Class- XII Chemistry Chapter-10 Haloalkane and...

Class- XII

Chemistry

Chapter-10

Haloalkane and Haloarene

Haloalkane

The replacement of hydrogen atom(s) in an aliphatic hydrocarbon by halogen atom(s)

results in the formation of alkyl halide (haloalkane).

Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an

alkyl group.

Haloarene

The replacement of hydrogen atom(s) in an aromatic hydrocarbon by halogen atom(s)

results in the formation of aryl halide (haloarene).

Haloarenes contain halogen atom(s) attached to sp2hybridised carbon atom(s) of an

aryl group.

Classification of Haloalkanes and Haloarenes

On theBasis ofNumber ofHalogenAtoms

According to the hybridisation of the carbon atom to which the halogen is

bonded.

Compounds Containing sp3 C—X Bond (X= F,Cl,Br,I)

(a) Alkyl halides or haloalkanes (R—X)

In alkyl halides, the halogen atom is bonded to an alkyl group (R). They form a

homologous series represented by CnH2n+1X. They are further classified as primary,

secondary or tertiary.

(b) Allylic halides

These are the compounds in which the halogen atom is bonded to a sp3-hybridised

carbon atom adjacent to carbon-carbon double bond (C=C) i.e. to an allylic carbon.

(c) Benzylic halides


carbon atom attached to an aromatic ring.

* Compounds Containing sp2 C—X Bond

(a) Vinylic halides


carbon atom of a carbon-carbon double bond (C = C).

(b) Aryl halides

These are the compounds in which the halogen atom is directly bonded to the sp2

hybridised carbon atom of an aromatic ring.

a) From Alcohols

b) From Hydrocarbons

(I) From alkanes by free radical halogenation

are the compounds in which the halogen atom is directly bonded to the sp2

hybridised carbon atom of an aromatic ring.

Preparation of Haloalkanes

(I) From alkanes by free radical halogenation

are the compounds in which the halogen atom is directly bonded to the sp2-

Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric

mono- and polyhaloalkanes, which is difficult to separate as pure compounds.

(II) From alkenes

i)Addition of hydrogen halides

by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide.

(II)Addition of halogens:

In the laboratory, addition of bromine in CCl

reddish-brown colour

of bromine constitutes an important method

molecule. The addition results in the synthesis of Vic

Vic-dibromide

chlorination or bromination of alkanes gives a complex mixture of isomeric

and polyhaloalkanes, which is difficult to separate as pure compounds.

Addition of hydrogen halides: An alkene is converted to corresponding alkyl halide


In the laboratory, addition of bromine in CCl4 to an alkene resulting in discharge of

of bromine constitutes an important method for the detection of double bond in a

molecule. The addition results in the synthesis of Vic-dibromides, which are colourless.

chlorination or bromination of alkanes gives a complex mixture of isomeric

and polyhaloalkanes, which is difficult to separate as pure compounds.

: An alkene is converted to corresponding alkyl halide


to an alkene resulting in discharge of

for the detection of double bond in a

dibromides, which are colourless.

c)From Halogen Exchange

Preparation of Haloarenes

I)From hydrocarbons by electrophilic substitution

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

like iron or iron (III) chloride. The ortho and para isomers can be easily separated due

to large difference in their melting points.

(ii) From amines by Sandmeyer’s reaction

or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium

salt is formed. Mixing the solution of freshly prepared diazonium salt with cuprous

chloride or cuprous bromide results in the r

or –Br.

Replacement of the diazonium group by iodine does not require the presence of

cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.

Preparation of Haloarenes

I)From hydrocarbons by electrophilic substitution

ides and bromides can be easily prepared by electrophilic substitution of



e difference in their melting points.

(ii) From amines by Sandmeyer’s reactionWhen a primary aromatic amine, dissolved



chloride or cuprous bromide results in the replacement of the diazonium group by



ides and bromides can be easily prepared by electrophilic substitution of



When a primary aromatic amine, dissolved



eplacement of the diazonium group by –Cl



Melting and boiling points

For the same alkyl group, the boiling points of alkyl halides decrease in the order:

Thisis because with the increase in size and mass of halogen atom, the magnitude of

van der Waal forces increases.

The boiling points of isomeric haloalkanes decrease with increase in branching

The para-isomers are high melting as compared to their ortho

due to symmetry of para-isomers that fits in crystal lattice better as compared to

ortho- and meta-isomers.

Density

Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water. The

density increases with increase in number of carbon atoms, halogen atoms and atomic

mass of the halogen atoms.

Solubility

The haloalkanes are very slightly soluble in

dissolve in organic solvents.

Chemical Properties of Haloalkanes

The reactions of haloalkanes may be divided into the following categories:

1. Nucleophilic substitution

2. Elimination reactions

3. Reaction with metals.

Nucleophiles are electron rich species. Therefore, they attack at that part of the

substrate molecule which is electron deficient. Examples:

Physical Properties


RI> RBr> RCl> RF


van der Waal forces increases.

someric haloalkanes decrease with increase in branching

isomers are high melting as compared to their ortho-and meta

isomers that fits in crystal lattice better as compared to



The haloalkanes are very slightly soluble in water. However, haloalkanes tend to

Chemical Properties of Haloalkanes


ucleophiles are electron rich species. Therefore, they attack at that part of the

substrate molecule which is electron deficient. Examples:



someric haloalkanes decrease with increase in branching.

and meta-isomers. It is

isomers that fits in crystal lattice better as compared to



, haloalkanes tend to


ucleophiles are electron rich species. Therefore, they attack at that part of the

Ambident nucleophiles:

Groups which possess two nucleophilic centres are called ambident nucleophiles.

Actually, cyanide group is a hybrid of two contributing structures and therefore can act

as a nucleophile in two different ways

[-C≡N ↔: C=N-], i.e., linking through carbon atom resulting in alkyl cyanides [RCN] and

through nitrogen atom leading to isocyanides [RNC

Similarly, nitrite ion also represents an ambident

nucleophile with two different points of linkage

. The linkage through oxygen results in alkyl nitrites

NO2].

(1) Nucleophilic substitution

The reaction in which a nucleophile replaces already existing nucleophile in a molecule

is called nucleophilicsubstitution reaction. In this type of reaction, a nucleophile reacts

with haloalkane (the substrate) having a partial positive charge

bonded to halogen. A substitution reaction takes place and halogen atom, called

leaving group departs as halide ion. Since the substitution reaction is initiated by a

nucleophile, it is called nucleophilic substitution reaction.


cyanide group is a hybrid of two contributing structures and therefore can act

as a nucleophile in two different ways

], i.e., linking through carbon atom resulting in alkyl cyanides [RCN] and

through nitrogen atom leading to isocyanides [RNC].

Similarly, nitrite ion also represents an ambident

nucleophile with two different points of linkage

. The linkage through oxygen results in alkyl nitrites

while through nitrogen atom, it leads to nitroalkanes [R

reactions:



with haloalkane (the substrate) having a partial positive charge on the carbon atom



nucleophile, it is called nucleophilic substitution reaction.


cyanide group is a hybrid of two contributing structures and therefore can act

], i.e., linking through carbon atom resulting in alkyl cyanides [RCN] and

while through nitrogen atom, it leads to nitroalkanes [R



on the carbon atom



Mechanism:

Substitution nucleophilic bimolecular (SN2)

The term ‘SN2’ stands for – Substitution Nucleophilic Bimolecular. This type of

reaction is also referred to as bimolecular nucleophilic substitution, associative

substitution, and interchange mechanism.

The SN2 reaction is a nucleophilic substitution reaction where a bond is broken and

another is formed simultaneously in a single step and no intermediate is formed. Two

reacting species are involved in the rate determining step of the reaction.

Reaction Kinetics: Since an SN2 Reaction is a second-order reaction, the rate-

determining step is dependent on the concentration of nucleophile as well as the

concentration of the substrate”.

SN2 reaction mechanism requires the attack of nucleophile from the back side of the

carbon atom. So, the product assumes a stereochemical position opposite to the

leaving group originally occupied. This is called inversion of configuration.

Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because

there are only three small hydrogen atoms. Tertiary halides are the least reactive

because bulky groups hinder the approaching nucleophiles. Thus, the order of

reactivity followed is:

Primary halide > Secondary halide > Tertiary halid

Substitution nucleophilic unimolecular (SN1)

SN1 reactions are generally carried out in polar protic solvents (like water, alcohol,

acetic acid, etc.). The reaction between tert-butyl bromide and hydroxide ion yields

tert-butyl alcohol and

follows the first order kinetics, i.e., the rate of reaction depends upon the

concentration of only one reactant, which is tert- butyl bromide.

It occurs in two steps. In step I, the polarised C—Br bond undergoes slow cleavage to

produce a carbocation and a bromide ion. The carbocation thus formed is then

attacked by nucleophile in step II to complete the substitution reaction.

Step 1 is the slowest and reversible. It involves the C–Br bond breaking for which the

energy is obtained through solvation of halide ion with the proton of protic solvent.

Since the rate of reaction depends upon the slowest step, the rate of reaction depends

only on the concentration of alkyl halide and not on the concentration of hydroxide

ion. Further, greater

the stability of carbocation, greater will be its ease of formation from alkyl halide and

faster

will be the rate of reaction. In case of alkyl halides, 30 alkyl halides undergo SN1

reaction

very fast because of the high stability of 30 carbocations.

For the same reasons, allylic and benzylic halides show high reactivity towards the SN1

reaction. The carbocation thus formed gets stabilised through resonance as shown

below:

Difference between SN1 and SN2

SN1 SN2

The rate of reaction is

unimolecular.

The rate of reaction is bimolecular

It follows the first order kinetics It follows the second order kinetics

It is a twostep mechanism It is only a one step mechanism

Carbocation is formed as an

intermediate part of the

reaction.

No carbocation is formed during the

reaction.

There is no partial bond formed

with the carbon during this

reaction.

Carbon forms a partial bond with the

nucleophile and the leaving group.

There are many steps in this

reaction which start with the

removal of the group while

attacking the nucleophile.

The process takes place in only one

cycle, with a single intermediate

stage.

the rate of reaction depends

only on the concentration of

substrate and not on the

concentration of nucleophile

the rate of reaction depends on the

concentration of both the substrate

and the nucleophile

Some basic stereochemical principles and notations

1)Optical Activity/Optically Active Substances: -

When a plane polarised light is made to pass through certain substances, the plane of

polarisation of the emergent light is not the same as that of incident light, but it has

been rotated through some angle. This phenomenon is known as optical activity. The

substances which rotate the plane of polarisation are said to be optically active.

Examples: quartz, sugar crystals, turpentine oil, sodium chloride etc.

Optically active substances are of two types,

Dextro-rotatory (right-handed)

Laevo - rotatory (left-handed)

Dextro-rotatory (right-handed) which rotate the plane of polarisation in the clock wise

direction on looking towards the source.

Laevo - rotatory (left-handed) which rotate the plane of polarisation in the

anti-clockwise direction on looking towards the source.

2) Chirality

Chirality is defined as “an object which is asymmetric and cannot be superimposed

over its mirror image is known as chiral or stereo enter”. This property is known as

chirality.

For example- our Hand, legs etc.

The object which is symmetric in nature and can be superimposed over its mirror

image is known as achiral.

For example- cube, cone etc.

chirality

Optically active substances are of two types,

handed)

handed)

handed) which rotate the plane of polarisation in the clock wise

direction on looking towards the source.

handed) which rotate the plane of polarisation in the

clockwise direction on looking towards the source.

“an object which is asymmetric and cannot be superimposed

is known as chiral or stereo enter”. This property is known as

our Hand, legs etc.


handed) which rotate the plane of polarisation in the clock wise

handed) which rotate the plane of polarisation in the

“an object which is asymmetric and cannot be superimposed

is known as chiral or stereo enter”. This property is known as


Polarimeter:

The angle by which the plane polarised light is rotated is measured by an instrument

called polarimeter.

Enantiomers:

Enantiomers are a pair of molecules that exist in two forms that are mirror images of

one another but cannot be superimposed one upon the other.

Enantiomers possess identical physical properties namely, melting point, boiling point,

solubility, refractive index, etc. They only differ with respect to the rotation of plane

polarised light. If one of the enantiomers is dextro rotatory, the other will be laevo

rotatory.

Racemic mixture:

A mixture containing two enantiomers in equal proportions will have zero optical

rotation, as the rotation due to one isomer will be cancelled by the rotation due to t

other isomer. Such a mixture is known as racemic mixture or racemic modification. A

racemic mixture is represented by prefixing dl or (±) before the name, for example (±)

butan-2-ol. The process of conversion of enantiomer into a racemic mixture is know

as racemisation.

Retention: If during a reaction, no bond to the stereo centre is broken, the product will

have the same general configuration of groups around the stereo centre as that of

reactant. Such a reaction is said to proceed with retention of



er but cannot be superimposed one upon the other.



the enantiomers is dextro rotatory, the other will be laevo


rotation, as the rotation due to one isomer will be cancelled by the rotation due to t



ol. The process of conversion of enantiomer into a racemic mixture is know

If during a reaction, no bond to the stereo centre is broken, the product will


reactant. Such a reaction is said to proceed with retention of the configuration.





the enantiomers is dextro rotatory, the other will be laevo


rotation, as the rotation due to one isomer will be cancelled by the rotation due to the



ol. The process of conversion of enantiomer into a racemic mixture is known

If during a reaction, no bond to the stereo centre is broken, the product will


the configuration.

Inversion of configuration: A process in which the relative configuration of an atom is

changed. Since a molecule can form two enantiomers around a chiral centre, the

Walden inversion converts the configuration of the molecule from one enantiomeric

form to another.

If (A) is the only compound obtained, the process is called retention of configuration.

If (B) is the only compound obtained, the process is called inversion of configuration.

If a 50:50 mixture of the above two is obtained then the process is called racemisation

and the product is optically inactive, as one isomer will rotate light in the direction

opposite to another.

2. Elimination reactions:

When a haloalkane with β-hydrogen atom is heated with alcoholic solution of

potassium hydroxide, there is elimination of hydro

halogen atom from the α-carbon atom. As a result, an alkene is formed as a product.

Since β-hydrogen atom is involved in elimination, it is often called β

A process in which the relative configuration of an atom is


converts the configuration of the molecule from one enantiomeric

nly compound obtained, the process is called retention of configuration.



he product is optically inactive, as one isomer will rotate light in the direction

hydrogen atom is heated with alcoholic solution of

potassium hydroxide, there is elimination of hydrogen atom from β

carbon atom. As a result, an alkene is formed as a product.

hydrogen atom is involved in elimination, it is often called β

A process in which the relative configuration of an atom is


converts the configuration of the molecule from one enantiomeric

nly compound obtained, the process is called retention of configuration.



he product is optically inactive, as one isomer will rotate light in the direction

hydrogen atom is heated with alcoholic solution of

gen atom from β-carbon and a

carbon atom. As a result, an alkene is formed as a product.

hydrogen atom is involved in elimination, it is often called β-elimination.

Alexander Zaitsev (also pronounced as

can be summarised as “in dehydrohalogenation reactions, the preferred product is

that alkene which has the greater number of alkyl groups attached to the doubly

bonded carbon atoms.”

3. Reaction with metals Most organic chlorides, bromides and iodides react with

certain metals to give compounds containing carbon

are known as organo-metallic compounds

Grignard Reagents: Important class of organo

Victor Grignard in 1900 is alkyl magnesium halide, RMgX, referred as

Reagents.

These reagents are obtained by the reaction of haloalkanes with magnesium metal in

dry ether.

Wurtz reaction

Alkyl halides react with sodium in dry ether to give hydro

the number of carbon atoms present in the halide. This reaction is

known as Wurtz reaction.

1. Nucleophilic substitution:

(also pronounced as Saytzeff) who in 1875 formulated a rule which

“in dehydrohalogenation reactions, the preferred product is


Most organic chlorides, bromides and iodides react with

certain metals to give compounds containing carbon-metal bonds. Such compounds

metallic compounds.

Important class of organo-metallic compounds discovered by

ctor Grignard in 1900 is alkyl magnesium halide, RMgX, referred as


Alkyl halides react with sodium in dry ether to give hydrocarbons containing double

the number of carbon atoms present in the halide. This reaction is

Reactions of Haloarenes

75 formulated a rule which

“in dehydrohalogenation reactions, the preferred product is


Most organic chlorides, bromides and iodides react with

metal bonds. Such compounds

metallic compounds discovered by

ctor Grignard in 1900 is alkyl magnesium halide, RMgX, referred as Grignard


carbons containing double

Aryl halides are extremely less reactive towards nucleophilic substitution reactions

due to the following reasons:

(i) Resonance effect: In haloarenes, the electron pairs on halogen atom are in

conjugation with π-electrons of the ring and the following resona

are possible. C—Cl bond acquires a partial double bond character due to

resonance. As a result, the bond cleavage in haloarene is difficult than

haloalkane and therefore, they are less reactive towards nucleophilic

substitution reaction.

(ii) Difference in hybridisation of carbon atom in C

In haloalkane, the carbon atom attached to halogen is sp3 hybridised while in case of

haloarene, the carbon atom attached to halogen is sp2

The sp2 hybridised carbon with a greater s

hold the electron pair of C—X bond more tightly than sp3

haloalkane with less s-character. Thus, C

in haloarene is 169 pm. Since it is difficult to b

therefore, haloarenes are less reactive than haloalkanes towards nucleophilic


(iii) Instability of phenyl cation:

In case of haloarenes, the phenyl cation formed as a result of self

be stabilised by resonance and therefore, SN1 mechanism is ruled out.

(iv) Because of the possible repulsion

to approach electron rich arenes.

2. Electrophilic substitution reactions

(i) Halogenation


due to the following reasons:

In haloarenes, the electron pairs on halogen atom are in

electrons of the ring and the following resona

Cl bond acquires a partial double bond character due to




Difference in hybridisation of carbon atom in C—X bond:


haloarene, the carbon atom attached to halogen is sp2-hybridised.

The sp2 hybridised carbon with a greater s-character is more electronegative and can

X bond more tightly than sp3-hybridised carbon in

character. Thus, C—Cl bond length in haloalkane is 177pm while

in haloarene is 169 pm. Since it is difficult to break a shorter bond than a longer bond,


Instability of phenyl cation:

In case of haloarenes, the phenyl cation formed as a result of self-ionisation will n


(iv) Because of the possible repulsion, it is less likely for the electron rich nucleophile

to approach electron rich arenes.

2. Electrophilic substitution reactions


In haloarenes, the electron pairs on halogen atom are in

electrons of the ring and the following resonating structures

Cl bond acquires a partial double bond character due to




character is more electronegative and can

hybridised carbon in

Cl bond length in haloalkane is 177pm while

reak a shorter bond than a longer bond,


ionisation will not


, it is less likely for the electron rich nucleophile

(ii) Nitration

(iii) Sulphonation

(iv) Friedel-Crafts reaction

3. Reaction with metals

Wurtz-Fittig reaction A mixture of an alkyl halide and aryl halide gives an alkylarene

when treated with sodium in dry ether and is

A mixture of an alkyl halide and aryl halide gives an alkylarene

when treated with sodium in dry ether and is

A mixture of an alkyl halide and aryl halide gives an alkylarene

called Wurtz-Fittig reaction.

Fittig reaction

Aryl halides also give analogous compounds when treated with sodium in dry ether, in

which two aryl groups are joined together. It is called Fittig reaction.

NCERT INTEXT QUESTIONS:

Page no. 285

Answer 10.1


aryl groups are joined together. It is called Fittig reaction.


aryl groups are joined together. It is called Fittig reaction.

Page no. 289

Answer: 10.2

H2SO4 is an oxidising agent. It oxidises HI produced during the reaction to I

prevents the reaction between an alcohol and HI to form alkyl iodide. To prevent this,

a nonoxidizing acid like H3PO3

Answer:10.3

Four isomers are possible. These are:

is an oxidising agent. It oxidises HI produced during the reaction to I

reaction between an alcohol and HI to form alkyl iodide. To prevent this,

a nonoxidizing acid like H3PO3 is used.

Four isomers are possible. These are:

is an oxidising agent. It oxidises HI produced during the reaction to I2 and thus

reaction between an alcohol and HI to form alkyl iodide. To prevent this,

Answer:10.4

Answer:10.5

Page No. 291

Answer:10.6

(i) Chloromethane < Bromomethane < Dibromo methane < Bromoform

The reason is:

(a)for same alkyl group, BoilingPoint increases with size of halogen atom.

(b)Boiling Point increases as number of halogen atoms increase.

(ii) Isopropyl chloride < 1 – Chloropropane < 1 – Chlorobutane

Reason:

(a)For same halogen, Boiling Point increases as size of alkyl group increases.

(b)Boiling Point decreases as branching increases.

Page No.307.

Answer: 10.7

In SN2 mechanism, reactivity depends upon the steric hindrance around the C-atom

carrying the halogen. Lesser the steric hindrance, faster the reaction.

Answer:10.8

Answer: 10.9

NCERT EXRECISES

Answer:10.1

(i) 2-Chloro-3methylbutane, 2° alkyl halide

(ii) 3-Chloro-4methyl hexane, 2° alkyl halide

(iii) 1 -Iodo-2,2-dimethylbutane, 1 ° alkyl halide

(iv) l-Bromo-3, 3-dimethyl -1-phenylbutane, 2° benzylic halide

(v) 2-Bromo-3-methylbutane, 2° alkyl halide

(vi) 1-Bromo-2-ethyI-2-methylbutane, 1° alkyl halide

(vii)3-Chloro-3-methylpentane, 3° alkyl halide

(viii) 3-Chloro-5-methylhex-2-ene, vinylic halide

(ix)4-Bromo-4-methylpent-2-ene, allylic halide

(x)1-Chloro-4-(2-methylpropyl) benzene, aryl halide

(xi)1-Chloromethyl-3- (2,2-dimethylpropyl) benzene, 1 ° benzylic halide.

(xii)1-Bromo-2-(l-methyl propyl) benzene, aryl halide.

Answer:10.2

(i) 2-Bromo-3-chlorobutane

(ii) 1 Bromo-1 -chloro-1,2,2-trifluoroethane

(iii) l-Bromo-4-chlorobut-2-ene

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

(v)2-Bromo-3,3-bis-(4-chlorophenyl) butane

(vi)l-Chloro-l-(4-iodophenyl)-3,3- dimethyl but-l-ene.

Answer:10.3

Answer: 10.4

The three-dimensional structures of the three compounds along with the direction of

dipole moment in each of their bonds are given below:

CCl4 being symmetrical has zero dipole moment. In CHCl3, the resultant of two C – Cl

dipole moments is opposed by the resultant of C – H and C – Cl bonds. Since the dipole

moment of latter resultant is expected to be smaller than the former, CHCl3 has a finite

dipole (1.03 D) moment.

In CH2CI2, the resultant of two C – Cl dipole moments is reinforced by resultant of two

C – H dipoles, therefore, CH2CI2 (1 .62 D) has a dipole moment higher than that of

CHCl3. Thus, CH2CI2 has highest dipole moment.

Answer:10.5

The hydrocarbon with molecular formula C5H10 can either a cycloalkane or an alkene.

Since the compound does not react with Cl2 in the dark, therefore it cannot be an

alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl2 in the presence

of bright sunlight to give a single monochloro compound, C5H9Cl, therefore, all the ten

hydrogen atoms of the cycloalkanes must be equivalent. Thus, the cycloalkane is

cyclopentane.

Answer:10.6

Answer: 10.7

Answer: 10.8

Nucleophiles which can attack through two different sites are called ambident

nucleophiles. For example, cyanide ion is a resonance hybrid of the following two

structures:





It can attack through carbon to form cyanide and through N to form is isocyanide.

Answer: 10.9

(i)Since I– ion is a better leaving group than Br

in SN2 reaction with OH– ion.

(ii)On steric grounds, 1° alkyl halides are more reactive than tert

reactions. Therefore, CH3CI will react at a faster rate than (CH

with OH– ion.

Answer:10.10


ion is a better leaving group than Br- ion, therefore, CH3I reacts faster CH

(ii)On steric grounds, 1° alkyl halides are more reactive than tert-alkyl halides in SN

CI will react at a faster rate than (CH3)3CCl in a SN


I reacts faster CH3Br

alkyl halides in SN2

CCl in a SN2 reaction

Answer: 10.11

Answer: 10.12

(i)

The Cl- atom in chlorobenzene is linked to sp2

cyclohexyl chloride it is linked to sp3

carbon has more s-character, it is more electronegative than sp3

atom. Hence, the density of electrons of C

chlorobenzene than in cyclohexyl chloride.

Moreover, the electron density is reduced by the

chlorobenzene. Hence, chlorobenzene has a decreased polarity of C

dipole moment is lower than that of cyclohexyl chloride.

(ii)Alkyl halides are polar molecules; therefore, their molecules are held together by

dipole-dipole attraction. The molecules of H2O are hold together by H

new forces of attraction between water and alkyl halide molecules are weaker than

the forces of attraction already existing between alkyl halide

and water-water molecules, therefore, alkyl halides are immiscible (not soluble) in

water. Alkyl halide are neither able to form H

the H-bounding network of water.

(iii)Grignard reagents are very reactive. They react with moistu

apparatus to form alkanes

Thus, Grignard reagents must be prepared under anhydrous conditions.

Answer: 10.14

atom in chlorobenzene is linked to sp2 hybridized carbon atom, whereas in

cyclohexyl chloride it is linked to sp3 hybridized carbon atom. Now, as sp2

character, it is more electronegative than sp3

atom. Hence, the density of electrons of C-Cl bond near the Cl atom is less in

chlorobenzene than in cyclohexyl chloride.

Moreover, the electron density is reduced by the –R effect of the b

chlorobenzene. Hence, chlorobenzene has a decreased polarity of C

dipole moment is lower than that of cyclohexyl chloride.


attraction. The molecules of H2O are hold together by H


the forces of attraction already existing between alkyl halide – alkyl halide molecules

ter molecules, therefore, alkyl halides are immiscible (not soluble) in

water. Alkyl halide are neither able to form H- bonds with water nor are able to break

bounding network of water.

(iii)Grignard reagents are very reactive. They react with moistu



hybridized carbon atom, whereas in

hybridized carbon atom. Now, as sp2 hybridized

character, it is more electronegative than sp3 hybridized carbon

Cl bond near the Cl atom is less in

R effect of the benzene ring of

chlorobenzene. Hence, chlorobenzene has a decreased polarity of C-Cl bond, also the


attraction. The molecules of H2O are hold together by H-bonds. Since the


alkyl halide molecules

ter molecules, therefore, alkyl halides are immiscible (not soluble) in

bonds with water nor are able to break

(iii)Grignard reagents are very reactive. They react with moisture present in the



Answer: 10.15

KCN is a resonance hybrid of the following two contributing structures:

Thus, CN– ion is an ambident nucleophile. Therefore, it can attack the “carbon atom of

C-Br bond in n-BuBr either through C or N. Since C – C bond is stronger than C – N

bond, therefore, attack occurs through C to form n

Answer:10.16

The SN2 reactions reactivity depends upon

slower the reaction. Thus, the order of reactivity will be 1°> 2° >3°

Since in case of 1° alkyl halides steric hindrance increases in the order) n

alkyl halides with a substituent at any positio

substituent at the β-position, two substituents at the β

reactivity decreases in the same order. Thus, the reactivity of the given alkyl bromides

decreases in the order:

1-Bromobutane > l-Bromo-3-me

dimethyl propane.

bond, therefore, attack occurs through C to form n-butyl cyanide.

The SN2 reactions reactivity depends upon steric hindrance. More the steric hindrance

slower the reaction. Thus, the order of reactivity will be 1°> 2° >3°

Since in case of 1° alkyl halides steric hindrance increases in the order) n

alkyl halides with a substituent at any position other than the β-position, one

position, two substituents at the β-position, therefore, the


methylbutane > l-Bromo-2-methyjbutane> 1

steric hindrance. More the steric hindrance

Since in case of 1° alkyl halides steric hindrance increases in the order) n-alkyl halides,

position, one

position, therefore, the


methyjbutane> 1-Bromo-2,2-

Answer: 10.17

C6H5CH2Cl is 10 aryl halides while C6H5CH(CI)C6H5 is a 2° aryl halide. In SN1 reactions, the

reactivity depends upon the stability of carbocations.

Since the C6H5CHC6H5carbocation is more stable than C6H5CH2 carbocation, therefore,

C6H5CHCIC6H5 gets hydrolysed more easily than C6H5CH2Cl under SN1 conditions.

However, under SN2 conditions, the reactivity depends on steric hindrance, therefore,

under SN2conditions, C6H5CH2Cl gets hydrolysed more easily than C6H5CHClC6H5.

Answer: 10.18

The p-isomer being more symmetrical fits closely in the crystal lattice and thus has

stronger inter-

molecular forces of attraction than o- and m-isomers. Since during melting or

dissolution, the crystal lattice breaks, therefore, a large amount of energy is needed to

melt or dissolve the p-isomer than the corresponding o-and m-isomers. In other

words, the melting point of the p-isomer is higher and its solubility lower than the

corresponding o-and m-isomers.

Answer:10.19

Answer: 10.20

If aqueous solution, KOH is almost completely ionized to give OH– ions which being a

strong nucleophile brings about a substitution reaction on alkyl halides to form

alcohols. Further in the aqueous solution, OH– ions are highly solvated (hydrated). This

solvation reduces the basic character of OH

hydrogen from the P-carbon of the alkyl chloride to form alkenes. In contrast, an

alcoholic solution of KOH contai

base than OH– ions preferentially eliminate a molecule of HCl from an alkyl chloride to

form alkenes.

Answer: 10.21

There are two primary alkyl halides having the molecular formula, C4H9Br.

(ii) Since compound (a) when reacted with Na metal gave a compound (d) with

molecular formula C8H18 which was different from die compound obtained when n

butyl bromide was reacted with Na metal, therefore, (a) must be isobutyl bromide and

compound (d) must be 2,3-dimeth

(iii) If compound (a) is isobutyl bromide, than the compound (b) which it gives on

treatment with alcoholic KOH must be 2

(iv) The compound (b) on treatment with HBr gives compound (c) in accordance with

Markownikov’s rule. Therefore, compound (c) is tert

of compound (a) ,i.e., isobutyl ‘ bromide

solvation reduces the basic character of OH– ions which, therefore, fails to abstract a

carbon of the alkyl chloride to form alkenes. In contrast, an

alcoholic solution of KOH contains alkoxide (RO–) ion which being a much stronger

ions preferentially eliminate a molecule of HCl from an alkyl chloride to


pound (a) when reacted with Na metal gave a compound (d) with

which was different from die compound obtained when n


dimethylhexane.


treatment with alcoholic KOH must be 2-methyl-1-propane.


rule. Therefore, compound (c) is tert-butyl bromide which is an isomer

of compound (a) ,i.e., isobutyl ‘ bromide.

ions which, therefore, fails to abstract a

carbon of the alkyl chloride to form alkenes. In contrast, an

) ion which being a much stronger

ions preferentially eliminate a molecule of HCl from an alkyl chloride to


pound (a) when reacted with Na metal gave a compound (d) with

which was different from die compound obtained when n-




butyl bromide which is an isomer

Thus

(a)is isobutyl bromide,

(b)is 2-methyl-1 -propane,

(c)is tert-butyl bromide, and

(d)is 2,5-dimethylhexane.

Answer: 10.22

Class- XII Chemistry Chapter-10 Haloalkane and...

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