any of the above two depending upon the temperature of the reaction
(d) Saytzeff rule
Answer:

(a) Reaction of $C_6{H}_{5}C{H}_{2}Br$ with aqueous sodium hydroxide follows the $S_{N}1$ mechanism since the carbocation formed $C_6{H}_{5}C{H}_2^{⨁}$ is resonance stabilized cation.
Benzylic halides are highly reactive in nature towards the $S_{N}1$ 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 four valencies are attached to 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 $O{H}^{-}$ 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 $S_{N}1$ 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) $(-N{O}_2)$ 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 much 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 the 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 the number of electron-releasing groups, the lower 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) The 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 the above reaction?

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

Answer:

(i), (iii)
$O{H}^{-}$ and $C{l}^{-}$are nucleophiles. In (c), the C atom is sp² hybridized due to the formation of the C – OH bond and the breaking of the 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 configurations.
(iii) (b) and (d) have the 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 this carbon is attached to 5 atoms.
(ii) Intermediate (c) is unstable because the carbon atom is $s{p}^2$ hybridized.
(iii) Intermediate (c) is stable because the carbon atom is $s{p}^2$ hybridized.
(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) $O{H}^{-}$ will attach the substrate (b) from one side and $C{l}^{-}$ will leave it simultaneously from the other side.
(iii) An unstable intermediate will be formed in which $O{H}^{-}$ and $C{l}^{-}$ will be attached by weak bonds.
(iv) Reaction proceeds through $S_{N}1$ mechanism.
Answer:

(i), (iv) are correct. As the given reaction is a tertiary halide, therefore $S_{N}1$ 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 the concentration of both (a) and (b).
(iii) Molecularity of the reaction is one.
(iv) Molecularity of the 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 the same product on treatment with alcoholic KOH.
(ii) Both the compounds form the same product on treatment with aq.NaOH.
(iii) Both the compounds form the 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) $(C{H}_3{)}_{2}CHBr$
(b) $(C{H}_3{)}_{3}CC{H}_{2}Br$
(c) $C{H}_{3}CH(Br)C{H}_{2}C{H}_3$
(d) $(C{H}_3{)}_{2}CBrC{H}_{2}C{H}_3$
Answer:

(i), (iii) are secondary bromides.

Question:41

Which of the following compounds can be classified as aryl halides?
(i) $p-Cl{C}_6{H}_{4}C{H}_{2}CH(C{H}_3{)}_2$
(ii) $p-C{H}_{3}CHCl(C_6{H}_4)C{H}_{2}C{H}_3$
(iii) $o-Br{H}_{2}C-C_6{H}_{4}CH(C{H}_3)C{H}_{2}C{H}_3$
(iv) $C_6{H}_{5}Cl$
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) $HCl+ZnC{l}_2$
(ii) $RedP+B{r}_2$
(iii) $H_{2}S{O}_4+KI$
(iv) all the above
Answer:

(i), (ii) are correct options

Question:43

Alkyl fluorides are synthesized by heating an alkyl chloride/bromide in
presence of ____ or ____
$$\begin{gathered} \text{(}i)Ca{F}_2 \\ (ii)Co{F}_2 \\ \text{(}iii)H{g}_2{F}_2 \\ \text{(}iv)NaF \end{gathered}$$
Answer:

(ii), (iii) 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 $HI{O}_4.$

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 $S_{N}1$ reaction with the $O{H}^{-}$ ion?
$C{H}_3-C{H}_2-Clor{C}_6{H}_5-C{H}_2-Cl$
Answer:

$C_6{H}_5-C{H}_2-Cl$ will react faster in $S_{N}1$ reaction with the $O{H}^{-}$ ion. This is because there is enough stability of the carbocation in this compound. Here, due to resonance, the $C_6{H}_5$ group is already stable and it is attached with $C{H}_2$ and makes the whole structure stable.

Question:47

Why does iodoform have appreciable antiseptic properties?
Answer:

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 the 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 $H_{2}S{O}_4$. Explain why?
(a) $C{H}_{3}C{H}_{2}C{H}_{2}OH$
(b)
Answer:

If we mix NaBr and $H_{2}S{O}_4$, we will get $B{r}_2$ gas. Here, the product (b) will not react with $B{r}_2$ 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 attraction 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 higher.

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 the compound and KOH.
(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 the compound ‘A’ only depends upon the concentration of ‘A’, therefore, the reaction mechanism is $S_{N}1$ 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 $S_{N}2$
(ii) Because of the $S_{N}2$ 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 the molecular formula, $C_7{H}_8$ is treated with $C{l}_2$ in the presence of $FeC{l}_3$.
Answer:

Toluene$(C_6{H}_{5}C{H}_3)$ is the compound that has a molecular formula $C_7{H}_8$. When toluene is treated with $C{l}_2$ in the presence of $FeC{l}_3$, we get o-chlorotoluene and p-chlorotoluene where the p-isomer predominates. This is because of the $-C{H}_3$ group is o-, p-directing.

Question:57

Identify the products A and B formed in the following reaction:
(a) $C{H}_3-C{H}_2-CH=CH-C{H}_3+HCl\to A+B$
Answer:

Question:58

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

Answer:

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

Question:59

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

1-Bromo-2,2-dimethylpropane is the IUPAC name for neo-pentyl bromide.

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 $(C_5{H}_{12})$ has a molecular mass of 72 g/ mol. Therefore, the isomer of pentane with a single monochloride derivative should consists 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 by giving a reason.
(i) 1-Bromobutane
(ii) 2-Bromobutane
(iii) 2-Bromo-2-methylpropane
(iv) 2-Chlorobutane
Answer:

(iii) 2-Bromo-2-methylpropane
This complex 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 the reaction of phenol with HCl in the presence of $ZnC{l}_2$ ?
Answer:

In the presence of $ZnC{l}_2$, phenol will not react with HCl. This is because there is a partial double bond characteristic present between the benzene ring and O, which will be produced between the 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 when Cl is removed, carbocation is formed and stabilized by resonance.

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

Question:65

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

Allyl chloride becomes highly reactive in nature due to the formation of carbocation by hydrolysis, which is very much stable due to resonance. In the 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 necessary 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.
$RMgX+H_{2}O\to RH+Mg(OH)X$

Question:67

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

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

Question:68

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

If a molecule contains a double bond, then it can be detected easily by the bromine water test and Baeyer’s test. The bromine water or aqueous $KMn{O}_4$ becomes colorless.

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 chemically stable and soluble in fat. 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 benzene hexachloride? 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-biodegradable 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 the case of alkyl halides. Specify the reagents used in both cases.

Answer:

Alkyl halides respond in nucleophilic substitution as well as elimination ($\beta$-elimination) reaction.
However, by setting the reaction condition and the right choice of reagent, a particular product can be obtained. Normally, elimination reaction suits the strong and larger bases along with high temperature. Whereas substitution reaction is suitable for the weaker and smaller bases in lower temperatures.
Ex:
$C{H}_{3}C{H}_{2}Br\underset{alc.KOH}{\overset{473-573K}{\to }}C{H}_2=C{H}_2$ (Elimination)
$C{H}_{3}C{H}_{2}Br\underset{aq.KOH}{\overset{373K}{\to }}C{H}_3-C{H}_{2}OH$ (Substitution)
Nucleophilic Subsitution : Reagents used are nucleophilies like ${}^{-}OH,N{H}_3{,}^{-}C\equiv N,O=N-O{,}^{-}OR$, 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 to the carbon carrying -Cl, due to resonance a stabilised intermediate compound is found. As the $-N{O}_2$ is electron withdrawing in nature, the nucleophile will get attached to the benzene ring easily. When the molecule contains more number of $-N{O}_2$ 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 $S_{N}1$ mechanism while n-butylbromide reacts by $S_{N}2$ mechanism. Why?
Answer:

In general, the $S_{N}1$ reaction goes forward by the formation of carbocation. The tert-butyl bromide will form $3^{\circ }$ carbocation by losing the $B{r}^{-}$ ion, which is actually stable. Therefore, it reacts with aqueous KOH by $S_{N}1$ mechanism as:

On the other hand, due to its unstable in nature, n-butyl bromide does not form 1° n-butyl carbocation by ionization. Therefore, it follows the mechanism of $S_{N}2$ mechanism. This happens through a transition state where $O{H}^{-}$ ion form (nucleophilic attack) in the remote side with simultaneous expulsion of $B{r}^{-}$ ion from the front side.

Question:75

Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.
Answer:

The major product is 2-chloro-2-methylpropane

Question:76

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

Resonance effect: The C – X bond holds the nature of a partial double bond character and as a result, the C – X bond is very difficult to break.
(i) The C – X bond in haloarenes is extremely less reactive towards nucleophilic
(ii) In C – X bond, the C atom attached to the halogen is sp² hybridized. Carbon that is sp² hybridized 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 hybridized 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 $HCl+ZnC{l}_2$ and Cl in the chloroethane can be replaced by I with the help of NaI.
$C_2{H}_{5}OH+HCl\stackrel{ZnC{l}_2}{\to }{C}_2{H}_{5}Cl\stackrel{NaI}{\to }{C}_2{H}_{5}I$

Question:78

Cyanide ion acts as an ambident nucleophile. From which end it act as a stronger nucleophile in the aqueous medium? Give a 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 (a bond between two similar atoms) than a 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.

Answer:

$(i\to c);(ii\to d);(iii\to a);(iv\to b)$
(i) Chloramphenicol is a broad-spectrum antibiotic that is used to treat typhoid fever.
(ii) Thyroxine is a hormone secreted by the 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 $(CHC{l}_3)$. It is used as an anesthetic.

Question:80

Match the items of Column I and Column II.

Answer:

$(i\to c),(ii\to e),(iii\to a),(iv\to b),(v\to d)$
(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 racemization. When alkyl halide follows S_N1 mechanism then racemization 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 the case of gem-dihalides.
(v) Elimination of HX from alkyl halide follows the 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.

Answer:

$(i\to b),(ii\to d),(iii\to a),(iv\to c)$
(i) A halogen atom in alkyl halide is bonded to a carbon atom that is sp3 hybridized, where further addition of alkyl group may happen. $C{H}_3-CH(X)-C{H}_3$ is alkyl halide.
(ii) A halogen atom in allyl halide is bonded to a carbon atom next to $C=C$ double bond that is $s{p}^3$ hybridized. $C{H}_2=CH-C{H}_2-X$ is allyl halide.
(iii) Halogen atoms in aryl halide are bonded to a carbon atom that is $s{p}^2$ hybridized aromatic ring. $C_6{H}_{5}X$ is aryl halide.
(iv) Halogen atoms in vinyl halide is bonded to a $s{p}^2$ hybridized carbon atom of a $C=C$ double bond. $C{H}_2=CH-X$ is vinyl halide.

Question:82

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

Answer:

$(i\to b),(ii\to d),(iii\to e),(iv\to a),(v\to c)$
(i) Here, substitution takes place when an electrophile $C{l}^{+}$ 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 a secondary halide in nature and the hydroxyl ion will substitute the halogen. Here the reaction mechanism is $S_{N}1$ 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 $O{H}^{-}$ group here, due to nucleophilic substitution.
(v) It follows Saytzeff's rule, i.,e elimination reaction.

Question:83

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

Answer:

$(i\to a),(ii\to c),(iii\to b),(iv\to d)$
(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:

$(i\to b),(ii\to a),(iii\to d),(iv\to c)$
(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 the 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 the 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 the reason is not the 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 a gaseous state. As a result, a 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 the reason is the correct explanation of the assertion.
(ii) Both the assertion and the reason both are wrong statements.
(iii) The assertion is correct but the reason is a wrong statement.
(iv) The assertion is wrong, but the reason is the correct statement.
(v) Assertion and reason both are correct statements but the reason is not the correct explanation of the 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 the reason is the correct explanation of the assertion.
(ii) Assertion and reason both are wrong statements.
(iii) The 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 the reason is not the correct explanation of assertion.
Assertion (A): KCN reacts with methyl chloride to give methyl isocyanide
Reason (R):$C{N}^{-}$ is an ambident nucleophile.
Answer:

(iv) $R-Cl+KCN\to R-CN+KCl$
Alkyl Cyanide
Methyl cyanide is also obtained as $C{N}^{-}$ 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 the 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 the reason is not the correct explanation of assertion.
Assertion (A): Tert-butyl bromide undergoes a Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane.
Reason (R): In the 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 containing double the number of carbon atoms present in the halide.
tert-Butyl bromide undergoes a 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 the 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 the reason is not the correct explanation of assertion.
Assertion (A): The presence of a nitro group at the ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution.
Reason (R): The Nitro group, being an electron-withdrawing group decreases the electron density over the benzene ring.

Answer:

(i) The nitro group is 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 the 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 the reason is not the correct explanation of assertion.
Assertion (A): In mono haloarenes, further electrophilic substitution occurs at ortho and para positions.
Reason (R): The halogen atom is a ring deactivator.
Answer:

(v) Due to the (+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 the 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 the reason is not the correct explanation of assertion.
Assertion (A): Aryl iodides can be prepared by a reaction of arenes with iodine in the presence of an oxidizing agent.
Reason (R): Oxidising agent oxidizes $I_2$ into $HI$.
Answer:

(iii) HI converts into $I_2$ by the oxidising agent such as $HI{O}_3$. Although, HI can convert aryl halides into arenes.
$5HI+HI{O}_3\to 3H_{2}O+3I_2$

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 the 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 the 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 the 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 the 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 the reason is not the 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, $S_{N}2$ 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 the 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 the reason is not the correct explanation of assertion.
Assertion (A): Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene.
Reason (R): $-N{O}_2$ 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 that are responsible for this difference.
Answer:

Primary alkyl halides prefer S_N2 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 S_N2 mechanism, substitution takes place. Tertiary alkyl halides follow S_N1 mechanism. In this case, there alkyl halides form 3^? carbocations. If the reagent used is a weak base then substitution occurs while if it is a strong base then instead of substitution elimination occurs.

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 the 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 the 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 a solvent for iodine, alkaloids, and other substances.
Triiodomethane is used as an antiseptic.
But some compounds of this class are responsible for the exposure of flora and fauna to more and more UV light which causes destruction to a 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 into 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, and 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 the reasons mentioned as follows:
(i) In the case of haloarenes, the benzene ring is attached to the lone pair electron of the halogen in resonance. As a result, the 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 the case of haloarenes, the $s{p}^2$ hybridized carbon atom is attached to a halogen. An $s{p}^2$ 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$(N{O}_2)$ at the ortho and para positions. 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 the case of o- and p-chlorobenzenes, the carbon atom bearing $(N{O}_2)$ group has a negative charge due to the resonating structure. Therefore, the $(N{O}_2)$ 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 $-N{O}_2$ group, in the case of m-nitrochlorobenzene. Therefore, the negative charge does not stabilize by the nitro group at the 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.