Chemistry

9. COORDINATION COMPOUNDS

9.1 Introduction : A coordination compound ligands depending on the number of electron
consists of central metal atom or ion surrounded donor atoms they have.
by atoms or molecules. For example, a
chemotherapy drug, cisplatin, Pt(NH3)2Cl2, is 9.2.1 Monodentate ligands : A monodentate
a coordination compound in which the central ligand is the one where a single donor atom
platinum metal ion is surrounded by two shares an electron pair to form a coordinate
ammonia molecules and two chloride ions. The bond with the central metal ion. For example:
species surrounding the central metal atom or the ligands Cl , OH or CN attached to metal
ion are called ligands. The ligands are linked have electron pair on Cl, O and N, respectively
directly to central metal ion through coordinate which are donor atoms :
bonds. A formation of coordinate bond occurs
when the shared electron pair is contributed Cl O - H C ≡ N
by ligands. A coordinate bond is conveniently
represented by an arrow →, where the arrow Use your brain power
head points to electron acceptor. The central Draw Lewis structure of the
metal atom or ion usually an electron deficient following ligands and identify the
species, accepts an electron pair while the donor atom in them : NH3, H2O
ligands serve as electron donors. Coordination
compounds having a metal ion in the centre are 9.2.2 Polydentate ligands : A polydentate
common. In cisplatin two ammonia molecules ligand has two or more donor atoms linked
and two chloride ligands utilize their lone pairs to the central metal ion. For example,
of electrons to form bonds with the Pt(II). ethylenediamine and oxalate ion. Each of these
ligands possesses two donor atoms. These are
bidentate ligands.

Cl NPHt 3 NH3 i. Ethylenediamine binds to metal using
Cl electron pair on each of its two nitrogens.

The donor nitrogen and chlorine atoms of H2N CH2 - CH2 NH2
the ligands are directly attached to and form Similarly oxalate ion (C2O4)2 utilizes
bonds with platinum. electron pair on each of its negatively charged
oxygen atoms upon linking with the metal.
Can you recall ?

What are Lewis acids and bases ? OO
C-C
Formation of a coordination compound
can be looked upon as the Lewis acid-base OO
interaction. The ligands being electron pair
donors are Lewis bases. The central metal ion ii. Ethylenediaminetetraacetate ion
being electron pair acceptor serves as Lewis (EDTA)4 binds to metal ion by electron pairs
acid. of four oxygen and two nitrogen atoms. It is a
hexadentate ligand.
9.2 Types of ligands : The ligands can
be classified as monodentate and polydentate OO

O - C - H2C N - CH2 - CH2 - N CH2 - C - O

O - C - H2C CH2 - C - O
O O

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9.2.3 Ambidentate ligand : The ligands which [Fe(CN)6]4 has charge number of -4. It can be
have two donor atoms and use the electron utilised to calculate O.S. of Fe. Thus,
pair of either donor atoms to form a coordinate
bond with the metal ion, are ambidentate charge number of complex = -4
ligands. For example, the ligand NO2 links to
metal ion through nitrogen or oxygen. = (O.S. of Fe + charge of ligands)

O = (O.S. of Fe + 6 × charge of CN ion)

MN or M O - N = O = (O.S. of Fe + 6 × (-1))

O Therefore, O.S. of Fe = -4 + 6 = +2.

Similarly, SCN has two donor atoms Can you tell ?
nitrogen and sulfur either of which links to
metal depicted as M ← SCN or M← NCS. A complex is made of Co(III)
and consists of four NH3 molecules
9.3 Terms used in coordination chemistry: and two Cl ions as ligands. What is the
The following terms are used for describing charge number and formula of complex ion ?
coordination compounds.
9.3.3 Coordination number (C.N.) of central
9.3.1 Coordination sphere : The central metal metal ion:Lookatthecomplex[Co(NH3)4Cl2]⊕.
ion and ligands linked to it are enclosed in a Here four ammonia molecules and two chloride
square bracket. This is called a coordination ions, that is, total six ligands are attached to
sphere, which is a discrete structural unit. When the cobalt ion. All these are monodentate since
the coordination sphere comprising central each has only one donor atom. There are
metal ion and the surrounding ligands together six donor atoms in the complex. Therefore,
carry a net charge, it is called the complex the coordination number of Co3⊕ ion in the
ion. The ionisable groups shown outside the complex is six. Thus, the coordination number
bracket are the counter ions. For example, of metal ion attached to monodentate ligands
the compound K4[Fe(CN)6] has [Fe(CN)6]4- is equal to number of ligands bound to it.
coordination sphere with the ionisable K⊕
ions representing counter ions. The compound Consider the bidentate ligand C2O42
ionizes as : or ethylenediamine (en). The complexes,
[Fe(C2O4)3]3 and [Co(en)3]3⊕, have three
K4[Fe(CN)6] 4K⊕ + [Fe(CN)6]4 bidentate ligands each. The total donor atoms
in three of ligands is six and the C.N. of Fe3⊕
Try this... and Co3⊕ in these complexes is six.

Can you write ionisation of C.N. of metal ion in a complex is the
[Ni(NH3)6]Cl2? Identify coordination number of ligand donor atoms directly
sphere and counter ions. attached to it or the number of electron
pairs involved in the coordinate bond.
9.3.2 Charge number of complex ion and
oxidation state of metal ion : Use your brain power
Coordination number used
The net charge residing on the complex
ion is its charge number. It is algebraic sum in coordination of compounds in
of the charges carried by the metal ion and somewhat different than that used in
the ligands. The charge carried by the metal solid state. Explain
ion is its oxidation state (O.S.). The complex

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Can you tell ? The secondary valencies for a metal ion are
fixed and satisfied by either anions or neutral
What is the coordination ligands. Number of secondary valencies is
number of Co in [CoCl2(en)2]⊕, equal to the coordination number.
of Ir in [Ir(C2O4)2Cl2]3⊕ and of Pt in
[Pt(NO2)2(NH3)2]? Postulate (iv) The secondary valencies have
a fixed spatial arrangement around the metal
9.3.4 Double salt and coordination complex ion. Two spheres of attraction in the complex
[Co(NH3)6]Cl3 are shown.
Combination of two or more stable
compounds in stochiometric ratio can give two H3N NH3 NH3 Cl3
types of substances, namely, double salt and H3N Co
coordination complexes.
NH3 NH3
Double salt : A double salt dissociates in water

completely into simple ions. For example (i)

Mohr's salt, FeSO4(NH4)2SO4.6H2O dissociates coordination ionization
as :

FeSO4(NH4)2SO4.6H2O water Fe2⊕(aq) (secondary) sphere (primary) sphere

+ 2NH4⊕(aq) +2SO42 (aq) Remember...

ii. Carnalite KCl.MgCl2.6H2O dissociates as When a complex is brought
into solution it does not dissociate
KCl.MgCl2.6H2O water K⊕(aq) + into simple metal ions. When [Co(NH3)6]
Mg2⊕ (aq) + 3Cl (aq) Cl3 is dissolved in water it does not give the
test for Co3⊕ or NH3. However, on reacting
Coordination complex : A coordination with AgNO3 a curdy white precipitate of
complex dissociates in water with at least AgCl corresponding to 3 moles is observed.
one complex ion. For example, K4[Fe(CN)6]
dissociates as the complex ion and counter ion.

K4[Fe(CN)6] 4K⊕(aq) + [Fe(CN)6]4 Problem 9.1 : One mole of a purple coloured
(counter ion) (complex ion) complex CoCl3 and NH3 on treatment with
excess AgNO3 produces two moles AgCl.
9.3.5 Werner theory of coordination Write the formula of the complex if the
complexes : The first attempt to explain nature coordination number of Co is 6.
of bonding in coordination compounds was
put forth by Werner. The postulates of Werner Solution : One mole of the complex gives
theory are as follows. 2 moles of AgCl. It indicates that two Cl
ions react with Ag⊕ ions. The complex has
Postulate (i) Unlike metal salts, the metal in two ionisable Cl ions. The formula of the
a complex possesses two types of valencies complex is then [Co(NH3)5Cl]Cl2.
: primary (ionizable) valency and secondary
(nonionizable) valency. Can you tell ?
One mole of a green coloured
Postulate (ii) The ionizable sphere consists complex of CoCl3 and NH3 on
of entities which satisfy the primary valency treatment with excess of AgNO3
of the metal. Primary valencies are generally produces 1 mole of AgCl. What is the
satisfied by anions. formula of the complex ? (Given : C.N. of
Co is 6)
Postulate (iii) The secondary coordination
sphere consists of entities which satisfy the
secondary valencies and are non ionizable.

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A complex with coordination number six has anionic sphere complex. For example,
octahedral structure. When four coordinating [Ni(CN)4]2 and K3[Fe(CN)6] have anionic
groups are attached to the metal ion the coordination sphere; [Fe(CN)6]3 and
complex would either be with square planar three K⊕ ions make the latter electrically
or tetrahedral structure. neutral.
iii. Neutral sphere complexes : A neutral
9.4 Classification of complexes: The coordination complex does not possess cationic
coordination complexes are classified or anionic sphere. [Pt(NH3)2Cl2] or [Ni(CO)4]
according to types of ligands and sign of have neither cation nor anion but are neutral
charge on the complex ion. sphere complexes.

9.4.1 Classification on the basis of types of Use your brain power
ligands
Classify the complexes
i. Homoleptic complexes : Consider
[Co(NH3)6]3⊕. Here only one type of ligands as cationic, anionic or neutral.
surrounds the Co3⊕ ion. The complexes in
which metal ion is bound to only one type of Na4[Fe(CN)6], Co(NH3)6Cl2,
ligands are homoleptic. Cr(H2O)2(C2O4)23 , PtCl2(en)2 and
Cr(CO)6.
ii. Heteroleptic complexes : Look at the
complex [Co(NH3)4Cl2]⊕. There are two types 9.5 IUPAC nomenclature of coordination
of ligands, NH3 and Cl attached to Co3⊕ compounds : Tables 9.1, 9.2 and 9.3 summarize
ion. Such complexes in which metal ion is the IUPAC nomenclature of coordination
surrounded by more than one type of ligands compounds.
are heteroleptic.
Rules for naming coordination compounds
Use your brain power recommended by IUPAC are as follows:

Classify the complexes 1. In naming the complex ion or neutral
as homoleptic and heteroleptic molecule, name the ligand first and then the
[Co(NH3)5Cl]SO4, [Co(ONO)(NH3)5]Cl2 metal.
[CoCl(NH3)(en)2]2⊕ and [Cu(C2O4)3]3 .
2. The names of anionic ligands are obtained
9.4.2 Classification on the basis of charge on by changing the ending -ide to -o and -ate to
the complex -ato.

i. Cationic complexes : A positively charged 3. The name of a complex is one single word.
coordination sphere or a coordination There must not be any space between
compound having a positively charged different ligand names as well as between
coordination sphere is called cationic sphere ligand name and the name of the metal.
complex.
For example: the cation [Zn(NH3)4]2⊕ and 4. After the name of the metal, write its
[Co(NH3)5Cl]SO4 are cationic complexes. The oxidation state in Roman number which
latter has coordination sphere [Co(NH3)5Cl]2⊕; appears in parentheses without any space
the anion SO42 makes it electrically neutral. between metal name and parentheses.
ii. Anionic sphere complexes : A
negatively charged coordination sphere or a 5. If complex has more than one ligand of the
coordination compound having negatively same type, the number is indicated with
charged coordination sphere is called prefixes, di-, tri-, tetra-, penta-, hexa- and
so on.

6. For the complex having more than one type
of ligands, they are written in an alphabetical
order. Suppose two ligands with prefixes

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Table 9.1 : IUPAC names of anionic and neutral ligands

Anionic ligand IUPAC name Anionic ligand IUPAC name

Br , Bromide Bromo CO32 , Carbonate Carbonato
Cl , Chloride Chloro OH , Hydroxide Hydroxo

F , Fluoride Fluoro C2O42 , Oxalate Oxalato
I Iodide Iodo NO2 , Nitrite
Cyano ONO , Nitrite Nitro (For N - bonded ligand)
CN , Cyanide
SO42 , Sulphate Sulphato SCN , Thiocyanate Nitrito(For O-bonded ligand)

NO3 , Nitro Nitrato NCS , Thiocyanate Thiocyanato (For ligand do-
Neutral ligand nor atom S)
IUPAC name Neutral ligand Isothiocyanato (For ligand
NH3, Ammonia Ammine (Note the
CO, Carbon monoxide donor atom N)
spelling) IUPAC names
Cabonyl
H2O, water Aqua
en, Ethylene diamine Ethylenediamine

Table 9.2 : IUPAC names of metals in anionic complexes

Metal IUPAC name Metal IUPAC name

Aluminium, Al Aluminate Chromium, Cr Chromate

Cobalt, Co Cobaltate Copper, Cu Cuprate

Gold, Au Aurate Iron, Fe Ferrate

Manganese, Mn Maganate Nickel, Ni Nickelate

Platinum, Pt Platinate Zinc, Zn Zincate

Table 9.3 : IUPAC names of some complexes

Complex IUPAC name

i. Anionic complexes :

a.[Ni(CN)4]2 Tetracyanonickelate(II) ion
b. [Co(C2O4)3]3 Trioxalatocobaltate(III) ion
c. [Fe(CN)6]4 Hexacyanoferrate(II) ion

ii. Compounds containing complex anions and metal cations :

a. Na3[Co(NO2)6] Sodium hexanitrocobaltate(III)
b. K3[Al(C2O4)3] Potasium trioxalatoaluminate(III)
c. Na3[AlF6] Sodium hexafluoroaluminate(III)
iii. Cationic complexes :
a. Cu(NH3)42⊕ Tetraamminecopper(II) ion
b.[Fe(H2O)5(NCS)]2⊕ Pentaaquaisothiocyanatoiron(III) ion,
c.[Pt(en)2(SCN)2]2⊕ Bis(ethylenediamine)dithiocyanatoplatinum(IV).

iv. Compounds containing complex cations and anion : Tetraamminedibromoplatinum(IV) bromide,
a. [PtBr2(NH3)4]Br2 Pentaamminecarbonatocobalt(III) chloride,
b. [Co(NH3)5CO3]Cl Pentaammineaquacobalt(III) iodide.
c. [Co(H2O)(NH3)5]I3
Triamminetrinitrocobalt(III)
v. Neutral complexes : Pentacarbonyliron(0)
a. [Co(NO2)3(NH3)3] Triamminetrithiocyanatorhodium(III)
b. Fe(CO)5
c. [Rh(NH3)3(SCN)3]

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are tetraaqua and dichloro. While naming rule states that a metal ion continues to accept
in alphabetical order, tetraaqua is first and electrons pairs till it attains the electronic
then dichloro. configuration of the next noble gas. Thus if the
7. If the name of ligand itself contains EAN is equal to 18 (Ar), 36 (Kr), 54 (Xe), or
numerical prefix then display number 86 (Rn) then the EAN rule is obeyed.
by prefixes with bis for 2, tris for 3, EAN can be calculated with the following.
tetrakis for 4 and so forth. Put the ligand Formula
name in parentheses. For example, EAN = number of electrons of metal ion + total
(ethylenediamine)3 or (en3) would appear
as tris(ethylenediamine) or tris(ethane-1, number of electrons donated by ligands
2-diamine). = atomic number of metal (Z) - number
8. The metal in cationic or neutral complex
is specified by its usual name while in the of electrons lost by metal to form the
anionic complex the name of metal ends ion (X) + number of electrons donated
with 'ate'. by ligands (Y).
=Z-X+Y
Try this... Cosider Co[NH3]63⊕
Oxidation state of Cobalt is +3, six ligands
Write the representation of donate 12 electrons.
Z = 27; X = 3; Y = 12
• Tricarbonatocobaltate(III) ion. EAN of Co3⊕ = 27 - 3 + 12 = 36.

• Sodium hexacyanoferrate(III). Try this... of

• Potassium hexa cyanoferrate (II) Find out the EAN
Zn(NH3)42⊕,[Fe(CN)6]4
• Aquachlorobis(ethylenediamine)
cobalt(III). Cr(CO)6 and [Fe(CN)6]4 are some examples
of coordination compounds which obey
• Tetraaquadichlorochromium(III) the EAN rule. Certain other coordination
chloride. compounds however do not obey the EAN
rule. For example, [Fe(CN)6]3 and Cu[NH3]42⊕
• Diamminedichloroplatinum(II). have EAN 35.

9.6 Effective Atomic Number (EAN) Rule : Use your brain power
An early attempt to explain the stability Do the following complexes follow
the EAN rule ? Cr(CO)4, Ni(CO)4,
of coordination compounds was made by Mn(CO)5, Fe(CO)5.
Sidgwick who proposed an empirical rule
known as effective atomic number (EAN) rule.
EAN equals total number of electrons around
the central metal ion in the complex. EAN

Isomers

Stereo isomers Constitutional or structural
isomers

Geometric or Optical isomers or
Distereoisomers enantiomers

Linkage isomers Coordination isomers Ionization isomers Solvate isomers

Fig. 9.1 : Classification of isomers in coordination compounds

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9.7 Isomerism in coordination compounds : Here the cis isomer is more soluble in
One of the interesting aspects of water than the trans isomer. The cis isomer
named cisplatin is an anticancer drug while
coordination chemistry is existence of isomers. the trans isomer is physiologically inactive.
Isomers are different compounds that have The cis isomer is polar with non-zero dipole
the same molecular formula. Their chemical moment. The trans isomer has zero dipole
reactivities and physical properties such moment as a result of the two opposite Pt - Cl
as colour, solubility and melting point are and two Pt-NH3 bond moments, which cancel
different. each other.

Broadly speaking, isomers are classified [Pt(NH3)(H2O)Cl2] (MA2BC type)
into two types namely stereoisomers and
constitutional (or structural) isomers as Cl H2O Cl H2O
displayed in Fig. 9.1.
Pt Pt
9.7.1 Stereoisomers : Stereoisomers have the
same links among constituent atoms however Cl NH3 H3N Cl
the arrangements of atoms in space are cis isomer trans isomer
different.
Four coordinate tetrahedral complexes do
not show cis and trans isomers.

There are two kinds of stereoisomers in iii. Cis and trans isomers in octahedral
coordination compounds: (a) geometric
isomers or distereoisomers and (b) enantiomers complexes : The octahedral complexes of
or optical isomers.
the type MA4B2, MA4BC and M(AA)2B2 exist
a. Geometric isomers or distereoisomers : as cis and trans isomers. (AA) is a bidentate
These are nonsuperimposable mirror image
isomers. These are possible in heteroleptic ligand.
complexes. In these isomers, there are cis and
trans types of arrangements of ligands. H3N[Co(CNCoHl 3)C4Cll2]⊕,⊕(MA4HB32NtypeCC)ol ⊕

Cis-isomers : Identical ligands occupy NH3
adjacent positions.
H3N NH3 NH3 H3N Cl NH3
cis isomer trans isomer

Trans-isomer : Identical ligands occupy the [Pt(NH3)4ClBr]2⊕, (MA4BC type)
opposite positions.

Cis and trans isomers have different NH3 2⊕ Cl 2⊕
properties. Cis trans isomerism is observed in Pt Pt
square planar and octahedral complexes. Cl NH3 H3N NH3

i. Cis and trans isomers in square planar Br NH3 NH3 H3N Br NH3
complexes : The square planar complexes of
MA2B2 and MA2BC type exist as cis and trans cis isomer trans isomer
isomers, where A, B and C are monodentate
ligands, M is metal. For example : Pt(NH3)2Cl2, [Co(en)2Cl2]⊕, (M(AA)2B2 type)
(MA2B2 type)
Cl Cl ⊕ Cl ⊕
Cl NH3 Cl NH3 en Co en Co en
Pt Pt
en Cl
Cl NH3 H3N Cl trans isomer
cis isomer cis isomer
trans isomer

198

Try this... Cl 2⊕ Cl Cl 2⊕

Draw structures of the cis and en Pt Cl
trans isomers of [Fe(NH3)2(CN)4]
Pt en

b. Optical isomers (Enantiomers) : en en
The complex molecules or ions that are d cis isomer l
nonsuperimposable mirror images of each
other are enantiomers. The nonsuperimposable Cl 2⊕
mirror images are chiral. (A more elaborate
discussion on chirality and optical isomerism en Pt en
is included in Chapter 10.)
Cl
trans isomer

Enantiomers have identical properties Square planar complexes do not show
however differ in their response to the plane- enantiomers since they have mirror plane and
polarized light. The enantiomer that rotates axis of symmetry.
the plane of plane-polarized light to right
(clockwise) is called the dextro (d) isomer, Try this...
while the other that rotates the plane to left 1. Draw enantiomers of
(anticlockwise) is called laevo (l) isomer. [Cr(ox)3]3

i. Optical iomers in octahedral complexes 2. Draw enantiomers and cis and trans
isomers of [Cr(H2O)2(ox)2] (where
[Co(en)3]3⊕ ox = C2O42 )
en 3⊕
en 3⊕ 9.7.2 Structural isomers (Constitutional
isomers) : Structural isomers possess different
en Co Co en linkages among their constituent atoms and
en have, their chemical formulae to be the same.
en
d They can be classified as linkage isomers,
mirror l ionization isomers, coordination isomers and
solvate isomers.
Remember...
a. Linkage isomers : These isomers are formed
Our hands are non when the ligand has two different donor atoms.
superimposable mirror images. It coordinates to the metal via different donor
When you hold your left hand upto a atoms. Thus the nitrite ion NO2 having two
mirror the image looks like right hand. donor atoms show isomers as :

ii. Octahedral complexes existing as both [Co(NH3)5(NO2)]2⊕ and [Co(NH3)5(ONO)]2⊕
geometric and optical isomers
The nitro complex has Co-N bond and the
[PtCl2(en)2]2⊕ nitrito complex is linked through Co-O bond.
In this type of complex, only the cis isomer These are linkage isomers.
exists as pair of enantiomers
Can you tell ?

Write linkage isomers of
[Fe(H2O)5SCN]⊕. Write their IUPAC
names.

199

b. Ionization isomers : Ionization isomers the free solvent molecule. I and II represent
involve exchange of ligands between solvate (hydrate) isomers.
coordination and ionization spheres. For
example: 9.8 Stability of the coordination compounds:

[Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 The stability of coordination compounds
(I) (II) can be explained by knowing their stability
constants. The stability is governed by metal-
In compound I, anion SO42 , bonded to ligand interactions. In this the metal serves
Co is in the coordination sphere while Br is as electron-pair acceptor while the ligand as
in the ionization sphere. In compound II, anion Lewis base (since it is electron donor). The
Br is in the coordination sphere linked to Co metal-ligand interaction can be realized as the
while SO42 is in the ionisation sphere. These Lewis acid-Lewis base interaction. Stronger
complexes in solution ionize to give different the interaction greater is stability of the
ions. complex.

[Co(NH3)5SO4]Br [Co(NH3)5SO4]⊕ + Br Consider the equilibrium for the metal
ligand interaction :
[Co(NH3)5Br]SO4 [Co(NH3)5Br]2⊕+ SO42
Ma⊕ + nLx [MLn]a⊕ + nx
I and II are examples of ionization isomers.
where a, x, [a⊕ + nx ] denote the charge on the
Can you tell ? metal, ligand and the complex, respectively.
Now, the equilibrium constant K is given by
Can you write IUPAC names of
isomers I and II? K = [MLn]a⊕ + nx
[Ma⊕][Lx ]n

c. Coordination isomers : Coordination Stability of the complex can be explained in
isomers show interchange of ligands between terms of K. Higher the value of K larger is the
cationic and anionic spheres of different metal thermodynamic stability of the complex.
ions. For example :
The equilibria for the complex formation
[Co(NH3)6] [Cr(CN)6] [Cr(NH3)6] [Co(CN)6] with the corresponding K values are given
(cationic) (anionic) (cationic) (anionic) below.

(I) (II) Ag⊕ + 2CN [Ag(CN)2] K = 5.5 ×1018

In isomer I, cobalt is linked to ammine Cu2⊕ + 4CN [Cu(CN)4]2 K = 2.0 ×1027
ligand and chromium to cyanide ligand. In
isomer II the ligands coordinating to metals Co3⊕ + 6NH3 [Co(NH3)6]3⊕ K = 5.0 ×1033
are interchanged. Cobalt coordinates with
cyanide ligand and chromium to NH3 ligand. From the above data, [Co(NH3)6]3⊕ is more
I and II are examples of coordination isomers. stable than [Ag(CN)2] and [Cu(CN)4]2 .

d. Solvate isomers (Hydrate isomers when 9.8.1 Factors which govern stability of the
water is solvent) : These are similar to
ionization isomers. Look at the complexes. complex : Stability of a complex is governed

[Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2 .H2O by (a) charge to size ratio of the metal ion and
(I) (II)
In compound I the solvent water is directly (b) nature of the ligand.

bonded to Cr. In compound II, H2O appears as a. charge to size ratio of the metal ion

Higher the ratio greater is the stability.
For the divalent metal ion complexes their
stability shows the trend : Cu2⊕ > Ni2⊕ > Co2⊕ >
Fe2⊕ > Mn2⊕ > Cd2⊕. The above stability order

200

is called Irving-William order. In the above list iii. The number of vacant hybrid orbitals
both Cu and Cd have the charge +2, however, formed is equal to the number of ligand
the ionic radius of Cu2⊕ is 69 pm and that of donor atoms surrounding the metal ion
Cd2⊕ is 97 pm. The charge to size ratio of Cu2⊕ which equals the coordination number of
is greater than that of Cd2⊕ . Therefore the Cu2⊕ metal.
forms stable complexes than Cd2⊕.
iv. Overlap between the vacant hybrid
b. Nature of the ligand. orbitals of the metal and filled orbitals of
the ligand leads to formation of the metal-
A second factor that governs stability ligand coordinate bonds.
of the complexes is related to how easily the
ligand can donate its lone pair of electrons to v. The hybrid orbitals used by the metal ion
the central metal ion that is, the basicity of the point in the direction of the ligand.
ligand. The ligands those are stronger bases
tend to form more stable complexes. vi. The (n-1)d or nd orbitals used in
hybridisation allow the complexes to be
Use your brain power classified as (a) outer orbital and (b) inner
orbital complexes.

The stability constant K of the vii. For hybridisation in the outer orbital
[Ag(CN)2] is 5.5 × 1018 while that complex nd orbitals are used, whereas
for the corresponding [Ag(NH3)2]⊕ is 1.6 × in the inner orbital complexes (n-1)d
107. Explain why [Ag(CN)2]2 is more stable. orbitals are used.

9.9 Theories of bonding in complexes : Type of hybridisation decides the structure
of the complex. For example when the
The metal-ligand bonding in coordination hybridisation is d2sp3 the structure is octahedral.
compounds has been described by Valence
bond theory (VBT) and Crystal field theory Steps to understand the metal-ligand bonding
(CFT). include :

i. Find oxidation state of central metal ion

9.9.1 Valence bond theory (VBT) ii. Writevalenceshellelectronicconfiguration

Can you recall ? of metal ion.
What is valence bond theory and
the concept of Hybridisation? iii. See whether the complex is low spin or
high spin. (applicable only for octahedral
complexes with d4 to d8 electronic

The hybridized state is a theoretical step that configurations).

describes how complexes are formed. VBT iv. From the number of ligands find the

is based on the concept of hybridization. The number of metal ion orbitals required for

hybrid orbitals neither exist nor can be detected bonding.

spectroscopically. These orbitals, however, v. Identify the orbitals of metal ion
help us to describe structure of coordination available for hybridisation and the type of
compounds. The steps involved in describing hybridisation involved.
bonding in coordination compounds using the
VBT are given below. vi. Write the electronic configuration after
i. Metal ion provides vacant d orbitals hybridisation.

for formation of coordinate bonds with vii Show filling of orbitals after complex
ligands. formation.

ii. The vacant d orbitals along with s and viii.Determine the number of unpaired
p orbitals of the metal ion take part in electrons and predict magnetic behaviour
hybridisation. of the complex.

201

Remember... vi. Six orbitals available for hybridisation
are two 3d, one 4s, three 4p orbitals
Complete the missing entries.
3d 4s 4p
Coordination Geometry Hybridization

number of complex

2 sp d2sp3

4 Tetrahedral The orbitals for hybridization are decided
from the number of ammine ligands which
4 Square is six. Here (n-1)d orbitals participate in
planar hybridization since it is the low spin complex.
vii. Electronic configuration after complex
6 d2sp3/ sp3d2
formation.
Try this...
3d 4s 4p
Give VBT description of
bonding in each of following d2sp3
complexes. Predict their magnetic
behavior. 3⊕

a. [ZnCl4]2 H3N NH3 NH3
b. [Co(H2O)6]2⊕ (high spin) H3N Co NH3
c. [Pt(CN)4]2 (square planar) NH3
d. [CoCl4]2 (tetrahedral)
e. [Cr(NH3)6]3⊕

9.9.2 Octahedral, complexes viii. As all electrons are paired the complex is
a. [Co(NH3)6]3 ⊕ low spin diamagnetic.
i. Oxidation state of Cobalt:3⊕ b. [CoF6]3 high spin
ii. Valence shell electronic configuration i.Oxidation state of central metal Co is 3+
ii.Valence shell electronic configuration of
of Co3⊕ is represented in box diagram as Co3⊕ is
shown below :
3d 4s 4p
3d 4s 4p

iii. Number of ammine ligands is 6, number iii. Six fluoride F ligands, thus the number of
of vacant metal ion orbitals required for vacant metal ion orbitals required for bonding
bonding with ligands must be six. with ligands would be six.
iv. Complex is high spin, that means pairing
iv. Complex is low spin, so pairing of electrons of electrons will not take place prior to
will take place prior to hybridisation. hybridisation. Electronic configuration would
remain the same as in the free state shown
v. Electronic configuration after pairing would above.
be

3d 4s 4p

202

v. Six orbitals available for the hybridisation. one 4s, three 4p. The complex is tetrahedral.
3d 4s 4p
Those are one 4s, three 4p, two of 4d orbitals

3d 4s 4p 4d

sp3d2 sp3

Six metal orbitals after bonding with six The four metal ion orbitals for bonding with
F ligands led to the sp3d2 hybridization. The d Cl ligands are derived from the sp3
orbitals participating in hybridisation for this hybridization.
complex are nd. vi. Four vacant sp3 hybrid orbital of Ni2⊕
vi. Six vacant sp3d2 hybrid orbital of Co3+
overlap with six orbitals of fluoride forming overlap with four orbitals of Cl ions.
Co - F coordinate bonds. vii. Configuration after complex formation
vii. Configuration after complex formation.
would be.
3d 4s 4p 4d 3d 4s 4p

sp3d2 sp3

viii. The complex is octahedral and has four viii.The complex has four unpaired electrons
unpaired electrons and hence, is paramagnetic. and hence, paramagnetic.

3 Cl 2

F Cl Ni Cl
FF
Cl
Co
FF 9.9.4 Square planar complex
[Ni(CN)4]2
F i. Oxidation state of nickel is +2
ii. Valence shell electronic configuration of
9.9.3 Tetrahedral complex
[Ni(Cl)4]2 Ni2⊕
i. Oxidation state of nickel is +2 3d 4s 4p
ii. Valence shell electronic configuration of
iii. Number of CN ligands is 4, so number
Ni2+ of vacant metal ion orbitals required for
3d 4s 4p bonding with ligands would be four.

iii. number of Cl ligands is 4. Therefore iv. Complex is square planar so Ni2⊕ ion uses
number of vacant metal ion orbitals dsp2 hybrid orbitals.
required for bonding with ligands must be
four. v. 3d electrons are paired prior to the
hybridisation and electronic configuration
iv. Four orbitals on metal available for of Ni2⊕ becomes :
hybridisation are
3d 4s 4p

203

vi. Orbitals available for hybridisation are In an isolated gaseous metal ion the five d
orbitals, d x2-y2 d, z2, dxy,dyzd,zx have the same
one 3d, one 4s and two 4p which give dsp2 energy i.e. they are degenerate.

hybridization. ii. When ligands approach the metal ion they
create crystal-field around the metal ion. If
vii. Four vacant dsp2 hybrid orbitals of Ni2⊕ it were symmetrical the degeneracy of the d
orbitals remains intact.
overlap with four orbitals of CN ions to form
Usually the field created is not symmetrical
Ni - CN coordinate bonds. and the degeneracy is destroyed. The d orbitals
thus split into two sets namely, (dxy, dyz, dxz)
vii. Configuration after the complex formation usually refered by t2g and ( d x2-y2 ,dz2) called
as eg. These two sets of orbitals now have
becomes. 4s 4p different energies. A separation of energies of
3d these two sets of d orbitals is the crystal field
splitting parameter. This is denoted by Δo (O
dsp2 for octahedral).

viii.The complex has no unpaired electrons
and hence, dimagnetic.

2

NC CN iii. The Δo depends on strength of the ligands.
Ni The ligands are then classified as (a) strong
field and (b) weak field ligands. Strong field
NC CN ligands are those in which donor atoms are
C,N or P. Thus CN , NC , CO, NH3, EDTA,
Try this... en (ethylenediammine) are considered to be
strong ligands. They cause larger splitting of
Based on the VBT predict d orbitals and pairing of electrons is favoured.
structure and magnetic behavior of These ligands tend to form low spin complexes.
the [Ni(NH3)6]3⊕ complex. Weak field ligands are those in which donor
atoms are halogens, oxygen or sulphur. For
9.9.5 Limitations of VBT example, F , Cl , Br , I , SCN , C2O42 . In
i. It does not explain the high spin or low spin case of these ligands the Δo parameter is
nature of the complexes. In other words, strong smaller compared to the energy required for
and weak field nature of ligands can not be the pairing of electrons, which is called as
distinguished. electron pairing energy. The ligands then can
ii. It does not provide any explanation for the be arranged in order of their increasing field
colour of coordination compounds. strength as
iii. The structure of the complexes predicted
from the VBT would not always match I < Br < Cl < S2 < F < OH < C2O42
necessarily with those determined from the <H2O<NCS <EDTA< NH3,< en< CN < CO.
experiments. Let us understand splitting of d orbitals and
To overcome these difficulties in VBT, the formation of octahedral complexes
Crystal field theory has been proposed which
has widely been accepted. In octahedral environment the central metal
9.9.6 Crystal Field theory (CFT) ion is surrounded by six ligands.

C.F.T. is based on following assumptions Ligands approach the metal ion along the x, y,
z axes. As the ligands approach the metal ion
i. The ligands are treated as point charges. the degeneracy of d orbitals is resolved.
The interaction between metal ion and ligand
is purely electrostatic or there are no orbital
interactions between metal and ligand.

204

eg

eg - The higher energy set of orbitals (dz2 and dx2 - y2 )

Energy 3/5 Δ0 t2g - The lower energy set of orbitals (dxy,dyz and dzx)
2/5 Δ0
Δ0 Δ0 or 10 Dq - The energy separation between the two
levels

The eg orbitals are repelled by an amount of 0.6 Δ0.

t2g The t2g orbitals to be stabilized to the extend of 0.4 Δ0.

Fig. 9.2 : Crystal field Splitting in an octahedral complex

With closer approach of ligands along the 9.9.7 Factors affecting Crystal Field Splitting
axes, the doubly degenerate dx2-y2 and dz2 (eg) parameter (∆0)
orbitals experience larger repulsion than the
a. The magnitude of crystal field splitting
triply degenerate, t2g orbitals. As shown in Fig. depends on strength of the ligands.
9.2, the eg set has higher energy than the t2g set The strong ligands those appear in
by the amount Δo. The Δo parameter is equal spectrochemical series approach closer to
to 10 Dq units of splitting of t2g and eg levels. the central metal which results in a large
For electronic configurations d1, d2, d3 the crystal field splitting.

electrons occupy t2g orbitals and obey the b. Oxidation state of the metal : A metal with
Hund’s rules. For electronic configurations d1, the higher positive charge is able to draw
ligands closer to it than that with the lower
d2, d3 and d8, d9, d10 the high spin and low spin one. Thus the metal in higher oxidation
state results in larger separation of t2g and
configurations cannot be distinguished. Only eg set of orbitals. The trivalent metal ions
cause larger crystal field splitting than
the electronic configurations d4 to d7 render corresponidng divalent ones.

the high and low spin complexes.These are 9.9.8 Colour of the octahedral complexes

depiciated in Fig 9.5. As discussed above, the formation of
Table 9.5 : d orbital diagrams for high spin an octahedral complex is accompanied by
splitting of d orbitals into t2g and eg sets. A
and low spin complexes. separation of these two sets of orbitals is Δo,
which can be measured from experiments.
d orbital High spin Low spin The Δo corresponds to a certain frequency
electronic of electromagnetic radiation usually in the
configuration visible region. A colour complementary to the
absorbed frequency is thus observed. Consider
d4 eg the [Ti(H2O)6]3⊕ complex. The central metal
ion titanium has electronic configuration 3d
t2g and the electron occupies one of the t2g orbitals
d5 eg (Figure 9.3).

t2g
d6 eg

t2g

d7 eg

t2g

205

eg Energy eg Try this...

t2g t2g Sketch qualitatively crystal
field d orbital energy level diagrams
Ground state Excited state for each of the following complexes:

Fig. 9.3 : d - d transition in d' system a. [Ni(en)3]2⊕ b. [Mn(CN)6]3

The absorption of the wavelength of light c. [Fe(H2O)6]2⊕
corresponding to Δo parameter promotes an
electron from the t2g level. Such energy gap in Predict whether each of the complexes is
case of the [Ti(H2O)6]3⊕ complex is 20,300 cm-1 diamagnetic or paramagnetic.
(520 nm, 243 kJ/mol) and a complimentary
colour to this is imparted to the complex. A The dxy, dyz, dzx orbitals with their lobes
violet color of the [Ti(H2O)6]3⊕ complex arises lying in between the axes point toward the
from such d-d transition. ligands. On the other hand, dx2-y2 and dz2
orbitals lie in between metal-ligand bond
9.9.9 Tetrahedral complexes axes. The dxy, dyz and dzx orbitals experience
more repulsion from the ligands compared to
A pattern of splitting of d orbitals, that by dx2-y2 and dz2 orbitals.

which is a key in the crystal field theory, is Due to larger such repulsions the dxy, dyz
and dzx orbitals are of higher energy while the
dependent on the ligand field environment. dx2-y2 and dz2 orbitals are of relatively lower
energy.
This is illustrated for the tetrahedral ligand
Each electron entering in one of the
field environment. dxy,dyz and dzx orbitals raises the energy by
z 4 Dq whereas that accupying d x2-y2 and dz2
orbitals lowers it by 6 Dq compared to the
Mx energy of hypothetical degenerate d orbitals
in the ligand field.
y
Fig. 9.4 : Tetrahedral structure A splitting of d orbitals in tetrahedral
crystal fields (assumed to be 10 Dq) thus is
The tetrahedral structure having the much less (typically 4/9) compared to that
metal atom M at the centre and four ligands for the octahedral environment. The crystal
occupying the corners is displayed along with field splitting of d orbitals in a tetrahedral
in Fig. 9.4. ligand field is compared with the octahedral
one in Fig. 9.5. Thus the pairing of electrons
Tetrahedral Octahedral is not favoured in tetrahedral structure. For
eg example, in d4 configuration an electron
t2g Δtet would occupy one of the t2g orbitals. The low
eg Δo spin tetrahedral complexes thus are not found.

Free-ion t2g Typically metal complexes possessing
the cetral metal ion with d8 electronic
configuration, for example, Ni(CO)4, favours
the tetrahedral structure.

Fig. 9.5 : Splitting of d orbitals in tetrahedral
and octahedral complexes

206

9.10 Applications of coordination c. To estimate hardness of water : Hardness
compounds of water is due to the presence of Ca2+ and
Mg2⊕ ions. The ligand EDTA forms stable
a. In biology : Several biologically important complexes with Ca2⊕ and Mg2⊕. It can,
natural compounds are metal complexes. They therefore, be used to estimate hardness.
play important role in a number of processes
occuring in plants and animals. d. In electroplating : Usually stable
coordination complexes on dissolution
For example, chlorophyll present in dissociate to small extent and furnish a
plants is a complex of Mg. Haemoglobin controlled supply of metal ions. The metal
present in blood is a complex of iron. ions when reduced clump together to form
the clusters or nanoparticles. When the
b. In medicines coordination complexes are used the ligands
in the complex keep the metal atoms well
i. Pt complex cisplatin is used in the treatment separated from each other. These metal atoms
of cancer. tend to form a protective layer on the surface.
Certain cyanide complexes K[Ag(CN)2]
ii. EDTA is used for treatment of lead and K[Au(CN)2] find applications in the
poisoning. electroplating of these noble metals.

Exercises

1. Choose the most correct option.

i. The oxidation state of cobalt ion in the 3. [PtCl2Br2]2 (square planar)

complex [Co(NH3)5Br]SO4 is 4. [FeCl2(NCS)2]2 (tetrahedral)
a. +2 b. +3
a. 1 and 3 b. 2 and 3
c. +1 d. +4

ii. IUPAC name of the complex c. 1 and 3 d. 4 only
[Pt(en)2(SCN)2]2+ is
a. bis (ethylenediamine v. Which of the following complexes are
dithiocyanatoplatinum (IV) ion chiral ?
b. bis (ethylenediamine)
dithiocyantoplatinate (IV) ion 1. [Co(en)2Cl2]⊕ 2. [Pt(en)Cl2]
c. dicyanatobis (ethylenediamine)
platinate IV ion 3. [Cr(C2O4)3]3 4. [Co(NH3)4Cl2]⊕
d. bis (ethylenediammine)dithiocynato
platinate (IV) ion a. 1 and 3 b. 2 and 3

c. 1 and 4 d. 2 and 4

iii. Formula for the compound sodium vi. On the basis of CFT predict the number of
hexacynoferrate (III) is unpaired electrons in [CrF6]3 .

a. 1 b. 2 c. 3 d. 4

a. [NaFe(CN)6] b. Na2[Fe(CN)6] vii. When an excess of AgNO3 is added to the
complex one mole of AgCl is precipitated.
c. Na[Fe(CN)6] d. Na3[Fe(CN)6] The formula of the complex is

iv. Which of the following complexes exist a. [CoCl2(NH3)4]Cl
as cis and trans isomers ? b. [CoCl(NH3)4] Cl2

1. [Cr(NH3)2Cl4] 2. [Co(NH3)5Br]2⊕

207

c. [CoCl3(NH3)3] v. Write formulae of the following complexes

d. [Co(NH3)4]Cl3 a. Potassium amminetrichloroplatinate
(II)
viii. The sum of coordination number and
oxidation number of M in [M(en)2C2O4]Cl b. Dicyanoaurate (I) ion
is
vi. What are ionization isomers ? Give an
a. 6 b. 7 c. 9 d. 8 example.

2. Answer the following in one or two vii. What are the high-spin and low-spin
sentences. complexes ?

i. Write the formula for viii. [CoCl4]2 is tetrahedral complex. Draw its
tetraammineplatinum (II) chloride. box orbital diagram. State which orbitals
participate in hybridization.
ii. Predict whether the [Cr(en)2(H2O)2]3+
complex is chiral. Write structure of its ix. What are strong field and weak field
enantiomer. ligands ? Give one example of each.

iv. Name the Lewis acids and bases in the x. With the help of crystal field energy-
complex [PtCl2(NH3)2]. level diagram explain why the complex
[Cr(en)3]3⊕ is coloured ?
v. What is the shape of a complex in which
the coordination number of central metal 4. Answer the following questions.
ion is 4 ?
i. Give valence bond description for the
vi. Is the complex [CoF6] cationic or anionic bonding in the complex [VCl4] . Draw
if the oxidation state of cobalt ion is +3 ? box diagrams for free metal ion. Which
hybrid orbitals are used by the metal ? 
vii. Consider the complexes [Cu(NH3)4][PtCl4] State the number of unpaired electrons.
and [Pt(NH3)4] [CuCl4]. What type of
isomerism these two complexes exhibit? ii. Draw qualitatively energy-level
diagram showing d-orbital splitting in
viii. Mention two applications of coordination the octahedral environment. Predict
compounds. the number of unpaired electrons in the
complex [Fe(CN)6]4 . Is the complex
3. Answer in brief. diamagnetic or paramagnetic? Is it
coloured? Explain.
i. What are bidentate ligands ? Give one
example. iii. Draw isomers in each of the following

ii. What is the coordination number and a. Pt(NH3)2ClNO2
oxidation state of metal ion in the complex b. Ru(NH3)4Cl2
[Pt(NH3)Cl5] ? c. [Cr(en2)Br2]⊕
iv. Draw geometric isomers and enantiomers
iii. What is difference between a double salt
and a complex ? Give an example. of the following complexes.

iv. Classify following complexes as a. [Pt(en)3]4⊕ b. [Pt(en2)ClBr]2⊕
homoleptic and heteroleptic

[Cu(NH3)4]SO4, [Cu(en)2(H2O)Cl]2⊕,
[Fe(H2O)5(NCS)]2⊕, tetraammine zinc (II)
nitrate.

208

v. What are ligands ? What are their types ? vii. How stability of the coordination
Give one example of each type. compounds can be explained in terms of
equilibrium constants ?
vi. What are cationic, anionic and neutral
complexes? Give one example of each. viii. Name the factors governing the
equilibrium constants of the coordination
compounds.

Activity :

1. The reaction of chromium metal with H2SO4 in the absence of air gives blue
solution of chromium ion.

Cr(s) + 2H⊕(aq) Cr2⊕(aq) + H2(s)

Cr2⊕ forms octahedral complex with H2O ligands.

a. Write formula of the complex

b. Describe bonding in the complex using CFT and VBT.

Draw crystal field splitting and valence bond orbital diagrams.

2. Reaction of complex [Co(NH3)3(NO2)3 with HCl gives a complex
[Co(NH3)3H2OCl2]⊕ in which two chloride ligands are trans to one another.
a. Draw possible stereoisomers of starting material

b. Assuming that NH3 groups remain in place, which of two starting isomers would give
the observed product ?

209

10. HALOGEN DERIVATIVES

Can you recall ? CH3 - CH2 - X CH2 = CH - X
Identify the functional group (Haloalkane) (Haloalkene)

in the following compounds. HC ≡ C - X X

i. Br ii. CCl2F2 (Haloalkyne)
(Haloarene)
(Benzyl bromide) (Freon - 12)
b. On the basis of number of halogen atoms,
iii. Cl iv. Cl - CH = CCl2 halogen derivatives are classified as mono,
Cl Cl di, tri or poly halogen compounds.
(Westrosol)
Cl Cl X
Cl CH3 - CH2 - X

(Hexachlorobenzene) Monohalogen compounds

v. Cl CH3 - CH - X CH2 - CH2 X
Cl Cl X XX X

Cl Cl Dihalogen compounds
Cl
X X
((HBeexnazcehnleorhoecxyacclholhoerxidaen)e)
CH2 - CH - CH2
The parent family of organic compounds X XX
is hydrocarbon. Replacement of hydrogen
atom/s in aliphatic or aromatic hydrocarbons X
by halogen atom/s results in the formation of Trihalogen compounds
halogen derivatives of hydrocarbons.
We will consider classification of mono
In this chapter we will study halogen halogen derivatives in more detail.
derivatives in a systematic way.
10.1.1 Classification of monohalogen
Internet my friend compounds : Monohalogen compounds are
further classified on the basis of position of
Find out the structures halogen atom and the type of hybridization of
of two thyroid hormones T3 carbon to which halogen is attached.
(triiodothyronine) and T4 (thyroxine).
How do these help our body ? a. Alkyl halides or haloalkanes : In alkyl
halides or haloalkanes the halogen atom is
10.1 Classification of halogen derivatives : bonded to sp3 hybridized carbon which is
a part of saturated carbon skeleton. Alkyl
Halogen derivatives of hydrocarbons are halides may be primary, secondary or tertiary
classified mainly in two ways. depending on the substitution state of the
carbon to which halogen is attached : (Refer
a. On the basis of hydrocarbon skeleton to to Std. XI Chemistry Textbook, section 14.3).
which halogen atom is bonded, the halogen
derivatives are classified as haloalkanes,
haloalkenes, haloalkynes and haloarenes.

210

R - CH2- X R - CH- R R d. Vinylic halides : In vinylic halides halogen
R−C−X atom is bonded to a sp2 hybridized carbon
Primary X atom of aliphatic chain. Vinylic halide is a
halide R haloalkene.
(10 halide) Secondary Tertiary
halide halide CH2= CH - X X
(30 halide)
(20 halide)

b. Allylic halides : In allylic halides, halogen e. Haloalkyne : When a halogen atom is
atom is bonded to a sp3 hybridized carbon bonded to a sp hybridized carbon atom it is
atom next to a carbon-carbon double bond. a haloalkyne.

CH2= CH - CH2 - X X CH ≡ C - X

c. Benzylic halide : In benzylic halides f. Aryl halides or haloarenes : In aryl
halides, halogen atom is directly bonded to
halogen atom is bonded to a sp3 hybridized the sp2 hybridized carbon atom of aromatic
ring.
carbon atom which is further bonded to an
CH3
aromatic ring. CH3 XX
CH2 - X CX
CH3

Table 10.1 Names of some halogen derivatives

Formula Common name IUPAC name
CH2Cl2 Methylene chloride Dichloromethane
CH3CH2Br
CH3CH(Cl)CH3 Ethyl bromide Bromoethane
(CH3)2 CH - CH2Br
(CH3)3 C Br Isopropyl chloride 2-Chloropropane
(CH3)3 C CH2Cl
Isobutyl bromide 1-Bromo-2-methylpropane
CH2 = CH - Cl
CH2 = CH - CH2Br Tert-butyl bromide 2-Bromo-2-methyl-propane
CH ≡ C - Cl
Neopentyl chloride 1-Chloro-2, 2-dimethyl pro-
CH2I Vinyl chloride pane

1-Chloroethene

Allyl bromide 3-Bromopropene

Chloro acetylene Chloroethyne

Benzyl iodide Iodophenylmethane

I p-Iodotoluene 1-Iodo-4-methyl benzene or
m-dichlorobenzene 4-Iodotoluene
H3C
Cl 1, 3-dichlorobenzene

Cl

211

10.2 Nomenclature of halogen derivatives a wide variety. The hydroxyl group may be
replaced by halogen atom using (a) halogen
Can you recall ? acid, (b) phosphorous halide or (c) thionyl
chloride.
• In IUPAC system of
nomenclature does the a. By using halogen acid or hydrogen
functional group 'halogen' appear halide (HX) : The conditions for reaction
as a suffix or prefix ? of alcohol with halogen acid (HX) depend
on the structure of the alcohol and particular
• What are the trivial names of laboratory halogen acid used. The order of reactivity of
solvents CHCl3 and CCl4 ? alcohols with a given haloacid is 30>20>10.
(Refer to section 11.2.1 a)
The common names of alkyl halides are
derived by naming the alkyl group followed R - OH + HX suitable R - X + H2O
by the name of halogen as halide. For condition
example, methyl iodide, tert-butyl chloride.
According to IUPAC system of nomenclature (Alcohol) (Alkyl halide)
(Std. XI Chemistry Textbook Chapter 14,
section 14.4.7) alkyl halides are named as Hydrogen chloride is used with zinc
haloalkanes. Aryl halides are named as chloride (Grooves' process) for primary and
haloarenes in common as well as IUPAC secondary alcohols, but tertiary alcohols
system. For dihalogen derivative of an arene, readily react with concentrated hydrochloric
prefix o-, m-, p- are used in common name acid in absence of zinc chloride.
system but in IUPAC system the numerals
1,2 ; 1,3 and 1,4 respectively are used. R - OH + HCl anhydrous R - Cl + H2O
Common and IUPAC names of some halogen ZnCl2
derivatives are given in Table 10.1.
Do you know ?
Use your brain power Zinc chloride is a Lewis acid

Write IUPAC names of the and consequently can coordinate
following. with the alcohol, weakening R - O bond.
Mixture of concentrated HCl and anhydrous
ZnCl2 is called Lucas reagent.

i. CH3 - CH - CH3 ii. CH3 - CH - CH2I Constant boiling hydrobromic acid
Br CH3 (48%) is used for preparing alkyl bromides.
Primary alkyl bromides can also be prepared
iii. CH3 - CH = CH - CH2Cl by reaction with NaBr and H2SO4. Here HBr
is generated in situ.

iv. CH3 - C ≡ C - CH2 - Br R - OH + HBr NaBr, H2SO4 R - Br + H2O
v. Br vi. heat

Br Good yield of alkyl iodides may be

Br obtained by heating alcohols with sodium or
10.3 Methods of preparation of alkyl halides potassium iodide in 95 % phosphoric acid.
Here HI is generated in situ.

R - OH + HI NaI/H3PO4 R - I + H2O

10.3.1 From alcohol : The most widely used Can you tell ?
method of preparation of alkyl halide is
replacement of hydroxyl group of an alcohol Why phosphoric acid is preferred to
by halogen atom. Alcohols are available in H2SO4 to prepare HI in situ ?

212

b. By using phosphorous halide : An Addition of hydrogen halide to alkene
alkyl halide may be prepared by action of
phosphorous halide on alcohol. Phosphorous Alkyl halides are formed on addition
tribromide and triiodide are usually generated of hydrogen halide to alkenes. Refer to Std
in situ (produced in the reaction mixture) by XI Chemistry Textbook Chapter 15, section
the action of red phosphorous on bromine and 15.2.4 for all the details including order of
iodine respectively. Phosphorous pentachloride reactivity of HX, Markownikov rule and
reacts with alcohol to give alkyl chloride. peroxide effect.

3R - OH + PX3 3R - X + H3PO3 Problem 10.1 : How will you obtain
R - OH + PCl5 R - Cl + HCl + POCl3 1-bromo-1-methylcyclohexane from alkene?

Do you know ? Write possible structures of alkene and the
reaction involved.

Some times during replacement Solution : CH3 Br
of -OH by -X, alcohols tend to CH3 Br
undergo rearrangement. This tendency CH3
can be minimized by use of phosphorous
halides. Straight chain primary alcohols + HBr
react with phosphorous trihalide to give
unrearranged alkyl halides. CH2

c. By using thionyl chloride : Thionyl + HBr
chloride reacts with straight chain primary
alcohols to give unrearranged alkyl chloride. Use your brain power
The byproducts obtained are gases. There
is no need to put extra efforts for its Rewrite the following reaction
separation. Therefore this method is preferred by filling the blanks :
for preparation of alkyl chloride.
• CH3 - CH = CH2 + HBr +
R - OH + SOCl2 ∆ R - Cl + SO2↑+ HCl↑
(major)(minor)

Can you recall ? • (CH3)2C=CHCH3+HBr peroxide +

Identify the products of the (major)(minor)
following reactions.
• CH3 - CH = CH2+HBr peroxide +

i. CH4 + Cl2 hν ? (major)(minor)

ii. CH3 - CH = CH2 HCl ? Do you know ?

iii. CH3 - CH = CH2 + HBr Peroxide ? Alkenes form additon product,
vicinal dihalide, with chlorine or
iv. CH2 = CH - CH3 + Br2 CCl4 ? bromine usually in inert solvent like CCl4
at room temperature.
10.3.2 From hydrocarbon

Alkyl halides are formed from saturated C =C + X2 C-C
as well as unsaturated hydrocarbons by (X = Cl, Br) XX
various reactions. Halogenation of alkanes is
not suitable for preparation of alkyl halides
as a mixture of mono and poly halogen
compounds is formed.

213

Do you know ? When toluene is brominated in presence
When alkenes are heated of iron, a mixture of ortho and para bromo
toluene is obtained.
with Br2 or Cl2 at high temperature,
hydrogen atom of allylic carbon is CH3 CH3 Br CH3
substittued with halogen atom giving + Br2 +
allyl halide. Fe
dark

CH2 = CH - CH3 + Cl2 Br
CH2 = CH - CH2Cl + HCl
(o - Bromotoluene) (p-Bromotoluene)

+ HBr

10.3.3 Halogen exchange : Alkyl iodides Aromatic electrophilic substitution
are prepared conveniently by treating alkyl with iodine is reversible. In this case use
chlorides or bromides with sodium iodide in of HNO3/HIO4 removes HI by oxidation to
methanol or acetone solution. The sodium I2, equilibrium is shifted to right and iodo
bromide or sodium chloride precipitates from product is formed. F2 being highly reactive,
the solution and can be separated by filtration. fluoro compounds are not prepared by this
method.
R - Cl + NaI acetone R - I + NaCl ↓

The reaction is known as Finkelstein reaction. 10.3.5 Sandmeyer's reaction : Aryl halides
are most commonly prepared by replacement
Alkyl fluorides are prepared by heating of nitrogen of diazonium salt. (For details
refer to Chapter 13 section 13.6).
alkyl chlorides or bromides with metal

fluorides such as AgF, Hg2F2, AsF3, SbF3 etc. 10.4 Physical properties : Physical properties
of alkyl halides are considerably different
R - Cl + AgF R - F + AgCl ↓ from those of corresponding alkanes. The
boiling point of alkyl halides is determined
The reaction is known as Swartz reaction. by polarity of the C-X bond as well as the
size of halogen atoms.
10.3.4 Electrophilic substitution :

Can you recall ? 10.4.1 Nature of intermolecular forces:
• Identify the product of the Halogens (X = F, Cl, Br and I) are more
electronegative than carbon.
following reaction.

+ Cl2 anhyd. Carbon atom that carries halogen
AlCl3 develops a partial positive charge while the
halogen carries a partial negative charge.
• Name the type of halide produced in Thus carbon-halogen bond in alkyl halide is a
the above reaction. polar covalent bond. Therefore alkyl halides
are moderately polar compounds.
• What type of reactions are shown by
benzene ?

Aryl chlorides and bromides can be Cδ⊕ Xδ
prepared by direct halogenation of benzene
and its derivatives through electrophilic Size of the halogen atom increases from
substitution. It may be conveniently carried fluorine to iodine. Hence the C-X bond length
out in dark at ordinary temperature in presence increases. The C-X bond strength decreases
of suitable Lewis acid catalyst like Fe, FeCl3 with an increase in size of halogen. This is
or anhydrous AlCl3. because as the size of p-orbital of halogen
increases the p-orbital becomes more diffused

214

and the extent of overlap with orbital of Haloalkane Boiling point (K)
carbon decreases. Some typical bond lengths,
bond enthalpies and dipole moments of C-X CH3CH2CH2CH2Br 375
bond are given in Table 10.2. 364
CH3 - CH - CH2 - CH3
Table 10.2 : Bond parameters of C-X bond Br 346

Bond Bond Bond en- Dipole CH3
length/ thalpy/ moment/ CH3− C − CH3
(pm) (kJ mol-1) debye
Br

CH3 - F 139 452 1.847 10.4.3 Solubility : Though alkyl halides
are moderately polar, they are insoluble in
CH3 - Cl 178 351 1.860 water. It is due to inability of alkyl halides to
CH3 - Br 193 293 1.830 form hydrogen bonds with water. Attraction
CH3 - I 214 234 1.636 between alkyl halide molecules is stronger
than attraction between alkyl halide and
10.4.2 Boiling point : Boiling points of alkyl water. Alkyl halides are soluble in non-polar
halides are considerably higher than those of organic solvents.
corresponding alkanes due to higher polarity
and higher molecular mass. Within alkyl Aryl halides are also insoluble in water
halides, for a given alkyl group, the boiling but soluble in organic solvents. If aryl halides
point increases with increasing atomic mass are not modified by presence of any other
of halogen, because magnitude of van der functional group, they show properties similar
Waals force increases with increase in size to corresponding alkyl halides. The isomeric
and mass of halogen. dihalobenzenes have nearly the same boiling
points, but melting points of these isomers
Thus boiling point of alkyl halide show variation. Melting point of para isomer
decreases in the order RI > RBr > RCl > RF is quite high compared to that of ortho or meta
isomer. This is because of its symmetrical
For example, : structure which can easily pack closely in
the crystal lattice. As a result intermolecular
Haloal- CH3F CH3Cl CH3Br CH3I forces of attraction are stronger and therefore
kane 194.6 248.8 276.6 315.4 greater energy is required to overcome its
Boiling lattice energy.
point (K)

For the given halogen, boiling point rises Cl Cl Cl
with increasing carbon number. Cl

For example,

Haloalkane Boiling point (K) Cl

CH3Cl 248.8 b.p./K 453 446 Cl
CH3CH2Cl 285.5 m.p./K 256 249 448
CH3CH2CH2Cl 320.0
CH3CH2CH2CH2Cl 351.5 323

For isomeric alkyl halides, boiling point
decrease with increased branching as surface
area decreases on branching and van der
Waals forces decrease. For example :

215

Problem 10.2 Arrange the following Let us, now, jot down the atoms/groups
compounds in order of increasing boiling attached to each carbon in 2 - chlorobutane.
points : bromoform, chloromethane,
dibromomethane, bromomethane. 12 34
CH3 - CHCl - CH2 - CH3
Solution : The comparative boiling points
of halogen derivatives are mainly related C-1 : - H, -H, -H, -C2 HCl-C3 H2-C4 H3
with van der Waals forces of attraction
which depend upon the molecular size. In C-2 : -H, -Cl, -C1 H3, -C3 H2-C4H3
the present case all the compounds contain C-3 : -H, -H, -C4 H3, -C2HCl-C1 H3
only one carbon. Thus the molecular size
depends upon the size of halogen and C-4 : -H, -H, -H, -C3 H2-C2 HCl-C1 H3
number of halogen atoms present.
It can be seen that the four groups bonded
Thus increasing order of boiling point is, to C-2 are all different from each other.
Carbon atom in a molecule which carries
CH3Cl < CH3Br < CH2Br2 < CHBr3 four different groups/atoms is called chiral
carbon atom. Thus, the C-2 in 2-chlorobutane
10.5 Optical isomerism in halogen is a chiral carbon. Chiral atom in a molecule
derivatives : is marked with asterisk (*). For example,
CH3-*CHCl-CH2-CH3.
Can you recall ?
• What is the relationship between When a molecule contains one chiral
atom, it acquires a unique property. Such a
two compounds having the same molecule can not superimpose perfectly on its
molecular formula? mirror image. It is called chiral molecule. A
chiral molecule and its mirror image are not
• What is meant by stereoisomerism ? identical (see Fig. 10.1).

Isomers having the same bond connectivities, CH3 CH3
that is, structural formula are called
stereoisomers. Knowledge of optical C* Cl *C
isomerism, which is a kind of stereoisomerism C2H5
will be useful to understand nucleophilic Cl H
substitution reactions of alkyl halides (see H C2H5
10.6.3). mirror plane

10.5.1 Chiral atom and molecular chirality CH3 CH3
C* *C
Try this... H Cl Cl H
C2H5 C2H5
• Make a three - dimensional
model of 2 - chlorobutane. Fig. 10.1 : Nonsuperimposable mirror images

• Make another model which is a mirror A chiral molecule and its mirror image
image of the first model. both have the same structural formula and,
of course, the same molecular formula. The
• Try to superimpose the two models on spatial arrangement of the four different
each other. groups around the chiral atom, however, is
different. In other words, a chiral molecule
• Do they superimpose on each other and its mirror image are stereoisomers of each
exactly ?

• Comment on whether the two models
are identical or not.

216

other. (Refer to Std. XI Chemistry Textbook, Do you know ?
Chapter 14). Nicol prism is a special type of

The relationship between a chiral prism made from pieces of calcite,
molecule and its mirror image is similar to a crystalline form of CaCO3, arranged
the relationship between left and right hands. in a specific manner. Nicol prism is also
Therefore it is called handedness or chirality. called polarizer.
(Origin : Greek word : Cheir means hand)
10.5.3 Optical activity : When an aqueous
The stereoisomerism in which the isomers solution of certain organic compounds like
have different spatial arrangements of groups/ sugar, lactic acid is placed in the path of
atoms around a chiral atom is called optical plane polarized light, the transmitted light
isomerism. The optical isomers differ from has oscillations in a different plane than
each other in terms of a measurable property the original. In other words, the incident
called optical activity. light undergoes rotation of its plane of
polarization. The plane of polarization rotates
To understand optical activity, we must either to the right (clockwise) or to the left
know what is plane polarized light. (anticlockwise). This property of a substance
by which it rotates plane of polarization
Remember... of incident plane polarized light is known
as optical activity. The compounds which
The phenomenon of optical rotate the plane of plane polarized light are
isomerism in organic compounds called optically active compounds and those
was observed first and its origin in which do not rotate it are optically inactive
molecular chirality was recognized later. compounds. Optical activity of a substance
is expressed numerically in terms of optical
10.5.2 Plane polarized light : An ordinary rotation. The angle through which a substance
light consists of electromagnetic waves having rotates the plane of plane polarized light on
oscillations of electric and magnetic field in passing through it is called optical rotation.
all possible planes perpendicular to direction In accordance with the direction of optical
of propagation of light. rotation an optically active substance is either
dextrorotatory or laevorotatory. A compound
When ordinary light is passed through which rotates the plane of plane polarized
Nicol's prism, oscillations only in one plane light towards right is called dextrorotatory
emerge out. Such a light having oscillations and designated by symbol d- or by (+)
only in one plane perpendicular to direction sign. A compound which rotates plane of
of propagation of light is known as plane plane polarized light towards left is called
polarized light. laevorotatory and designated by symbol l-
or by (-) sign.

Isomerism in which isomeric compounds
have different optical activity is known as
optical isomerism. French scientist Louis
Pasteur first recognized that optical activity is
associated with certain type of 3-dimensional
structure of molecules. Pasteur introduced
the term enantiomers for the optical isomers
having equal and opposite optical rotation.

217

Figure 10.2 indicates a few objects in our Enantiomers have identical physical
day to day life which exhibit superimposable properties (Such as melting point, boiling
and non-superimposable mirror image points, densities, refractive index) except the
relationship. sign of optical rotation. The magnitude of
their optical rotation is equal but the sign
Non superimposable Superimposable of optical rotation is opposite. They have
identical chemical properties except towards
Fig. 10.2 : Superimposable and optically active reagent.
nonsuperimposable mirror image
An equimolar mixture of enantiomers
(dextrorotatory and laevorotatory) is called
racemic modification or racemic mixture.
A racemic modification is optically inactive
because optical rotation due to molecules
of one enatiomer is cancelled by equal and
opposite optical rotation due to molecules of
the other enantiomer. A racemic modification
is designated as (dl) or by (±) sign.

10.5.5 Representation of configuration of
molecules :

Remember... Can you recall ?

• Optical activity is an • Identify the type of following
experimentally observable 3-D representation (I) and
(II) of a molecule and state
property of compounds. Chirality is significance of the lines drawn.

a description of molecular structure. WW

Optical activity is the consequence of

chirality. XY
Z
• Molecules which contain one chiral XC
atom are chiral, that is, they are YZ
nonsuperimposable on their mirror
image. (I) (II)

• The two non-superimposable mirror a. Fischer projection formula (cross
image structures are called pair of formula) : Two representations are used to
enantiomers. represent configuration of chiral carbon and
the 3-dimensional structure of optical isomers
• Enantiomers have equal and opposite on plane paper. These are (a) wedge formula
optical rotation. Thus, enantiomers are and (b) Fischer projection formula (also called
a kind of optical isomers. cross formula) (Std. XI Chemistry Textbook
Chapter 14 section 14.2.3).

10.5.4 Enantiomers : The optical isomers Cl Chiral carbon Cl Bonds below
which are non-superimposable mirror image
of each other are called enantiomers or the plane
enantiomorphs or optical antipodes. For
example, 2 - chlorobutane exists as a pair of I Br I C Br
enantiomers (Fig. 10.1).
H Bonds above H
the plane
Fischer projection
Convention of vertical
and horizontal lines

Fig. 10.3 Fischer projection formula

218

a. Wedge formula : When a tetrahedral Can you recall ?
carbon is imagined to be present in the plane • What is meant by substitution
of paper all the four bonds at this carbon
cannot lie in the same plane. The bonds in reaction ?
the plane of paper are represented by normal
lines, the bonds projecting above the plane • Can you identify substitution reaction
of paper are represented by solid wedges from the following ?
(or simply by bold lines) while bonds going
below the plane of paper are represented by (i) CH3 - CH2 - OH + HCl ZnCl2
broken wedges (or simply by broken lines). CH3 - CH2 - Cl + H2O

below the plane Br (ii) CH2 = CH2 + HI
CH3 - CH2 - I
I C In the plane
Cl • Is the carbon carrying halogen in alkyl
halide, an electrophilic or a nucleophilic
H centre ?

Above the plane hybridization of that carbon the reaction is
called substitution reaction. The C-X bond
Try this... in alkyl halides is a polar covalent bond
and the carbon in C-X bond is positively
1. Draw structures of enantiomers polarized. In other words, the C-X carbon is
of lactic acid (CH3-CH-COOH) an electrophilic centre. It has, therefore, a
OH tendency to react with a nucleophile. (Refer to
using Fischer projection formulae. Std. XI Chemistry Textbook Chapter 14.) Alkyl
halides react with a variety of nucleophiles
2. Draw structures of enantiomers of to give nucleophilic substitution reactions
2-bromobutane using wedge formula. (SN). The reaction is represented in general
form as shown below.
10.6 Chemical properties :

10.6.1 Laboratory test of haloalkanes : Nu + − C δ⊕ Xδ − C − Nu + X
Haloalkanes are of neutral type in aqueous −
medium. On warming with aqueous sodium
or potassium hydroxide the covalently bonded When a substrate reacts fast it is said to
halogen in haloalkane is converted to halide be reactive. The reactivity of alkyl halides
ion. in SN reaction depends upon two factors,
namely, the substitution state (10, 20 or 30)
R - X + OH ∆ R - OH + X of the carbon and the nature of the halogen.
The order of reactivity influenced by these
When this reaction mixture is acidified two factors is as shown below.
by adding dilute nitric acid and silver nitrate tertiary alkyl halide (30) > secondary alkyl
solution is added a precipitate of silver halide halide (20) >primary alkyl halide (10) and
is formed which confirms presence of halogen R - I > R - Br > R - Cl
in the original organic compound. Examples of some important nucleophilic
substitution reactions of alkyl halides are
Ag⊕ (aq) + X (aq) AgX↓ (s) shown in Table 10.3.

10.6.2 Nucleophilic substitution reactions of
haloalkanes :

When a group bonded to a carbon in
a substrate is replaced by another group to
get a product with no change in state of

219

10.3 Nucleophilic substitution reactions of alkyl halides

Sr. No. Alkyl halide Reagent Substitution product
1. R-X + NaOH(aq) ∆
2. R - OH + NaX
R-X + (or KOH)
3. (alcohol) (or KX)
N⊕ aOR' ∆
4. R - O - R' + NaX
5. (sodium alkoxide)
6. (ether)
7. O
R-X + R' - C - OA⊕ g ∆ O

(silver carboxylate) R' - C - OR + AgX ↓

NH3(a(elxcc.e)ssp)re∆ssure (ester)

R-X + KCN (alc.) ∆ R - NH2 + HX
R-X +
R-X + AgCN (alc.) ∆ (primary amine)
R-X +
K⊕ O - N = O R - CN + RX

(potassium nitrite) (nitrile)(alkyl cyanide)

R - N C + AgX ↓

(isocyanide)

R - O - N = O + KX

(alkyl nitrite)

8. R-X + Ag - O - N = O R N⊕ O + AgX ↓
O
(silver nitrite)

(nitroalkane)

Do you know ? Can you tell ?

Cyanide ion is capable of Alkyl halides when treated with
attacking through more than one alcoholic solution of silver nitrite give
site (atom). nitroalkanes whereas with sodium nitrite
they give alkyl nitrites Explain.

C≡N C=N 10.6.3 Mechanism of SN reaction :

Such nucleophiles are called ambident Can you recall ?
nucleophiles. KCN is predominantly ionic
(K⊕C ≡ N) and provides cyanide ions. Both • What is meant by order and
carbon and nitrogen are capable of donating molecularity of a reaction ?
electron pair. C-C Bond being stronger than
C-N bond, attack occurs through carbon atom • What is meant by mechanism of
of cyanide group forming alkyl cyanides as chemical reaction ?
major product. However AgCN (Ag-C ≡ N)
is mainly covalent compound and nitrogen It can be seen from the Table 10.3 that in
is free to donate pair of electron. Hence a nucleophilic substitution reactions of alkyl
attack occurs through nitrogen resulting in halides the halogen atom gets detached from
formation of isocyanide. the carbon and a new bond is formed between
that electrophilic carbon and nucleophile.
Another ambident nucleophile is nitrite The covalently bonded halogen is converted
ion, which can attack through ‘O’ or ‘N’. into halide ion (X ). It means that the two
electrons constituting the original covalent
O-N=O bond are carried away by the halogen along
with it. The halogen atom of alkyl halide
is, therefore, called ‘leaving group’ in the

220

context of this reaction. Leaving group is H [ [1 H 1
HO C HO2 C B2 r
the group which leaves the carbon by taking
H HH
away the bond pair of electrons. The substrate H Br
Transition state
undergoes two changes during a SN reaction. (T.S.)
The original C-X bond undergoes heterolysis
H
and a new bond is formed between the carbon

and the nucleophile using two electrons of the

nucleophile. These changes may occur in one

or more steps. The description regarding the HO C + Br
H
sequence and the way in which these two
H
changes take place in SN reaction is called
mechanism of SN reaction. The mechanism is Fig. 10.4 : SN2 mechanism
deduced from the results of study of kinetics Salient features of SN2 mechanism :
i. Single step mechanism with simultaneous
of SN reactions. Two mechanisms are observed bond breaking and bond forming.
in various SN reactions. These are denoted as
SN1 and SN2 mechanisms. ii. Backside attack of nucleophile : The
nucleophile attacks the carbon undergoing
a. SN2 Mechanism : The reaction between substitution from the side opposite to that
methyl bromide and hydroxide ion to give of the leaving group. This is to avoid steric
repulsion (repulsion due to bulkyness of
methanol follows a second order kinetics, the groups) and electrostatic repulsion
between the incoming nucleophile and
that is, the rate of this reaction depends on the leaving group.

concentration of two reacting species, namely,

methyl bromide and hydroxide. Hence it is

called subtitution nucleophilic bimolecular,

SN2. iii. In the transition state (T.S.) the nucleophile
and leaving groups are bonded to the
CH3Br + OH CH3OH + Br carbon with partial bonds and carry
partial negative charge. (Thus, the total
rate = k [CH3Br] [OH ] negative charge is diffused.)

Rate of a chemical reaction is influenced by

the chemical species taking part in the slowest iv. The T.S. contains pentacoordinate
carbon having three σ (sigma) bonds in
step of its mechanism. In the above reaction one plane making bond angles of 1200
with each other and two partial covalent
only two reactants are present and both are bonds along a line perpendicular to this
plane.
found to influence the rate of the reaction.

This means that the reaction is a single step

reaction which can also be called the slow

step. This further implies that the two changes,

namely, bond breaking and bond forming at v. When SN2 reaction is brought about at
chiral carbon (in an optically active
the carbon take place simultaneously. This substrate), the product is found to have
opposite configuration compared to
SN2 mechanism is represented as shown in that of the substrate. In other words,
Fig. 10.4. SN2 reaction is found to proceed with
inversion of configuration. This is like
flipping of an umbrella (See Fig. 10.4).
It is known as Walden inversion. The
inversion in configuration is the result of
backside attack of the nucleophile.

221

b. SN1 Mechanism : The reaction between iv. When SN1 reaction is carried out at chiral
tert-butyl bromide and hydroxide ion to give carbon in an optically active substrate,
tert-butyl alcohol follows a first-order kinetics, the product formed is nearly racemic.
that is the rate of this reaction depends on This indicates that SN1 reaction proceeds
concentration of only one species, which is mainly with racemization. This means
the substrate molecule, tert-butyl bromide. both the enantiomers of product are formed
Hence it is called substitution nucelophilic in almost equal amount. Racemization in
unimolecular, SN1. SN1 reaction is the result of formation
of planar carbocation intermediate (Fig.
CH3 CH3 10.5). Nucleophile can attack planar
carbocation from either side which results
CH3− C − Br + OH CH3− C − OH+ Br in formation of both the enantiomers of
the product.
CH3 CH3

rate = k [(CH3)3CBr]

It can be seen in this reaction that Use your brain power

concentration of only substrate appears • Draw the Fischer projection
formulae of two products
in the rate equation; concentration of the obtained when compound (A) reacts
with OH by SN1 mechanism.
nucleophile does not influence the reaction
C2H5
rate. In other words, tert-butyl bromide reacts H3C− C − Br (A)

with hydroxide by a two step mechanism. n-C3H7

In the slow step C-X bond in the substrate • Draw the Fischer projection formula of
the product formed when compound (B)
undergoes heterolysis and in the subsequent reacts with OH by SN2 mechanism.

fast step the nucleophile uses its electron

pair to form a new bond with the carbon

undergoing change. This SN1 mechanism is
represented as shown in Fig. 10.5.

Step I CH3 CH3 (B)
(CH3)3C - Br slow C ⊕ + Br H− C − Cl

H3C CH3 C2H5
(carbocation intermediate)

Step II 10.6.4 Factors influencing SN1 and SN2
CH3 mechanism :

C⊕ + OH (CH3)3C - OH a. Nature of substrate : SN2 : The T.S. of
H3C CH3 SN2 mechanism is pentacoordinate and thus
crowded (See Fig. 10.4). As a result SN2
Salient Fig. 10.5 : SSNN11 mechanism : mechanism is favoured in primary halides
features of mechanism and least favoured in tertiary halides.

i. Two step mechanism. SN1 : A planar carbocation intermediate
is formed in SN1 reaction. It has no steric
ii. Heterolyis of C-X bond in the slow crowding. Bulky alkyl groups can be easily
and reversible first step to form planar accommodated in planar carbocation See
carbocation intermediate. (Fig. 10.5). As a result SN1 mechanism is
most favoured in tertiary halides and least
iii. Attack of the nucleophile on the favoured in primary halides. (Formation of
carbocation intermediate in the fast planar carbocation intermediate results in a
second step to form the product.

222

H H H H
HCH HCH HCH
11
22 11 11 11
22 22 22
(a) Nu C X
Nu C X Nu C X Nu C X
HH
HH HC H HC CH
HH H HH H

crowding and destabilization increases H
H Cα H
H H H
HCH H Cα H
(b)
C⊕ C⊕
HH C⊕ C⊕ H C α Cα H
HH
H Cα H H HH H
HH

steric relief, stabilization by +I and hyperconjugation of
α - hydrogens increases

Fig. 10.6 : Influence of substrate in SN1 and SN2 (a) Transition states (T.S.) in SN2
(b) Carbocation intermediates in SN1

relief from steric crowding present in the Problem 10.4 : Primary allylic and
tertiary halide substrate). primary benzylic halides show higher
reactivity by SN1 mechanism than
Secondly the carbocation intermediate is other primary alkyl halides. Explain.
stabilized by +I effect of alkyl substituents Solution : SN1 reaction involves
and also by hyperconjugation effect of alkyl formation of carbocation
substituents containing α-hydrogens. As a intermediate. The allylic and benzylic
result, SN1 mechanism is most favoured carbocation intermediate formed are
in tertiary halides and least favoured in resonance stabilized, and hence SN1
primary halides. This can be represented mechanism is favoured.
diagramatically as shown below.

SN1 rate increases CH2 = CH - C⊕H2 C⊕ H2 - CH = CH2

CH3 - X 10 20 30

Resonance stabilization of allylic carbocation

SN2 rate increases C⊕ H2 CH2 CH2
C⊕ H2
Tertiary halides undergo nucleophilic ⊕ ⊕
substitution by SN1 mechanism while primary
halides follow SN2 mechanism. Secondary ⊕ CH2
halides react by either of the mechanism or
by mixed mechanism depending upon the
exact conditions.

Resonance stabilization of benzylic carbocation

223

b. Nucleophilicity of the reagent : carbocation is relatively poor and solvation
of anion is particularly important. Anions are
Can you recall ? solvated by hydrogen bonding solvents, that
mis,oprerortaicpisdollyveinntsp.oTlahrusprSoNt1icresaoclvtieonntsprtohcaenedins
• Give some examples of aprotic solvents.
nucleophiles that are electrically
neutral. Polar protic solvents usually decrease the
rsaatsetepnuoocffleSoNSp2Nh2rielemaceitsciohnian.nviIosnlmvtehdes.urbAastterpaodtleeatrearsmsoilnvwieennlgtl
• Give some examples of anionic stabilizes nucleophile (one of the reactant)
nucleophiles. by solvation. Thus solvent deactivates the
nucleophile by stabilizing it. Hence aprotic
• What is the difference between a base solvents or solvents of low polarity will
and a nucleophile ? favour SN2 mechanism.

A nucleophile is a species that uses its Problem 10.5 : Which of the following
electron pair to form a bond with carbon. two compounds would react faster by SN2
Nucleophilic character of any species is mechanism and Why ?
expressed in its electron releasing tendency,
which can be corelated to its strength as CH3-CH2-CH2-CH2Cl CH3-CH-CH2-CH3
Lewis base.
Cl
A more powerful nucleophile attacks the
sTthuhebesntrraaatttueeroeffaosStfeN1rnumacneldecohfpaahnviiolseum.rsNisSuiNcn2ldemoeppeehcnihldeaennditsomeosf. 1-Chlorobutane 2-Chlorobutane
not react in slow step of SN1. It waits till
the carbocation intermediate is formed, and Solution : In SN2 mechanism, a
reacts fast with it. pentacoordinate T.S. is involved. The order
of reactivity of alkyl halides towards SN2
Do you know ? mechanism is,
Primary > Secondary > Tertiary, (due to
1. A negatively charged increasing crowding in T.S. from primary
to tertiary halides. 1-Chlorobutane being
nucleophile is more powerful primary halide will react faster by SN2
mechanism, than the secondary halide
than its conjugate acid. For example 2-chlorobutane.

R-O is better nucleophile than R-OH. Can you recall ?

2. When donor atoms are from same • How are alkenes prepared
period of periodic table, nucleophilicity from alkyl halides ?
decreases from left to right in a period.
For example H2O is less powerful • Which is stronger base from the
nucleophile than NH3. following ?

3. When donor atoms are from same group i. aq. KOH ii. alc. KOH
of the periodic table, nucleophilicity
increases down the group. For example,
I is better nucleophile than Cl .

c. Solvent polarity : SN1 mechanism proceeds
via formation of carbocation intermediate.
A good ionizing solvent, polar solvent,
stabilizes the ions by solvation. Solvation of

224

10.6.5 Elimination reaction : The different products of elimination do
Dehydrohalogenation not form in equal proportion. After studying
a number of elimination reactions, Russian
When alkyl halide having at least one chemist Saytzeff formulated an empirical rule
β-hydrogen is boiled with alcoholic solution of given below.
potassium hydroxide, it undergoes elimination
of hydrogen atom from β-carbon and halogen In dehydrohalogenation reaction, the
atom from α - carbon resulting in the formation preferred product is that alkene which has
of an alkene. greater number of alkyl groups attached to
doubly bonded carbon atoms.
Remember...
Therefore, in the above reaction but-2-ene
The carbon bearing halogen is is the preferred product, and is formed as the
commonly called α-carbon (alpha major product. It turned out that more highly
carbon) and any carbon attached to α-carbon substituted alkenes are also more stable alkenes.
is β-carbon (beta carbon). Hydrogens Hence Saytzeff elimination is preferred
attached to β-carbon are β-hydrogens. formation of more highly stabilized alkene
during an elimination reaction. The stability
This reaction is called β-elimination order of alkyl substituted alkenes is :

(or 1,2 - elimination) reaction as it involves R2C = CR2 > R2C = CHR > R2C = CH2,

elimination of halogen and a β - hydrogen RCH = CHR > RCH = CH2

atom. Do you know ?

H Elimination versus substitution:

B + βC αC alc. KOH C = C + B⊕H +X Alkyl halides undergo sunstitution as
∆ well as elimination reaction. Both reactions
are brought about by basic reagent, hence
X there is always a competition between these
two reactions. The reaction which actually
(base) (alkyl halide) (alkene) predominates depends upon following
factors.
As hydrogen and halogen is removed
in this reaction it is also known as a. Nature of alkyl halides : Tertiary alkyl
dehydrohalogenation reaction. halides prefer to undergo elimination
reaction where as primary alkyl halides
If there are two or more non-equivalent prefer to undergo substitution reaction.
β-hydrogen atoms in a halide, then this
reaction gives a mixture of products. Thus, b. Strength and size of nucleophile :
2-bromobutane on heating with alcoholic Bulkier electron rich species prefers to act
KOH gives mixture of but-1-ene and but-2- as base by abstracting proton, thus favours
ene. elimination. Substitution is favoured in the
case of comparatively weaker bases, which
β1 α β2 prefer to act as nucleophile

CH3 − CH2 − CH − CH3
Br

(2-bromobutane)

alc. KOH ∆

loss of β2 - hydrogen loss of β1 -hydrogen

HC3 − CH2 − CH = CH2 c. Reaction conditions : Less polar solvent,
high temperature fovours elimination where
(But-1-ene) as low tempertaure, polar solvent favours
substitution reaction.
CH3− CH = CH - CH3

(But-2-ene)

225

10.6 Reaction with active metals b. Wurtz reaction : Alkyl halides react with
metallic sodium in dry ether as solvent, and
Active metals like sodium, magnesium form higher alkanes containing double the
cadmium readily combine with alkyl chlorides, number of carbon atoms present in alkyl halide.
bromides and iodides to form compounds This reaction is called Wurtz reaction. (Refer
containing carbon-metal bonds. These are to Std. XI Chemistry Textbook sec. 1.5.3)
known as organometallic compounds.
2 R-X + 2 Na dry R - R + 2 NaX
a. Reaction with magnesium : When alkyl ether
halide is treated with magnesium in dry ether
as solvent, it gives alkyl magnesium halide. It 2C2H5 Br + 2Na dry CH3-CH2-CH2-CH3
is known as Grignard reagent. ether

(Ethyl bromide) (Butane)

R-X + Mg dry ether R - Mg - X + 2 NaBr
When a mixture of two different alkyl halides is
alkyl magnesium halide
(Grignard reagent) used, all the three possible alkanes are formed.

For example : C(eHth3-aCneH)3

Grignard reagents are very reactive compounds.

They react with water or compounds containing CH3Br + C2H5Br Na CH3-CH2-CH3
(Methyl (Ethyl dryether (propane)
hydrogen attached to electronegative element.
CH3-CH2-CH3-CH3
X bromide) bromide) (butane)

R - Mg -X + CH3OH R - H + Mg 10.6.1 Reaction of haloarenes :
R - Mg -X + NH3 (Hydrocarbon) OCH3
a. Reactions of haloarene with metals
X
R - H + Mg The reaction of aryl halide with alkyl
(Hydrocarbon) NH2
halide and sodium metal in dry ether to give
Do you know ?
substituted aromatic compounds is known as
Carbon-magnesium bond in
Grignard reagent is a polar covalant Wurtz- Fittig reaction. This reaction is an
bond. The carbon pulls electrons from the extension of Wurtz reaction and was carried
electropositive magnesium. Hence carbon out by Fittig. This reaction allows alkylation
in Grignard reagent has negative polarity of aryl halides.
and acts as a nucleophite
Br dry CH3
Rδ- Mδ⊕g X ether
+ CH3-Br + 2Na + 2NaBr
Victor Grignard received Nobel Prize
in 1912 for synthesis and study of organo- (Bromobenzene) (Toluene)
magnesium compounds. Grignard reagent
is a very versatile reagent used by organic In case only aryl halide takes part in the
chemist. Vinyl and aryl halides also form reaction, the product is biphenyl and the
Grignard reagent.
reaction is known as Fittig reaction.

Cl

+ 2Na dry + 2NaCl
ether

(Chlorobenzene) (Biphenyl)

226

b. Nucleophilic substitution SN of haloarenes: Thus nuclophilic substitution reaction
involving cleavage of C-X bond in haloarene
Can you recall ? proceeds with difficulty. However, the presence
• What is resonance? of certain groups at certain positions of the
ring, markedly activate the halogen of aryl
• Draw resonance structures of halides towards substitutuion. For example,
bromobenzene. presence of electron withdrawing group at
ortho and/or para postion greatly increases the
• Identify the type of hybridization of reactivity of haloarenes towards subsitution of
carbon to which halogen is attached in halogen atom. Greater the number of electron
haloarene. withdrawing groups at o/p position, greater
is the reactivity. Electron withdrawing group
Aryl halides show low reactivity at meta position has practically no effect on
towards nucleophilic substitution reactions. reactivity.
The low reactivity of aryl halides is due to :

i. Resonance effect and i. Cl OH
ii. sp2 hybrid state of C .

i. One of the lone pairs of electrons (i)NaOH 433 K
on halogen atom is in conjugation with (ii)H3O⊕
π -electrons of the ring. For example the
following different resonance structures can NO2 NO2
be written for chlorobenzene.
(p-nitrochlorobenzene) (p-nitrophenol)

: Cl: : + Cl : +Cl : ii. Cl NO2 OH NO2
:
: (i)aq.Na2CO3 403 K
:
: (ii)H3O⊕

I II III NO2 NO2

(2,4-dinitrochlorobenzene) (2,4 - dinitrophenol)

+ Cl : : Cl : iii. Cl OH
NO2 NO2
NO2 NO2

warm

IV V NO2 H2O NO2

Resonance structures II, III and IV show (2,4,6-trinitrochlorobenzene) (2,4,6 - trinitrophenol)
double bond character to carbon-chlorine bond.
Thus carbon-chlorine bond in chlorobenzene Can you tell ?
is stronger and shorter than chloroalkane
molecule, C-Cl bond length in chlorobenzene Conversion of chlorobenzene
is 169 pm as compared to C-Cl bond length to phenol by aqueous sodium
in alkyl chloride 178 pm. Hence it is difficult hydroxide requires high temperture of
to break. Phenyl cation produced due to self- about 623K and high pressure. Explain.
ionization of haloarene will not be stabilised
by resonance, which rules out possibility of SN1 Cl OH
mechanism. Back side attack of nucleophile is
blocked by the aromatic ring, which rules out (i) 623K, OH 300 atm Phenol
SN2 mechanism. (ii) H3O⊕

Chlorobenzene

227

Do you know ? • Identify the product A of following
reaction.
Occurrence of nucleophilic
substitution in p-nitrochlorobenzene + HNO3 conc. H2SO4 A
can be explained on the basis of resonance
stabilization of the intermediate. Aryl halides undergo electrophilic
substitution reaction slowly as compared to
Cl Cl OH benzene.

+ OH slow In resonance structures of chlorobenzene
N step (see section 10.6.5) elelctron density is
O⊕ O relatively more at ortho and para position.
N Therefore incoming electrophilic group is
Cl OH O ⊕O more likely to attack at these positions. But
due to steric hinderance at ortho position, para
(I) product usually predominates. In haloarenes,
Cl OH halogen atom has strong electron withdrawing
inductive effect (-I). This deactivates the ring
N N and electrophilic substitution reaction occurs
O ⊕O O ⊕O slowly.

(II) (III)

Cl OH OH Remember...
The -I effect of Cl is more
N fast + Cl
O ⊕O step powerful than its +R effect.
Therefore Cl is o-/p- directing but ring
(IV) N deactivating group.
O⊕O
i. Halogenation : It is carried out by reacting
The resonance structure (III) shows haloarene with halogen in presence of ferric
that the electron withdrawing nitro group salt as Lewis acid catalyst.
(-NO2) in the p-position extends the
conjugation. As a result, the intermediate Cl
carbanion is better stabilized which favours
nucleophilic substitution reaction. + Cl2

c. Electrophilic substitution (SE) in (Chlorobenzene)
arylhalides

Can you recall ? anhydrous Cl2
• What is an electrophile? FeCl3
• Give some examples of
Cl Cl
electrophiles + Cl
• What type of reactions are observed in + HCl

benzene? Cl (1,2 - Dichlorobenzene)
(minor)
228 (1,4 - Dichlorobenzene)
(major)

ii. Nitration : It is carried out by heating Cl O
haloarene with conc. HNO3 in presence of CH3-C -Cl
conc. H2SO4. +

Cl (Chlorobenzene)

+ HNO3 anhyd. AlCl3 Cl
conc.
Cl O
(Chlorobenzene)
C CH3 +
(4 -ChlorOoacCetopChHen3one)
conc. H2SO4 ∆ (1-Chloroacetophenone)

Cl Cl 10.7 Uses and Environmental effects of some
+ NO2 polyhalogen compounds

NO2 10.7.1 Dichloromethane/ methylene chloride
(CH2Cl2) : It is a colourless volatile liquid
(1-Chloro-4-nitrobenzene) (1-Chloro-2-nitrobenzene) with moderately sweet aroma. It is used as a
solvent, and used as a propellant in aerosols.
iii. Sulfonation : It is carried out by heating
haloarene with fuming H2SO4. Over exposure to dichloromethane causes
dizziness, fatigue, nausea, headaches,
Cl numbness, weakness. It is highly dangerous
if it comes in direct contact with eyes as it
+ H2SO4 damages cornea.
(fuming)
(chlorobenzene) 10.7.2 Chloroform / trichloromethane
(CHCl3) : It is a colourless liquid with
∆ peculiar sweet smell. It is used to prepare
chlorofluromethane, a refrigerant R-22. It
Cl Cl is used as a solvent for extraction of natural
SO3H products like gums, fats, resins. It is used as a
source of dichlorocarbene. Chloroform causes
+ depression of central nervous system. Inhaling
chloroform for a short time causes fatigue,
(1 - Chlorobenzene SO3H dizziness and headache. Long exposure to
sulfonic acid) chloroform may affect liver. Chloroform when
(4 - Chlorobenzene sulfonic exposed to air and light forms a poisonous
acid) compound phosgene so it is stored in dark
coloured air tight bottles.
iv. Friedel Craft’s reaction : It is carried out
by treating haloarene with alkyl chloride or 10.7.3 Carbon tetrachloride /
acyl chloride in presence of anhydrous AlCl3 tetrachloromethane (CCl4) :
as a catalyst.
It is a colourless liquid with sweet smell. It is
Cl Cl Cl very useful solvent for oils, fats, resins. It serves
CH3 as a source of chlorine. It is used as a cleaning
+ CH3-Cl anhyd. agent. It is highly toxic to liver. Exposure to
AlCl3 + high concentration of CCl4 can affect central

(Chlorobenzene) CH3
(1-Chlorotoluene) (4-Chloro-

toluene)

229

nervous system and it is suspected to be 10.7.6 Dichlorodiphenyltrichloroethane
carcinogenic. Prolonged exposure may cause (DDT) : It is colourless, tasteless and odorless
death. It is a green house gas. crystalline compound having insecticidal
property.
10.7.4 Idoform or triiodomethane (CHI3):
It kills insects such as houseflies,
It is a yellow crystalline substance with mosquitoes and body lice. It was used for
disagreeable smell. It is used in medicine as controlling maleria and typhus.
a healing agent and antiseptic in dressing of
wounds, however its use is limited. Exposure to high doses of DDT may cause
vomiting, tremors or shakiness. Laboratory
It causes irritation to skin and eyes. It may animal studies showed adverse effect of DDT
cause respiratory irritation or breathing on liver and reproduction. DDT is a pressistent
difficulty, dizziness, nausea, depression of organic pollutant, readily absorbed in soils
central nervous system, visual disturbance. and tends to accumulate in the ecosystem.
When dissolved in oil or other lipid, it is
10.7.5 Freons : These are organic compounds readily absorbed by the skin. It is resistant to
of chlorine and fluorine, chlorofluorocarbons, metabolism. It accumulates in fatty tissues.
CFC's are commonly used as refrigerants. There is a ban on use of DDT due to all these
The most common representative is adverse effects .
dichlorodifluromethane (Freon-12) others
include chlorodifluromethane (R-22), Cl Cl Cl
trichlorofluromethane (R-11) and so on.
Cl Cl
They are used as refrigerants in fridge and
airconditioning, propellants in aerosol and Do you know ?
solvents. They are used as blowing agents in
making foams and packing materials. DDT, the first chlorinated
organic insecticides, was originally
Chloroflurocarbons are responsible for ozone prepared in 1873, but it was not until 1939
depletion of ozone in stratosphere. Regular that Paul Muller of Geigy Pharmaceuticals
large inhalation of freons results in breathing in Switzerland discovered the effectiveness
problems, organ damage, loss of consciousness. of DDT as an insecticide. Paul Muller
was awarded the Nobel Prize in Medicine
Do you know ? and Physiology in 1948 for this discovery.
The use of DDT increased enormously
How do CFC distroy the ozone on a worldwide basis after World War
II, primarily because of its effectiveness
layer in the atmosphere ? against the mosquito that spreads malaria
and lice that carry typhus. Many species
When ultraviolet radiation (UV) of insects developed resistance to DDT,
and it was also discovered to have a
strikes CFC (CFCl3) molecules in the high toxicity towards fish. DDT is not
metabolised very rapidly by animals;
upper atmosphere, the carbon-chlorine instead, it is deposited and stored in the
fatty tissues. The use of DDT was banned
bond breaks and produces highly reactive in the United States in 1973, although it is
still in use in some other parts of the world.
chlorine atom (Cl).

CFCl3 CFCl2 + Cl
This reactive chlorine atom

decomposes ozone (O3) molecule into

oxygen molecule (O2).

O3 + Cl O2 + ClO
ClO + O O2 + Cl

One atom of chlorine can destroy upto

100,000 ozone molecules.

230

Exercises

1. Choose the most correct option. iv. The best method for preparation of
alkyl fluorides is
i. The correct order of increasing
reactivity of C-X bond towards a. Finkelstein reaction
nucleophile in the following
compounds is b. Swartz reaction

XX c. Free radical fluorination

(CH3)3C-X (CH3)2CH-X d. Sandmeyer's reaction

NO2 (III) (IV) v. Identify the chiral molecule from the
(I) (II) following.

a. 1-Bromobutane

a. I < II < III < IV b. 1,1- Dibromobutane

b. II < I < III < IV c. 2,3- Dibromobutane

c. III < IV < II < I d. 2-Bromobutane

d. IV < III < I < II vi. An alkyl chloride on Wurtz reaction
gives 2,2,5,5-tetramethylhexane.
ii. CH3-CH=CH2 HI The same alkyl chloride on reduction
peroxide with zinc-copper couple in alchol
give hydrocarbon with molecular
The major product of the above formula C5H12. What is the structure
reaction is, of alkyl chloride

a. I-CH2-CH=CH2 b. CH3-CH2-CH2I CH3 CH3
a. CH3-C-CH2Cl b. CH3-C-CH2CH3
c. CH3-CH-CH3 d. CH3-CH-CH2
CH3 Cl
I I OH

iii. Which of the following is likely to c. CH3-CH2-CH-Cl d. CH3-CH-CH-CHCl
undergo racemization during alkaline
hydrolysis ? CH3 CH3 CH3

CH3-CH-C2H5 CH2Cl vii. Butanenitrile may be prepared by
(II) heating
Cl
a. propanol with KCN
(I)

Cl CH3 b. butanol with KCN
(III)
CH3-CH c. n-butyl chloride with KCN

CH2Cl d. n-propyl chloride with KCN
(IV)

a. Only I b. Only II

c. II and IV d. Only IV

231

viii. Choose the compound from the c. CH3-CH-CH=CH2 + HBr peroxide
following that will react fastest by SN1 CH3
mechanism. OH

a. 1-iodobutane d. + SOCl2

b. 1-iodopropane CH3
e. + Cl2
c. 2-iodo-2 methylbutane

d. 2-iodo-3-methylbutane dark
Fe

ix. Cl + Mg dry A H2O iii. Identify chiral molecule/s from the
ether B following.
a. CH3-CH-CH2-CH3
The product 'B' in the above reaction OH
sequence is,
b. CH3-CH2-CH-CH2-CH3
a. Mg b. Mg-Cl
Br
c. Cl Mg d.
c. CH3-CH2-CH2-CH2Br
x. Which of the following is used as
source of dichlorocarbene d. CH3-CH-CH3-CH3

a. tetrachloromethane CH3

b. chloroform iv. Which one compound from the
following pairs would undergo SN2
c. iodoform faster from the?

d. DDT

2. Do as directed. a. CH2Cl and Cl

i. Write IUPAC name of the following b. CH3CH2 CH2I and CH3CH2CH2Cl
compounds v. Complete the following reactions

a. CH3-CH=C-CH-Br giving major product.

H3C CH3

b. CH3-CH-CH-CH2-CH3 a. CH3-CH=CH2 HBr A alc. KOH B
peroxide
Cl CH3
b. CH3-CH=CH2 Red P/Br2 A Ag2O/H2O B
OH
Cl Cl

c. d. CH3
c. CH3-C-CH2-Cl
CH3 Na/dry ether A
Cl CH3
C2H5

ii. Write structure and IUPAC name Cl
of the major product in each of the d. AMg
following reaction.
dry ether

a. CH3-CH-CH2Cl + NaI Acetone
CH3

b. CH3-CH2Br + SbF3

232

vi. Name the reagent used to bring about iv. But-1-ene to n-butyl iodide
the following conversions. v. 2-Chloropropane to propan-1-ol
vi. tert-Butyl bromide to isobutyl bromide
a. Bromoethane to ethoxyethane vii. Aniline to chlorobenzene
viii. Propene to 1-nitropropane
b.1-Chloropropane to 1 nitropropane
7. Answer the following
c. Ethyl bromide to ethyl
isocyanide

d. Chlorobenzene to biphenyl i. HCl is added to a hydrocarbon 'A'
(C4H8) to give a compound 'B' which
vii. Arrange the following in the increase on hydrolysis with aqueous alkali
order of boiling points forms tertiary alcohol 'C' (C4H10O).
Identify 'A' , 'B' and 'C'.
a. 1-Bromopropane

b. 2- Bromopropane ii. Complete the following reaction
sequences by writing the structural
c. 1- Bromobutane formulae of the organic compounds
'A', 'B' and 'C' .
d. 1-Bromo-2-methylpropane

viii. Match the pairs. a. 2-Bromobutane alc.KOH A Br2 B NaNH2 C

Column I Column II b. Isopropyl alcohol ∆ A NH3 excess B
PBr3
a. CH3CH-CH3 i. vinyl halide

X iii. Observe the following and answer the
questions given below.
b. CH2=CH-CH2X ii. alkyl halide

c. CH2=CH-X iii. allyl halide CH2=CH-X CH2-CH=X⊕

iv. benzyl halide a. Name the type of halogen
derivative
v. aryl halide

3. Give reasons b. Comment on the bond length of
C-X bond in it
i. Haloarenes are less reactive than halo
alkanes. c. Can react by SN1 mechanism?
Justify your answer.
ii. Alkyl halides though polar are
immiscible with water. Activity :
1. Collect detailed information
iii. Reactions involving Grignard reagent
must be carried out under anhydrous about Freons and their uses.
condition. 2. Collect information about DDT as a

iv. Alkyl halides are generally not persistent pesticide.
prepared by free radical halogenation Reference books
of alkanes. i. Organic chemistry by Morrison, Boyd,

4. Distinguish between - SN1 and SN2 Bhattacharjee, 7th edition, Pearson
mechanism of substitution reaction ? ii. Organic chemistry by Finar, Vol 1, 6th

5. Explain - Optical isomerism in edition, Pearson
2-chlorobutane.

6. Convert the following.
i. Propene to propan-1-ol
ii. Benzyl alcohol to benzyl cyanide
iii. Ethanol to propane nitrile

233

11. ALCOHOLS, PHENOLS AND ETHERS

Can you recall ? Do you know ?

1. What is the name and formula Epoxide are cyclic ethers in which
of 2nd member of homologous the ethereal oxygen is a part of a
series of alcohols? three membered ring.

2. What is the structural formula of CC H2C CH2
functional group of ether? O O

3. What is the name of the compound (Ethylene oxide) (1,2 - Epoxyethane)
having -OH group bonded to benzene
ring?

11.1 Introduction : Alcohols are organic 11.2 Classification : Let us first consider
compounds whose molecules contain hydroxyl classification of alcohols, phenols and then
group, (-OH) attached to a saturated carbon ethers.
atom.
11.2.1 Mono, di, tri and polyhydric
C OH compounds : Alcohols and phenols are
Hydroxyl group can also be present in aromatic classified as mono, di-, tri, or polyhydric
compounds. There are two types of aromatic compounds on the basis of one, two, three
hydroxy compounds: phenols and aromatic or more hydroxyl groups present in their
alcohols. Phenols contain a hydroxyl group molecules as :
directly attached to the carbon atom of benzene
ring. When the hydroxyl group is present in the CH3 - CH2 - OH H2C - OH H2C - OH
side chain of aromatic ring, the compound is H2C - OH HC - OH
termed as aromatic alcohol.
OH H2C - OH
OH OH OH
OH
OH OH
Phenols
OH

Monohydric Dihydric alcohols/ Trihydric

alcohols/phenols phenols alcohols/phenols

CH2-OH Monohydric alcohols are further classified on
the basis of hybridisation state of the carbon
Aromatic alcohol atom to which hydroxyl group is attached.
a. Alcohols containing sp3C - OH bond : In
Ethers are compounds which contain these alcohols -OH group is attached to a sp3
an oxygen atom bonded to two alkyl groups hybridised carbon atom of alkyl group. These
or two aryl groups or one alkyl and one aryl alcohols are further classified as primary
group. Ethers are organic oxides. Ethers are (10), secondary (20) and tertiary (30) alcohols
considered as anhydrides of alcohols. in which -OH group is attached to primary,
secondary and teriary carbon atom respectively.
R-O-R', R-O-Ar, Ar-O-Ar'. (see Fig. 11.1), also refer to Std. XI Chemistry
Textbook Chapter 14, sec. 14.3.2)

234

H Primary R Secondary b. Alcohols contianing sp2C -OH bonds :
carbon carbon
In these alcohols -OH group is attached
R C OH R C OH to a sp2 hybridised carbon atom which is part of
a carbon-carbon double bond. These alcohols
H H are known as vinylic alcohols. For example
10 alcohol 20 alcohol
CH2 = CH - OH (Vinyl alcohol)
R tertiary 11.2.2 Classification of Ethers : Ethers are
carbon classified as symmetrical ethers (simple
ethers) or unsymmetrical ethers (mixed
R C OH ethers) depending on whether the two alkyl/
aryl groups bonded to oxygen atom are same
R or different respectively. For example :
30 alcohol
R - O - R/Ar - O - Ar
Fig. 11.1 : Primary, secondary and tertiary
alcohols CH3 - O - CH3

Each of these three types of alcohols can also symmetrical ethers (simple ethers)
be either allylic or benzylic if the sp3 carbon
carrying -OH is further bonded to sp2 carbon. R - O - R'/Ar - O - Ar'

• Allylic alcohols : In this type of alcohols CH3 - O - C2H5, C6H5 - O - CH3
-OH group is attached to sp3 hybridised
carbon atom which is further bonded to a unsymmetrical ethers (mixed ethers)
carbon-carbon double bond. Allylic alcohol
may be primary, secondary or tertiary. 11.3 Nomenclature :

• Benzylic alcohols : In this type of alcohols 11.3.1 Alcohols : There are three systems of
-OH group is attached to sp3 hybridised nonmenclature of monohydric alcohols.
carbon atom which is further bonded to an
aromatic ring. Benzylic alcohol may be a. Common/trivial names : The common
primary, secondary or tertiary. or trivial names of alcohols are obtained by
adding word alcohol after the name of alkyl
Use your brain power group bonded to -OH. Names of higher alkyl
groups also include prefixes like normal, iso,
Classifiy the following secondary, tertiary (see. Table 11.1).

alcohols as 10 / 20 / 30 and allylic/ b. Carbinol system : In this system alcohols
are considered as derivatives of methyl
benzylic - CCHH2 3- OH, alcohol which is called carbinol. The alkyl
H2C = CH group attached to the carbon carrying -OH
group are named in alphabetical order. Then
H2C = CH - CH - OH, the suffix carbinol is added. For example :

CH3 H Carbinol carbon
CH3
H3C C OH
H2C = CH - C - OH, H
CH3
Methyl carbinol
CH3
CH2 OH CH OH

235

Use your brain power 11.3.2 Nomenclature of phenols : The
hydroxyl derivative of benzene is called

Name t-butyl alcohol using phenol. The IUPAC system name of phenol is

carbinol system of nomenclature. benzenol. The common name phenol is also

c. IUPAC system : Accoriding to IUPAC accepted by IUPAC. The common names have
system (Std XI Chemistry Text book, Chapter prefixes ortho, meta and para in subsituted
14), alcohols are named as alkanols. The phenols. IUPAC system uses the locant 2-, 3-,
ending ‘e’ in the name of the parent alkane, 4-, etc. to indicate the positions of substituents
alkene or alkyne is replaced by the suffix (see Table 11.2).

‘ol’. For naming polyhydric alcohol, ‘e’ in the 11.3.3 Nomenclature of Ethers : In the

ending of alkane is retained, the ending ‘ol’ is common system of nomenclature, the ethers

added and number of -OH groups is indicated are named by writing names of the alkyl groups

by prefix di, tri, etc., before ‘ol’. The positions attached to the oxygen atom in alphabetical

of -OH groups are indicated by appropriate order and word ether is added. If two alkyl

locants. For example ethane -1,2-diol (see. groups are same, prefix di- is used. According

Table 11.1). Similarly cyclic alcohols are to the IUPAC system of nomenclature, ethers

named by using prefix cyclo to the parent are named as alkoxyalkanes (see Table 11.3).

alkane and considering -OH group attached to The larger alkyl group is considered to be the

C-1 carbon atom. parent alkane. The name of the smaller alkane

is prefixed by the name of alkoxy group and its

OH locant. For example :

Cyclobutanol C2H5 -12CCHH3- Some alkoxy group
OH
CH3 - O 3 CH3-CH2O- Ethoxy
2- Ethylcyclopentanol CH3-CH2-CH2-O- n-Propoxy
CH3 Isopropoxy
CH3-CH-O-
2-Methoxypropane CH3

Table 11.1 Common/Trivial and IUPAC Names of some alcohols

Structural formula Common/ Trivial Name IUPAC Name

H3C-OH Methyl alcohol Methanol
H3C-CH2-OH Ethyl alcohol Ethanol
n -Propyl alcohol Propan -1-ol
H3C-CH2-CH2-OH sec-Butyl alcohol Butan -2-ol
H3C-CH2-CH-OH
Isobutyl alcohol 2- Methylpropanol
CH3
H3C-CH-CH2-OH tert-Butyl alcohol 2-Methylpropan-2-ol

CH3 Ethylene glycol Ethane-1, 2-diol
CH3

H3C C OH

CH3
H2C CH2

OH OH

H2C − CH − CH2 Propylene glycerol Propane-1,2,3-triol
OH OH OH

H3CCH = CHCH2OH Crotonyl alcohol But-2-en-1-ol

236

Table 11.2 Common and IUPAC names of some phenols

Structural formula Common name IUPAC Name
OH Phenol Benzenol/Phenol

OH o- Cresol 2-Methylphenol
CH3

HO p-Nitrophenol 4-Nitrophenol
Catechol Benzene-1,2-diol
NO2
OH

OH

OH Resorcinol Benzene-1,3-diol
Hydroquinone/ Benzene -1,4-diol
OH Benzene-1,3,5-triol
HO quinol Benzene-1,2,3-triol
Phloroglucinol
OH
OH Pyrogallol

HO OH
OH

HO OH

Table 11.3 Common and IUPAC Names of some Ethers

Structural formula Common Name IUPAC Name
Dimethyl ether Methoxymethane
H3C - O - CH3 Ethyl methyl ether Methoxyethane
H3C - O - CH2 - CH3 Methyl n-propyl ether 1-Methoxypropane
H3C - O - CH2 - CH2 - CH3 Methyl phenyl ether (Anisole) Methoxybenzene

C6H5 - O - CH3 Phenyl n-propyl ether 1- Propoxybenzene
O - CH2 - CH2 - CH2

CH3 OCH3 - 2- Methoxy-1,1-
H3C dimethylcyclobutane

237

Problem 11.1 : Draw structures of alcohol. (Refer to Std. XI Chemistry Textbook
following compounds. section 15.2.4). This is an antimarkownikoff
hydration of alkene.
i. 2,5-Diethylphenol ii. Prop-2-en-1-ol

iii. 2-methoxypropane iv. Phenylmethanol Do you know ?
The mechanism of acid catalyzed
Solutuion : hydration of alkene involves the
following three steps:
i. OH ii. 3 2 1
1 2 C2H5 H2C = CH - CH2 - OH
6 4 Step 1: Formation of carbocation
5 3 intermediate.

H5C2 H
H-O-H+ C=C
iii. CH3-CH-CH3 iv. CH2 - OH H2O + C-C
O-CH3 ⊕ ⊕
H

Step 2: Nucleophilic attack of H2O on C⊕

Try this... H C-C
Write IUPAC names of the C-C +H-O H O⊕
following compounds.
⊕
H

i. OH OH ii. H3C-CH-CH2- CH3 HH
OCH2- CH3
Step 3: Deprotonation

C-C H2O C - C + H3O⊕
H OH
iii. CH2 - CH2 - OH iv. H3C OH H O⊕
HH

11.4 Alcohols and Phenols : Use your brain power

11.4.1 Prepartion of alcohols : Predict the major product of the
following reactions :
a. From alkyl halide by hydrolysis with
aqueous alkali or moist silver oxide (refer to • CH3 - CH = CH2 (i)B2H6 - THF ?
section 10.6.2) (ii)H2O2, OH

b. By acid catalyzed hydration of alkenes : • CH ?3
(i) con. H2SO4
Alkene reacts with sulfuric acid to produce (ii) H2O
alkyl hydrogen sulfate, which on hydrolysis
gives alcohol (Refer to Std XI Chemistry d. By reduction of carbonyl compounds :
Textbook, section 15.2.4). This reaction i. By reduction of aldehydes and ketones :
follows Markownikoff’s rule. Aldehydes on reduction by H2/Ni or
LiAlH4 give primary alcohols (10). Similarly
c. Hydroboration - Oxidation of alkenes : ketones on reduction with H2/Ni or LiAlH4
give secondary alcohols (20).
With diborane (B2H6) alkene undergoes
addition reaction (Hydroboration) to give to H2/Ni or Pd
trialkylborane (R3B), which on oxidation with
hydrogen peroxide in alkaline medium gives R - CHO ∆ R- CH2 - OH
(i) LiAlH4

(ii) H3O⊕ 10 alcohol

238

O H2/Ni or Pd R CH OH Problem 11.2 : Predict the products for the
RCR following reaction.
∆ R
(i)LiAlH4 20 alcohol ?
(ii) H3O⊕ ?

ii. By reduction of carboxylic acids : H2/Ni
Caboxylic acids require strong reducing agent
LiAlH4 to form primary alcohols. CH3 - CH = CH - CH2 - CHO
(A) (i) LiAlH4
(ii) H3O⊕

Solution : The substrate (A) contains an

O isolated C = C and an aldehyde group.
H2/Ni can reduce both these functional
R C OH (i) LiAlH4 R - CH2 - OH groups while LiAlH4 can reduce only -CHO
(ii) H3O⊕ of the two, Hence

However LiAlH4 is an expensive reagent. (A) H /Ni CH3-CH2-CH2-CH2-CH2-OH
Therefore, commercially acids are first CH3-CH=CH-CH2-CH2-OH
transformed into esters which on catalytic 2
hydrogenation give primary alcohols.
(i(i)i)HL3iOA⊕lH4

R - COOH + R'OH H⊕ R - COOR' + H2O This reaction is useful in synthesis of a variety
of alcohols (see Table 11.4).
RCOOR' + 2H2 Ni/Pd R - CH2OH + R'OH
∆

Remember... Table 11.4 Preparation of alcohols by Grignard
reagent
The advantage of LiAlH4 over
H2/Ni is that it does not reduce the Aldehyde/ Grignard Final product Type of
isolated olefinic bond and hence it can ketone reagent alcohol
reduce unsaturated aldehyde and ketones to
unsaturated alcohols. H - CHO R - Mg Br R - CH2OH 10
(formaldehyde) 20

R' - CHO R - Mg Br R - CH - OH 30
(aldehyde)
R'
e.By addition of Grignard reagent to
aldeheydes and ketones : Grignard reagent R' - CO - R'' R''
reacts with aldehyde or ketone to form an (ketone) R - Mg Br R - C - OH
adduct which on hydrolysis with dilute acid
gives the corresponding alcohols. R'

δO OMgX Do you know ?
δ⊕C -C-
δ δ⊕ dry Epoxide reacts with Girgnard
+ R - Mg - X ether R reagent followed by acidic hydolysis
to give primary alcohols
(adduct)
H2C - CH2 + R Mg X dry
OH ether
- C - + Mg
H3O⊕ X O

R OH [R - CH2- CH2 - OMgX] H3O⊕
X

R-CH2CH2-OH + Mg OH

239

11.4.2 Preparation of phenol : d. From aniline : Aniline is treated with nitrous
acid [NaNO2 + HCl] at low temparature to
a. From chlorobenzene (Dow Process) : obtain benzene diazonium chloride, which on
hydrolysis gives phenol (Also refer to chapter
Chlorobenzene is fused with NaOH at high 13 for this reaction).

temperature and pressure (623K and 150atm)

followed by treatment with dilute HCl to

obtain phenol. O N⊕a OH NH2 ⊕N2Cl
Cl
+ HNO2 (Benzene diazonium
NaOH H3O⊕ NaNO2 chloride)
623 K/150atm (Aniline)
HCl OH
273 K

(Chloro (Sodium (Phenol)
benzene) phenoxide)

b. From Cumene : This is the commercial

method of preparation of phenol. Cumene H2O + N2
∆
(isopropylbenzene) on air oxidation in

presence of Co-naphthenate gives cumene (Phenol)

hydroperoxide, which on decomposition with

dilute acid gives phenol with acetone as a 11.4.3 Physical Properties of alcohols and
phenols
valuable by product.
Try this...
CH3 CH3 Arrange O-H, C-H and N-H
H3C CH H3C C O O H
bonds in increasing order of their
Co-naphthenate bond polarity.

+ O2 423K a. Nature of intermolecular forces : Alcohols
and phenols are very polar molecules due to
(Cumene) (air) (Cumene hydroperoxide) presence of -OH groups. The polar -OH groups
are held together by the strong intermolecular
dil. HCl OH CH3 forces, namely hydrogen bonding.
∆ + C=O

(Phenol) CH3

(Acetone)

c. From benzene sulfonic acid : Benzene δH⊕- Oδ - δH⊕- δ - δH⊕- Oδ - δH⊕- δ -
O O
sulfonic acid on neutralization by NaOH

gives sodium benzene sulfonate, which on RR

fusion with solid NaOH at 573 K gives sodium

phenoxide, followed by reaction with dilute (alcohol) Intermolecular (phenol)
hydrogen bonding
acid gives phenol. SO3N⊕a
SO3H b. Physical State : Lower alcohols are
colourless, toxic liquids having characterstic
NaOH alcoholic odour. Pure phenol is colourless,
toxic, low melting solid having characterstic
(Benzene sulphonic acid) (Sodium benzene sulphonate) carbolic or phenolic odour.

O ⊕Na OH c. Boiling Points : The boiling points of
alcohols and phenols increase with increase in
NaOH dil. HCl their molecular mass (Table 11.5).
523 K

(Sodium phenoxide) (Phenol)

240

Table 11.5 M.P/B.P and solubilities of some alcohols and phenols

Name Formula M.P. (0C) B.P. (0C) Solubility (g/100g H2O)
Methyl alcohol H3C-OH -97 65 0.793
Ethyl alcohol H3C-CH2-OH -115 78
n-Propyl alcohol H3C-CH2-CH2-OH 0.789
Isopropyl alcohol H3C-CH-OH -126 97
CH3 -86 83 0.804
n-Butyl alcohol H3C-CH2-CH2-CH2-OH
Isobutyl alcohol H3C-CH-CH2-OH 0.789
CH3
sec-Butyl alcohol H3C-CH2-CH-OH -90 118 0.810
-108 108 0.802
tert-Butyl alcohol CH3
H3C -114 99.5 0.806
Phenol H3C C-OH
H3C 25.5 83 0.789

OH 41 182 9.3

p-Cresol H3C OH 35 202 2.3
o-Nitrophenol 45 217 0.2
OH 114 1.7
p-Nitrophenol O2N NO2 -

OH

Problem 11.3 : The boiling point of n-butyl d. Solubility : Phenols and lower alcohols
alcohol, isobutyl alcohol, sec-butyl alcohol (having upto three carbons) show appreciable
and tert-butyl alcohol are 1180C, 1080 C. solubility in water due to their ability to form
990C and 820C respectively. Explain. intermolecular hydrogen bonding with water
molecule (See Table 11.5).
Solution : As branching increases
intermolecular van der Waal’s force become δ- δ⊕ O H δ- δ⊕ O H
weaker and the boiling point decreases. H H
Therefore n-butyl alcohol has highest
boiling point 1180C and tert-butyl alcohol R-O Hδ ⊕Oδ -H Ar-O Hδ ⊕Oδ - H
has lowest boiling point 830C. Isobutyl
alcohol is a primary alcohol and hence its Hydrogen H Hydrogen
boiling points is higher than that of sec- bond H
butyl alcohol.
bond Intermolecular hydrogen bonding

of R-OH and Ar-OH with water

241