ROUTERA


Chapter 10 Haloalkanes and Haloarenes

Class 12th Chemistry NCERT Exemplar Solution



Multiple Choice Questions I
Question 1.

The order of reactivity of following alcohols with halogen acids is ___________.

(A) CH3CH2 —CH2—OH

(B)

(C)

A. (A) > (B) > (C)

B. (C) > (B) > (A)

C. (B) > (A) > (C)

D. (A) > (C) > (B)


Answer:

Haloalkanes are prepared from alcohols and halogen acids where the hydroxyl group of the alcohol is replaced by the halogen. Options (A) (B) and (C) are primary, secondary, tertiary alcohols respectively. Tertiary alcohols are more reactive than secondary and primary alcohol, and they form haloalkanes from haloacids at room temperature without catalysts. The order of reactivity of alcohols is 3°>2°>1°. Hence, the correct option is (ii).


Question 2.

Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?
A. CH3CH2—CH2—OH

B.

C.

D.




Answer:

Alkyl halides can be obtained by treating alcohols with haloacids. The reactivity of tertiary alcohols with the haloacids is the highest and primary alcohols is the lowest. Formation of alkyl chloride using primary and secondary alcohols requires the usage of a catalyst ZnCl2, where dry hydrogen chloride gas is passed through a solution of alcohol or by heating a mixture of alcohol and concentrated aqueous halogen acid. Reaction with tertiary alcohols does not require a catalyst and can be carried out by using tertiary alcohols where they can be shaken with concentrated HCl at room temperature. Out of the given alcohols, option (iv) is the tertiary alcohol and will yield alkyl chloride at room temperature.


Question 3.

Which of the following compounds will give racemic mixture on nucleophilic substitution by OH ion?

(a)



(b)



(c)



A. (a)

B. (a), (b), (c)

C. (b), (c)

D. (a), (c)


Answer:

A mixture containing two enantiomers in equal proportions but with zero optical activity because the opposite optical rotations of the two enantiomers cancel out each other, is called a racemic mixture. For a racemic mixture to occur after nucleophilic substitution, the optically active reactant undergoes SN1 reaction. Options (a) is a chiral carbon atom and it will undergo SN1 mechanism to give a racemic mixture. Option (b) is not an asymmetric carbon, and option (c) contains a secondary carbon asymmetric atom, which has less reactivity towards an SN1 substitution. The correct option is (i).


Question 4.

Identify the compound Y in the following reaction.



(i)



(ii)



(iii)



(iv)




Answer:

Haloarenes can be prepared from amines by Sandmeyer’s reaction. In this process, a primary aromatic amine which is dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite and a diazonium salt is formed. This freshly prepared salt when mixed with cuprous chloride replaces the diazonium group with -Cl, forming the aryl chloride. The compound Y is option (i) which is the aryl chloride. The correct option is (i).


Question 5.

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:

Toluene is an aromatic hydrocarbon (C6H5—CH3) and can be treated with halogens and used to produce aryl halides in the presence of Lewis acid catalyst iron (III) chloride by an electrophilic substitution reaction. The reaction is carried out with chlorine or bromine in the absence of light, and the products are o- and p- haloarenes.


Question 6.

Which of the following is halogen exchange reaction?
A. R X + NaI → RI + NaX

B.



C.

D.




Answer:

The reaction in option (i) is an example of Finklestein reaction, which is a halogen exchange reaction, where alkyl iodide is produced by treating alkyl halides with NaI in dry acetone. The reaction in option (ii) is an addition reaction, where an alkene is converted to corresponding alkyl halide. The reaction in option (iii) is a substitution, and the reaction in (iv) is an electrophilic substitution. Hence option (i) is correct.


Question 7.

Which reagent will you use for the following reaction?

CH3CH2CH2CH3→ CH3CH2CH2CH2Cl + CH3CH2CHClCH3

A. Cl2/UV light

B. NaCl + H2SO4

C. Cl2 gas in dark

D. Cl2 gas in the presence of iron in dark


Answer:

Production of alkyl chlorides from alkanes can be carried out by chlorination under the presence of UV light. Chlorine molecule under the influence of UV light forms free radicals, which react with alkanes to form a mixture of isomeric mono- and polyhaloalkanes. The correct option is (i).


Question 8.

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

(a)



(b)



(c)



(d)



A. (a) < (b) < (c) < (d)

B. (a) < (c) < (d) < (b)

C. (d) < (c) < (b) < (a)

D. (b) < (d) < (c) < (a)


Answer:

Alkyl halides are heavier in density than water. Their densities are based on the masses of the halogen atoms and the number of halogen atoms, and also the carbon atoms. Simply put, the atomic mass of Br is 79 and that of Cl is 35. From the molecules, (d) will have the heaviest mass, followed by (c), (b), and (a). Density is directly proportional to mass, hence the order will be the same in terms of reducing densities. The correct option is (i).


Question 9.

Arrange the following compounds in increasing order of their boiling points.

(a)



(b) CH3CH2CH2CH2Br

(c)



A. (b) < (a) < (c)

B. (a) < (b) < (c)

C. (c) < (a) < (b)

D. (c) < (b) < (a)


Answer:

The boiling points of isomeric alkyl halides are proportional to their branching, with a decrease in B.P. with increase in branching. Hence, boiling point of the tertiary isomer is the lowest, and that of primary isomer is the highest. The order of reducing boiling points is thus (b) > (a) > (c). The correct option is (iii).


Question 10.

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

(a)



(b)



(c)



(d)



A. (a), (b), (c), (d)

B. (a), (b), (c)

C. (b), (c), (d)

D. (a), (c), (d)


Answer:

The molecules (a), (b), (c) have non-superimposable mirror images. Molecule (d) cannot have an asymmetric carbon atom because all four species attached to the carbon are not different from each other, and the mirror image of (d) when rotated 180° is superimposable on the original image. The correct option is (ii).


Question 11.

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



A.



B.



C.



D.




Answer:

The stereoisomers of a compound which are non-superimposable on each other are known as enantiomers. The molecule (A) has an asymmetric carbon atom. The mirror image of the molecule (i) and the 180° rotation are not superimposable on (A). The molecules (ii), (iii) and (iv) are superimposable on (A). The 180° rotation of the molecules (iii) and (iv) are superimposable on (A). Hence, the correct option is (i).


Question 12.

Which of the following is an example of vic-dihalide?
A. Dichloromethane

B. 1,2-dichloroethane

C. Ethylidene chloride

D. Allyl chloride


Answer:

Dihaloalkanes having the same halogen are classified as geminal halides or gem-dihalides and vicinal halides or vic-dihalides. Gem-dihalides are molecules where Vic-dihalides are those dihaloalkanes where the halogen atoms are present on two adjacent carbon atoms. In common naming system, the gem-dihalides are named as alkylidene halides, vic-dihalides are named as alkylene dihalides. Dichloromethane contains only one carbon, so adjacent halogen atoms cannot occur in the molecule. 1,2-dichloroethane contains two carbon atoms with adjacent halogen atoms. Ethylidene chloride, as its common name states, is a gem-dihalide. Allyl chloride contains only one chlorine atom. The correct option is (iv).


Question 13.

The position of –Br in the compound in CH3CH==CH—C(Br)(CH3)2 can be classified as ____________.
A. Allyl

B. Aryl

C. Vinyl

D. Secondary


Answer:

Allyl halides occur when the halogen atom is bonded to an sp3-hybridised carbon atom adjacent to an allylic carbon, that is, a carbon-carbon double bond.


Aryl halides are formed when the halogen atom is bonded directly to the sp2-hybridised carbon atom of an aromatic ring.


Vinylic bonds are formed when the halogen atom is bonded to an sp2-hybridised carbon atom of a carbon-carbon double bond. Simply put, it is the ethylene molecule with one of the hydrogen atoms replaced by another –R group.


Secondary bonds occur when the halogen is bound to a secondary carbon atom in the molecule. In the given molecule, the –Br is attached to a tertiary carbon atom which is in turn attached an sp3-hybridised carbon-carbon double bond. Hence this –Br is in an allylic position. Option (i) is correct.


Question 14.

Chlorobenzene is formed by reaction of chlorine with benzene in the presence of AlCl3. Which of the following species attacks the benzene ring in this reaction?
A. Cl

B. Cl+

C. AlCl3

D. [AlCl4]


Answer:

Aluminum chloride is a Lewis acid catalyst and works in the same way as FeCl3 does. Benzene is converted into chlorobenzene by chlorination of benzene in the presence of AlCl3. The reaction occurs by electrophilic substitution. Cl2 forms a coordination complex with AlCl3, forming Cl+AlCl4- complex, which gives a slight positive charge to Cl and AlCl4- is negatively charged. This Cl+ then reacts with the aromatic double bond of the benzene ring to form an addition product, followed by deprotonation to form chlorobenzene and AlCl3 and HCl as side products. The correct option is (ii).


Question 15.

Ethylidene chloride is a/an ______________.
A. vic-dihalide

B. gem-dihalide

C. allylic halide

D. vinylic halide


Answer:

Gem-dihalides are dihaloalkanes which have two halogen atoms of the same type attached to the same carbon atom in a molecule. The common naming system of gem-dihalides is alkylidene dihalides. Ethylidene dichloride thus is a gem-dihalide. The correct option is (ii).


Question 16.

What is ‘A’ in the following reaction?



A.



B.



C.



D.




Answer:

The products obtained can be explained by Markonikov’s rule. According to the rule, the hydrogen from HCl will get added to the carbon directly bonded to the most hydrogen atoms and the –Cl will get bonded to the carbon directly bonded to the least hydrogens. Hence, the correct answer is (iii).


Question 17.

A primary alkyl halide would prefer to undergo _____________.
A. SN1 reaction

B. SN2 reaction

C. α–Elimination

D. Racemisation


Answer:

An alkyl halide with β-hydrogen atoms when reacting with a base or a nucleophile can undergo a particular type of reaction depending on the nature of the alkyl halide, the strength of the base or nucleophile, the size of the molecules and the conditions of the reaction. An alkyl halide can follow substitution or elimination reaction, and can also follow the two types of substitution, SN1 or SN2. A primary alkyl halide will prefer a SN2 reaction as the incoming nucleophile cannot interact with the bulky substituents on or near the carbon atom thus having an inhibiting effect. Hence, primary alkyl halides and methyl halides have the highest tendency towards an SN2 reaction. The correct answer is (ii).


Question 18.

Which of the following alkyl halides will undergo SN1 reaction most readily?
A. (CH3)3C—F

B. (CH3)3C—Cl

C. (CH3)3C—Br

D. (CH3)3C—I


Answer:

SN1 reactions are carried out mainly in polar protic solvents like H2O and they follow first order kinetics. This means that the rate of reaction depends only on one reactant. This reaction favours tertiary alkyl halides because of the high stability of the formed carbocation. The carbocation forms when the molecule is polarized in water to form a carbocation and halide ion. The reactivity of the halides are R–I> R–Br>R–Cl>>R–F. Hence, (CH3)3C—I will undergo the reaction most readily.


Question 19.

Which is the correct IUPAC name for:

A. 1-Bromo-2-ethylpropane

B. 1-Bromo-2-ethyl-2-methylethane

C. 1-Bromo-2-methylbutane

D. 2-Methyl-1-bromobutane


Answer:

First we need to identify the longest carbon chain. Once we do that, the actual structure should read CH3—CH2—CH(CH3)—CH2—Br. –Br, the functional halide group is attached to the first carbon atom, so we start the numbering from that position. The methyl group branch is bond to the second carbon atom in the chain. The number of carbons in the unbranched parent chain are four, thus giving the name butane. The molecule is named 1-Bromo-2-methylbutane. Option (ii) is correct.


Question 20.

What should be the correct IUPAC name for diethylbromomethane?
A. 1-Bromo-1, 1-diethylmethane

B. 3-Bromopentane

C. 1-Bromo-1-ethylpropane

D. 1-Bromopentane


Answer:

From the common name of the given compound, we can deduce there are two –C2H5 groups, one –Br group bound to a CH4 molecule. Bromomethane would appear as CH3—Br. For diethylbromomethane, two ethyl groups can be substituted for the hydrogens in the methyl molecule, forming (C2H5)CH(C2H5)Br. To provide an IUPAC name, we need to identify the longest parent chain. CH3—CH2—CH(Br)—CH2—CH3 is identified as the parent chain, and the –Br group from either side is attached to the third carbon atom. The number of carbons in the parent chain are five, so the compound is called 3-bromopentane. The correct answer is (ii).


Question 21.

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

A.



B.



C.



D. Mixture of (ii) and (iii)


Answer:

Reaction of aromatic arenes with chlorine in the presence of Lewis acid catalysts like iron (III) chloride gives ortho and para isomers of haloarenes by electrophilic substitution reaction. Both (ii) and (iii) are products in the reaction. Cl2 forms a coordination complex with FeCl3, forming Cl+FeCl4- complex, which gives a slight positive charge to Cl and FeCl4- is negatively charged. This Cl+ then reacts with the aromatic double bonds of the toluene molecule to form an addition product, followed by deprotonation to form a mixture of o- p- and m- isomers of the chlorotoluene. The m- isomer is very unstable, so the product is not available as o- and p- are.


Question 22.

Chloromethane on treatment with excess of ammonia yields mainly
A. N, N-Dimethylmethanamine

B. N–methylmethanamine (CH3—NH—CH3)

C. Methanamine (CH3NH2)

D. Mixture containing all these in equal proportion


Answer:

Ammonia molecule is a nucleophile in nature as it has unpaired electrons. This nucleophile attacks the chloromethane molecule and form methyl amine or methanamine by nucleophilic substitution reaction. The carbon atom is partially positive in the molecule, due to electronegativity of the halide attached which is partially negative. The electron-rich nucleophile attacks the positive ion, causing the halide ion to be separated from the molecule. The correct answer is (iii).


Question 23.

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:

The structure of 2-Bromobutane is shown below. All the groups attached to the central carbon atom are different, making the mirror image of molecule non superimposable over original molecule.



The structure of 1-Bromobutane is given below. There are only three groups different from each other, hence the molecule is not chiral in nature.



The structure of 1-Bromopropane is given below. Again, there are only three groups different from each other, and the molecule is not chiral.



The structure of 2-Bromopropan-2-ol, given below, again has two similar species and thus cannot be chiral.



Question 24.

Reaction of C6H5CH2Br with aqueous sodium hydroxide follows ____________.
A. SN1 mechanism

B. SN2 mechanism

C. Any of the above two depending upon the temperature of reaction

D. Saytzeff rule


Answer:

When benzyl chloride is treated with aqueous sodium hydroxide, where OH is the nucleophile, nucleophilic substitution reaction takes place forming benzyl alcohol. Here, the benzene ring is resonance stabilized and its stability is extended to the methylene group attached, giving a positive charge to –CH2, making the whole carbocation stable when the bond between benzyl and bromide is broken. This reaction is SN1 reaction, it has two steps, and is followed due to the stability of carbocation. This reaction is concerted and occurs in two steps. First the halide group leaves the carbocation, and the nucleophile then attaches itself to the cation, forming an alcohol. The correct option is (i).


Question 25.

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



A. a, b, c, d

B. b, c

C. a, d

D. a, b, c


Answer:

Chiral molecules contain one carbon atom surrounded by four nonidentical species. All straight chain molecules cannot be chiral because of the presence of two or more identical groups like hydrogens. Even the carbons with double or triple bonds to a group cannot be considered as a chiral carbon. An asymmetric carbon needs to be surrounded by four species different from each other through covalent bonds. Hence, the atoms b and c are asymmetric. Option (ii) is the correct answer.


Question 26.

(a)



(b)



(c)



A. (a) < (b) < (c)

B. (c) < (b) < (a)

C. (a) < (c) < (b)

D. (c) < (a) < (b)


Answer:

The reactivity of aryl halides to nucleophilic substitution is extremely low due to the resonance stabilization of the benzene ring. The –Cl bond also acquires a partial double bond due to resonance. Presence of an electron withdrawing –NO2 group at ortho or para positions on the ring increases the reactivity of aryl halides. The presence of –NO2 at a closer position to the C—Cl makes the molecule more reactive. The order of reactivity from highest to lowest should be (b) > (c) > (a). The correct option is (iii).


Question 27.

(a)



(b)



(c)



A. (a) < (b) < (c)

B. (a) < (c) < (b)

C. (c) < (b) < (a)

D. (b) < (c) < (a)


Answer:

The –CH3 group is an electron releasing group, thus reducing the reactivity of aryl halides when present at ortho and para positions. Hence the aryl halides without the presence of electron releasing groups are more reactive. Hence the order of reactivity is (a) > (c) > (b). The correct option is (iv).


Question 28.

(a)



(b)



(c)



A. (c) < (b) < (a)

B. (b) < (c) < (a)

C. (a) < (c) < (b)

D. (a) < (b) < (c)


Answer:

As explained in question 26, –NO2 are electron withdrawing groups and presence at the o- and p- position increase reactivity of the halides. The more the groups, the higher the reactivity, hence the order is (c) > (b) > (a). The correct answer is (iv).


Question 29.

(a)



(b)



(c)



A. (a) < (b) < (c)

B. (b) < (a) < (c)

C. (c) < (b) < (a)

D. (a) < (c) < (b)


Answer:

As explained in question 27, presence of electron releasing groups like –CH3 cause reduction in reactivity of aryl halides. Hence the increase in methyl groups reduces the reactivity. The reactivity order is (a) > (b) > (c). The correct option is (iii).


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:

Due to the polar nature of alkyl halides and the increase in molecular weight compared to their parent alkanes, the boiling points of alkyl halides are higher than that of their parent alkanes. The boiling points of alkyl halides depend on the molecular mass and the size of the halogen atom. With the increase in size, mass, and number of electrons in halogen atoms, the Van Der Waals forces increase and the boiling point also increases. The boiling point of alkyl halides reduces in the order RI > RBr > RCl > RF.


Therefore, the order of increasing order of boiling points should be Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane. The correct option is (i).


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:

If the halogen atom is the same type in all the given alkyl halides, the boiling point of a compound increases on the increase of addition of alkyl groups. The attractions between the molecules and electrons get stronger as their number and size increase. Hence the higher the molecular weight of the alkyl halide, the higher the boiling point. The increasing order of boiling points of the given compounds are 1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene. The correct option is (iv).



Multiple Choice Questions Ii
Question 1.

Which of the statements are correct about above reaction?
A. (a) and (e) both are nucleophiles.

B. In (c) carbon atom is sp3 hybridised.

C. In (c) carbon atom is sp2 hybridised.

D. (a) and (e) both are electrophiles.


Answer:

The given reaction where CH3Cl reacts with hydroxide ion to give CH3OH is an SN2 nucleophilic substitution reaction. In this reaction, OH- and Cl- are both nucleophiles with an excess of electrons. In the transition state (c) the carbon atom is bound to only 3 hydrogens as the binding of OH- and leaving of Cl- is simultaneous. Hence, the carbon atom is sp2 hybridised. The correct statements are (i) and (iii).


Question 2.

Which of the following statements are correct about this reaction?
A. The given reaction follows SN2 mechanism.

B. (b) and (d) have opposite configuration.

C. (b) and (d) have same configuration.

D. The given reaction follows SN1 mechanism.


Answer:

The given reaction is an SN2 reaction, a nucleophilic substitution. Alkyl halides undergoing SN2 reaction undergo inversion of configuration, as the incoming nucleophile causes the groups surrounding the carbon atom move in a direction opposite to that of the nucleophile. Hence, (b) and (d) will have opposite configuration. This reaction is also SN2 because it has only one step where addition of OH- and removal of Cl- is simultaneous. The correct options are (i) and (ii).


Question 3.

Which of the following statements are correct about the reaction intermediate?
A. Intermediate (c) is unstable because in this carbon is attached to 5 atoms.

B. Intermediate (c) is unstable because carbon atom is sp2 hybridised.

C. Intermediate (c) is stable because carbon atom is sp2 hybridised.

D. Intermediate (c) is less stable than the reactant (b).


Answer:

The intermediate (c) is unstable because the carbon atom is transiently attached to five atoms, and is less stable than reactant (b) which is attached to four groups.


Question 4.

Which of the following statements are correct about the mechanism of this reaction?
A. A carbocation will be formed as an intermediate in the reaction.

B. OH will attach the substrate (b) from one side and Cl will leave it

simultaneously from other side.

C. An unstable intermediate will be formed in which OH and Cl will be

attached by weak bonds.

D. Reaction proceeds through SN1 mechanism.


Answer:

The given reaction is a nucleophilic substitution that follows SN1 mechanism as it is a tertiary alkyl halide. It is a two-step reaction. The first step is the slow cleavage of polarised C—Br bond which produces a carbocation and a bromide ion. This stable intermediate carbocation is then attacked by nucleophile OH- to complete the substitution. The correct statements are (i) and (iv).


Question 5.

Which of the following statements are correct about the kinetics of this reaction?
A. The rate of reaction depends on the concentration of only (b).

B. The rate of reaction depends on concentration of both (a) and (b).

C. Molecularity of reaction is one.

D. Molecularity of reaction is two.


Answer:

The reaction is SN1 nucleophilic substitution type. This type of reaction follows second order kinetics, where the rate of reaction only depends on the concentration of the reactant. Molecularity of a reaction is defined as the number of molecules participating in the rate determining step. Molecularity of this reaction is one because of second order kinetics. The correct options are (i) and (iii).


Question 6.

Haloalkanes contain halogen atom (s) attached to the sp3 hybridised carbon atom of an alkyl group. Identify haloalkane from the following compounds.
A. 2-Bromopentane

B. Vinyl chloride (chloroethene)

C. 2-chloroacetophenone

D. Trichloromethane


Answer:

Option (i) and (iv) have halogen atom attached to sp3 hybridized carbon which is attached to other groups in single bonds. The other options are attached to carbon atoms which themselves are attached to other groups in double bonds, which means they are sp2 hybridized. Options (i) and (iv) are correct.


Question 7.

Ethylene dichloride and ethylidene chloride are isomers. Identify the correct statements.
A. Both the compounds form same product on treatment with alcoholic KOH.

B. Both the compounds form same product on treatment with aq. NaOH.

C. Both the compounds form same product on reduction.

D. Both the compounds are optically active.


Answer:

Ethylidene dichloride CH3—CHCl2 and Ethylene chloride ClCH2—CH2Cl are isomers. Both molecules are not optically active as the carbon atoms are not surrounded by different groups.


When haloalkanes are treated with alcoholic KOH, they undergo elimination reaction where there is an elimination of hydrogen atom from β-carbon atom and a halogen atom from the α-carbon atom. Both these compounds lose hydrogen and chlorine atoms to give ethyne molecule.


In case of treatment with aqueous NaOH, the molecules undergo nucleophilic substitution, where the –Cl groups are replaced with –OH molecules. Due to the positions of the halides on different carbon atoms, the products will also be different.


Reduction of the two compounds is carried out with Zn dust in alcohol, and the end product is of these is the corresponding alkene, hence both compounds on reduction give alkene.


Options (i) and (iii) are correct.


Question 8.

Which of the following compounds are gem-dihalides?
A. Ethylidene chloride

B. Ethylene dichloride

C. Methylene chloride

D. Benzyl chloride


Answer:

Gem-dihalides are dihaloalkanes which have two halogen atoms attached to the same carbon atom in a molecule. They are also known as alkylidene halides. Ethylene dichloride as a molecule is a vicinal dihalide, which has chlorine atoms attached to adjacent carbon atoms. Benzyl chloride has only one chlorine atom. Ethylidene chloride is a gem-dihalide as indicated by the name of the molecule. Methylene chloride is known as dichloromethane by IUPAC nomenclature. It contains two chlorine atoms attached to the same single carbon atom. The correct statements are (i) and (iii).


Question 9.

Which of the following are secondary bromides?
A. (CH3)2 CHBr

B. (CH3)3C CH2Br

C. CH3CH(Br)CH2CH3

D. (CH3)2CBrCH2CH3


Answer:

A method to determine whether the given compound has a secondary bromide is to check with carbon atom it is attached to. A secondary carbon group in a molecule is attached to one hydrogen atom and three groups which are not hydrogen. Hence from the above options, the –Br group attached to –CH groups are secondary bromides. The correct options are (i) and (iii).


Question 10.

Which of the following compounds can be classified as aryl halides?
A. p-ClC6H4CH2CH(CH3)2

B. p-CH3CHCl(C6H4)CH2CH3

C. o-BrH2C-C6H4CH(CH3)CH2CH3

D. C6H5-Cl


Answer:

Aryl halides are those halides in which the halogen atom is directly attached to the sp2 hybridised carbon atom of an aromatic ring. From the given options, only (i) and (iv) have halogen atoms directly attached to aryl rings. Hence, they are the aryl halides. The correct options are (i) and (iv).


Question 11.

Alkyl halides are prepared from alcohols by treating with
A. HCl + ZnCl2

B. Red P + Br2

C. H2SO4 + KI

D. All the above


Answer:

Alcohols on treatment with various reagents give corresponding alkyl halides as products. Alcohols on treatment with HCl + ZnCl2 give alkyl chlorides and so does red phosphorous and bromine treated with alcohol give alkyl bromide. H2SO4 and KI cannot be used for converting an alcohol to alkyl iodide as this converts KI to corresponding acid, HI which is then oxidised by it to I2. Hence the correct statements are (i) and (ii).


Question 12.

Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ____________ or ____________.
A. Ca F2

B. CoF2

C. Hg2F2

D. NaF


Answer:

Alkyl fluorides can be synthesized by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3. The reaction is termed as Swarts reaction. Out of the given options, CoF2 and Hg2F2, work as reagents. NaF cannot be used because the end products of NaF + RCl → RF + NaCl are soluble in the solvent i.e. water, making it difficult to separate. Hence the correct options are (ii) and (iii).



Short Answer
Question 1.

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


Answer:

Iodination of arenes is reversible in nature due to formation of HI. To move the reaction forward, an oxidizing agent like HNO3 or HIO4 oxidises HI, thus stabilizing the product.



Question 2.

Out of o-and p-dibromobenzene which one has higher melting point and why?


Answer:

Out of the two compounds, p-dibromobenzene has the higher melting point. This is because the symmetry of p-dibromobenzene which makes the molecule fit better in crystal lattice. Due to this, it requires a higher temperature to break the bonds between the molecules and thus has a higher melting point.



Question 3.

Which of the compounds will react faster in SN1 reaction with the OH ion?

CH3— CH2— Cl or C6H5— CH2— Cl


Answer:

C6H5— CH2— Cl will react faster in an SN1 reaction with the OH- ion. This happens due to the stability of the carbocation in the compound. C6H5 group is already stable due to resonance, and the CH2 attached will gain that stability, thus forming a stable C6H5CH2+ carbocation after the cleavage in the first step of the SN1 reaction. The same type of carbocation does not occur in CH3—CH2—Cl.



Question 4.

Why iodoform has appreciable antiseptic property?


Answer:

The chemical name of Iodoform is Triiodomethane. It has an antiseptic property because of its property of release of free iodine.



Question 5.

Haloarenes are less reactive than haloalkanes and haloalkenes. Explain.


Answer:

The major reason haloarenes are less reactive than haloalkanes and haloalkenes is the resonance stabilization of the aryl ring. For example, in C6H5-Cl, the electron pairs on halogen atom are in conjugation with ϖ-electrons of the ring. Due to resonance, the C—Cl bond acquires a partial double bond character, making it less reactive to nucleophilic substitution than haloalkanes and haloalkanes.



Question 6.

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 from arenes by electrophilic substitution. This reaction is carried out by treating the arene with chlorine or bromine in the presence of iron (III) chloride in the absence of light. Iron (III) chloride, that is, FeCl3 is a Lewis acid, which generates the electrophile required to take the reaction forward. FeCl3 forms a coordination compound with Cl2, making the complex Cl+[FeCl4-]. The chloride ion gains a partial positive charge, acting as an electrophile, and attacking the ϖ bonds. The final products are o-arylhalide and p-arylhalide.



Question 7.

Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4. Explain why?

(a) CH3CH2CH2OH

(b)




Answer:

A mixture of NaBr and H2SO4 gives Br2 gas as a product. Molecule (b) will not react with Br2 gas because of the stable molecule that is formed due to resonance stabilization.



Question 8.

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

Explain.

CH3CH=CH2 + HI → CH3CH2CH2I + CH3CHICH3

(A) (B)


Answer:

This addition reaction is carried out by following Markovnikoff’s rule, where in a double bond, the hydrogen from the hydrogen halide is added to the carbon atom with the most hydrogen atoms attached to it, while the halogen atom is attached to the carbon atom with lesser hydrogens. The molecule that follows this rule will be the major product in the mixture. Hence, the molecule (B) will be the major product in the reaction.



Question 9.

Why is the solubility of haloalkanes in water very low?


Answer:

Haloalkanes are very slightly soluble in water because in order to dissolve a haloalkane in water, energy is required to overcome the attractions between the haloalkane molecules and also to break the hydrogen bonds between water molecules. On the other hand, less energy is released when new attractions are created between the haloalkane and the water molecules as these are not as strong as the original hydrogen bonds in water.



Question 10.

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 other resonance structures are as follows:


The functional group present in these molecules are ortho-para directing as electron density is more at ortho and para positions.



Question 11.

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:

The given compounds can be classified as primary, secondary and tertiary halides by knowing which carbon the halogen atom is attached to. In 1-Bromobut-2-ene, the –Br is attached to the first carbon in the chain, which is a primary carbon atom. The structure is H3C—HC=CH—H2C—Br and this is a primary halide

In 4-Bromopent-2-ene, the bromine is attached to the secondary carbon atom, hence this a secondary halide. The structure is CH3—CH(Br)CH=CH—CH3.


In 2-Bromo-2-methylpropane, the bromine is attached to the tertiary carbon atom, making it a tertiary halide. The structure is CH3—C(Br)(CH3)—CH3.



Question 12.

Compound ‘A’ with molecular formula C4H9Br 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 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 inverted configuration?


Answer:

The compound ‘A’, C4H9Br when treated with aq. KOH, proceeds with the rate of the reaction depends on the concentration of ‘A’ only. This means that the reaction follows first order kinetics, which is the characteristic of SN1 reactions. This means C4H9Br is a tertiary halide, because tert-halides undergo SN1 reactions. On the other hand, the optically active isomer ‘B’, when subjected to treatment with aq. KOH, undergoes SN2 reaction due to the rate of reaction depending on both the reactants. This means that the isomer is a secondary halide.

(i) The structure of ‘A’ is



The structure of ‘B’ is .


(ii) Compound ‘B’ will undergo inversion of configuration and give inverted product because it undergoes SN2 reaction.



Question 13.

Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, C7H8 is treated with Cl2 in the presence of FeCl3.


Answer:

When an arene is treated with Cl2 in presence of FeCl3, it undergoes chlorination to give aryl chloride by electrophilic addition reaction. Thus the compound given is C6H5—CH3, or toluene. On chlorination by electrophilic substitution, the products are o-chlorotoluene and p-chlorotoluene.



Question 14.

Identify the products A and B formed in the following reaction:

(a) CH3—CH2—CH=CH—CH3+HCl → A + B


Answer:

The alkene undergoes addition reaction to give two products, as shown below.



Question 15.

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

(I)



(II)



(III)




Answer:

Compound (II) has both methyl groups as well as chlorine atoms are placed symmetrically at para-positions, due to this, these molecules fit in the crystal lattice better than other isomers, hence it has the highest melting point.



Question 16.

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


Answer:

The IUPAC name for this compound is 1-Bromo-2,2-dimethylpropane,



Question 17.

A hydrocarbon of molecular mass 72 g mol–1 gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.


Answer:

The hydrocarbon that has molecular mass of 72g mol-1 is the alkane C5H12. The tert-isomer of pentane that on photo-chlorination gives a single monochloro derivative and two dichloro derivatives. The structure of the alkane and the chloride derivatives are given as follows.



Question 18.

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


Answer:

There are two compounds which can yield 1-chloro-1-methylcyclohexane, methylenecyclohexane and 1-methylcyclohex-1-ene.



Question 19.

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:

Haloalkanes will react with aqueous KOH to undergo nucleophilic substitutions. Out of the given compounds, the tertiary halide 2-Bromo-2-methylpropane will react with aq. KOH most easily because the first step will involve cleavage of the alkyl group and the halide group, leading to formation of the very stable carbocation.



Question 20.

Why can aryl halides not be prepared by reaction of phenol with HCl in the presence of ZnCl2?


Answer:

Phenol contains the phenyl ring bonded to the hydroxyl group. Due to the resonance stabilization of the ring, the hydroxyl group attached to it gains a partial double bond, which makes it difficult for the cleavage of this bond in order to be subject to chlorination.



Question 21.

Which of the following compounds would undergo SN1 reaction faster and why?

(A)



(B)




Answer:

The compound (B) would undergo SN1 reaction faster than (A) due to the resonance stabilization of the benzyl ring. After the cleavage of the –Cl from the ring, the carbocation C6H5CH2Cl+ is highly stable and it is driven to carry out the SN1 reaction.



Question 22.

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


Answer:

Allyl Chloride, CH2=CH—CH2Cl, is more readily hydrolysed than n-propyl chloride, CH3—CH2—CH2Cl, because of the presence of the ϖ bond which gives the property of resonance stabilization to the CH3—CH2—CH2+ carbocation to be very stable. Due to the formation of carbocation, the molecule undergoes hydrolysis more readily.



Question 23.

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


Answer:

RMgX + H2O → RH + Mg(OH)X

Grignard’s reagent are highly reactive and on contact with even slight traces of water, form the corresponding hydrocarbon.



Question 24.

How do polar solvents help in the first step in SN1 mechanism?


Answer:

SN1 reaction, a type of nucleophilic substitution, is carried out in the presence of a polar protic solvent like water, alcohol and acetic acid. The polar solvents aid in the cleavage of the R—X bond leading to formation of stable carbocation. The energy for this cleavage is obtained through the solvation energy of the halide with the protons of the polar solvent.



Question 25.

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


Answer:

There are two tests to detect the presence of unsaturated double bond in an organic molecule, the bromine water test and Bayer’s test.

In the bromine water test, when the molecule is added to bromine water, bromine is added to the carbon atoms across the double bond. The orange-brown colour of bromine water changes and becomes colourless at the end of the reaction.



In Bayer’s test, alkaline KMnO4 is used to detect unsaturated carbon with double bond. The chemical gives rise to [O], which hydrolyses the carbons across double bond, and the solution containing the two compounds goes from purple colour to colourless.


2KMnO4 + H2O → 2KOH + 2MnO2 + 3[O]




Question 26.

Diphenyls are potential threat to the environment. How are these produced from arylhalides?


Answer:

Diphenyls such as p,p-dichlorodiphenyl trichloroethane (DDT), which have been used as organic pesticides, are a potential threat to the environment because in the early years of its extensive usage, it was shown to have high toxicity towards fish and many insects also developed resistance towards it. The chemical stability of DDT and its fat solubility compounded the problem. In the chemical aspect, DDT is not metabolised very rapidly or solubilized effectively by animals. It is deposited and stored in the fatty tissues. If ingestion continues at a steady rate, DDT builds up within the animal over time, damaging the environment. Diphehyls are produced from arylhalides from two methods, Fittig reaction and Ullman biaryl synthesis.

In Fittig reaction, the aryl halides are treated with sodium metal in the presence of dry ether to produce substituted aryl compounds, in this case, biphenyl compound.


In Ullmann synthesis, which is a coupling reaction, aryl halides are heated in the presence of Cu to form biphenyl/diphenyl compounds.




Question 27.

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


Answer:

The IUPAC name of DDT is 2,2-bis(4-chlorophenyl)-1,1,1-trichloroethane. The IUPAC name of bezenehexachloride is 1,2,3,4,5,6-hexachlorocyclohexane.

Their use is banned in India and other countries because DDT is not metabolised very rapidly or solubilized effectively by animals. It is deposited and stored in the fatty tissues. If ingestion continues at a steady rate, DDT builds up within the animal over time, damaging the environment. It was also shown to have high toxicity towards fish and many insects also developed resistance towards it.



Question 28.

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


Answer:

Alkyl halides can undergo both beta-elimination as well as substitution reactions, the desired product can be obtained using an appropriate reagent and environmental conditions of the system. Consider the case of CH3CH2Br, which can undergo β-elimination as well as substitution. On treatment with alcoholic KOH, at 473-523°K, ethyl bromide undergoes elimination to give ethane. On the other hand, treatment with aqueous KOH at a lower temperature of 373°K gives ethyl alcohol by substitution reaction.



Question 29.

How will you obtain monobromobenzene from aniline?


Answer:

Monobromobenzene can be prepared from aniline, C6H5NH2 by Sandmeyer’s reaction. When aniline suspended in cold mineral acid is treated with sodium nitrite, a diazonium salt is formed. Mixing the diazonium salt with cuprous bromide results in replacement of the diazonium ion by –Br, forming monobromobenzene.



Question 30.

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

(I)



(II)



(III)




Answer:

Aryl halides are very less reactive towards nucleophilic substitution because of resonance stabilization. The presence of an –NO2, an electron-withdrawing group at ortho or para position increases the reactivity of the aryl halide towards substitution. The more positions filled with electron withdrawing groups, the more reactive the aryl halide. According to this concept, the decreasing order of reactivity is III > II > I.



Question 31.

tert-Butylbromide reacts with aq. NaOH by SN1 mechanism while n-butylbromide reacts by SN2 mechanism. Why?


Answer:

Tert-butylbromide undergoes substitution by SN1 mechanism because it is able to form a stable carbocation in the first step after cleavage of the halide group. The carbocation then reacts with the nucleophile OH-. On the other hand, primary halide n-butylbromide cannot form a stable carbocation so it undergoes SN2 mechanism, which is a one-step substitution which involves attack of OH- and simultaneous leaving of X- to form n-butyl alcohol.




Question 32.

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


Answer:

The reaction of HCl and isobutylene is given as:


The mechanism of the reaction is given as follows:




Question 33.

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


Answer:


Haloarenes are extremely less reactive to nucleophilic substitutions due to resonance stabilization of the molecule. The electron pairs on halogen atom are conjugated with ϖ-electrons of the aryl ring and the above resonating structures are possible. The C—X bond gains a partial double bond in resonance and thus it becomes difficult for cleavage of the bond, making is less reactive. There is also a difference in the hybridization of the carbon atom in the C—X bond. The carbon atom attached to halogen is sp2 hybridised unlike halloalkanes which are sp3 hybridised. The sp2 hybridised carbon holds the halogen more tightly, with a smaller bond length compared to the carbon atom attached to the halogen in sp3 hybridisation in haloalkane.



Question 34.

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


Answer:

Ethanol can be reacted with HCl gives chloroethane in the presence of ZnCl2. Chloroethane on reacting with NaI gives iodoethane.



Question 35.

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:

Groups like cyanide and nitrile are known as ambident nucleophiles, that is, groups which possess two nucleophilic centers. In the cyanide group, the linking can occur from the carbon end as well as the nitrogen end. In an aqueous medium, the carbon end acts as a stronger nucleophile because it leads to the formation of a C—C bond which is stronger than a N—C bond in the same compound.




Matching Type
Question 1.

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



Answer:

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

The halogen derivatives are matched to their pharmaceutical applications. Chloramphenicol is an antibiotic used in treatment of typhoid fever, thyroxine is a hormone whose dysregulation causes goiter, chloroquine is a drug used in treatment of malaria, and chloroform is used as an anesthetic in some cases.



Question 2.

Match the items of Column I and Column II.



Answer:

(i) → (c) (ii) → (e) (iii) → (a) (iv) → (b) (v) → (d)

The products of SN1 reaction undergo racemization, which is the process of conversion of enantiomers into a racemic mixture, a mixture containing two optically active enantiomers in equal proportions. Chlorobromocarbons are a chemical fire extinguisher. Bromination of alkenes will lead to formation of vic-dibromides, in which halogen atoms of the same type are attached to adjacent carbon atoms of the molecule. Alkylidene dihalides, also known as gem-dihalides are dihalides which contain halogen atoms attached to the same carbon atom. Elimination of HX from alkylhalide follows the Saytzeff rule, that is, “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 3.

Match the structures of compounds given in Column I with the classes of compounds given in Column II.



Answer:

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

The first molecule is an alkyl halide because it contains a halogen atom bound to an sp3 hybridised carbon atom, which is further bound to other alkyl groups.


The second molecule is an allyl halide which are compounds where the halogen atom is attached to a carbon atom is bonded to an sp3 hybridised carbon atom which is attached to a carbon-carbon double bond.


The third molecule is an aryl halide which is a compound where the halogen atom is attached to an aromatic ring carbon atom which is sp2 hybridised.


The fourth molecule is a vinyl halide in which the halogen atom is bound to an sp2 hybridised carbon atom in a carbon-carbon double bond.



Question 4.

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



Answer:

(i) → (b) (ii) → (d) (iii) → (e) (iv) → (a) (v) → (c)

The first reaction is an electrophilic substitution reaction where Cl+ from Cl2 and FeCl3 complex attacks the benzene ring and carry out the substitution.


The second reaction is an electrophilic addition reaction where HBr is added across the double bond following Markownikoff’s rule.


The third reaction is a nucleophilic substitution reaction following the SN1 mechanism where the –I group is replaced by –OH group.


The fourth reaction is a nucleophilic aromatic substitution reaction where the –Cl group is replaced by –OH.


The fifth reaction is a dehydrohalogenation reaction where the alkyl halide follows Saytzeff elimination reaction.



Question 5.

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



Answer:

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

The given names are IUPAC names of the given structures.



Question 6.

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



Answer:

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

The first reaction is a Wurtz-Fittig reaction where an alkyl halide and an aryl halide react in the presence of Na metal in dry ether to form alkylarene.


The second reaction is the Fittig reaction where a coupling reaction between two arylhalides takes place in the presence of Na metal in dry ether, forming diaryl compounds.


The third reaction is Sandmeyer reaction where aryl halides are prepared from primary amines by conversion into diazonium salt and treatment with cuprous bromide.


The fourth reaction is the Finkelstein reaction where alkyl iodides are prepared by reaction of alkyl chlorides or bromides with NaI in dry acetone.




Assertion And Reason
Question 1.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : Phosphorus chlorides (tri and penta) are preferred over thionyl chloride for the preparation of alkyl chlorides from alcohols.

Reason : Phosphorus chlorides give pure alkyl halides.


Answer:

Thionyl chloride is preferred over phosphorous chlorides because along with the alkyl halides formed, the by-products are SO2 and HCl, which are gaseous and hence can escape the reaction leaving pure halides.


Question 2.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : The boiling points of alkyl halides decrease in the order :

RI > RBr > RCl > RF

Reason : The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.


Answer:

The boiling points of the alkyl halides decrease in the given order because of the size of the halogen atom. Iodide having the highest atomic number, has more electrons, which means increase in Van Der Waals forces and the higher boiling point.


Question 3.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : KCN reacts with methyl chloride to give methyl isocyanide

Reason : CN is an ambident nucleophile.


Answer:

Haloalkanes react with AgCN to form alkyl isocyanides as main product while KCN forms alkyl cyanides as the chief product.


Question 4.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane.

Reason: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.


Answer:

Tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane because in the tertiary butyl bromide reacts with NaI in dry ether and thus form 2, 2, 3, 3-tetramethylbutane.


Question 5.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : Presence of a nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution.

Reason : Nitro group, being an electron withdrawing group decreases the electron density over the benzene ring.


Answer:

Nitro Group present on ortho and para position of haloarenes act as an electron-withdrawing group and thus there are the deficiency of electrons makes it more reactive towards nucleophilic substitution reaction.


Question 6.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : In monohaloarenes, further electrophilic substitution occurs at ortho and para positions.

Reason : Halogen atom is a ring deactivator.


Answer:

Further electrophilic substitution occurs at ortho and para positions because halogen atoms are ortho and para directing, not because they are ring deactivators.


Question 7.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : Aryl iodides can be prepared by reaction of arenes with iodine in the presence of an oxidising agent.

Reason : Oxidising agent oxidises I2 into HI.


Answer:

Oxidising agents like HIO3 oxidise HI to I2 because in their absence, the presence of HI makes the aryl iodides go back to being arenes.


Question 8.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : It is difficult to replace chlorine by –OH in chlorobenzene in comparison to that in chloroethane.

Reason : Chlorine-carbon (C—Cl) bond in chlorobenzene has a partial double bond character due to resonance.


Answer:

In chlorobenzene, due to resonance partial double bond character appears in bond between C & Cl atom and we all know that replacing of partial double bond character is bit difficult than replacing single bond as present in C-Cl bond in chloroethane.


Question 9.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : Hydrolysis of (–)-2-bromooctane proceeds with inversion of configuration.

Reason : This reaction proceeds through the formation of a carbocation.


Answer:

Hydrolysis of alkyl halides with inversion of configuration is an example of SN2 mechanism. This mechanism is a one-step process and does not involve formation of carbocation.


Question 10.

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
A. Assertion and reason both are correct and reason is correct explanation of assertion.

B. Assertion and reason both are wrong statements.

C. Assertion is correct but reason is wrong statement.

D. Assertion is wrong but reason is correct statement.

E. Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Assertion : Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene

Reason : —NO2 group is a m-directing group.


Answer:

The —NO2 group is a meta-directing group by m-nitrochlorobenzene is not a stable compound, and the products of the reactions contain nitro groups at o- and p- positions.


Question 11.

Some alkylhalides 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:

Some alkylhalides undergo substitution or elimination, and this is determined by the structure of the alkyl halide as well as the reagents used in the reaction. To understand the structural features of alkyl halides, their reactivity towards substitution mechanisms can be observed. Primary alkyl halides prefer to undergo substitution reaction by SN2 mechanism which is a one-step reaction involving cleavage of the halide from the carbon atom and simultaneous attachment of the attacking nucleophile. On the other hand, tertiary halides undergo elimination reaction due to the formation of stable carbocation. Tertiary halides also prefer to undergo SN1 reaction which is a two-step process mediated by the formation of a stable carbocation after cleavage of the halide atom. Secondary halides have intermediate reactivity towards both elimination and substitution reactions, depending on the solvent and temperature conditions.

The given reaction is for a primary alkyl halide undergoing SN2 substitution.



The given reaction is for a tertiary alkyl halide undergoing SN1 substitution. The mechanism is determined by the strength of the base used. In case of usage of weak base, aqueous KOH, the reaction undergone is substitution while in case of strong base, alcoholic base, the reaction undergone is elimination.






Long Answer
Question 1.

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 minimise harmful effects of these compounds.


Answer:

These carbon compounds contain multiple halogen groups and have wide application in industry and have wide application in industry. They are commonly known as polyhalogen compounds. The various compounds and their effects on life as follows:

1. Dichloromethane or methylene chloride: This is used as a solvent as paint remover, as a propellant in aerosols, as a process solvent in the manufacture of drugs and also as a metal cleaning and finishing solvent. Dichloromethane has been known to harm the central nervous system. Exposure to low levels of this compound can lead to slightly impaired hearing and vision, while higher levels in air cause dizziness, nausea, tingling and numbness in the fingers and toes. In humans, direct skin contact with methylene chloride causes intense burning and mild redness of the skin. Direct contact with the eyes can burn the cornea.


2. Trichloromethane or chloroform: This is mainly used as a solvent for fats, alkaloids, iodides and others and mainly used in production of a Freon R-22. It used to have applications in medicine as an anaesthetic but lesser toxic versions like ether are used. Chronic exposure to chloroform is fatal as it gets converted to phosgene, causing liver and kidney damage and skin sores.


3. Tetrachloromethane or carbon tetrachloride was widely used in the manufacture of cleaning liquids, chlorofluorocarbons and propellants for aerosols. CCl4 usage has caused depletion of ozone layer, which protects the earth from UV rays, which has led to higher cases of skin cancer and other disorders.


4. p,p’-dichlorodiphenyltrichloroethane or DDT: DDT was widely used as insecticide. They are chlorinated organic insecticides whose usage increased greatly after World War II because they were highly effective against mosquitoes causing malaria and lice carrying typhus. The serious effects of this started with insects gaining resistance to DDT, and a high toxicity towards fish. The chemical stability of DDT and its fat solubility compounded the problem. In the chemical aspect, DDT is not metabolised very rapidly or solubilized effectively by animals. It is deposited and stored in the fatty tissues. If ingestion continues at a steady rate, DDT builds up within the animal over time, damaging the environment.



Question 2.

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 than alkyl halides due to following reasons:

(i) The electrons on the halogen atom are in conjugation with the ϖ-electrons of the aryl ring, giving the C—X bond a partial double bond character. The entire molecule is then resonance stabilized and the following structures are possible. Due to this bond formation, C—X bond cleavage in haloarenes is more difficult than haloalkanes.



(ii) In haloalkanes, the carbon attached to halogen is sp3 hybridised while in haloarene, the carbon attached to the halogen is sp2 hybridised. The sp2 hybridised carbon holds the halogen atom tightly and tighter than a bond between sp3 carbon atom and halogen atom in haloalkanes. The C—X bond length in haloarenes is shorter than haloalkanes and thus more difficult to break.


(iii) The phenyl cation does not get resonance stabilized and hence cannot follow SN1 mechanism of substitution.


(iv) Aryl halides are less reactive than alkyl halides because of possible repulsion between electron rich arenes and approaching electron rich nucleophiles.


Reactivity of aryl halides can be increased with the presence of electron withdrawing group at ortho- and para- positions. No effect is observed at the meta- position. Presence of nitro group, an example of electron withdrawing group, at ortho- and para- positions withdraws the electron density from the benzene ring and thus facilitates the attack of the nucleophile on haloarene. The carbanion thus formed is stabilised through resonance. The negative charge present at ortho- and para- positions with respect to the halogen substituent is stabilised by –NO2 group while in case of meta-nitrobenzene, none of the resonating structures bear the negative charge on carbon atom bearing the –NO2 group. Hence no effect on reactivity of aryl halide is observed on presence of meta- position electron withdrawing group.