(3) By catalytic reduction of styrene : C6H5CH = CH2 + H2 → C6H5CH2CH3
(4) By alkyl benzene synthesis : C6 H5 H + H2C = CH2 AlCl3 , HCl → C6 H 5 CH 2CH 3
95°C,Pressure
It undergoes electrophilic substitution reactions in the same way as toluene. When oxidised with dil. HNO3 or
alkaline KMnO4 or chromic acid it forms benzoic acid.
C6H5C2H5 [O] → C6H5COOH
Styrene (C6H5CH=CH2).
It is present in storax balsam and in coal-tar traces.
(1) Preparation
(i) Dehydrogenation of side chain of ethylbenzene : Dehydrogenation of side chain is affected by
heating ethylbenzene to high temperature in presence of a catalyst.
CH2CH3 CH = CH2
+ CH2 = CH2 AlCl3 600°C
Cr2O3 / Al2O3
Benzene Ethylbenzene Styrene
(ii) Decarboxylation of cinnamic acid : This is the laboratory preparation. It involves heating of cinnamic
acid with a small amount of quinol.
C6H5CH = CHCOOH Quinol → C6H5CH = CH2 + CO2
(iii) Dehydration of 1-phenyl ethanol with H2SO4 : C6 H5CHOHCH3 H2SO4 → C6 H5CH = CH2
−H2O
(iv) Dehydration of 2-phenyl ethanol with ZnCl2 : C6 H5CH2CH2OH ZnCl2 ,heat → C6 H5CH = CH2
−H2O
(v) Dehydrohalogenation of 1-phenyl-1-chloro ethane : On heating with alcoholic potassium
hydroxide, a molecule of hydrogen chloride is eliminated by the chloroderivative.
C6 H5CHClCH 3 Alc.KOH → C6 H5CH = CH2
Heat
(2) Properties : It is a colourless liquid, boiling point 145°C. On keeping, it gradually changes into a solid
polymer called metastyrene. The polymerisation is rapid in sunlight or when treated with sodium. It shows
properties of benzene ring (Electrophilic substitution) and unsaturated side chain (Electrophilic addition). However,
the side chain double bond is more susceptible to electrophilic attack as compared to benzene ring.
At lower temperature and pressure, it reacts with hydrogen to produce ethylbenzene and at higher
temperature and pressure, it is converted into ethyl cyclohexane.
CH = CH2 CH2CH3 CH2CH3
H2 / Ni H2 / Ni
20°C, 3 atm 125°C, 110 atm
Styrene Ethyl benzene Ethyl cyclohexane
With bromine, it gives the dibromide. CHBr.CH2Br
CH = CH2
+ Br2
Styrene Styrene dibromide
Halogen acids add to the side chain. C6H5CH = CH2 + HX → C6H5CHXCH3
Preparation of ring substituted styrenes is not done by direct halogenation but through indirect route.
CH2CH3 CH2CH3 CHClCH3 CH = CH2
+ Cl2 FeCl3 Cl2 Alc.. KOH
hv Heat
Cl Cl Cl
When oxidised under drastic conditions, the side chain is completely oxidised to a carboxyl group.
CH = CH2 COOH
[O]
KMnO4
Styrene
In presence of peroxides, styrene undergoes free radical polymerisation resulting in the formation of
polystyrene – an industrially important plastic.
Peroxide →−
nC6 H5CH = CH 2 CH − CH 2 −
|
C6H5
n
Co-polymers of styrene with butadiene and other substances are also important since many of them are
industrially useful products such as SBR ( A rubber substitute).
Bi-phenyl (C6H5 – C6H5).
It occurs in coal-tar. It is the simplest example of an aromatic hydrocarbon in which two benzene rings are
directly linked to each other.
(1) Methods of formation
(i) Fittig reaction : It consists heating of an ethereal solution of bromobenzene with metallic sodium.
Br + 2Na + Br + 2NaBr
(ii) Ullmann biaryl synthesis : Iodobenzene, on heating with copper in a sealed tube, forms biphenyl. The
reaction is facilitated if a strong electron wihtdrawing groups is present in ortho or para position.
I + 2Cu + I + 2CuI
(iii) Grignard reaction : Phenyl magnesium bromide reacts with bromo benzene in presence of CoCl2 .
MgBr + Br CoCl2 + MgBr2
(2) Properties : It is a colourless solid, melting point 71°C. It undergoes usual electrophilic substitution
reactions. Since aryl groups are electron withdrawing , they should have deactivating and m-orientating effect. But,
it has been experimentally shown that presence of one benzene ring activates the other for electrophilic substitution
and directs the incoming group to o- and p- positions. It has been shown that monosubstitution in the bi-phenyl
results in the formation of para isomer as the major product.
Another special feature of the biphenyl is the behaviour towards second substitution in a monosubstituted
biphenyl. The second substituent invariably enters the unsubstituted ring in the ortho and para position no matter
what is the nature of substituent already present.
HNO3 / H2SO4 NO HNO3 / H2SO4 O2N NO2
2
Diphenyl methane.
(1) Methods of preparation
(i) Friedel-craft's reaction : C6 H5CH 2Cl+ C6 H6 AlCl3 → C6 H5CH 2C6 H5 + HCl or
Benzyl chloride Benzene Diphenyl methane
2C6H6 + CH2Cl2 AlCl3 → C6H5CH2C6H5 + 2HCl
Dichloromethane
(ii) By action of formaldehyde on benzene in presence of conc. sulphuric acid
2C6 H6 + O = CH2 Conc. H2SO4 → C6 H5CH2C6 H5 + H2O
(iii) By Grignard reaction : Phenyl magnesium bromide reacts with benzyl bromide to from diphenyl
methane. C6H5MgBr + BrCH2C6H5 → C6H5CH2C6H5 + MgBr2
(iv) By reduction of benzophenone : Reduction can be done with LiAlH4 or P and HI.
C6H5COC6H5 4[H] → C6H5CH2C6H5 + H2O
(2) Properties : It is a colourless solid, melting point 26°C.Like biphenyl, it also easily undergoes electrophilic
substitution reactions.
CH2 HNO3 CH2 NO2 HNO3 O2N CH2 NO2
H2SO4 H2SO4
The methylene hydrogens of diphenylmethane are situated on carbon atom linked by two electron attracting
benzene rings. Thus, these are somewhat acidic in nature.
C6 H5CH 2C6 H5 + Br2 → C6 H5CHBrC6 H5 + HBr
When oxidised with K2Cr2O7 / H 2SO4 mixture, it forms benzophenone.
C6 H5CH 2C6 H5 [O] → C6 H5CC6 H5
||
O
It forms fluorene when its vapours are passed through a red hot tube.
CH2 Heat
HH Fluorene
Polynuclear hydrocarbons.
Compounds having two or more benzene rings fused together in ortho positions are termed as fused
polynuclear hydrocarbons. These hydrocarbons also called fused ring hydrocarbons.
Naphthalene Anthracene Phenanthren
(1) Naphthalene
Naphthalene is the largest single constituent of coal-tar (6-10%). It is obtained in the middle oil fraction of
coal-tar distillation. It is recovered as crude product when the middle oil fraction is cooled. The crude crystalline
product is separated by centrifugation and purified by washing successively with dilute H 2SO4 (to remove basic
impurities), sodium hydroxide solution (to remove acidic impurities) and water. Finally, the solid is sublimed to get
pure naphthalene.
Aromatic hydrocarbons
Word aromatic is now reserved for benzene and for the carbocyclic derivatives which resembles with benzene in
chemical behaviour. These are also known as benzenoid compounds. All aromatic hydrocarbons (benzene,
naphthalene, anthracene etc.) have been given a new name “Arenes”.
Monocyclic aromatic hydrocarbon Arenes Polycyclic aromatic hydrocarbon
Example : Benzene, Toluene Example : Naphthalene Anthracene
Source of Arenes.
Source of arenes is coal. It contains benzene, xylene, naphthalene etc. Arenes are obtained by destructive
distillation of coal.
(1) Distillation of coal COAL
Heated to 1273-1373 K
(Destructive distillation)
COKE HOT VAPOURS AND GASES
(Solid residue nearly 70%).
It is used as a fuel and as reducing Cooled and passed through water
agent in metallurgy.
Condensed liquid COAL GAS
(Mixture of uncondensed gases nearly 17%).
Allowed to settle,
two layers are separated It is used as a fuel.
Upper layer Lower layer
AMMONIACAL LIQUOR (nearly 8-10%) COAL TAR (nearly 4-5%)
It is used for the preparation of Black, viscous liquid having unpleasant odour
(NH4)2SO4 to be used as a fertilizer.
Coal tar is a mixture of large numbers of arenes.
(2) Distillation of coal tar : Arenes are isolated by fractional distillation of coal tar,
Name of the fraction Temperature range (K) Main constituents
Light oil (or crude oil) fraction Upto 443 Benzene, toluene, xylene
Middle oil fraction (Carbolic oil) 443-503 Phenol, naphthalene, pyridine
Heavy oil fraction (Creosote oil) 503-543 Naphthalene, naphthol and cresol
Green oil (Anthracene oil) 543-633 Anthracene, phenanthrene
Pitch (left as residue) Non-volatile Carbon
Note : The residue left after fractional distillation of coal-tar is called pitch.
(3) Isolation of benzene
Light oil coldH2SO4 → Basic impurities removed NaOH → Phenols removed distillation → Benzene (255 - 257K)
[Like pyridene] Toluene (383 K)
[Acidic impurities]
General characteristics of arenes.
(1) All arenes have general formula [Cn H 2n − 6y] . Where y is number of benzene rings and n is not less than 6.
(2) Arenes are cyclic and planar. They undergo substitution rather than addition reactions.
(3) Aromaticity or aromatic character : The characteristic behaviour of aromatic compounds is called
aromaticity. Aromaticity is due to extensive delocalisation of π-electrons in planar ring system. Huckel (1931)
explained aromaticity on the basis of following rule.
Huckel rule : For aromaticity the molecule must be planar, cyclic system having delocalised
(4n + 2)π electrons where n is an integer equal to 0, 1, 2, 3,------.
Thus, the aromatic compounds have delocalised electron cloud of 2,6,10 or 14 π electrons.
Example : 4n + 2 = 6 ; 4n = 4 ;n = 4 =1
4
Naphthalene Anthracene
14π electrons
10π electrons
n=2 n=3
Similarly cyclolpentadienyl anion or tropylium ion are also aromatic because of containing 6π electrons (n=1).
HH H
HH
H .. H ⊕
H ⊕H Cyclopropenyl cation (n = 0)
H HH
Cyclopentadienyl anion 6π electrons
Tropyllium ion 6 π electrons
Hetrocyclic compounds also have 6π electrons (n = 1).
.. .. ..
N FuOran ThioSphene PyrNidine
H H ..
Pyrrole
Molecules do not satisfy huckel rule are not aromatic.
Cyclopentadiene H⊕ Cyclooctatetraene Cyclopropenyl anion
4π electrons 8π electrons 4π electrons
Cyclopentadienyl cation
4π electrons
(4) Antiaromaticity : Planar cyclic conjugated species, less stable than the corresponding acyclic unsaturated
species are called antiaromatic. Molecular orbital calculations have shown that such compounds have 4nπ
electrons. In fact such cyclic compounds which have 4nπ electrons are called antiaromatic compounds and this
characteristic is called antiaromaticity.
Example : 1,3-Cyclobutadiene, It is extremely unstable antiaromatic compound because it has 4nπ electrons
(n = 1).
4n = 4 ; n = 4 =1
4
Thus, cyclobutanediene shows two equivalent contributing structures and it has n = 1 .
In terms of Huckel rule antiaromatic compounds have cyclic, planar structure with 4nπ electrons. They are
destabilised by resonance. Some other examples are,
⊕
Cyclopropenyl anion Cyclopentadienyl Cycloheptatrienyl anion Cycloctatetraene
4π electrons 4π electrons 8π electrons 8π electrons
Comparison of aromatic and aliphatic hydrocarbons
Characteristic Benzene and its homologous Aliphatic hydrocarbons
Composition
These are closed ring compounds. These are open chain compounds.
Carbon percentage These are represented by general These are represented by general
formula CnH2n−6 . formulae; CnH2n+ 2 (Alkanes), CnH2n
Combustion (Alkenes) and CnH2n−2 (Alkynes).
Nature These contain high percentage of
Physical state carbon. In benzene C6H6 , the carbon These have low percentage of carbon
Addition reactions percentage is 92.3. in comparison to aromatic
hydrocarbons. In hexane, C6 H14 , the
Substitution reactions These burn with smoky flame. carbon percentage is 83.7.
These have high unsaturation. For These burn with non smoky flame.
Stability example, benzene molecule consists These are saturated as well as
(4n + 2) rule three double bonds. unsaturated.
Oxidation These are colourless liquids or solids.
They have characteristic odour A few lower members are colourless
(Aromatic). gases while higher members are
liquids or solids. Generally no
Inspite of the fact that these are characteristic odour exists.
unsaturated, generally resist addition The unsaturated hydrocarbons show
reactions. These do not react with addition reactions.
HCl, HBr, HI or HClO .
Generally exhibit substitution The saturated hydrocarbons show
substitution reactions such as
(Electrophilic) reactions such as halogenation. The unsaturated
hydrocarbons resist substitution
halogenation, nitration, sulphonation, reactions. Nitration and sulphonation
occur with difficulty in higher alkanes.
Friedel-craft's reaction etc. Friedel-craft's reaction is not shown by
aliphatic hydrocarbons.
Highly stable.
The unsaturated hydrocarbons are less
Follow (4n + 2) rule, i.e., contain stable.
where
(4n + 2)π electrons (4n + 2) rule does not apply to
aliphatic unsaturated hydrocarbons.
n = 0,1,2,3,..........
Except benzene, all oxidise easily. Alkanes do not oxidise easily while
unsaturated hydrocarbons oxidise
easily.
Benzene (C6H6).
Benzene is the first member of arenes. It was first discovered by Faraday (1825) from whale oil. Mitscherllich
(1833) obtained it by distillating benzoic acid with lime. Hofmann (1845) obtained it from coal tar, which is still a
commercial source of benzene.
(1) Structure of benzene : Benzene has a special structure, which is although unsaturated even then it
generally behave as a saturated compound.
(i) Kekule's structure : According to Kekule, in benzene 6-carbon atoms placed at corner of hexagon and
bonded with hydrogen and double bond present at alternate position.
(a) Evidence in favour of Kekule's structure
• Benzene combines with 3 molecules of hydrogen or three molecules of chlorine. It also combines with 3
molecules of ozone to form triozonide. These reactions confirm the presence of three double bonds.
• Studies on magnetic rotation spectroscopy show the presence of three double bonds.
• The synthesis of benzene from three molecule of acetylene also favour's Kekule's structure.
3CH ≡ CH ∆ →
• Benzene gives cyclohexane by reduction by hydrogen. C6 H6 + 3H 2O Ni →
(b) Objections against Kekule's formula Cyclohexane
• Unusual stability of benzene.
• According to Kekule, two ortho disubstituted products are possible. But in practice only are ortho
disubstituted product is known.
• Heat of hydrogenation of benzene is 49.8 kcal/mole, whereas theoretical value of heat of hydrogenation of
benzene is 85.8 kcal/mole. It means resonance energy is 36 kcal/mole.
• C − C bond length in benzene are equal, although it contains 3 double bonds and 3 single bonds.
Kekule explained this objection by proposing that double bonds in benzene ring were continuously oscillating
between two adjacent positions.
(ii) Some other structures of benzene
(a) Ladenberg's prism formula : This formula shows benzene three dimensional structure where X-ray studies
of benzene molecule indicate a planar.
(b) Claus diagonal formula :
(c) Dewar's parallel formula :
(d) Armstrong and Baeyer's centric formula :
(e) Thiele's formula :
(iii) Valence bond theory [Resonance theory] : According to this theory, benzene can not be represented by only
one structural formula but as a hybrid of structure.
⇌≡ Resonance hybrid
(a) (b) (c)
The resonance hybrid structure of benzene explain all the properties of benzene. The resonance structure of
benzene is supported by the following facts,
(a) The C − C bond length in benzene is 139 pm which is intermediate between bond lengths for C − C bond
(154 pm) and C = C (134 pm).
(b) Due to resonance the π electron charge in benzene get distributed over greater area. As a result of
delocalisation the energy of resonance hybrid decrease as compared to contributing structure by about 50 kJ/mole.
The decrease in energy is called resonance energy.
(iv) M.O.T. [Modern concept] : According to the orbital concept each H 1.39Å H
carbon atom in benzene is sp 2 hybridised and one orbital remains σ(sp2–sp2) CC 1.09Å
unhybridised. Out of three hybrid orbitals two overlap with neighbouring carbon H–C 120o C–H
atoms and third hybrid orbital overlap with hydrogen atom for σ bonds. Thus
benzene has a planar structure with bond angle of 120° each. 120o CC σ(sp2–sp2)
H 120o H
(2) Methods of preparation of benzene
(i) Laboratory method : COONa CaO + Na2CO3
+ NaOH heat
(ii) From benzene deriSvoadtiuimvebsenzoate Benzene
OH
(a) From phenol : + Zn distill ZnO
Phenol Cl Benzene
(b) From chlorobenzene : + 2H Ni-Al alloy + HCl
NaOH
Chlorobenzene Benzene
(c) By preparing grignard reagent of chlorobenzene and then hydrolysed
C6 H5Cl Mg → C6 H5 MgCl H2O → C6 H6 + Mg OH
dry ether Cl
Chlorobenzene Phenyl magnesium Benzene
chloride
SO3H
(d) From benzene sulphonic acid : + HOH 150°-200°C +H2SO4
HCl,pressure +N2 +HCl
Steam
Benzene sulphonic acid Benzene
N2Cl
(e) From benzene diazonium chloride : + 2H SnCl2
NaOH
Benzene
(f) From acetylene :
+ HC CH red hot tube
HC + 1500-2000°C
CH
HC Benzene
+ HC
Three molecules of acetylene
Note : Cyclic polymerisation takes place in this reaction.
(g) Aromatisation : C6 H14 Cr2O3 /Al2O3 → C6 H6 + 4H2
500°C
n− Hexane at high pressure Benzene
(3) Properties of benzene
(i) Physical properties
(a) Benzene is a colourless, mobile and volatile liquid. It's boiling point is 80°C and freezing point is 5.5°C. It
has characteristic odour.
(b) It is highly inflammable and burns with sooty flame.
(c) It is lighter than water. It's specific gravity at 20°C is 0.8788.
(d) It is immiscible with water but miscible with organic solvents such as alcohol and ether.
(e) Benzene itself is a good solvent. Fats, resins, rubber, etc. dissolve in it.
(f) It is a non-polar compound and its dipole moment is zero.
(g) It is an extremely poisonous substance. Inhalation of vapours or absorption through skin has a toxic effect.
(ii) Chemical properties : Due to the presence of π electron clouds above and below the plane benzene
ring, the ring serves as a source of electrons and is easily attacked by electrophiles (Electron loving reagents). Hence
electrophilic substitution reaction are the characteristic reactions of aromatic compounds.
Substitution reactions in benzene rather than addition are due to the fact that in the former reactions
resonance stabilised benzene ring system is retained while the addition reactions lead to the destruction of benzene
ring. Principal reactions of benzene can be studied under three heads,
(a) Addition reactions (b) Substitution reactions (c) Oxidation reactions
(a) Addition reactions : In which benzene behaves like unsaturated hydrocarbon.
• Addition of hydrogen : Benzene reacts with hydrogen in the presence of nickel (or platinum) catalyst at
150°C under pressure to form cyclohexane.
+ 3H2 Ni or C6H12
150°C,pressure
Cl H
Benzene H Cyclohexane
C
H C Cl
Cl H
• Addition of halogen : HC CH hv C C
HC CH + 3Cl2
H C C Cl
Cl H
C
C
H H Cl
Benzene Benzene hexachloride (BHC)
• Addition of ozone : H OO
HC C CH O CH O 3H2O CHO
HC C +3O3 → C CH O Zn 3 + 3H2O2
CHO
CH O C CH O Glyoxal
C
H OO
Benzene triozonide
(b) Substitution reactionsBenzene
• Nucleophilic mechanism
• Unimolecular : Mostly uncommon in aromatic substitution, there is only one example which obtain in
benzene diazonium dichloride. HOH ArOH
X–
Ar − N + (Slow) → N 2 + Ar + (Fast) Phenol
2
Arenediazonium ArX
cation
Aryl halide
• Bimolecular
ZZ YZ YZ YZ YZ Y
Y + . . + Z
Y + or (Fast)
Y attaches to the positively
charged carbon atom (slow)
Cl OH Cl OH (Resonating structure of the
hexadienyl anion)
Example : (Slow) (Fast) + Cl
OH +
Phenol
• Elimination-addition mechanism (Benzyne mechanism)
*Cl * ⊕ * *NH2 *H
– HCl N⊕ H3 H+ NH2
H NH3 * NH3
+ (47%) (53%)
(Benzene)
• Electrophilic substitution reaction : Benzene undergoes this reaction because it is an electron rich system due to
delocalized π- electrons. ⊕H ⊕H HH
+ E⊕ (Slow) E; E EE
⊕⊕
Carbonium ion Resonance forms of carbonium ion (Arenium ion)
⊕H (σ - complex) E
E + Nu: (Fast) + H – Nu
Substitution product
Electrophile (E⊕) Name Source Name of substitution
reaction
Cl + Chloronium Cl2 + AlCl3 or FeCl3
Br + Bromonium Br2 + AlBr3 or FeBr3 Chlorination
NO2+ Nitronium HNO3 + H2SO4
SO3 Sulphur trioxide Conc. H2SO4 , Fuming sulphuric acid Bromination
R+ Alkyl carbonium RX + AlX3 (X = Cl or Br), ROH + H +
Nitration
+ Acyl carbonium RCOCl + AlCl3
Sulphonation
R−C =O
Friedel-Craft's
(Alkylation)
Friedel-Craft's
(Acylation)
• Free radical aromatic substitution : The aromatic substitution reactions which follow free radical mechanisms are
very few and have limited synthetic value. But some typical example of these reactions are:
⋅⋅
(CH 3 )3 COOC − (CH 3 )3 heat → 2(CH 3 )3 C O → 2CH 3 + 2CH 3COCH3
X X XX
. H .+ . H
+ CH3 . CH3 +
o- m-IntermediateCH3
H CH3
p
X XX
CH3 + .
+ +H
m CH3
O
CH3
Products
The mechanism of chlorination of benzene at high temperature is similar to that of the free radical aliphatic
substitution
..
Cl2 → Cl + Cl (Chain initiation)
..
C6 H6 + Cl → C6 H5 + HCl (H- abstraction)
..
C6 H5 + Cl2 → C6 H5Cl + Cl (Chain propagation)
(c) Oxidation : 2C6H6 + 15O2 →12CO2 + 6H2O ∆H = 6530 kJ/mole
When vapours of benzene and air are passed over vanadium pentoxide at 450 – 500°C, maleic anhydride is
obtained.
C6 H6 + 9[O] V2O5 → CHCO O+ 2CO2 + 2H 2O
450 − 500°C ||
CHCO
Maleic anhydride
Note : Strong oxidising agents converts benzene slowly into CO2 and water on heating.
(d) Reduction : 2 + 12HI + CH3 + 6I2
Benzene Cyclohexane Methylcyclopentane
(iii) Uses : (a) In dry cleaning (b) As a motor fuel when mixed with petrol. (c) As a solvent. (d) In the
manufacture of gammexane (As insecticide). (e) In the preparation of nitrobenzene, chlorobenzene, benzene
sulphonic acid, aniline, styrene, etc. Many of these are employed for making dyes, drugs, plastics, insecticides, etc.
Directive effect in substituted benzene derivatives.
(1) Directive effect in mono substituted benzene derivatives : The substituent already present on the
benzene ring directs the incoming substituent to occupy ortho (2 or 6), meta (3 or 5) or para (4) position. This
direction depends on the nature of the first substituent and is called directive or the orientation effect.
The substituent already present can increase or decrease the rate of further substitution, i.e., it either activates
or deactivates the benzene ring towards further substitution. These effects are called activity effects.
There are two types of substituents which produce directive effect are,
(i) Those which direct the incoming group to ortho- and para-positions simultaneously (Neglecting meta all
together).
(ii) Those which direct the incoming group to meta-position only (Neglecting ortho- and para-positions all together).
Ortho-para directors Meta directors
Moderately deactivating
Strongly activating − .. .. .. – .. – ..O..:– −C ≡ N,−SO3H, −COOH,−COOR, −CHO, COR
NH2,− NHR,− NR2, .O.H,
Strongly deactivating − NO2,−NR3⊕,−CF3,−CCl3
.. .. .. ..
Moderately activating − NHCOCH3, − NHCOR, – .O.CH3, – .O.R
Weakly dacetaicvtaitviantgin−gC–H..F3..,:−,C–2C..H..l5: ,−R.,−. C6H. .5 ,
Weakly ,– B. .r: ,–.I.:
Theory of ortho – para directing group
:S ⊕S ⊕S ⊕S :S
.. ..
,i.e.,
δ ⊕S Ortho attack .. S
δ Ortho S
δ E
+
+ E⊕
δ Ortho product E
Para attack Para product
The above mechanism is followed when S is − OH, − NH 2,−Cl, − Br,−I,−OR,−NR2,−NHCOR etc.
CH3 CH3 CH3 CH3
.. ..
..
In methyl or alkyl group, the +I effect of the methyl group or alkyl group initiates the resonance effect.
Thus, methyl or alkyl group directs all electrophiles to ortho and para positions.
Theory of meta directing group : The substituent, S withdraws electrons from ortho and para positions.
Thus, m-position becomes a point of relatively high electron density and further substitution by electrophile occurs
at meta position. For example, − NO2 group is a meta directing (Electron withdrawing). Its mechanism can be
explained as :
O N⊕ O O N⊕ O O N⊕ O O N⊕ O O N⊕ O
⊕ .⊕.
⊕
All meta-directing groups have either a partial positive charge or a full positive charge on the atom directly
attached to the ring.
Aromatic hydrocarbons
(2) Directive effect in disubstituted benzene
(i) If the directive effects of two substituents reinforce, then a single product is formed.
CH3 OH OH
Example : Nitration NO2
; +NO2
NO2 (m) NO2 NO2
Thus, both (CH3, NO2) direct further substitution to the same position (Orth).
(ii) If the directing effect of two groups oppose each other strongly activating groups win over deactivating or
weakly activating group. The sequence of directing power is
− NH 2 > −OH > −OCH3 − > NHCOCH3 > −C6 H5 > CH 3 > meta directors
OH OH OH Directs OH OH
(Powerful Br
Directs
activator)
Example : Br2
; FeBr3
CH3 CH3 CH3 CH3 CH3
Directs Directs
(iii) There is normally little substitution when the two groups CH3 Too hindered position
are meta to each other. Aromatic rings with three adjacent substituents
are generally prepared by same other routes.
Cl
Toluene, methyl benzene or phenyl methane.
Toluene is the simplest homolouge of benzene. It was first obtained by dry distillation of tolubalsam and
hence named toluene. It is commercially known as tolual.
(1) Methods of preparation + CH3Cl AlCl3 CH3 HCl
(i) From benzene [Friedel-craft's reaction] : +
Note : Alkyl halide employed may undergoBeanzneniesomeric change Toluene
C6 H6 + ClCH2CH2CH3 AlCl3 → C6 H5CH CH 3 + HCl CH3 + 2NaBr
n− Propyl chloride CH 3
Isopropyl benzene (65 −70%)
Catalysts can be used in place of anhydrous AlCl3 are,
AlCl3 > SbCl3 > SnCl4 > BF3 > ZnCl2 > HgCl2
(ii) Wurtz fitting reaction : Br + 2Na + BrCH3 Ether
Bromobenzene Methyl bromide Toluene
(iii) Decarboxylation : C6 H4 CCOHO3 Na+ NaOH Soda lime → C6 H5CH 3 + Na 2 CO3
Toluene
(o-,m- or p-)
Sodium toluate
(iv) From cresol : CH3 heat CH3 ZnO
OH +
+ Zn
o-Cresol Toluene
CH3 CH3
+ H2SO4
(v) From toluene sulphonic acid : + HOH
Toluene
SO3H
p-Toluene sulphonic acid
CH3 CH3 CH3
+ N2 + CH3CHO + HCl
(vi) From toluidine : NaNO2 C2H5OH
HCl
NH2 N2Cl Toluene
p-Toluidine p-Toluene diazonium chloride CH3
+ MgBr2
(vii) From grignard reagent : MgBr
+ CH3Br
Phenyl magnesium bromide Toluene
(viii) Commercial preparation
From coal tar : The main source of commercial production of toluene is the light oil fraction of coal-tar. The
light oil fraction is washed with conc. H 2SO4 to remove the bases, then with NaOH to remove acidic substances
and finally with water. It is subjected to fractional distillation. The vapours collected between 80 − 110°C is 90%
benzol which contains 70 − 80% benzene and 14 − 24% toluene. 90% benzol is again distilled and the portion
distilling between 108 − 110°C is collected. It is toluene. CH3 CH3
(ix) From n- heptane and methyl cyclohexane |
CH2 Cr2O3 / Al2O3
H2C
CH3
H2C CH2 500-550°C Toluene
150 atms
(2) Physical properties n-HCepHta2ne
(i) It is a colourless mobile liquid having characteristic aromatic odour.
(ii) It is lighter than water (sp. gr. 0.867 at 20°C).
(iii) It is insoluble in water but miscible with alcohol and ether in all proportions.
(iv) Its vapours are inflammable. It boils at 110°C and freezes at –96°C.
(v) It is a good solvent for many organic compounds.
(vi) It is a weak polar compound having dipole moment 0.4D.
(3) Chemical properties : Toluene shows the behavior of both
CH3 Side chain (Aliphatic)
Benzene ring (Aromatic)
(i) Electrophilic substitution reactions : Aromatic character (More reactive than benzene) due to electron
releasing nature of methyl group. CH3 CH3 CH3
E
+ E⊕ +
Electrophile
o-Derivative E
Note : E+ may be chlorine, HNO3, H2SO4,CH3Cl . p-Derivative
(ii) Reactions of side chain CH3 CH2Cl CHCl2 CCl3
(a) Side chain halogenation : Cl2 Cl2 Cl2
UV UV UV
Toluene Benzyl chloride Benzal chloride Benzo trichloride
Note : Benzyl chloride on hydrolysis with aqueous caustic soda forms benzyl alcohol.
C6H5CH2Cl + NaOH → C6H5CH2OH + NaCl
(Phenyl methyl chloride)
Benzal chloride on hydrolysis forms benzaldehyde.
C6 H5CHCl2 + 2NaOH → C6 H5CH(OH)2 + 2NaCl
(Benzylide chloride)
C6 H5CH↓O+ H 2O
Benzo trichloride on hydrolysis forms benzoic acid.
C6H5CCl3 + 3NaOH → C6H5C(OH)3 + 3NaCl
(Benzylidyne chloride)
C6H5C↓OOH + H2O
(b) Oxidation : CH3 COOH
• With hot acidic KMnO4 : + H2O
KMnO4 / H+
3[O]
Toluene Benzoic acid CH3 CHO
+ H2O
• With acidic manganese or chromyl chloride (Etards reaction) : + 2[O] CrO2C2
Toluene Benzaldehyde
Note : All alkyl benzenes on oxidation with hot acidic KMnO4 or Na2Cr2O7 form benzoic acid. The
length of the side chain does not matter.
R R
(c) Hydrogenation : + 3H2 Na / liquid NH3 – C2H5OH
Birch reduction
Alkyl benzene Alkyl cyclohexane
CH3 CH3
| |
C CH
HC CH
+ 3H2 Ni H2C CH2
HC CH 200°C
CH H2C CH2
Methyl benzene CH2
Methylcyclohexane
(d) Combustion : C6H5CH3 + 9O2 →7CO2 + 4H2O
(e) Ozonolysis : H OO
HC C CH O CH O Zn CH–C=O CHO
HC C +3O3 → C CH O +2 + 3H2O2
HOH H – C=O CHO
CH O C CH O
C Methyl glyoxal Glyoxal
H OO
(4) Uses Toluene Triozonide
(i) In the manufacture of benzyl chloride, benzal chloride, benzyl alcohol, benzaldehyde, benzoic acid,
saccharin, etc.
(ii) In the manufacture of trinitrotoluene (TNT), a highly explosive substance.
(iii) As an industrial solvent and in drycleaning.
(iv) As a petrol substitute.
(v) In the manufacture of certain dyes and drugs.
T.N.T. (Tri-nitro toluene)
CH3 CH3
NO2
Preparation : H2SO4 O2N
+
+ 3FHumNinOg 3 3H2O
Toluene NO2
Properties : It is pale yellow crystalline solid (M.P. = 81°C).
Uses : • It is used as an explosive in shells, bombs and torpedoes under the name trotyl.
• When mixed with 80% ammonium nitrate it forms the explosive amatol.
• TNT is also used as a mixture of aluminium nitrate, alumina and charcoal under the name ammonal.
CH3 T.N.B. (Tri-nitro benzene)
Preparation : CH3 COOH
NO2
O2N NO2 O NK2Cr2O7 2 NO2 Soda O2N
H2SO4 lime
HNO3 H2SO4
Toluene NO2 NO2 NO2
T.N.B.
Properties and uses: It is colourless solid (M.P. = 122°C). It is more explosive than T.N.T. and used for
making explosive.
Xylenes (Dimethyl benzene) C6H4(CH3)2.
The molecular formula, C8H10 represents four isomers.
CH3 CH3 CH3 C2H5
CH3
Ethyl benzene
o-Xylene CH3 CH3
m-Xylene p-Xylene
These are produced along with benzene, toluene and ethylbenzene when aromatisation of C6 − C8 fraction of
petroleum naphtha is done. The xylenes are isolated from the resulting mixtrue (BTX) by fractional distillation.
These can be prepared by Wurtz – Fittig reaction. A mixture of bromotoluene and methylbromide is treated
with sodium in dry ethereal solution to form the desired xylene.
CH3 CH3 CH3 CH3
Br CH3
+ 2Na + BrCH3 + 2Na + BrCH3 + 2NaBr
+ 2NaBr ; Br CH3
o-Bromotoluene
o-Xylene m-Bromotoluene m-Xylene
CH3 CH3
+ 2Na + BrCH3 + 2NaBr
Br p-XCylHen3e
p-Bromotoluene
• These can also be obtained by Friedel – craft's synthesis,
CH3 CH3 CH3
+ CH3Cl CH3 +
AlCl3
o-Xylene CH3
• m-Xylene can be obtained from mesitylene. p-Xylene
CH3 CH3 Soda lime CH3
heat
H3C CH3 [O] CH3 + CO2
CH3
HOOC
Mesitylene Mesitylenic acid m-Xylene
Xylenes are colourless liquids having characteristic odour. The boiling points of three isomers are,
o-Xylene = 144°C; m-Xylene = 139°C; p-Xylene = 138°C.
Xylenes undergo electrophilic substitution reactions in the same manner as toluene. Upon oxidation with
KMnO4 or K2Cr2O7 , Xylenes form corresponding dicarboxylic acids. COOH
COOH
COOH ,
COOH , COOH
Phthalic acid Isophthalic acid COOH
Terephthalic acid
Xylenes are used in the manufacture of lacquers and as solvent for rubber. o-Xylene is used for the
manufacture of phthalic anhydride.
Ethyl benzene (C6H5C2H5).
It can be prepared by the following reactions,
(1) By Wurtz-Fittig reaction : C6H5Br + 2Na + BrC2H5 → C6H5C2H5 + 2NaBr
(2) By Friedel-craft's reaction : C6 H5 H + BrC2 H5 AlCl3 → C6 H5C2 H5 + HBr
(3) By catalytic reduction of styrene : C6H5CH = CH2 + H2 → C6H5CH2CH3
(4) By alkyl benzene synthesis : C6 H5 H + H2C = CH2 AlCl3 , HCl → C6 H 5 CH 2CH 3
95°C,Pressure
It undergoes electrophilic substitution reactions in the same way as toluene. When oxidised with dil. HNO3 or
alkaline KMnO4 or chromic acid it forms benzoic acid.
C6H5C2H5 [O] → C6H5COOH
Styrene (C6H5CH=CH2).
It is present in storax balsam and in coal-tar traces.
(1) Preparation
(i) Dehydrogenation of side chain of ethylbenzene : Dehydrogenation of side chain is affected by
heating ethylbenzene to high temperature in presence of a catalyst.
CH2CH3 CH = CH2
+ CH2 = CH2 AlCl3 600°C
Cr2O3 / Al2O3
Benzene Ethylbenzene Styrene
(ii) Decarboxylation of cinnamic acid : This is the laboratory preparation. It involves heating of cinnamic
acid with a small amount of quinol.
C6H5CH = CHCOOH Quinol → C6H5CH = CH2 + CO2
(iii) Dehydration of 1-phenyl ethanol with H2SO4 : C6 H5CHOHCH3 H2SO4 → C6 H5CH = CH2
−H2O
(iv) Dehydration of 2-phenyl ethanol with ZnCl2 : C6 H5CH2CH2OH ZnCl2 ,heat → C6 H5CH = CH2
−H2O
(v) Dehydrohalogenation of 1-phenyl-1-chloro ethane : On heating with alcoholic potassium
hydroxide, a molecule of hydrogen chloride is eliminated by the chloroderivative.
C6 H5CHClCH 3 Alc.KOH → C6 H5CH = CH2
Heat
(2) Properties : It is a colourless liquid, boiling point 145°C. On keeping, it gradually changes into a solid
polymer called metastyrene. The polymerisation is rapid in sunlight or when treated with sodium. It shows
properties of benzene ring (Electrophilic substitution) and unsaturated side chain (Electrophilic addition). However,
the side chain double bond is more susceptible to electrophilic attack as compared to benzene ring.
At lower temperature and pressure, it reacts with hydrogen to produce ethylbenzene and at higher
temperature and pressure, it is converted into ethyl cyclohexane.
CH = CH2 CH2CH3 CH2CH3
H2 / Ni H2 / Ni
20°C, 3 atm 125°C, 110 atm
Styrene Ethyl benzene Ethyl cyclohexane
With bromine, it gives the dibromide. CHBr.CH2Br
CH = CH2
+ Br2
Styrene Styrene dibromide
Halogen acids add to the side chain. C6H5CH = CH2 + HX → C6H5CHXCH3
Preparation of ring substituted styrenes is not done by direct halogenation but through indirect route.
CH2CH3 CH2CH3 CHClCH3 CH = CH2
+ Cl2 FeCl3 Cl2 Alc.. KOH
hv Heat
Cl Cl Cl
When oxidised under drastic conditions, the side chain is completely oxidised to a carboxyl group.
CH = CH2 COOH
[O]
KMnO4
Styrene
In presence of peroxides, styrene undergoes free radical polymerisation resulting in the formation of
polystyrene – an industrially important plastic.
Peroxide →−
nC6 H5CH = CH 2 CH − CH 2 −
|
C6H5
n
Co-polymers of styrene with butadiene and other substances are also important since many of them are
industrially useful products such as SBR ( A rubber substitute).
Bi-phenyl (C6H5 – C6H5).
It occurs in coal-tar. It is the simplest example of an aromatic hydrocarbon in which two benzene rings are
directly linked to each other.
(1) Methods of formation
(i) Fittig reaction : It consists heating of an ethereal solution of bromobenzene with metallic sodium.
Br + 2Na + Br + 2NaBr
(ii) Ullmann biaryl synthesis : Iodobenzene, on heating with copper in a sealed tube, forms biphenyl. The
reaction is facilitated if a strong electron wihtdrawing groups is present in ortho or para position.
I + 2Cu + I + 2CuI
(iii) Grignard reaction : Phenyl magnesium bromide reacts with bromo benzene in presence of CoCl2 .
MgBr + Br CoCl2 + MgBr2
(2) Properties : It is a colourless solid, melting point 71°C. It undergoes usual electrophilic substitution
reactions. Since aryl groups are electron withdrawing , they should have deactivating and m-orientating effect. But,
it has been experimentally shown that presence of one benzene ring activates the other for electrophilic substitution
and directs the incoming group to o- and p- positions. It has been shown that monosubstitution in the bi-phenyl
results in the formation of para isomer as the major product.
Another special feature of the biphenyl is the behaviour towards second substitution in a monosubstituted
biphenyl. The second substituent invariably enters the unsubstituted ring in the ortho and para position no matter
what is the nature of substituent already present.
HNO3 / H2SO4 NO HNO3 / H2SO4 O2N NO2
2
Diphenyl methane.
(1) Methods of preparation
(i) Friedel-craft's reaction : C6 H5CH 2Cl+ C6 H6 AlCl3 → C6 H5CH 2C6 H5 + HCl or
Benzyl chloride Benzene Diphenyl methane
2C6H6 + CH2Cl2 AlCl3 → C6H5CH2C6H5 + 2HCl
Dichloromethane
(ii) By action of formaldehyde on benzene in presence of conc. sulphuric acid
2C6 H6 + O = CH2 Conc. H2SO4 → C6 H5CH2C6 H5 + H2O
(iii) By Grignard reaction : Phenyl magnesium bromide reacts with benzyl bromide to from diphenyl
methane. C6H5MgBr + BrCH2C6H5 → C6H5CH2C6H5 + MgBr2
(iv) By reduction of benzophenone : Reduction can be done with LiAlH4 or P and HI.
C6H5COC6H5 4[H] → C6H5CH2C6H5 + H2O
(2) Properties : It is a colourless solid, melting point 26°C.Like biphenyl, it also easily undergoes electrophilic
substitution reactions.
CH2 HNO3 CH2 NO2 HNO3 O2N CH2 NO2
H2SO4 H2SO4
The methylene hydrogens of diphenylmethane are situated on carbon atom linked by two electron attracting
benzene rings. Thus, these are somewhat acidic in nature.
C6 H5CH 2C6 H5 + Br2 → C6 H5CHBrC6 H5 + HBr
When oxidised with K2Cr2O7 / H 2SO4 mixture, it forms benzophenone.
C6 H5CH 2C6 H5 [O] → C6 H5CC6 H5
||
O
It forms fluorene when its vapours are passed through a red hot tube.
CH2 Heat
HH Fluorene
Polynuclear hydrocarbons.
Compounds having two or more benzene rings fused together in ortho positions are termed as fused
polynuclear hydrocarbons. These hydrocarbons also called fused ring hydrocarbons.
Naphthalene Anthracene Phenanthren
(1) Naphthalene
Naphthalene is the largest single constituent of coal-tar (6-10%). It is obtained in the middle oil fraction of
coal-tar distillation. It is recovered as crude product when the middle oil fraction is cooled. The crude crystalline
product is separated by centrifugation and purified by washing successively with dilute H 2SO4 (to remove basic
impurities), sodium hydroxide solution (to remove acidic impurities) and water. Finally, the solid is sublimed to get
pure naphthalene.
Aromatic hydrocarbons
(i) Methods of preparation
(a) Haworth synthesis O
O || CH2CH2
C CH2
|| C CH2 CH2
C – CH2 | ||
+O | AlCl3 HO – C CH2 Zn(Hg) Conc. H2SO4
HCl HO – C CH2 heat (–H2O) O
C – CH2 |
Benzene || α-Tetralone
CH2
Zn(Hg)
O || || HCl
Succinic O
anhydride O
γ-phenylbutyric acid
β-Benzoylpropionic
acid CH2
Se CH2
heat TetraliCn H2CH2
(b) By 4-phenyl-1-butene : CHC2H2 Naphthalene + 2H2
(ii) Structure | CaO
heat
CH Naphthalene
H2C
(a) In naphthalene all carbon atoms are sp2 - hybridized. sp2 - hybrid orbital overlap with s- orbital of
hydrogen atoms forming C − C and C − H sigma bond.
(b) All carbon and hydrogen atoms lie in one plane in naphthalene.
H H sp2- sp2
H C-C
H σ-bond
H sp2-s
H
H C-H
H σ-bond
(c) According to resonance theory. It is a resonance hybrid.
81
72
63 (II) (III) Hybrid
54
(I)
(d) Position : 1,4,5,8 = α ; 2,3,6,7 = β
α α 8 1
β β 7 2
β β 6 3
α α 5 4
(e) Resonance energy of naphthalene is 61 kcal/mol. Which is less than of benzene. So that naphthalene is
less aromatic i.e. more reactive than benzene.
(f) In naphthalene C1 − C2 bond length is shorter (1.36Å) i.e. C = C and C2 − C3 is 1.40Å i.e. single bond.
(iii) Physical properties : It is a colourless crystalline compound. It melts at 80.2°C. It is very volatile and
sublimes slowly even at room temperature. It has strong characteristic odour. It is insoluble in water but very soluble
in ether, benzene and hot alcohol. It burns with smoky flame.
(iv) Chemical properties : It undergoes usual aromatic electrophilic substitution reactions. The product of
monosubstitution is either α or β -depending on conditions, but the α -product always predominates.
E E
+ H+
+ E+ and/or
α-product β-product
Substitution at β -position occurs only when the reaction is carried at high temperatures or when bulkier
solvents are used. Cl COC
Cl2 Cl CH3CO
250°C + AlCl3 in CS2 at
Cl CH2O + CH2Cl
ZnCl2 O
S2O52°Cl
S1O020C- l Cl CrO3
HN5O0°3/CH2S CH3COOH,
Cl
hHigNhOt3e/mH2pS. NO Na2Cr2O7/H2SO4 O
H2SO or acid COO
40°- Naphthalen COO
H2SO
160°C NO NO NO O
C2H5B +
AlCl3 O2 /V2O5, ||
NO or conc. H2SO4,
SO3 C
SO3 O
C2H
C
Naphthalene ||
Pthalic aOnhydride
OO
+O CH
O CH
DiozoOnide O Pthalaldehyd
Na/C-
Reflux
(Dialine)
1, 4-Dihydro
Na/Isopentyl
Reflux
(Tetralin)
(1, 2, 3, 4-
H2/Ni
Reflux
(Decalin)
(Decahydronaphthale
(v) Uses
(a) As moth repellent. It is, however, now being replaced by more powerful insecticides such as p-
dichlorobenzene and DDT.
(b) For commercial production of phthalic anhydride, α -naphthol, β -naphthol, etc.
(c) For manufacture of dyes, explosives and synthetic resins.
(d) For increasing the illuminating power of coal gas.
(2) Anthracine
The hydrocarbon derives its name from the world anthrax (Greek = coal) as coal is the chief source from
which it is manufactured. It is present in coal-tar (Less than 0.5%) and is obtained from the anthracene oil or green
oil fraction (because of its dark green fluorescence) formed during coal-tar distillation. This fraction is collected
between 270 - 360°C.
The anthracene oil fraction is cooled when crude anthracene crystallizes out. The crude product consists
phenanthrene and carbazole as impurities. The crude product is successively washed with solvent naphtha as to
remove phenanthrene and pyridine to remove carbazol. Finally, the solid is sublimed to get pure anthracene.
(i) Methods of preparation
Haworth synthesis
O AlCl3 O Conc. H2SO4 O Zn dust
(heat) (–H2O) distill.
|| || ||
C C C
O+ C
C COOH || Anthracene
|| o-Benzoyl benzoic acid O
O 9, 10-Anthraquinone
Phthalic anhydride
(ii) Structure
(a) Anthracine is tricyclic aromatic hydrocarbon.
(b) All carbon atoms in anthracene are sp 2 hybridized.
(c) resonance hybrid are as follows
(I) (II) (III) (IV) Hybrid form
(d) It gives both addition and electrophilic substitution reaction.
(e) In anthracene the numbering of carbon atom is
891 αγα
72 ββ
63 ββ
5 10 4 αγα
Anthracene Anthracene
(iii) Physical properties : Anthracene is a colourless solid. It melts at 217°C. It is insoluble in water but
soluble in alcohol and ether in small amounts. It is comparatively more soluble in hot benzene. With picric acid, it
forms a red colour picrate. Br
(4) Chemical properties
Br2
CCl4
9-bromoanthracene
NO2
HNO3
AC2O
NO2
9, 10-dinitroanthracene
SO3H
H2SO4
low temp.
H2SO4 SO3H
High temp.
Anthracene
Na/C2H5OH
HH
(v) Uses : Anthracene is used HH
(a) For manufacture of anthraquinone.
(b) For making dyes (Alizarin). 9, 10-dihydroanthracene
(c) In smoke screens.
O
[O] + H2O
Na2Cr2O7 + H2SO4 O
9, 10-anthraquinone
Heterocyclic compounds.
These are cyclic compounds in which the ring includes in addition to carbon atoms at least one atom of
another element (Hetero = other, different). The common hetero atoms present in the carbon rings are O,N and S.
(1) Thiophene
It is found in the benzene fraction of coal-tar and petroleum. The benzene fraction is shaken with cold
concentrated sulphuric acid. Thiophene present in the fraction combines with sulphuric acid more readily than
benzene to form thiophene sulphonic acid which is separated with water being soluble. Thiophene sulphonic acid is
treated with super-heated steam to recover thiophene.
HC CH
or
S HC S CH
Properties : It is a colourless liquid. Its boiling point is 84°C. Its odour is similar to that of benzene. It is
insoluble in water but soluble in organic solvents. It is flammable and toxic in nature.
Its resonance energy is 31k cal mol −1 . Hence, it is more stable and resembles benzene more closely than
furan (23 k cal mol −1) and pyrrole (25 k cal mol −1). It does not show basic properties and does not undergo
Diels-Alder reaction. .–. .–.
.. S+ S+ .–. .–. S+ S+
S
CH Thiophene as a resonance hybrid (Resonance energy is 31 k cal /mol
2 CH +
Al2O3 Cl2 Cl
400°C 2-
S
Br
2- S Br
I2
HgO 2-IodothSiopheIne
F.HNO3, –10°C S NO
Acetic anhydride
2-
Cold conc.
CH2 – CH2 + 4S 650°C S 2-ThiopheSne SO3H
| | Thiophene sulphonic acid
CH3 CH3 HgCl2
CuO
2-ChloroSmercHurigthCiolphene
CH3COCl 2- S COCH3
AlCl3
HCHO + HCl S CH2Cl
2-
CH2COON Pd/H2
a or Na-Hg/C2H5OH
| S
Tetrahydrothiophene (Thiophan)
P2S3 Raney CH3CH2CH2CH3 +
[Laboratory heat NiS
(2) Furan sp2 sp2
Furan derives its name from furfur meaning bran in Greek which is the 3 ↓ ↓
source of its aldehyde, furfural. It is present in pine-wood tar and may be 4 H –σ– C C –σ– H
extracted from it. 5 O 2 H –σ– C .. C –σ– H
s↑p2 O s↑p2
Furan shows aromatic behaviour because resulting π-molecular orbital 1 s↑p2
satisfies the Huckel rule ((4n + 2)π where, n = 1 .
Furan is also considered as a resonance hybrid of the canonical forms. Out of which the first three are the
main contributing structures. .–. .–.
Furan has resonance energy about
.. O+ .–. .–. O+ O+ O+
23 k cal mol −1 which is less than benzene. However it O II III
is less aromatic and more reactive than benzene. IV V
I
Properties : Furan is a colourless liquid. Its
Canonical forms of furan
boiling point is 32°C. It is insoluble in water but soluble in orgainic solvents. It is a reactive compound. It is a weak
base. In Furan electrophilic substitution reactions take place preferably at 2 and 5 position where electron density is
high. If these positions are occupied, substitution occurs at 3 and 4 positions. It undergo Diels-Alder reaction.
Pine wood distillation Br2 O Br
Dioxane, 0°C
2-Bromofuran
O
|| O NO2
CH3 – C – ONO2 or HNO3
(CH3CO)2O, 10°C 2-Nitrofuran
[O] COOH 200°C SO3
CHO –CO2 Pyridine
O O
Furfural
Furoic acid
O SO3H
2-Furan sulphonic acid
steam O (CH3CO)2O O
Ag2O BF3
Furan ||
O CHO
– C – CH3
O
2-Acetylfuran
CH2 – CHO CH = CHOH P2O5 2H2/Ni H2C –– CH2 HCl H2C –––– CH2
∆ heat
|| ||
| ⇌ | = CHOH
H2C CH2 H2C – Cl CH2 – OH
CH2 – CHO CH O
Tetra methylene chloroform
(4-chloro-1-butanol)
Tetrahydro furan (THF)
Br2 O Br
or X2 O COR
CHOH –– CHOH HgCl2
||
Dry distill. CH3COON HgCl
H – C – OH H – C – OH – CO2
| | – 3H2O CH3OH + HCl O
COOH COOH 2-Chloro
Mucic acid or Saccharic acid or Glucaric RCOCl
acid
CH2 – CH(OCH3)2
|
CH2 – CH(OCH3)2
Diacetal of succinic dialdehyde
(i) HCN + HCl, anhy. AlCl3 O
||
–C–H
O
(2-Formyl furan)
Furfural
NH3+Al2O3
C4H9 Li N
H
Pyrrol
(i) O COO
O Li (ii) Furoic acid
2-lithium furan
(3) Pyridine
Pyridine is a six membered aromatic heterocycle with one nitrogen atom in the ring. It may be supposed to
have been derived by replacement of = CH − group of benzene by = N − . Hence, it is isoster of benzene. Its
systematic name is azabenzene. (Prefix aza stands for nitrogen). The hybrid structure of pyridine is represented as:
sp2-s H sp2-sp2
(C-C, σ bond) (C-C, σ bond)
CH H H
HC CH
1.37Å 1.40Å
N HC CH
N HN H
Completely filled sp2-sp2
sp2-hybrid orbital (C-N, σ bond)
(i) Properties : It is a colourless liquid having an unpleasant odour. It boils at 115°C. It is miscible with water
and is hygroscopic. It is a good solvent for many organic compounds and inorganic salts.
(ii) Aromatic character : Each carbon atom and nitrogen atom in the ring have sp 2 hybridized and one
unhybrid p-orbital containing one electron. These orbitals overlap to form π molecular orbital consisting six
electrons. The π molecular orbital satisfies Huckel's rule (4n + 2) and thus aromatic properties are observed in
pyridine.
The resonance energy is 43 k cal/mol and bond length of C − C bond is 1.40 Å and C − N 1.39 Å.
••• N: π electron cloud
••
.–.
N• : N..
4n + 2 = 6 ; n = 1 .–. ..
.–. N.. + N
(iii) Resonance structure : + N..
+
Canonical form of pyridine
(iv) Basic nature : Pyridine is basic in nature due to presence of lone pair nitrogen atom. It is more and
more basic than pyrrole and less basic than aliphatic amine.
Example : + H⊕Cl
.N.
⊕
[NH]Cl
Pyridinium hydrochloride
(v) Chemical properties :
CH2CH2NH2HCl N KNO3/H2SO4 NO2
H 300°C 3-NitroNpyridine
CH2 Piperidine H2SO4/300°C
CH2CH2NH2HCl
Penta methylene diamine dihydrochloride
F.H2SO4/HgSO4 SO3H
250°C
3-Pyridine Nsulphonic acid Electrophilic
substitution
2C2H2 + HCN Br2 Br
300°C 3-BromNopyridine
+ CH2I2 CH3ONa NaNH2 N NH2
N 200°C 100°C
H 2-Aminopyridine
C6H5Li
N 100°C N C6H5 Nucleophilic
substitution
Pyridine 2-Phenylpyridine
KOH N OH
300°C
2-Hydroxypyridine
2CH3CHO + HCHO + NH3 Al2O3
Na/C2H5OH or Red
H2/Ni
N
H
Piperidine
2CH2 = CH. CHO NH3, K2Cr2O7/H+ HI CH3CH2CH2CH2CH3 + NH3
∆ [O] 300°C
n-Pentane
(vi) Uses
(a) To denature alcohol.
(b) As a basic solvent in organic reactions.
(c) For preparing sulpha-pyridine and vitamin B6 .
(d) As a catalyst in many reactions, e.g., in the formation of Grignard reagent, in Perkin and Knoevenagel
reactions.
(4) Pyrrole
It occurs in coal-tar and bone oil and is found in many natural products including chlorophyll, haemoglobin
and alkaloids.
N HC CH
or
|
HC N CH
H
|
Pyrrole
H
sp2-sp2
(C-C, σ-bond)
HH .–. .–.
sp2-s N+
H H (C-H, σ-bond) .. N+ N+ .–. .–. N+ |
N N
sp2-s | | | H
Hsp2-sp2 | (N-H, σ-bond) |
H H H
(C-N, σ-bond) H
(i) Preparation : C4H4NK + H2O steam
distillation
2C2H2 + NH3 –H2
H2C CH2 Zn dust N
distillation
O=C N C=O |
H 200°C
Glycerol H
CHOHCHOHCOONH4
Pyrrole
|
CHOHCHOHCOONH4
+ NH3 Al2O3
400°C
O
Note : Furan needed for the process is obtained from agricultural waste materials which are rich in
pentosans. The pantosans on acid hydrolysis yields furfural which is decarbonylated.
(ii) Properties : Pyrrole is a colourless liquid. Its boiling point is 131°C. It is slightly soluble in water but
highly soluble in alcohol and ether. Its odour is similar to chloroform. It rapidly becomes brown when exposed to
air. Vapours of pyrrole turn a pine splint moistened with HCl red. Pyrrole derives its name from this property.
Chemical properties : Br Br KOH C4H4N–K+
Br HN Br HCl
Br2 CH3I Pyrrole potassium
C2H5OH Tetrabromo pyrrole
CH3COCl C4H4NH.HCl
I2/KI II
Pyrrole hydrochloride
I HN I
N
Tetraiodo pyrrole (Iodole)
|
SOCl2
CH3
HN Cl
N-Methyl pyrrole
2-Chloro pyrrole
N
SO3 HN SO3H
Pyridine, 100°C |
2-Pyrrole sulphonic acid
COCH3
N-Acetyl pyrrole
HN HN HONO
Pyrrole HNO3 Pyrrole N
(CH3CO)2O, 5°C
HN NO2 |
2-Nitro pyrrole NO
N-Nitroso pyrrole
(CH3CO)2O, 200°C Zn/CH3COOH
AlCl3 (Mild reduction)
HN COCH3 HN
2-Acetyl pyrrole 2, 5-Dihydro pyrrole
C6H5N2Cl HN N=NC6H5 H2/Pt HN
diazotisation in acid heat, pressure
2-Phenylazo pyrrole Cr2O3/CH3COOH Pyrrolidine
solution
3[O] OC HN CO
CHCl3/KOH
Reimer-Tiemann HN CHO Maleic imide
2-Formyl pyrrole
(Pyrrole-2-aldehyde)
(iii) Uses : It is used as a commercial solvent, as an intermediate in the production of nylon and for making
pharmaceuticals.
Nuclear chemistry
“The branch of chemistry which deals with the study of composition of atomic nucleus and the nuclear trans-
formations”
The discussion of nuclear science with special emphasis on its chemical aspects is termed Nuclear
chemistry. It has become a very important and fascinating branch of science due to the tremendous amount of
energy liberated during nuclear reaction, which led to the discovery of atom bomb, hydrogen bomb, etc. The
controlled release of nuclear energy promises to lead us into a new era, in which an unlimited storehouse of energy
is available to man. It is rightly said that we are now living in the nuclear age and the future of humanity is at the
mercy of the nuclear scientists. They can annihilate the whole world within a few minutes. They are the hope of
human happiness and prosperity as well. In this chapter we shall deal broadly with the various aspects of nuclear
chemistry
Nucleus.
Nucleus is found to be a source of tremendous amount of energy, which has been utilised for the destructive
as well as constructive purposes. Hence the study of nucleus of an atom has become so important that it is given a
separate branch of chemistry under the heading of nuclear chemistry. According to an earlier hypothesis, the
nucleus is considered as being composed of two building blocks, proton's and neutron's, which are collectively
called nucleons.
(1) Nuclear forces : Since the radius of nucleus is very small≈ 10 −15 m, two protons lying in the nucleus
are found to repel each other with an electrostatic force of about 6 tonnes. The forces, which hold the nucleons
together means stronger proton – proton, neutron – neutron and even proton – neutron attractive forces, exist in the
nucleus. These attractive forces are called nuclear forces. Unlike electrostatic forces which operate over long ranges,
but the nuclear forces operate only within small distance of about 1 × 10 −15 m or 1 fermi (1 fermi = 10 −13 cm ) and
drops rapidly to zero at a distance of 1×10−13 cm. Hence these are referred to as short range forces. Nuclear forces
are nearly 10 21 times stronger than electrostatic forces.
(2) Nuclear stability : Nucleus of an atom contains all the protons and neutrons in it while all electrons are
in the outer sphere. Nuclides can be grouped on the basis of nuclear stability, i.e. stable and unstable nucleus. The
most acceptable theory about the atomic nuclear stability is based upon the fact that the observed atomic mass of
all known isotopes (except hydrogen) is always less from the sum of the weights of protons and neutrons
(nucleons) present in it. Other less important (or unusual) fundamental particles of the nucleus are electron,
antiproton, positron, neutrino, photon, graviton, meson and γ - particles are considered as created by stresses in
which energy is converted into mass or vice versa, e.g. an electron (β- particle) from a radio active nucleus may be
regarded as derived from a neutron in the following way.
Neutron → Proton + Electron
Similarly, photons are produced from internal stresses within the nucleus.
A list of elementary particles is given below :
LEPTONS Name of particle Symbol Anti – particle Mass Spin Charge
symbol
MEONS Photon hν – 0 1 0
Electron e– e+ 1 1/2 –1
HARDONS Neutrino Ve 0 1/2 0
BARYONS Muon Ve µ+ 207 1/2 –1
Muon –neutrino Vµ 0 1/2 0
Tauon µ− τ+ 3500 1/2 –1
Vµ 264 0 0
Pions τ− π−
π0 273 0 +1
Kaons k−
π+ k0 966 0 +1
Etameson – 974 0 0
Proton k+ p– 0 0
Neutron k0 n– 1074 1/2 +1
Lambda hyperon n0 1836.6 1/2 0
p λ0 1836.6
Sigma hyperons n 2183 1/2 0
λ0 Σ+ 1/2 +1
Σ0 2328 1/2 0
Σ+ Σ −1 2334 1/2 –1
Σ0 2343 1/2 0
3/2 –1
Σ−
Xi hyperons 1.1, 1.1 – 2573
Omega hyperons Ω− Ω− 3273
Some common important elementary particles are listed below :
Name Symbol Mass Charge Discoverer
Electron e− 9.1 × 10−31kg − 1.602 × 10−19 C J.J. Thomson (1896)
Proton p 1.673 × 10−27 kg + 1.602 × 10−19 C E. Goldstein (1886)
Neutron n 1.675 × 10−27 kg Zero J. Chadwick (1932)
Neutrino V 3.64 × 10−32kg Zero Pauli
Mesons µ 275 – 300 times mass of +ve,0 or −ve Yukawa (1935)
electron
Positron e+ 9.1 × 10−31kg +ve Anderson (1932)
The stability of nucleus may be discussed in terms of any one of the following.
(i) Nuclear Binding Energy and Mass defect : The mass of hydrogen atom is equal to the sum of the
masses of a proton and an electron. For other atoms, the atomic mass is less than the sum of the masses of protons,
neutrons and electrons present. This difference in mass termed as, mass defect, is a measure of the binding energy
of protons and neutrons in the nucleus. The mass energy relationship postulated by Einstein is expressed as :
∆E = ∆mc 2 , Where ∆E is the energy liberated, ∆m the loss of mass and c is the speed of light.
Consider the helium nucleus, which contains 2 protons and 2 neutrons; the mass of helium nucleus on
12 C = 12mu , scale is 4.0017mu . The masses of individual isolated proton and neutron are 1.0073 and 1.0087 mu
respectively. The total mass of 2 protons and 2 neutrons is (2 × 1.0073)+ (2 × 1.0087) = 4.0320mu . The loss in mass
or mass defect for helium nucleus is, 4.0320mu − 4.0017mu = 0.0303mu
1mu = 1.66057 × 10−27 kg and c = 2.998 × 108 ms −1
∆E = 0.0303 × 1.66057 × 10−27 × 6.02 × 10 23 × (2.998 × 108 )2 kg m2 s −2mol −1 = 2.727 × 1012 J mol −1
Thus, the molar nuclear binding energy of helium nucleus, 2 He 4 , is 2.73 × 1012 J mol −1 . Binding energy of a
nucleus is generally quoted as energy in million electron volts (MeV) per nucleon. One million electron volts are
equivalent to 9.6 × 1010 J mol −1 . Thus, the formation of helium nucleus results in the release of
2.7 × 1012 / 9.6 × 1010 MeV = 28 MeV (approximately). In comparing the binding energies of different nuclei, it is
more useful to consider the binding energy per nucleon. For example, helium nucleus contains 4 nucleons (2
protons and 2 neutrons), the binding energy per nucleon in this case is 28/4 = 7 MeV.
Binding energies of the nuclei of other atoms can be calculated in a similar manner. When we plotted binding
energies of the nuclei of atoms against their respective mass number. Three features may be noted. First, nuclei with
mass number around 60 have the highest binding energy per nucleon. Second, species of mass numbers 4, 12, and
16 have high binding energy per nucleon implying that the nuclei 4 He, 12C and 16 O , are particularly stable. Third
the binding energy per nucleon decreases appreciably above mass number 100. The form of relationship between
binding energy per nucleon and mass number indicates that heavy nuclei would release mass (and therefore
energy) on division (or fission) into two nuclei of medium mass and that the light nuclei would release mass (and
therefore energy) on fusion to form heavier nuclei. These processes called fission and fusion are described later in
this Unit.
The average binding energy for most of the nuclei is in the vicinity of 8 MeV. Nuclei having binding energy
per nucleon very near to 8 MeV are more or less stable.
Iron has the maximum average binding energy (8.79 MeV) and thus its nucleus is thermodynamically most
stable.
The isotopes with intermediate mass numbers 40 to 100 are most stable. The elements with Low Mass numbers
or High Mass numbers tend to become stable by acquiring intermediate mass number. Evidently, nuclei of
lighter elements combine together to form a heavier nucleus of intermediate mass number (nuclear fusion);
while the nuclei of heavy elements split into two lighter nuclei of intermediate mass numbers (nuclear fission).
In either case, energy is released and hence the stability is enhanced.
Note : Relation between different units of energy 1cal = 4.2J ; 1 J = 107 ergs; 1eV = 1.622 × 10−19 J
(ii) Relative stability of isotopes and binding energy: Value of binding energy predicts the relative
stability of the different isotopes of an element. If the value of binding energy is negative, the product nucleus or
nuclei will be less stable than the reactant nucleus. Thus the relative stability of the different isotopes of an element
can be predicted by the values of binding energy for each successive addition of one neutron to the nucleus.
2 He 3 + 0n1 → 2 He 4 + 20.5MeV ; 2 He4 + 0n1 → 2He5 − 0.8MeV
Therefore, 2 He 4 is more stable than 2 He 3 and 2 He 5 .
(iii) Packing fraction: The difference of actual isotopic mass and the mass number in terms of packing
fraction is defined as:
Packing fraction = Actual isotopic mass − Mass number × 10 4
Mass number
The value of packing fraction depends upon the manner of packing of the nucleons with in the nucleus. Its
value can be negative, positive or even zero.
Note : Actual isotopic mass is not a whole number whereas, mass number is a whole number.
(a) Zero packing fraction: Carbon12 has zero packing fraction because it is taken as a reference on the
atomic scale and its actual isotopic mass (12) is equal to its mass number (12).
(b) Negative packing fraction : Negative value of the packing fraction means that the actual isotopic mass
is less to the mass number. This term indicates that some mass has been transformed into energy (binding energy)
during formation of nucleus. Such nuclei are, therefore more stable.
(c) Positive packing fraction : Positive packing fraction should imply the opposite, i.e., the nuclei of such
isotopes should be unstable. However, this generalisation is not strictly correct especially for elements of Low Mass
numbers. For these elements, though packing fraction is positive, yet they are stable. This is explained on the basis
that the actual masses of protons and neutrons (of which the nuclei are composed) are slightly greater than unity.
In general, lower the packing fraction, greater is the binding energy per nucleon and hence greater is the
stability the relatively low packing fraction of He, C and O implies their exceptional stability packing fraction is least
for Fe (negative) and highest for H (+78).
(iv) Meson theory of nuclear forces : Neutron is found to play a leading role in binding the nuclear
particles. It has been established that neutron proton attractions are stronger than the proton-proton or neutron –
neutron attraction. This is evident by the fact that the deutron, 1 H 2 having one proton and one neutron, is quite
stable.
Yukawa in 1935, put forward a postulate that neutrons and protons are held together by very rapid
exchange of nuclear particles called Pi-mesons (π-mesons have mass equal to 275 times of the mass of an electron
and a charge equal to +1, 0 or –1. There are designated as π+ π0 and π– respectively). The nuclear force which is
used in rapid exchange of Pi-mesons between nucleons are also called exchange forces.
• The binding forces between unlike nucleons (p and n) are explained by the oscillation of a charged π-
meson (π+ orπ − ) (a) p1 + n2 n1 + π + + n2 n1 + p2
(b) p1 + n2 n1 + π − + p2 n1 + p2
• Binding forces between like nucleons (p - p or n - n) result from the exchange of neutral mesons (π0) as
represented below.
(a) p1 p2 + π 0 or p1 + π 0 p2
(b) n1 n2 + π 0 or n1 + π 0 n2
(v) Nuclear shell model : According to this theory, nucleus of atom, like extra-nuclear electrons, also has
definite energy levels (shells). The shell structure is supported by the existence of periodicity in the nuclear
properties. For example, elements with even number of protons and neutrons are more abundant, more stable and
richer in isotopes. Nuclides with odd number of protons and neutrons are least abundant in nature (only 5 are
known 1 H 2 , 5 B10 , 7 N 14 and 73 Ta180 ).
Thus elements have a tendency to have even number of both protons and neutrons. This suggests that like
electrons, nucleon particles in the nucleus are paired. Magnetic fields of the two paired protons spinning in opposite
direction cancel each other and develop attractive forces, which are sufficient to stabilize the nucleus. Further nuclei
with 2, 8, 20, 28, 50, 82 or 126 protons or neutrons have been found to be particularly stable with a large number
of isotopes. These numbers, commonly known as Magic numbers are defined as the number of nucleons required
for completion of the energy levels of the nucleus. Nucleons are arranged in shells as two protons or two neutrons
(with paired spins) just like electrons arranged in the extra-nuclear part. Thus the following nuclei
2 He 4 , 8O16 ,20 Ca 40 and 82 Pb 208 containing protons 2, 8, 20 and 82 respectively (all magic numbers) and
neutrons 2, 8, 20 and 126 respectively (all magic numbers) are the most stable.
Magic numbers for protons : 2, 8, 20, 28, 50, 82, 114
Magic numbers for neutrons : 2, 8, 20, 28, 50, 126, 184, 196
When both the number of protons and number of neutrons are magic numbers, the nucleus is very stable.
That is why most of the radioactive disintegration series terminate into stable isotope of lead (magic number for
proton = 82, magic number for neutron = 126). Nuclei with nucleons just above the magic numbers are less stable
and hence these may emit some particles to attain magic numbers.
(vi) Nuclear fluid theory : According to this theory the nucleus is considered to resemble a liquid drop.
Nucleons are believed to be present in the nucleus as nuclear fluid of very high density equal to 130 trillion
tonnes/m3, which is about 100 trillion times the density of water. The density is uniform and does not vary from
atom to atom. Along with its almost unbelievable high density nuclear fluid possesses a correspondingly high
surface tension ( = 9.3 × 1019 Nm−1 , i.e., 1.24 × 1018 times the surface tension of water). A nuclear film attached to a
wire one centimetre long would support the mass of one billion tonnes. 130
This force of nuclear surface tension is, in fact, responsible for keeping the
nucleons bound together against the forces of repulsion. This is known as 120 n/p=1.5 200 Hg
the Nuclear fluid theory of the stability of the nuclei. Thus according to 80
this theory the nucleons are free to move with in the nucleus whereas
according to the nuclear shell structure theory the nucleons exist in definite 110
energy levels.
100 Stability
(vii) Neutron-proton ratio and nuclear stability or causes of belt
radioactivity : The nuclear stability is found to be related to the
neutron/proton (n/p) ratio. If for different elements the number of neutrons 90
is plotted against the number of protons, it is found that the elements with
stable nuclei (non-radioactive elements) lie within a region (belt) known as Number of neutrons (n) 80 n/p=1.4 120 Sn
50
70
60 n/p=1.25 90 Zr
40
50
40
n/p=1
30
zone or belt of stability. 20
10
0 10 20 30 40 50 60 70 80
Number of protons (P)
Nuclei lying above or below the stability belt are unstable
(a) For elements with low atomic number (less than 20), n/p ratio is 1, i.e., the number of protons is equal to
the number of neutrons. Remember that n/p ratio of 1 H 1 is zero as it has no neutron. Nuclide with highest n/p
ratio is 1 H 3 (n/p = 2.0)
(b) With the increase in atomic number although the number of protons increases but the number of neutrons
increases much more than the number of protons with the result the n/p ratio goes on increasing from 1 till it
becomes nearly equal to 1.5 at the upper end of the belt.
(c) When the n/p ratio exceeds 1.52 as in elements with atomic number 84 or higher, the element becomes
radioactive and undergoes disintegration spontaneously. Note that these elements lie outside the zone of stability.
The way an unstable nucleus disintegrates is decided by its position with respect to the actual n/p plot of stable
nuclei (the zone of stability)
• Neutrons to proton (n/p) ratio too high. If the n/p ratio is too high, i.e., when the nucleus contains too
many neutrons, it falls above the zone of stability. The isotope would be unstable and would tend to come within
the stability zone by the emission of a β-ray (electron). Electron, is produced in the nucleus probably by the
following type of decay of a neutron.
0 n1 →1 H 1 + −1e 0 (Beta particle)
The electron thus produced is emitted as a β-particle and thus the neutron decay ultimately increases the
number of protons, with the result the n/p ratio decreases and comes to the stable belt. Consider the example of
C12 and C14 . In C12 , the n/p ratio (6/6) is 1, hence its nucleus is quite stable. On the other hand, in C14 , the n/p
ratio (8/6) is 1.3, hence it should be unstable. In practice also it is found to be so and C14 decays in he following
way to give N 14 (n/p ratio = 1)
6 C14 → 7 N 14 + −1e 0 ; n U 238 → 90 Th234 + 2 He 4
p n 144
92 p = 90
n 8 n 7 146
p = 6 =1.33 p =7 =1.0 = 92 =1.587 =1.6
• Neutron to proton ratio (n/p) too low, (i.e., when the nucleus contains excess protons) : There are no
naturally occurring nuclides with n/p ratio less than 1, however there are many artificially nuclides in such cases, the
nucleus lies below the zone of stability, it would again be unstable and would tend to come within the zone of
stability by losing a positron.
6 C11 → 5 B11 + +1e 0 ; N 13 → C 13 + e0
n 5 n 7 6 +1
p 6 p 5
= =0.83 n = 6 =1.2 = 6 =0.83 n 7 =1.16
p 5 p =6
Such nuclides can increase n/p ratio by adopting any one of the following three ways :
By emission of an alpha particle : U 238 → 90 Th234 + 2 He 4
n 92 n 144
p 146 p = 90
= 92 =1.58 =1.60
By emission of a positron : N13 → 13 C + e0
6
7 n 7 +1
n 6 p =6
p = 7
By K-electron capture : 194 Au + e0 → 194 Pt
79 78
n 115 −1 n 116
p 79 p 78
= =
α-emission is usually observed in natural radioactive isotopes while emission of positron or K-electron capture
is observed in artificial radioactive isotopes. The unstable nuclei continue to emit α or β-particles. Until stable nuclei
comes into existence.
(3) Nuclear reactions : In a chemical reaction, only electrons (extra-nuclear particle) of the atom take part
while the nucleus of the atom remains unaffected. However, the reverse reactions (i.e., where only nuclei of atoms
take part in reactions) are also possible. Such reactions in which nucleus of an atom itself undergoes spontaneous
change or interact with other nuclei of lighter particles resulting new nuclei and one or more lighter particles are
called nuclear reactions.
(i) Some characteristics of nuclear reactions :
(a) Nuclear reactions are written like a chemical reaction : As in a chemical reaction, reactants in a nuclear
reaction are written on the left hand side and products on the right hand side with an arrow in between them.
(b) Mass number and atomic number of the elements are written in a nuclear reactions : Mass number and
atomic number of the element involved in a nuclear reaction are inserted as superscripts and subscripts respectively
on the symbol of the element. For example 27 Al or Al1237 or 13 Al 27 stands for an atom of aluminum with mass
13
number 27 and atomic number 13.
(c) Mass number and atomic number are conserved : In a nuclear reaction the total mass numbers and total
atomic numbers are balanced on the two sides of the reaction (recall that in an ordinary reaction the total number
of atoms of the various elements are balanced on the two sides)
(d) Energy involved in the nuclear reactions is indicated in the product as +Q or –Q of reactions accompanied
by release or absorption of energy respectively.
(e) Important projectiles are α-particles (2 He 4 ) , Proton (1 H 1 or p) , deutron ( 1 H 2 or 1 D 2 ), neutron (0 n1 ),
electron (β-particle or e0 or e–) and positron (+1 e 0 ) .
−1
(f) Representation of nuclear reactions : For example, 7 N 14 + 2 He 4 → 8O17 + 1H 1 + Q . Some times a short
hand notation is used, e.g., the above reaction can be represented as below. 7 N 14 (α, p)8 O17
(ii) Nuclear reactions Vs chemical reactions :
(a) As per definition, chemical reactions depend upon the number of extranuclear electrons while nuclear
reactions are independent upon the electrons but depend upon the nature of the nucleus.
(b) Chemical reactions involve some loss, gain or overlap of outer orbital electrons of the two-reactant atoms.
On the other hand, nuclear reactions involve emission of some light particles (α, β, positron, etc.) from the nucleus
of the atom to form another element.
(c) The chemical reactivity of the element is dependent on the nature of the bond present in the concerned
compound. On the other hand, the nuclear reactivity of the element is independent of its state of chemical
combination, e.g., radium, whether present as such or in the form of its compound, shows similar radioactivity.
(d) The energy change occuring in nuclear reactions is very high as compared to that in chemical reactions.
Again in chemical reactions the energy is expressed in kcal per mole while in nuclear reactions the energy is
expressed in MeV per nucleus. Nuclear reactions, which liberate energy are called exoergic reactions and which
absorb energy are called endoergic.
(e) A chemical reaction is balanced in terms of mass only while a nuclear reaction must be balanced in terms
of both mass and energy. In endoergic reactions, the mass of products is more than the mass of reactants. While in
exoergic reaction the mass of products is less than the mass of reactants.
(f) The chemical reactions are dependent on temperature and pressure while the nuclear reactions are
independent of external conditions.
(iii) Types of nuclear reactions : Nuclear reactions may broadly be divided into two types :
(a) Natural nuclear reactions : In these reactions, nucleus of a single atom undergoes a spontaneous
change itself.
(b) Artificial nuclear reactions : In these reactions, two nuclei of different elements are brought to interact
artificially. Bombarding a relatively heavier nucleus (non-radioactive) with a lighter nucleus, viz. proton, deutron
and helium, does this. Artificial nuclear reactions are divided as follows :
• Projectile capture reactions : The bombarding particle is absorbed with or without the emission of γ-
radiations.
92 U 238 + 0n1 → 92U 239 + γ ; 13 Al 27 + 0n1 → 13 Al 28 + γ
• Particle-particle reactions : Majority of nuclear reactions come under this category. In addition to the
product nucleus, an elementary particle is also emitted.
11 Na 23 + 1 H 1 → 12 Mg 23 + 0n1 ; 11 Na 23 + 1 H 2 → 11 Na 24 + 1 H 1
11 Na 23 + 2 He 4 → 12 Mg 26 + 1 H 1 ; 7 N 14 + 0n1 → 6 C14 + 1 H 1
• Spallation reactions : High speed projectiles with energies approximately 40 MeV may chip fragments
from a heavy nucleus, leaving a smaller nucleus.
29 Cu63 + 2 He 4 + 400 MeV → 17 Cl 37 + 14 1 H 1 + 160 n1
• Fission reactions : A reaction in which a heavy nucleus is broken down into two or more medium heavy
fragments. The process is usually accompanied with emission of neutrons and large amount of energy.
92 U 235 + 0n1 → 56 Ba141 + 36 Kr 92 + 3 0n1 + 200 MeV
• Fusion reactions : Light nuclei fuse together to reproduce comparatively heavier nuclei. A fusion
reactions is the source of tremendous amount of energy.
1 H 2 + 1 H 3 → 2 He 4 + 0n1 + 17.6 MeV
Example : 1 Sulphur-35 (34.96903 amu) emits a β-particle but no γ-ray. The product is chlorine-35 (34.96885 amu). The
Solution: (c)
maximum energy emitted by the β-particle is [CBSE 1999]
(a) 16.758 MeV (b) 1.6758 MeV (c) 0.16758 MeV (d) 0.016758 MeV
The mass converted into energy = 34.96903amu − 34.96885 amu = 1.8 × 10 −4 amu (1amu = 931.5 MeV )
Energy produced = 1.8 × 10−4 × 931.5 = 0.16758 MeV
Example : 2 If the atomic masses of lithium, helium and proton are 7.01823 amu, 4.00387 amu and 1.00815 amu
Solution: (a) respectively, calculate the energy that will be evolved in the reaction. Li7 + H 1 → 2He 4 + energy .
Example : 3 (Given that 1 amu = 931 MeV)
Solution: (a)
(a) 17.3 MeV (b) 17.8 MeV (c) 17.2 MeV (d) 17.0 MeV
Total mass of the reacting species (Li7 and H 1) = 7.01823 + 1.00815 = 8.02638 amu
The mass of the resulting species (2 He 4 ) = 2 × 4.00387 = 8.00774 amu
Mass of reacting species converted into energy, i.e., ∆m = 8.02638 − 8.00774 = 0.01864 amu
∴ Energy evolved in the reaction = 0.01864 × 931.5 = 17.363 MeV.
Calculate the mass defect and binding energy per nucleon for 27Co 59 . [The mass of Co 59 = 58.95 amu ,
mass of hydrogen atom = 1.008142 amu and mass of neutron = 1.008982 amu].
(a) 8.77 MeV (b) 8.25 MeV (c) 9.01 MeV (d) 8.00 MeV
Number of protons in 27Co 59 = 27
∴ Number of neutrons = 59 – 27 = 32
∆m = (1.008142 × 27 + 1.008982 × 32) − 58.95 = 0.556438 amu
The binding energy (EB ) per nucleon = ∆m × 931 = 0.556438 × 931 MeV = 8.77 MeV
Mass of cobalt 59
Radioactivity.
“Radioactivity is a process in which nuclei of certain elements undergo spontaneous disintegration without
excitation by any external means.’’
Henry Becquerel (1891) observed the spontaneous emission of invisible, penetrating rays from potassium
uranyl sulphate K2UO2(SO4 )2 , which influenced photographic plate in dark and were able to produce
luminosity in substances like ZnS.
Later on, Madam Marie Curie and her husband P. Curie named this phenomenon of spontaneous
emission of penetrating rays as, Radioactivity. They also pointed out that radioactivity is a characteristic
property of an unstable or excited nucleus, i.e., a nuclear property is independent of all the external conditions
such as pressure, temperature, nature of other atoms associated with unstable atom but depends upon the
amount of unstable atom.
Curies also discovered a new radioactive element Radium from pitchblende (an ore of U i.e. U3O8 ) which is
about 3 million times more radioactive than uranium. Now a days about 42 radioactive elements are known.
The elements whose atoms disintegrate and emit radiations are called radioactive elements.
Radioactivity can be detected and measured by a number of devices like ionisation chamber, Geiger Muller
counter, proportional counter, flow counter, end window counter, scintillation counter, Wilson cloud chamber,
electroscope, etc. The proper device depends upon the nature of the radioactive substance and the type of
radiation emitted. GM counter and proportional counter are suitable for solids and liquids, ionisation chamber is
most suitable for gases.
Lightest radioactive isotope is tritium ( 1H 3 ) ; other lighter radioactive nuclides are 14 C , 40 K and 99Tc .
(1) Nature of radioactive emissions : The nature of the radiations emitted from a radioactive substance
was investigated by Rutherford (1904) by applying electric and magnetic fields to the radiation as shown in
figure.
Photographic plate Photographic plate
γ Magnet γ
αβ Slit αβ
Radioactive
substance
Block of Lead
(as Shiled)
Study of the nature of radiations emitted form a radioactive substance
It is observed that on applying the field, the rays emitted from the radioactive substances are separated into
three types, called α, β, and γ-rays. The α-rays are deflected in a direction which shows that they carry positive
charge; the β-rays are deflected in the opposite direction showing that they carry negative charge and the γ-rays are
not deflected at all showing that they carry no charge.
(2) Characteristics of radioactive rays : Radioactive rays are characterised by the following properties :
(i) They blacken photographic plates.
(ii) They pass through thin metal foils.
(iii)They produce ionization in gases through which they passes.
(iv)They produce luminescence in zinc sulphide, barium platinocyanide, calcium tungstate, etc.
Radioactive radiations are composed of three important rays, namely α, β and γ − rays which differ very
much in their nature and properties, e.g. penetrating power, ionising power and effect on photographic plates.
Remember that γ-rays are not produced simultaneously with α and β-rays but are produced subsequently.
Comparison of α, β and γ-rays
α-Particle or α-Ray β-Particle or β-Ray γ-Ray
(1) Charge and mass : It carries 2 It carries 1 unit negative charge and These are electromagnetic rays with
units positive charge and 4 unit mass. no mass. very short wavelength (app. 0.05 Å)
(2) Nature : It is represented as helium It is represented as electron − 1e0 . It is represented as 0γ 0
nucleus or helium ions 2He4 or
He++ .
(3) Action of magnetic field : These These are deflected to anode. These are not deflected.
are deflected towards the cathode.
(4) Velocity : 2 ×109 cm / s or 2.36 to 2.83 ×1010 cm / s (2.36 to Same as that of light 3 ×1010 cm / s
2 ×107 m / sec (1/10th to that of light) 2.83 ×108 m / s ) (3 ×108 m / s)
(5) Ionizing power : Very high nearly Low nearly 100 times to that of γ-rays. Very low.
100 times to that of β-rays.
(6) Effect on ZnS plate : They cause Very little effect. Very little effect.
luminescence.
(7) Penetrating power : Low 100 times that of α-particles. 10 times that of β-particles.
(8) Range : Very small (8-12 cm.) More that of α-particles. More
(9) Nature of product : Product Product obtained by the loss of 1 β- There is no change in the atomic
obtained by the loss of 1 α-particle particle has atomic number more by 1 number as well as in mass number.
has atomic number less by 2 units and unit, without any change in mass
mass number less by 4 units. number.
Note : β-particles originates in the nucleus; they are not orbital electrons.
β-particles having their velocity almost equal to velocity of light are known as hard β-particles and the others
having their velocity ≈ 1 × 1010 cm sec −1 are called soft β-particles.
γ-radiation always accompany alpha or beta emissions and thus are emitted after α- and β-decay.
Only one kind of emission at a time is noticed. No radioactive substance emits both α- and β-particles
simultaneously.
Theory of radioactive disintegration.
Rutherford and Soddy, in 1903, postulated that radioactivity is a nuclear phenomenon and all the
radioactive changes are taking place in the nucleus of the atom. They presented an interpretation of the radioactive
processes and the origin of radiations in the form of a theory known as theory of radioactive disintegration.
The main points of this theory are as follows :
• The atomic nuclei of the radioactive elements are unstable and liable to disintegrate any moment.
• The disintegration is spontaneous, i.e., constantly breaking. The rate of breaking is not affected by external
factors like temperature, pressure, chemical combination etc.
• During disintegration, atoms of new elements called daughter elements having different physical and
chemical properties than the parent elements come into existence.
• During disintegration, either alpha or beta particles are emitted from the nucleus.
The disintegration process may proceed in one of the following two ways :
(1) α-particle emission : When an α-particle (2 He 4 ) is emitted from the nucleus of an atom of the parent
element, the nucleus of the new element, called daughter element possesses atomic mass or atomic mass number
less by four units and nuclear charge or atomic number less by 2 units because α-particle has mass of 4 units and
nuclear charge of two units.
Parent element -α → Daughter element
Atomic mass :W W−4
Atomic number : Z Z−2
Examples are:
88 Ra 226 → 86 Rn222 + 2 He 4 ; 92 U 238 → 90Th234 + 2 He 4
(Radium) (Radon) (Uranium) (Thorium)
83 Bi 213 → 81Tl 209 + 2 He 4 ; 84 Po 215 → 82 Pb 211 + 2 He 4
(Bismuth) (Thallium) (Polonium) (Lead)
(2) β-particle emission : β-particle is merely an electron which has negligible mass. Whenever a beta
particle is emitted from the nucleus of a radioactive atom, the nucleus of the new element formed possesses the
same atomic mass but nuclear charge or atomic number is increased by 1 unit than the parent element. Beta
particle emission is due to the result of decay of neutron into proton and electron.
0n1 → 1 p1 + −1e 0
The electron produced escapes as a beta-particle-leaving proton in the nucleus.
Parent element -β→ Daughter element
Atomic mass :W W
Atomic number : Z Z +1
Examples are:
82 Pb 214 → 83 Bi 214 + −1e 0 ; 90 Th234 → + 91 Pa 234 + −1e 0 ; 83 Bi 213 → 84 Po 213 + −1e 0
(Lead) (Bismuth) (Thorium) (protoactinium) (Polonium)
Special case : If in a radioactive transformation 1 alpha and 2 beta-particles are emitted, the resulting
nucleus possesses the same atomic number but atomic mass is less by 4 units. A radioactive transformation of this
type always produces an isotope of the parent element.
Z AW −α → Z −2BW −4 −β → Z −1CW −4 −β → Z DW −4
A and D are isotopes.
• γ-rays are emitted due to secondary effects. The excess of energy is released in the form of γ-rays. Thus γ-
rays arise from energy re-arrangements in the nucleus. As γ-rays are short wavelength electromagnetic radiations
with no charge and no mass, their emission from a radioactive element does not produce new element.
Group displacement law.
Soddy, Fajans and Russell (1911-1913) observed that when an α-particle is lost, a new element with
atomic number less by 2 and mass number less by 4 is formed. Similarly, when β-particle is lost, new element with
atomic number greater by 1 is obtained. The element emitting then α or β-particle is called parent element and the
new element formed is called daughter element. The above results have been summarized as Group
displacement laws as follows :
(1) When an α-particle is emitted, the new element formed is displaced two positions to the left in the
periodic table than that of the parent element (because the atomic number decreases by 2).
New element formed Original element
(daughter) (parent)
Period YA – 4 XA
Z–2 Z
Two position left of By loss of α - particle
the original element
For example, when
92U 238 → 90 Th234 + 2 He 4
(2) When a β-particle is emitted, the new element formed is displaced one position to the right in the periodic
table than that of the parent element (because atomic number increased by 1).
Original element New element formed
(parent) (daughter)
Period Y XA A
Z Z+1
By loss of α - particle One position right of
the original element
For example,
90Th234 → 91 Pa 234 + −1e 0 ; 6 C14 → 7 N 14 + −1e 0
Hence, group displacement law should be applied with great care especially in the case of elements of
lanthanide series (57 to 71), actinide series (89 to 103), VIII group (26 to 28; 44 to 46; 76 to 78), IA and IIA groups.
It is always beneficial to keep in mind the setup and skeleton of the extended form of periodic table.
IA IIA IIIB IVB VB VIB VIIB VIII IB IIB IIIA IVA VA VIA VIIA Zero
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
1st period IA IIA IIIB IVB Zero
2nd period 1 18
3rd period – – –
4th period At. No. – – – At. No.
5th period 1 – – – 2
6th period 3 – – – 10
7th period 11 – – – 18
19 56 57*–71 72 36
37 88 89!–103 104 54
55 86
87
*Lanthanides, ! Actinides
Important tips
α-Decay produces isodiapher i.e., the parent and daughter nuclide formed by α-decay have same isotopic number, i.e., difference
between the number of neutrons and protons is same. For example,
88Ra226 → 86Rn222
No. of neutrons 138 136
No. of protons 86
Difference 88 50
50
Thus note that an α-decay leads to
(i) Decrease in atomic weight, mass number and number of nucleons by four units.
(ii) Decrease in number of protons, neutrons, nuclear charge and atomic number by two units.
(iii) Increase in n/p ratio.
β-Decay results in the formation of an isobaric element i.e., parent and daughter nuclide have different atomic numbers but same mass
number. For example,
19 K 40 → 20Ca 40 + −1e 0
Thus note that a β-decay leads to
(i) No change in atomic weights, mass number and number of nucleons.
(ii) Decrease in number of neutrons by one unit.
(iii) Increase in nuclear charge, number of protons and atomic number by one unit.
(iv) Decrease in n/p ratio.
It is important to note that although β-particle (electron) is not present in the nucleus, even then it is emitted from the nucleus since a
neutron at first breaks down to a proton and electron.
0n1 → 1 p1 + −1e 0
The proton is retained by the nucleus while the electron is emitted as a β-particle.
Emission of 1 α-particle and 2 β-particles in succession produces an isotope of the parent element. For example,
92U 235 −α → 90Th231 −β → 91Pa 231 −β → 92U 231
This law helps to fix the position of the radioelements in the periodic table.
To determine the number of α- and β- particles emitted during the nuclear transformation. It can be done in following manner :
a X → b Y + x 4 He + y −1e 0 ,
c d 2
a = b + 4 x or x = a−b .......(i)
4
c = d + 2x − y ......(ii)
where x = no. of α-emitted, y = no. of β-emitted
substituting the value of x from eq. (i) in eq. (ii) we get c = d + a − b 2 − y ; y = d + a − b − c
4 2
Example : 4 Calculate the number of neutrons in the remaining atom after emission of an alpha particle from 238 U atom
Solution: (c) 92
Example : 5 [Roorkee 1978]
Solution: (c)
(a) 146 (b) 145 (c) 144 (d) 143
On account of emission of an alpha particle, the atomic mass is decreased by 4 units and atomic number by 2
units
So, Atomic mass of daughter element = 234
Atomic number of daughter element = 90
Number of neutrons = Atomic mass – Atomic number
= 234 – 90 = 144
Radioactive disintegration of 226 Ra takes place in the following manner into RaC
88
Ra −α → Rn −α → RaA −α → RaB −β → RaC , Determine mass number and atomic number of RaC.
(a) 214 and 84 (b) 214 and 86 (c) 214 and 83 (d) 214 and 85
Parent element is 226 Ra
88
Atomic mass = 226
Atomic number = 88
RaC is formed after the emission of 3 alpha particles. Mass of 3 alpha particles = 3 × 4 = 12
So Atomic mass of RaC = (226 − 12) = 214
With emission of one α-particle, atomic number is decreased by 2 and with emission of β-particle, atomic
number is increased by 1.
So Atomic number of RaC = 88 − (3 × 2) + 1 = 83
Example : 6 How many ‘α and β’ particles will be emitted when 90Th234 changes into 84 Po 218 [CBSE 1992]
Solution: (b)
(a) 2 and 6 (b) 4 and 2 (c) 2 and 4 (d) 6 and 2
The change is; 90 Th234 → 84 Po 218
Parent End product
Decrease in mass = (234 − 218) = 16 amu
Mass of 1 α-particle = 4 amu
Therefore, number of α-particles emitted = 16 = 4
4
Number of β-particles emitted = 2 × No. of αparticles emitted − (atomic no. of parent − At. no. of product )
= 2 × 4 − (90 − 84) = 2
Hence number of α-particles = 4 and number of β-particles = 2
Example : 7 An atom has atomic mass 232 and atomic number 90. During the course of disintegration, it emits 2 β-
Solution: (a)
particles and few α-particles. The resultant atom has atomic mass 212 and atomic number 82. How many α-
particles are emitted during this process [CBSE 1992]
(a) 5 (b) 6 (c) 7 (d) 8
The decrease in atomic mass = (232 – 212) = 20
Decrease in mass occurs due to emission of α-particles. Let x be the number of alpha particles emitted.
Mass of 'x' α-particles = 4x
So 4 x = 20 or x = 20 = 5
4
Alternative method : This can also be determined by the application of following equation :
No. of β-particles emitted = 2 × No. of α-particles emitted −(ZParent − ZEnd product )
2 = 2 × x – (90 – 82) or x = 5
Example : 8 An element X with atomic number 90 and mass number 232 loses one α- and two β-particles successively to
give a stable species Z. What would be the atomic number and atomic weight of Z [CPMT 1990]
(a) Z 228 (b) Z 235 (c) Z 235 (d) Z 238
90 91 90 91
Solution: (a) At. no. and At. wt. of the element (Y) produced by the loss of one α-particle ( 2 He 4 ) from 90 X 232 = 88 Y 228
At. no. and At.wt. of Z produced by the loss of 2 β-particles ( −1e 0 ) from Y = 90 Z 228
Example : 9 Find out the total number of α- and β-particles emitted in the disintegration of 90 Th232 to 82 Pb 208
Solution: (a)
[CPMT 1989]
(a) 6 and 4 (b) 8 and 4 (c) 9 and 6 (d) 2 and 4
Change in at. wt. = 232 – 208 = 24 amu (at. wt. unit)
Now since in one α-particle emission, at. wt. is decreased by 4 amu, the number of α-emissions for 24 amu =
24/4 = 6
Atomic number after 6α-emissions = 90 – 12 = 78 ( α = 2 He 4 )
Increase in atomic number from 78 to the given 82 = 82 – 78 = 4 ( β - particle = −1e 0 )
∴ No. of β-particle emissions = 4
Example : 10 92U 235 belongs to group III B of the periodic table. It loses one α-particle to form the new element. Predict the
position of the new element in the periodic table [MLNR 1984]
Solution: Since loss of an α-particle decreases the atomic number of the element by 2, the resulting product should lie
two groups to the left of the parent group. However, in the present case the element will remain in the same
group of the periodic table because it is an actinide element.
Example : 11 234 Th disintegrates to give 206 Pb as the final product. How many alpha and beta particles are emitted in this
90 82
process [IIT 1986]
(a) 6 (b) 8 (c) 9 (d) 2
Solution: (a) 234 Th → 206 Pb
90 82
Parent End product
Decrease in mass = (234 – 206) = 28
Mass of α-particle = 4
So Number of α-particles emitted = 28 = 7
4
Number of beta particles emitted = 2 × No. of α-particles – (At. no. of parent – At. no. of end product)
= 2 × 7 – (90 – 82) = 6
Rate of radioactive decay.
“According to the law of radioactive decay, the quantity of a radioelement which disappears in unit time (rate
of disintegration) is directly proportional to the amount present.”
The law of radioactive decay may also be expressed mathematically.
Suppose the number of atoms of the radioactive element present at the commencement of observation, i.e.
when t = 0 is N0 , and after time t, the number of atoms remaining unchanged is N t , then the rate of decay of
atoms is − dN t (the word ‘d’ indicates a very-very small fraction; the negative sign shows that the number of atoms
dt
N t decreases as time t increases)
Now since the change in number of atoms is proportional to the total number of atoms N t , the relation
becomes − dN t = λN t , where λ is a radioactive constant or decay constant.
dt
• Rate of decay of nuclide is independent of temperature, so its energy of activation is zero.
• Since the rate of decay is directly proportional to the amount of the radioactive nuclide present and as the
number of undecomposed atom decreases with increase in time, the rate of decay also decreases with the
increase in time. Various forms of equation for radioactive decay are,
N t = N 0 e −λt ; log N0 − log N t = 0.4343 λt ; log N0 = λt
Nt 2.303
λ = 2.303 log N0
t Nt
where N0 = Initial number of atoms of the given nuclide, i.e. at time 0
N t = Number of atoms of that nuclide present after time t.
λ = Decay constant
Note : This equation is similar to that of first order reaction, hence we can say that radioactive
disintegration are examples of first order reactions. However, unlike first order rate constant (K), the decay constant
(λ) is independent of temperature.
Decay constant (λ) : The ratio between the number of atoms disintegrating in unit time to the total number
of atoms present at that time is called the decay constant of that nuclide.
Characteristics of decay constant (λ) :
It is characteristic of a nuclide (not for an element).
Its units are time−1 .
Its value is always less than one.
Half life and Average life period.
(1) Half-life period (T1/2 or t1/2) : Rutherford in 1904 introduced a constant known as half-life period of
the radioelement for evaluating its radioactivity or for comparing its radioactivity with the activities of other
radioelements. The half-life period of a radioelement is defined, as the time required by a given amount of the
element to decay to one-half of its initial value.
Mathematically, t1/ 2 = 0.693
λ
Now since λ is a constant, we can conclude that half-life period of a particular radioelement is independent of
the amount of the radioelement. In other words, whatever might be the amount of the radioactive element present
at a time, it will always decompose to its half at the end of one half-life period.
Half-life period is a measure of the radioactivity of the element since shorter the half-life period of an element,
greater is the number of the disintegrating atoms and hence greater is its radioactivity. The half-life periods or the
half-lives of different radioelements vary widely, ranging form a fraction of a second to million of years.
Fraction and Percent of radioactive nuclides left after n-Half-Lives
No. of half-lives Fraction of mass Percent of mass
passed (n) Decayed
Decayed Left Left
0 0 100
0 1.0
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