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ELECTROPHILIC AROMATIC SUBSTITUTION REACTION
Introduction :
Aromatic hydrocarbons are known generally as arenes. An aryl group is one derived from an arene by removal
of a hydrogen atom and its symbol is Ar—. Thus, arenes are designated ArH just as alkanes are designated
RH.
The most characteristic reactions of benzenoid arenes are the substitution reactions that occur when they react
with electrophilic reagents. These reactions are of the general type shown below.
Ar – H + E – A Ar – E + H – A
+E – A E
+H– A
The electrophiles are either a positive ion (E+) or some other electron- deficient species with a large partial
positive charge. For example, benzene can be brominated when it reacts with bromine in the presence of
FeBr3. Bromine and FeBr3 react to produce positive bromine ions, Br+. These positive bromine ions act as
electrophiles and attack the benzene ring, replacing one of the hydrogen atoms in a reaction that is called an
electrophilic aromatic substitution (EAS).
Electrophilic aromatic substitutions allow the direct introduction of a wide variety of groups into an aromatic
ring, and because of this they provide synthetic routes to many important compounds. The five electrophilic
aromatic substitutions that we shall study in this chapter are outlined in figure.
X2, FeX3 X
(x=Cl,Br) + HX Halogenation
HONO2 NO2 + H2O Nitration
H2SO4
SO3 SO3H Sulfonation
H2SO4
(RCl, AlCl3) R +HCl Friedel -Crafts Alkylation
O
O (R can rearrange)
Cl C—R +HCl Friedel-Crafts Alkylation
R , AlCl3
Electrophilic Aromatic Substitution Reactions
( 1 ) A GENER AL MECHANISM FOR ELECTROPHILIC AROMATIC SUBSTITUTION :
Benzene is susceptible to electrophilic attack primarily because of its exposed -electrons. In this respect
benzene resembles an alkene, for in the reaction of an alkene with an electrophilic the site of attack is the
exposed bond.
We saw in however, that benzene differs from an alkene in a very significant way. Benzene's closed shell of
six electrons gives it a special stability. So although benzene is susceptible to electrophilic attack, it undergoes
substitution reactions rather than addition reactions. Substitution reactions allow the aromatic sextet of
electrons to be regenerated after attack by the electrophile was occurred. We can see how this happens of
we examine a general mechanism for electrophilic aromatic substitution.
Once the electrophilic, E+ is generated in the reaction, it enters into some kind of a weak interaction with the
cloud of benzene ring leading to the formation of a - complex. This - complex is a donor-acceptor type
of a complex, benzene being the donor and electrophile, the acceptor. These adducts are known as charge
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transfer complexes. In the complex that benzene forms with bromine, it has been shown that the halogen
molecule is located centrally and at right angles to the plane of the benzene ring.
+ E+ (Fast) E
+E E E
H H H
Step I (slow) + +
In step 1 the electrophile takes two electrons of the six- electron system to form a bond to one carbon
atom of the benzene ring. Formation of this bond interrupts the cyclic system of - electrons, because in the
formation of the arenium ion the carbon that forms a bond to the electrophile becomes sp3 hybridized and
therefore, no longer has an available p-orbital. Now only five carbon atoms of the ring are still sp2 hybridized
and still have p-orbitals. A calculated electrostatic potential map for the arenium ion formed by electrophilic
addition of bromine to benzene indicates that positive charge is distributed in the arenium ion ring (figure) just
as was shown in the contributing resonance structures.
In step 2 a proton is removed from the carbon atom of the arenium ion that bears the electrophile. The two
electrons that bonded this proton to carbon becomes a part of the system. The carbon atom that bears the
electrophile becomes sp2 hybridized again and a benzene derivative with six fully delocalized electrons is
formed. We can represent step 2 with any one of the resonance structures for the arenium ion.
+E (Fast) E
H + H–A
(The proton is removed by any of the bases present, for example, by the anion derived from the electrophile).
Kekul'e structure are more apropriate for writing mechanisms such as electrophilic aromatic substitution because
they permit the use of resonance theory, it can be described however, using the modern formula for benzene
in the following way.
Step I E
K1 + H + : A– (Slow step)
K 1
Arenium ion
E E
+ E – A (Fast step)
Step II + H – k2
Where k1 and k–1 is the rate constant of the forward and backward reactions in step-I, k2 is the rate constant
of the step-II.
There is firm experimental evidence that the arenium ion is a true intermediate in electrophilic substitution
reactions. It is not a transition state. This means that in a free energy diagram (figure shown below) the arenium
ion lies in an energy valley between two transition states.
The free energy of activation, G ) , for the reaction leading from benzene and the electrophile, E+, to the
(1
arenium ion has been shown to be much greater than the free energy of activation, G ) , leading from the
(2
arenium ion to the final product. This is consistent with what we would expect. The reaction leading from
benzene and an electrophile to the arenium ion is high endothermic, because the benzene ring loses its resonance
energy. The reaction leading from the arenium ion to the substituted benzene, by contrast, is highly exothermic
because in it the benzene ring regains its resonance energy.
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+
+ +
EH +
1
+ EH
2
Potential Energy
EH
G1 G2
E–A E–A
E
Reaction coordinate
Of the above two steps, step 1 the formation of the arenium ion – is usually the ratedetermining step in
electrophilic aromatic substitution.
Step 2, the removal of a proton, occurs rapidly relative to step 1 and has no effect on the overall rate of
reaction.
Isotope Effects :
The next question which comes up is how do we ascertain the presence of two discrete steps in the mechanism
and also that the formation of - complex is the rate determining step. Answer to this question can be obtained
from the study of kinetic isotope effect. If the rate of a reaction depends on a step which involves breaking
of a C—H bond, then a kinetic isotope effect (KH/KD) of 6 to 7 is expected. Absence of any significant isotope
effect in aromatic electrophilic substitution (except sulphonation) suggests that the proton is lost in the fast step,
subsequent to rds. We see that the isotope effect study has provided two pieces of important information
regarding the mechanism. Firstly, it has shown that the reaction takes place in two steps and secondly, that
the first step is slower than the second step.
( 2 ) Nitration :
Nitration reaction is generally carried out with a mixture of concentrated nitric acid and sulphuric acid. The
reagents which cause nitration are called nitrating agents.
The various nitrating agents which are commonly employed are :
(a) N2O5 in CCl4 in the presence of P2O5 is used when anhydrous conditions are required.
(b) Ethyl nitrate (C2H5ONO2) is used to carry out nitration in alkaline medium.
(c) In the case of polycyclic hydrocarbons N2O4 and nitronium salts such as N O BF4– , N O 2 P F6– , N O S O – ,
2 2 3
can be used. The electrophile involved in nitration reaction is nitronium ion N O .
2
Mechanism :
Generation of electrophile from nitrating agent.
(a) In a mixture of nitric acid and sulphuric acid, an acid base reaction takes place in which nitric acid acts
as the base.
HO–NO2 +H2SO4
Base H–O–NO2 +HSO4
NO2 +H2O
H– O – NO2
H2O + H2SO4 H3O+HSO4
HNO3+2H2SO4 NO2 +H3O+ 2 HSO4
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(b) N2O5 in CCl4 when used, results in a spontaneous dissociation reaction.
N2O5 N O N O –
2 3
(c) With concentrated HNO3 alone
2 HNO3 N O N O – H2O
2 3
The electrophile generated in this case is obtained by the behaviour of one nitric acid as the base and
other molecule as the acid, but the equilibrium lies in the reactant side.
( 3 ) Sulphonation :
Sulphonation is another synthetically important reaction. It is often accomplished with concentrated sulphuric
acid or fuming sulphuric acid containing excess of SO3 or chlorosulphonic acid, ClSO2OH.
It is believed that the electrophile varies with the reagent, though in all cases SO3, is involved either free or
along with a carrier, like in H2SO 4 (SO3 + H3O+) or H2S2O7. Sulphur trioxide is generated form sulphuric
acid as follows 2H2SO4 H SO – +H3O+ + SO3. The mechanism of sulphonation of benzene is given
4
below :
O H SO3 SO3H
K1 (Slow) + SO3 HSOI4–I,,SloHw2SO4 H3IOII,(f,asHt )2O
+S
O O k1
Sulphonation is different from other aromatic electrophilic substitution reactions, Firstly, it is reversible and
secondly, it shows some amount of isotope effect which is totally absent in other cases. Let us have a look
at the potential energy diagram, Fig. of sulphonation reaction to understand these anomalies.
Potential Energy H SO3
+ SO3 +
SO3
+H+
Reaction coordinate
We see that once the - complexed benzenium intermediate is formed ; the energy barriers on either side
of the intermediate are roughly of the same magnitude. This means that the intermediate can cross over to
the product and can also come back to the reactant. This accounts for a reversible nature. Now, if we have
the deuterated substrate, then the potential energy diagram gets slightly modified (dotted curve). The barrier
to step II becomes higher as it now involves the cleavage of C—D bond. The barrier for step I, on the other
hand, remains the same as it pertains to - complex formation. The rate of its reverting back to reactants
is higher than its crossover to the product. Therefore, there is a net decrease in the overall rate of sulphonation
for deuterated substrate-it shows a kinetic isotope effect. The loss of proton (Step II) is the slowest step (rds).
The equilibrium in step III lies to the left as aryl sulphonic acids are strong acids.
In the case of other electrophilic substitutions, in contrast, energy barrier for the first step is much higher than
that for the second step even for a deuterated substrate.
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( 4 ) Halogenation :
Halogenation of an aromatic ring is a synthetically important reaction. It takes place in the presence of varied
reaction condition depending on the reactivity of the aromatic ring. For very reactive aromatic compounds
in polar solvents, the molecular halogens themselves may acts as electrophiles. In the case of nonpolar solvents,
halogenation is catalysed by a Lewis acid like AlCl3, or FeCl3, Reactivity of halogens has the following order,
I2 < Br2 < Cl2
Let us take chlorination as a representative reaction to understand the mechanism of halogenation. Chlorine,
in the presence of AlCl3 or FeCl3 forms a complex, Cl2—AlCl3. This complex can itself be the reactive electrophile
or it may dissociate to give Cl+.
Step 1 : •• AlCl3
Cl—Cl + AlCl3 Cl—Cl — AlCl3
(anhydrous) H Cl
••
Step 2 : slow
Potential Energy
+ Cl — Cl— AlCl3 + AlCl4
Cl H Cl
+ AlCl3 + HCl
Step 3 : Fast
+ AlCl4
However, there is no significant evidence for the involvement of Cl+ as an electrophile and it is likely that the
complex itself attacks the substrate. In the Cl2—AlCl3 complex, role of the Lewis acid is to polarize the halogen
molecule and weaken the Cl—Cl bond. This lowers the activation energy for the formation of -complex.
H Cl---Cl H Cl---Cl-AlCl3
+
+ H Cl
+
intermediate
H Cl
+–
AlCl4
reactants Cl products
+ Cl2—AlCl3 + HCl + AlCl3
Reaction coordinate
Schematic potential energy diagram for chlorination
of benzene with Cl2(–)and Cl2 AlCl3 (---) as the electrophile
Bromination follows a similar mechanism. As said above, Iodine is weakest of the three halogens and even
in the presence of a Lewis acid, It can halogenate reactive aromatics only. Therefore, in most other cases
iodine-monochloride is used in the presence of Lewis acid, ZnCl2.
–
Halogenation may be effected by hypohalous acids, HO — X ,also. This is markedly slower than with molecular
–
halogens as HO– is a poorer leaving group from HO — X ,than, X– is from X — X . The reaction is speeded
in the presence of X–, however, as HO—X is then converted into the more reactive X2, e.g. :
OCl + Cl + 2H Cl2 + H2O
In the presence of strong acid, however, HO—Hal becomes a very powerful halogenating agent due to the
formation of a highly polarized complex :
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HO—X + H H2O + X
H2 O— X
The evidence is that this species is the effective electrophile under these conditions and does not support the
further conversion of complex into Hal, i.e. unlike the case with H2O —NO2. F2 reacts vigorously with
benzene, but C—C bond breaking occurs and the reaction is of no preparative significance.
( 5 ) Fridel-Crafts Alkylation :
Alkyl halides react with benzene in presence of aluminium chloride to yield alkyl benzenes. Alkylation of
benzene with alkyl halides in the presence of aluminium chloride was discovered by Charles Friedel and James.
M. Crafts in 1877.
A general equation for a friedel-Crafts alkylation reaction is the following
+R — X AlCl3 R
+HX
Alkyl halides by themselves are insufficiently electrophilic to react with benzene. AlCl3 serves as a Lewis acid
catalyst to enhance the electrophilicity of the alkylating agent. The mechanism for the reaction (shown in the
following steps, with isopropyl chloride as R—X) starts with the formation of carbocation (step I). The carbocation
then acts as an electrophile (step II) and attacks the benzene ring to form an arenium ion. The arenium ion
(Step III) then loses a proton to generate isopropyl benzene. Some times carbocation rearrange to a more
stable carbocation.
Mechanism for the Reaction :
H3C Cl H3C Cl H3C
CH–C••••l•• + Al CH + AlCl4
Step I +
CH—Cl—Al—Cl H3C
H3C Cl Cl H3C Cl
Step II CH3 CH3 other canonical forms
+ HC CH3
H
CH3
Step III CH3 Cl CH3
+ Cl—Al—Cl + HCl + AlCl3
H
CH3 Cl CH3
With R–X is a primary halide, a simple carbocation probably does not form. Instead, the AlCl3 forms a complex
with the alkyl halide and this complex acts as the electrophile. In the complex the carbon halogen bond is
nearly broken and one in which the carbon atom has a considerable +ve charge.
+ C••••l•• –
R—CH2 AlCl3
These complexes are so carbocation like that they also undergo typical carbocation rearrangements.
Friedel-Crafts alkylations are not restricted to the use of alkyl halides and AlCl3. Many other pairs of reagents
that form carbocations (or carbocation like species) may be used as well. These possibilities include the use
of a mixture of an alkene and an acid.
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CH(CH3)2
CH3 — CH CH2 O C
propene HF
Isopropyl benzene
A mixture of an alcohol and an acid may also be used.
+ HO 60C
B F3
( 6 ) Friedel-crafts Acylation :
The group is called an acyl group and a reaction whereby an acyl group is introduced into a compound is called
an acylation reaction.
The Friedel-Crafts acylation reaction is an effective means of introducing an acyl group into an aromatic ring.
The reaction is often carried out by treating the aromatic compound with an acyl halide.
+ H3C O O
Cl AlCl3 CH2–CH3
+ HCl
Acyl chlorides (also called acid chlorides) are early prepared by treating carboxylic acids with thionyl chloride
(SOCl2) or phosphorous pentachloride (PCl5).
H3C OH H3C Cl
+ SOCl2 80C + SO2 + HCl
O O
Friedel-Crafts acylations can also be carried cut using carboxylic acid anhydrides. eg.
O
O C CH3 O
O AlCl3
H3C + H3C OH
+ O
H3C
Mechanism of the Reaction :
In most Friedel-Crafts acylation reactions the electrophile appears to be an acylium ion formed from an acyl
halide in the following way.
Step I O O
R—C—C••••l••+ AlCl3
R—C—Cl – AlCl3
Step II O O••••+ AlCl4
R—C
H3C—C—Cl— AlCl3
R—C O
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Step III R
H
R—C O
O
+
Step IV R R
H O
O +HCl + AlCl3
+ + AlCl4
R R
O•••• O— AlCl3
–Cl –
+ AlCl3
Step V
In the last step, AlCl3 (a Lewis acid) forms a complex with the ketone (a Lewis base). After the reaction is over,
treating the complex with water liberates the ketone.
R R
C6H5
AlCl3+ 2H2O O + Al(OH)3+ 3HCl
O
C6H5
Limitations of Friedel - Crafts Reactions :
Several restrictions limit the usefulness of Friedel- Crafts reactions.
(a) When the carbocation formed from an alkyl halide, alkene, or alcohol can rearrange to a more stable
carbocation, it usually does so and the major product obtained from the reaction is usually the one from
the more stable carbocation.
When benzene is alkylated with butyl bromide, for example, some of the developing butyl cations rearrange
by a hydride shift-some developing 1° carbocations (see following reactions) become more stable 2°
carbocations. The benzene reacts with both kinds of carbocations to form both butylbenzene and sec-
butylbenzene. H
H3C
Br AlCl3 H3C + [–AlCl3 Br] H3C
CH2---Br---AlCl3
CH3
–AlCl3 –H
–HBr
H2C—CH2-CH2-CH3 H3C—CH–CH2-CH3
minor major
(b) An aromatic ring less reactive than that of halobenzene don't undergo Friedal Craft's reaction. Aromatic
ring containing - NH2, –NHR, –NR2, groups does not undergo friedal craft's alkylation due to formation
of anilinum complex which is meta directing and has more electron withdrawing power than halogen
in benzene ring.
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H H
+
H—N:
H—N—AlCl3
+ AlCl3
Meta directing group
and does not go
Friedal Craft's reaction
(c) Friedel-Crafts reactions do not occur when powerful electron-withdrawing groups are present on the
aromatic ring or when the ring bears an –NH2, –NHR or –NR2 group. This applies to both alkylations
and acylations.
NO2 +O C OH
N(CH3)3
These do not undergo Friedel -Crafts reactions
(d) Aryl and vinylic halides cannot be used as the halide component because they do not form carbocations
readily.
Cl
No reaction
AlCl3
No reaction
Cl
(e) Polyalkylations often occur - Alkyl groups are electron releasing groups, and once one is introduced into
the benzene ring it activates the ring towards further substitution.
CH(CH3)2 CH(CH3)2
H3C BF3 +
+ CHOH 60C
H3C 24% Isopropyl benzene
CH(CH3)2
14% diisopropylbenzene
Polyacylations are not a problem in Friedel-Crafts acylations however. The acyl group (RCO–) by itself
is an electron-withdrawing group, and when it forms a complex with AlCl3 in the last step of the reaction,
it is made even more electron withdrawing. This strongly inhibits further substitution and makes
monoacylation easy.
( 7 ) Orientation and reactivity in electrophilic aromatic substitution :
In the case of benzene we get just one monosubstituted product in electrophlic aromatic substitution as all the
positions are equivalent. However, for substitution in a compound which already has a group attached to the
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ring, three disubstituted products, (1, 2), (1, 3) (1, 4) commonly called ortho, meta and para meaning next
to, between and opposite respectively are possible. These are abbreviated as o, m and P.
Y
Y YY
E
E ++ E
1, 2 ortho E 1, 4
1, 3 meta
Para
Reactivity :
We have now seen that certain groups activate the benzene ring toward electrophilic substitutions, while other
groups deactivate the ring. We mean that an aromatic compound with an activating group reacts faster in
electrophilic substitutions than benzene. When we say that a group deactivates the ring, we mean that an
aromatic compound with a deactivating group reacts slower than benzene.
We have also seen that we can account for relative reaction rates by examining the transition state for the rate-
determining steps. Any factor that increases the energy of the transition state relative to that of the reactants
decreases the relative rate of the reaction. It does this because it increases the free energy of activation of the
reaction. In the same way, any factor that decreases the energy of the transition state relative to that of
activation and increases the relative rate of the reaction.
The rate-determining step in electrophilic substitutions of substituted benzene compounds is the step that
results in the formation of arenium ion. We can write the formula for a substituted benzene in a generalized
way if we use the letter S to represent any ring substituent including hydrogen, (if S is hydrogen, the compound
is benzene). We can also write the structure for the arenium ion in the way shown here. By this formula we
mean that S can be in any position -ortho, meta or para - relative to the electrophile, E. Using these conventions,
we can write the rate-determing step for electrophilic aromatic substitution in the following way.
When we examine this step for a large number of reactions, we find that the relative rates of the reactions
depend on whether S withdraws or release electrons. If S is an electron-releasing group (relative to hydrogen),
the reaction occurs faster than the corresponding reaction of benzene. If S is an electron withdrawing group,
the reaction occurs slower than that of beznene.
S S
Reaction is
+E+
+ faster
S releases HE
electrons Arenium ion
is stabilized
S
S
+E+
Reaction is
+ slower
S withdraws HE
electrons Arenium ion
is destabilized
It appears, that the substituent (S) must affect stability of the transition state relative to that of the reactants.
Electron-releasing groups apparently make the transition state more stable, while electron withdrawing grops
make it less stable. that is entirely reasonable, because the transition state resembles the arenium ion, and the
arenium ion is a delocalized carbocation.
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The effects of substituents on the reactivity of a benzene ring toward electrophilic substituion:
Strong activating groups : —NH2, —NHR, —NR2, —OH, —OR.
OO Ortho/para
Moderately activating groups : — NHCR, — OCR, directing
Weakly activating groups :—R, —Ar, —CH=CR2
Weakly deactivating groups :—F, —Cl, —Br, —I
Moderately deactivating groups : —CHO, —COR, —COOR,
—COOH, —COCl Meta
directing
Strong deactivating groups : —CN, —SO3H, — N H3 , — N H2R ,
— NHR2 , — NR3 ,—NO2
( 8 ) HETEROCYCLIC AROMATIC COMPOUND :
This is a class of compounds with aromatic properties, in which one or more of the ring carbons has been
replaced by a heteroatom (an atom other than carbon) such as nitrogen, oxygen or sulphur. Some of these
compounds preserve the six-membered ring and are therefore called benzenoid aromatic. In other, the (4n +2)
-electron reside in a five- membered ring, so these are called non-benzenoid. A key difference between these
two classes of aromatic compounds is that in non-benzenoid heteroaromatic compounds a formally non-
bonding pair of electrons on the heteroatom becomes part of the (4n +2) -electron of the aromatic -system,
whereas in benzenoid aromatic compounds, the non-bonding pairs on heteroatoms are normally separated
from the aromatic system. Here are a few examples :
Benzenoid N N N:
heteroaromatic Pyridine Quinoline
compounds N
Pyrimidine
:
::Benzenoid :N :O: :S:
heteroaromatic
:compoundsH Furan Thiophene
Pyrrole
::
Electrophilic Substitution in Pyridine :
:
::
:
::
In pyridine the attack may takes palce at "3" or "4" position in general.
"3" CH+ E E E E
E+ H H H
H N
N HC+ CH+ 3d
4 H 3a N N
53 3b 3c E
62 HE
CH+ HE HE N
1 HC+ 4d
N
N 4a N
4c
"4" N+
E+ 4b
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At first glance the two sets of resonance structures look quite similar. But notice how in 4b (one of the resonance
structure of the intermediate that results from attack at the 4th position) the valence of nitrogen is left unfilled,
an unfavourable situation not encountered in the intermediate formed by attack at the 3rd position. Again, it
is the electronegativity of the nitrogen that reduces the reactivity of pyridine (relative to benzene) toward
electrophiles.
( 9 ) REACTIONS OF ALKYL BENZENES :
Halogenation of the side chain :
The side chain halogenation of alkyl benzenes takes place in the presence of light or high temperatures. Side
chain halogenation is mostly carried out by using the reagent N-bromosuccinamide (NBS) in the presence of light.
CH3 O CH2Br
+ NBr hv
CCl4
O Benzyl bromide
An alkyl benzene with side chain other than methyl may lead to the formation of more than one products
CH2CH3 CH–CH3
Br2 Br
Heat, light
(I)
CH2–CH2Br
(II)
Product (I) is the only product obtained, and formation of such product can be attributed to mechanism governing
this reaction. Side chain halogenation has a similar mechanism as that of alkanes. It involves the formation
of free radical intermediates. Now if we observe the free radicals formed by attack of bromine free radical on
ethyl benzene, we will find that product (I) involves formation of benzyl free radical and (II) involves formation
of primary free radical.
• Br
CH–CH3
CH–CH3
CH2CH3 Br2 CH2CH2Br
hv •
CH2–CH2
hv
Br2
The benzyl free radical is more stable than primary free radical because its bond dissociation energy is 19 kcal/
mole less than that of primary free radical.
Order of stability of free-radials.
Benzyl > allyl > 3° > 2° > 1° >CH3 vinyl (from bond dissociation energy data)
Oxidation of side chain :
Although benzene itself is not susceptible to oxidation but side chain attached to the benzene ring undergoes
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oxidation and converts itself into a —COOH group. Oxidation of side chain takes place after substance is
heated for a long time with KMnO4.
R COOH
KMnO4
The above reaction can be used for identifying substitution pattern in aromatic compound. If any compound
gives pthalic acid on heating with KMnO4, then we can infer that it is ortho disubstituted benzene.
R COOH CH3
COOH H3C—C—CH3
R
KMnO4 No reaction
KMnO4
There is no reaction because of the absence of a benzylic hydrogen.
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SOLVED EXA MPLES
E x . 1 Which of the following is not correctly matched :
(A) Hydrolysis of phenyl magnesium iodide – benzene
(B) –Isomer of BHC – lindane
(C) (2n + 4) Rule – aromaticity
(D) Trimerisation of propyne – mesitylene
Sol. (C) The Huckel rule to account for aromaticity is closed ring of (4n + 2) electrons.
E x . 2 Ozonolysis of benzene in vigorous condition yields :
(A) Glyoxal (B) Glycerine (C) Glycol (D) Glycerol
S o l . (A ) Ozonolysis of benzene yields glyoxal. Benzene adds three molecules of ozone and forms benzene triozonide
which on decomposition with water gives three molecules of glyoxal.
C6H6 3O3 C6H6O9
benzene triozonide
benzene ozone
C6H6O9 H2O H C O
benzene trizonide C O
3
H
Glyoxal
E x . 3 Benzene undergoes subsitution reaction more easily than addition because :
(A) It has cyclic structure (B) It has three double bonds
(C) It has six hydrogen atoms (D) Of resonance
Sol. (D) If there were no resonance in benzne, electrons would have not been delocalised and hence easily
available to undergo addition reactions as in ethylene. Further the subtituted benzene is stable due to
resonance.
Ex.4 Benzene reacts with n-propyl chloride in the presence of anhydrous AlCl3 to give predominantly :
(A) n–Propylbenzene (B) Isopropylbenzene
(C) 3–Propyl–1–chlorobenzene (D) No reaction
Sol. (B) Propyl carbonium ion, CH3CH2 C H2 is primary carbonium ion, it rearranges to the more stable secondary
carbonium ion CH3 C HCH3, which then reacts to form isopropylbenzene.
+ +
CH3 CH CH2 CH3 CH CH3
Ex.5 H
Propyl carbonium ion(1°) Iso Propyl carbonium ion(2°)
Which of the following reactions of benzene does not account for the three ‘C = C’ bonds in the molecule–
[a] Benzene + Br2 FeBr3 bromobenzene + HBr
conc. H2SO4
[b] Benzene + HNO3 nitrobenzene + H2O
[c] Benzene + 3O3 Triozonide
[d] Benzne + 3H2 Ni cyclohexane
(A) a, c (B) b, d (C) b, c, d (D) a, b
Sol. (D) a and b are the electrophilic subsitution reactions and do not account for the C = C bond reaction.
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Ex.6 In which of the following reaction t–butylbenzene is formed :
(A) Benzene + t–butyl chloride AlCl3 (B) Benzene + (CH3)2C = CH2 BF3.HF
(C) Benzene + t–butyl alcohol H2SO4 (D) All of these
Sol. (D) + CH3 CH3 CH3 CH3
C Cl AlCl3 C CH3
CH3
+ HCl
CH3 CH3
C CH3
CH3 C CH2 BF3.HF
+
CH3
+ CH3 CH3 CH3 CH3
C OH C CH3
H2SO4
+ H2O
CH3
E x . 7 The order of reactivity of :
–CH3 (I), –CH2–CH3 (II), –CH(CH3)2 (III) and –C(CH3)3 (IV)
Where = C6H5
(A) I > II > III > IV (B) IV > III > II > I (C) II > I > III > IV (D) III > II > I > IV
Sol. (A) More are the number of –hydrogen present in the alkyl group attached to the benzene ring more
pronounced will be the hyperconjugation and the benzene ring will be more electron rich and easily
be attacked by an electrophile. –hydrogen in –CH3, –CH2 –CH3, –CH(CH3)2 and –C(CH3)3 respectively
are three, two one and zero.
Ex.8 Ethylbenzene + Cl2 Light (main) compound is :
(A) o– & p– Chloroethylbenzene (B) 1–Chloroethylbenzene
(C) 2–Chloroethylbenzene (D) m–Chloroethylbenzene
C2H5 Cl
CH CH3
+ Cl2 Light
Sol. (B)
1-Chloroethylbenzene
Ex.9 –CH3 KMnO4 A Soda Lime B
Compound B is :
(A) Toluene (B) Benzene (C) Cresol (D) Benzaldehyde
Sol. (B)
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E x . 1 0 Formation of which of the following compound confirms the unsaturation character of benzene :
(A) Cyclohexane (B) Gammexane (C) Triozonide (D) All the above
Sol. (D) Formation of all the three compounds are the result of addition reaction. Hence confirm the unsaturation
nature of benzene.
E x . 1 1 Toluene may be prepared by :
(A) Toluic acid (B) Cresol
(C) Toluene sulphonic acid (D) All the above
Sol. (D) Toluene may be prepared by all the above compounds descrbied earlier.
Ex. 12 Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH
gives :
(A) o–Cresol (B) p–Cresol (C) 2,4–Dihydroxy toluene (D) Benzoic acid
Sol. (D) C6H5CH3 Cl2 C6H5CH2Cl Cl2 C6H5CH2Cl2 Cl2 C6H5CCl3
hv
NaOH NaOH NaOH
C6H5CH2OH C6H5CHO C6H5COOH
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ISOMERISM
The name was given by Berzilius. Two or more than two organic compounds having the same molecular
formula and molecular weight but different physical and chemical properties are called isomers and the
phenomenon is called isomerism.
ISOMERISM
Structural Isomerism Stereoisomerism
Chain Position Ring Chain Functional Metamerism Tautomerism
Structural Isomerism :
S .No. Is om er s C ha r a cter is tics Conditions
1 Chain Isomers The y ha ve d iffe r e nt size of The y ha ve sa m e na tu r e of
2 Positional Isomers m a in cha in or sid e cha in locant
The y ha ve d iffe r e nt p osition The y shou ld ha ve sa m e size of
of locant m a in cha in a nd sid e cha in a nd
sa m e na tu r e of loca nt
3 Functional Isomers D iffe r e nt na tu r e of loca nt Cha in a nd p ositiona l
isom e r ism is not consid e r e d
4 Metamerism D iffe r e nt na tu r e of a lk yl The y shou ld ha ve sa m e na tu r e
group along a polyvalent of functional group chain &
fu nctiona l g r ou p p ositiona l isom e r is ig nor e d
Different position of hydrogen The two functional isomers
5 Tautomerism r e m a ins in d yna m ics
atoms
e q uilib r iu m with e a ch othe r
1 . Chain Isomerism (CI) :
The compounds which have same molecular formula, same functional group, same position of functional
group or multiple bond or substituent but different arrangement of carbon chain (different parent name of
compound) shows chain isomerism.
Example : CH3 CH2 CH2 CH3 CH3 CH CH3
Butane(4C)
CH3
2-Methyl propane (3C)
Example : CH3 CH2 CH CH2 CH2 C CH3
Example : 1–Butene(4C) CH3
2-Methyl-1-propene(3C)
CH3 CH2 CH2 CH2 OH CH3 CH CH2 OH
1-Butanol (4C) CH3
2-Methyl-1-propanol (3C)
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O O
Example : CH3 CH2 CH2 CH2 C OH CH3 CH CH2 C OH
Pentanoic acid CH3
3-Methyl butanoic acid
CH3 O
CH3 C C OH
CH3
2,2-Dimethyl propanoic acid
2 . Position Isomerism (PI) :
The compounds which have same molecular formula, same functional group, same parent carbon chain but
different position of functional group or multiple bond or substituents, shows position isomerism.
Example : CH2 CH CH2 CH3 CH3 CH CH CH3
But–1–ene But–2–ene
Example : CH3 CH2 CH2 CH2 OH CH3 CH2 CH CH3
1–Butanol
OH
2–Butanol
Example : CH3 CH2 CH2 CH2 Cl CH3 CH2 CH CH3
1–Chlorobutane Cl
Example of CI and PI : 2–Chlorobutane
( i ) C4H10 have two isomers : Both butane and isobutane are chain isomers.
Example : CH3 CH2 CH2 CH3 CH3 CH CH3
Butane CH3
Isobutane
( i i ) C5H12 have three isomers : All of three structures are chain isomers because only carbon chain (parent)
is different.
Example :
CH3 CH2 CH2 CH2 CH3 , CH3 CH2 CH CH3 , CH3
CH3 C CH3
Pentane CH3
2–Methyl butane CH3
2,2–Dimethylpropane
(iii) C6H14 has 5 isomers : (b) CH3 CH2 CH2 CH CH3
(a) CH3CH2CH2CH2CH2CH3 CH3
Hexane 2–Methyl pentane
(c) CH3 CH2 CH CH2 CH3 (d) CH3 CH CH CH3
CH3 CH3 CH3
3–Methyl pentane 2,3–Dimethyl butane
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CH3 a–b, b–d, a–c, c–d Chain Isomers
(e) H3C C CH2 CH3
b–c, d–e Position Isomers
CH3
2,2–Dimethyl butane
(i v) C7H16 has 9 isomers Heptane
1. CH3 CH2 CH2 CH2 CH2 CH2 CH3 2–Methylhexane
2. CH3 CH2 CH2 CH2 CH CH3
3–Methylhexane
CH3
3. CH3 CH2 CH2 CH CH2 CH3
CH3
CH3
4. CH3 CH2 CH2 C CH3 2,2–Dimethyl pentane
CH3
CH3
5. CH3 CH2 C CH2 CH3 3,3-Dimethylpentane
CH3
CH3
6. CH3 CH CH CH2 CH3 2,3–Dimethylpentane
CH3
CH3 CH3
7. CH3 CH CH2 CH CH3 2,4–Dimethylpentane
CH2 CH3
8. CH3 CH2 CH CH2 CH3 3–Ethylpentane
CH3
9. CH3 CH C CH3 2,2,3–Trimethylbutane
CH3 CH3
( v ) C3H6Cl2 has 4 isomers : Position of chlorine atom is different in all the structure, so these are position
Isomers.
Cl Cl
1 . H3C CH2 CH Cl 2 . H2C CH2 CH2 Cl
1,1–Dichloropropane 1,3–Dichloropropane
Cl Cl
3 . H3C C CH3 4 . H2C CH CH3
Cl Cl
2,2–Dichloropropane 1,2–Dichloropropane
(vi) C5H11Cl has 8 isomers
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(vii) C8H10 has 4 aromotic isomers CH2 CH3
CH3
CH3 ( o , m , p )
( v i i i ) C6H4X2 3 Aromatic isomers
C6H4XY 3 Aromatic isomers
C6H3X3 3 Aromatic isomers
C6H3XYZ 10 Aromatic isomers
E x . Structures CH3—CH2—CH CH2 and CH3 C CH2 are :–
CH3
S o l . Chain Isomers
Molecular formula No. of Isomers Molecular formula No. of Isomers
2 18
C4H10 3 C8H18 35
C5H12 5 C9H20 75
C6H14 9 C10H22
C7H16
3 . Ring chain isomerism (RCI) :
Same molecular formula but different mode of linking (open chain or closed chain) of carbon atoms.
CH3 CH CH2 [open chain]
C3H6 CH2 [closed chain or ring]
CH2 CH2
They have same molecular formula so they are Ring chain isomers.
E x . Relate a,b and c:– (a)
H3C C CH
C3H4
(b)
CH2 C CH2
(c) CH
CH2 CH
S o l . a–b Functional Isomers
a–c , b–c Ring-chain Isomers, Functional Isomers
Special points :
Alkenes with cycloalkane and alkynes (Alkadienes) with cycloalkenes show Ring-chain Isomerism.
Ring-chain Isomers are also Functional Isomers.
E x . Relate structures a,b,c and d. (a) CH3—CH2—C CH (b) CH2 C CH—CH3
CH3 (d)
(c)
S o l . a, b Functional Isomers
a, c. Ring-chain Isomers and Functional Isomers
b, c Ring-chain Isomers and Functional Isomers
a, d. Ring-chain Isomers and Functional Isomers
c, d. Chain Isomers
b, d. Ring-chain Isomers and Functional Isomers
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54 3 2 and 4 32
E x . CH3 CH2 CH2 CH2 CH3 CH2 CH CH3 are ?
1CN 1CN
Pentanenitrile 2–Methyl butanenitrile
S o l . Molecular formula same, Functional group same, position of Functional group same but different parent carbon
atom chain so both are Chain isomers
E x . How many minimum carbons required for Chain isomerism and Position isomerism in alkanes ?
Sol. 4, 6
E x . How many minimum carbons required for Chain isomerism and Position isomerism in alkenes ?
Sol. 4, 4
E x . How many minimum carbons required for Chain isomerism and Position isomerism in alkynes ?
Sol. 5, 4
4 . Functional Isomerism :-
Same molecular formula but different functional groups.
Following compounds show Functional isomerism, as they have same molecular formula and different functional
group.
(i) Alcohol and ether C H 3 — C H 2 — O H and CH3—O—CH3
(ii) Aldehydes and ketones CH3 CH2 C H and CH3 C CH3
OO
(iii) Acids and ester CH3 C OH and H C O CH3
O
O
CH3—CH2—CH2—NC
(iv) Cyanide and isocyanide C H 3— C H 2— C H 2— C N and
(v) Nitro and Nitrite O and CH3—CH2—O—N O
CH3 CH2 N
O
(vi) Keto and enol O and CH2 C CH3
CH3 C CH3 OH
(vii) Amide and Oxime CH3 C NH2 and CH3—CH NOH
(viii) 1°, 2°, 3° amines O
(i) CH3—CH2—CH2—NH2 (ii) CH3—NH—CH2—CH3 (iii) CH3
CH3 N CH3
(ix) Alcoholic and phenolic compounds : CH2OH OH
and CH3
(x) Alkyl halides do not show Functional isomerism.
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5 . Metamerism :
Same molecular formula, same polyvalent Functional group but different alkyl groups attached to polyvalent
Functional group.
Polyvalent Functional group [More than one valency] are :
—O— , —S—, C , C O , C NH , —NH—, N , O CN
O O O C O , O
C
O
Example : CH3—O—CH2—CH2—CH3 ; CH3–CH2–O–CH2–CH3
Both are metamers.
Example : CH3—CH2—NH—CH2—CH3 CH3—NH—CH2—CH2—CH3
N–Ethyl ethanamine N–Methyl propanamine
They are only metamers not CI
Example : CH3 C CH2 CH2 CH2 CH3 CH3 CH2 C CH2 CH2 CH3
O O
2–Hexanone 3–Hexanone
Both are metamers and Position isomer
Example : CH3 C CH2 CH2 CH2 CH3 CH3 CH2 C CH CH3
O O CH3
2–Hexanone 2–Methyl–3–pentanone
Both are Metamers, Chain and Position isomers
E x . Structures O and O are
C6H5 C O CH3
CH3 C O C6H5
S o l. Both are metamers.
6 . Tautomerism or De smotropism :
Tautomerism was introduced by "Laar". It's also called desmotropism.
Desmotroism means bond turning. [Desmos = Bond ; Tropos = Turn]
Tautomers have same molecular formula but different structural formula due to wandering nature of
active hydrogen between two atoms.
The tautomerism is also called kryptomerism or allotropism or desmotropism or dynamic
isomerism.
CH2 C H –H attached to carbonyl compound is active H
HO
–Hydrogen or active H
Example : H H
HC CH HC CH
HO OH
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Example : CH3 C CH2 CH3 C CH2
OH OH
keto ene + ol = enol
Note : Tautomers are also F. I. and exist in dynamic equilibrium is used to show tautomerism]
(i) By shifting of H–atom bond also change its position.
(ii) Tautomers always exist in dynamic equilibrium.
(iii) Number of electrons and lone pairs in both tautomers always remain the same.
(iv) It is a chemical phenomenon which takes place only in liquids and gaseous phase only. It never
(v) takes place in solid state.
The process can be catalyzed by the acid as well a bases.
(vi)
Condition for Tautomerism :
( a ) For carbonyl compounds :- Carbonyl compounds having atleast one –H show tautomerism
(i) CH3 C H 3 H, show tautomerism.
O
(ii) CH3 C CH3 6 H, show tautomerism
O
CH3 CH C H 1 H, show tautomerism
(iii) No H, No tautomerism
CH3 O
(iv) H C H
O
O
(v) C H No H, No Tautomerism
O 3 H, shows tautomerism (Aceto phenone)
(vi) C CH3
(vii) Ph C Ph No H, No tautomerism (Benzophenone)
O
(viii) Ph C CH2 C Ph 2 H, shows tautomerism
OO
O
H H 4 H, shows tautomerism
(ix) H H
O
HH
(x) H H H, attached sp2 carbon does not take part in tautomerism
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Ex . Which of the following show keto enol tautomerism.
O O H (C) CH CH OH
(A) (B) H (D)
O O
S o l . (A), (B) and (D)
( b ) For nitro compounds : Nitro compounds having atleast one – H show tautomerism
O O
CH2 N O
H CH2 N
OH
Nitro form Aci nitro form
(acidic form so soluble in base)
Note : Nitro compounds with at least one –H are soluble in NaOH.
Ex . Which of the following show tautomerism.
(A) CH3CH2—NO2 (B) (CH3)2CH—NO2 (C) (CH3)3C—NO2 (D) Ph—CH2—NO2
S o l . (A), (B) and (D) (D) Ph—CH2—NO2
Ex . Which of the following do not soluble in NaOH.
(A) CH3CH2—NO2 (B) (CH3)2CH—NO2 (C) (CH3)3C—NO2
Sol. Only (C)
( c ) H—C N and H—N C are tautomers [also Functional isomers] while R—C N and R—N C
are only Functional isomers.
H—C N C N—H
Active H
O and H—O—N O are tautomers.
(d) H N
CH2 C H
O OH
Enol Content : "enol" ( 1%)
1 . CH2 C H CH3 C CH2 Less stable or unstable
HO OH
"keto" ( 99%) 1%
2 . CH3 C CH2 OH
OH sp2
99%
O
3. H
"keto" ( 1%) "enol" (stable by resonance and aromatic nature) ( 99%)
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4 . CH3 C CH2 C OC2H5 CH3 C CH C OC2H5
OO OH O
Aceto acetic ester (AAE) keto : enol = 93 : 7
aqueous solution keto : enol = 25 : 75
AAE liquid state
In aqueous solution
CH3 C CH2 C O C2H5
OO
HOH HOH Keto from is stablised by intermolecular H–Bonding.
Enol form is stabilised by intramolecular H–Bonding.
In liquid state
CH3 C CH C OC2H5
OH O
(i) Enol content number of >C O group.
(ii) If number of >C O groups are equal then proportional to number of –H.
(iii) Group —OH attached to sp2 carbon or double bond is less stable or unstable.
(iv) More active H, more take part in tautomerism.
(v) Stability of enol form depnds on (i) Resonance and (ii) H – Bond.
E x . Arrange the following in correct order of enol content.
(A) CH3 C H (B) CH3 C CH3
O O
(C) CH3 C CH2 C H (D) CH3 C CH2 C CH3
OO OO
Sol. (A) < (B) < (C) < (D)
E x . Which have maximum stable enol form
O OH O OH
H H
(A) (C) H
O OH OH
(B) O
O OH (D)
S o l . (D) Stable by Resonance and is Aromatic
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STEREO ISOMERISM
Stereoisomers
Configurational Conformational
(non-interconvertible (Interconvertible
resolvable) non-resolvable)
Geometrical diastereomers Optical Isomers
(non-mirror image
Geo. isomers)
Enantiomers Optical
(Mirror-image (non-mirror image
stereoisomers)
optical isomers)
Two or more than two compounds having same molecular formula, same structural formula but different
arrangements of atoms or groups in space.
( A ) CONFIGUR ATIONAL ISOMERISM : Stereo isomers which have following characteristics.
Stereo isomer which cannot interconvert in each other at room temperature due to restricted rotation known
as Geometrical isomerism.
Stereo isomer which cannot super impose on each other due to chirality know as optical isomerism.
Geometrical isomerism (G. I) : Alkenes ( >C C<), oximes (>C N—) and azo compounds
[—N N—], show Geometrical Isomers due to restricted rotation about double bond while cycloalkanes show
Geometrical Isomers due to restricted rotation about single bond.
Geometrical Isomers in Alkenes :
Reason : Restricted rotation about double bond : It is due to overlapping of p–orbital.
Restricted Rotation Free Rotation
CH3 CH3
H
CH3 CH CH CH3 H
sp2 sp2
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CH3 CH3 CH3 H
Example : CC CC
HH H CH3
cis–2–butene trans–2–butene
Note : cis trans is possible only when bond break.
Condition for Geometrical isomerism :
Only those alkenes show G. I. in which "Each sp2 carbon have different atoms or groups"
aa ax
CC b C C y
bb
Geometrical isomers Geometrical isomers
ay pp
CC C
ay C
qq
Not Geometrical isomers Not Geometrical isomers
Ex . Which of the following show Geometrical isomerism –
(A) 1,1–diphenyl–1–butene (B) 1,1–diphenyl–2–butene
(C) 2,3–dimethyl–2–butene (D) 3-phenyl–1–butene
Sol. (B)
( B ) NOMENCLATURE SYSTEMS OF GEOMETRICAL ISOMERS :
( a ) Cis–Trans System : If same groups at same side then cis and if same groups at different side then
trans.
aa ba
CC C C
a b
bb
[Same groups, same side] [Same groups different side]
cis trans
Example : aa zz
CC CC
xy xy
cis cis
CH3 C H Cl Cl
C C
Example : H CH2 CH3
C
Br Br
trans-2–pentene It does not show Geometrical isomers So no cis–trans
Physical Proper ties of Cis–Trans Isomers :
S.No. Physical proper ties C om p a ri s on Remarks
1 D ipole moment cis > trans
cis-isomer has resulta nt of dipoles w hile in trans isomer dipole
2 B oiling point cis > trans m om e n ts c a n ce l o ut
3 S olubility (in H2O) cis > trans
4 Melting point trans > cis Molecules having higher dipole moment have hig he r boiling
point due to larger intermolecular force of attra ction
5 Stability trans > cis
M o re po la r m o le c ules a re m o re s o luble in H 2 O
M o re sym m e tri c i so m e rs h a ve hi gh e r m elti n g po in ts du e to
be tt e r p ac kin g i n cra s ta lli n e la tt ice & tran s is o m e rs a re m o re
sy m m etri c t ha n c is .
The molecule having more vander wall strain are less stable. In
cis isomer the bulky g roup are close r they have larger vander
w a a l s tra i n .
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Dipole moment [] :
H C CH3
Example : CH3 C H
Example :
Example :
CH3 C H CH3 C H
cis 0 trans = Zero
CH3 C H CH3 C H
H C CH2 CH3 Cl C H
cis 0
trans 0
Cl
CH3 C H
trans 0
H C Cl
E x . If dipole moment of chlorobenzene is , then dipole moment of is –
Cl Cl
Sol. Zero
( b ) E – Z System :
E (Entgegen) : When high priority groups are opposite side.
Z (Zussaman) : When high priority groups are same side.
LP C C HP HP C C HP
HP LP LP LP
'E' 'Z'
HP – High priority and LP – Low priority
Priority Rules :
Rule I : Priority is proportional to atomic number of atom which is directly attached to sp2 carbon.
Br C C F [HP] Cl C CH3
C N H2 [HP]
[HP] I Cl [HP]
Cl C
'Z' Cl Cl 'E'
Rule II : If rule-I is failed then consider next atom
CH3 CH3 [C, C, C]
[HP]
[HP] F C C CH3
C decreasing order of atomic number
C CH3
[LP] CH3H2C CH3 [C, C, H]
[LP]
H
'Z'
Rule III :– If multiple bond is present then consider them as :-
CC
C C C C C C C C
CC CC
C OC O N
O C N C N
N
O
[O, O, H]
[HP] NC CH
HNC C
H
C OH
O [O, O, O]
[HP]
'Z'
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Prioity order for some groups is :
Br > Cl > OH > NH2 > COOH > CHO > CH2OH > CN > C6H5 > C CH > C(CH3)3 > CH CH2 > CH(CH3)2
Rule IV : If isotopes are present then consider atomic weight.
H CH3
CC
CH2 CH3 [HP]
[HP] D
'Z'
CH3 H CH3 CH3
C C
Example : Cl CH3 [HP] ; C C
H Cl
[HP]
'Z' 'E'
NC C CBr3 [Br, Br, Br]
[HP] C CH2I
CH3 CH2 CH2Cl [Cl, H, H] C
CC [O, O, O] CH3 [I, H, H]
Example : CH C COOH ; CH3
[LP] CH3
'E' 'E'
GEOMETRICAL ISOMERS IN OXIMES [>C = N–OH] :
Oximes show G. I. due to restricted rotation about double bond.
Only those oximes show Geometrical isomerism in which sp2 carbon have two different groups.
[CH3—CH O + H2N—OH] CH3—CH N—OH (oxime)
Example : Acetaldoximes has two Geometrical isomers –
CH3 C H CH3 C H
N OH
syn HO N
When H and OH are on the same side
anti
When H and OH are on the opposite side
Example : Ph—CH N—OH Ph C H
Benzaldoxime N OH [syn.]
Ph C H [Anti]
HO N
Ex . Which of the following show Geometrical isomerism – (B) H2C N—OH
(A) CH3—CH2—CH N—OH (D) CH3 C CH2CH3
(C) CH3 C CH3
N OH N OH
Sol. (A), (D)
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GEOMETRICAL ISOMERS IN AZO COMPOUNDS : ( .. .. )
N N
Ph Ph
N N
N N
Ph Ph
(syn) (Anti)
Ph—N N—Ph (Azo benzene)
GEOMETRICAL ISOMERS IN CYCLOALKANES :
Cycloalkanes show Geometrical isomers due to restricted rotation about single bond. Only those cyclo alkanes
show Geometrical isomers in which atleast two different carbons have two different groups.
Two Geometrical isomers
Me Me Me Me cis
HH HH
Me H trans
H Me
Ex . Which of the following show Geometrical isomerism –
H Cl H Me
H Me
(A) H H (B) Cl Cl (C) Cl Br (D) Me
Cl Cl H H Me
Cl Br
Sol, (A), (B) and (D)
NUMBER OF GEOMETRICAL ISOMERS IN POLYENES :
R1—CH CH—CH CH .................................. CH CH—R2
(a) If R1 R2 then number of Geometrical isomers = 2n [n = number of double bonds.]
Example : CH3—CH CH—CH CH—CH CH—CH2CH3
As n = 3 number of Geometrical isomers = 23 = 8
(b) If R1 = R2 then number of Geometrical isomers = 2n – 1 + 2p – 1
where p n (when n is even) and p n 1 (n is odd)
2 2
Example : CH3—CH CH—CH CH—CH CH—CH3 [n = 3]
Number of Geometrical isomers = 2n – 1 + 2p – 1 [ p = 3 1 ]
2
= 22 + 21 = 4 + 2 = 6
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Special Point : CH2 CH—CH CH2 [1,3–butadiene] show Geometrical isomerism due to resonance.
CH2 CH CH CH2
H C CH2
CH2 CH—CH CH2 CH2 C C CH2 C H
HH CH2
cis
trans.
OPTICAL ISOMERISM :
Certain substances possess the property to rotate the plane of polarized light. Such substances are called
optically active substances and this phenomenon is called optical activity.
Optical activity :
Light is propagated by a vibratory motion of the 'ether' particles present in the atmosphere. Thus in ordinary
light vibrations occur in all planes at right angles to the line of propagation. In plane polarized light the
vibrations take place only in one plane, vibrations in other planes being cut off. Plane polarized light can be
obtained by passing ordinary light through a Nicol prism.
ordinary light Nicol prism
waves vibrating in
Plane polarized waves
all directions vibrating in one direction
plane rotated Plane polarized plane rotated
to the right light to the left
Certain organic compounds, when their solutions are placed in the path of a plane polarized light, have the
remarkable property of rotating its plane through a certain angle which may be either to the left (or) to the
right. This property of a substance of rotating the plane of polarized light is called optical activity and the
substance possessing it is said to be optically active.
The observed rotation of the plane of polarized light [determined with the help of polarimeter] produced by
a solution depends on :
(a) the amount of the substance in tube ;
(b) on the length of the solution examined ;
(c) the temperature of the experiment and
(d) the wavelength of the light used.
The instrument used to measure angle of rotation is called polarimeter. The measurement of optical rotation
is expressed in terms of specific rotation []Dt ; this is given by the following relation :
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[]Dt obs [where = observed angle of rotation]
C
[]Dt = specific rotation determined at t°C, using D-line of sodium light.
= length of solution in decimeters
C = concentration of the active compound in grams per millilitre.
For example, the specific rotation of amyl alcohol [2-methyl-1butanol] at 25°C for D-line of sodium light is
given by
[]2D5 –5.756
The sign attached with the angle of rotation signifies the direction of rotation. Negative sign (–) indicates that
the rotation is towards the left, while positive (+) sign means that the direction of rotation is toward right.
The rotation may be different in different solvents and this needs to be mentioned while reporting the specific
rotation. Thus,
[]2D5 = + 24.7° (in chloroform)
Asymmetric carbon (or) Chiral Carbon :
If all the four bonds of carbon are satisfied by four different atoms/groups, it is chiral. Chiral carbon is designated
by an asterisk (*).
Optical isomerism in bromo chloro iodo methane :
The structural formula of bromo chloroiodomethane is
Br
H C* Cl
I
The molecule has one chiral carbon as designated by star. So molecule is chiral. It is non-super imposable on
its mirror image.
According to Van't Hoff rule.
Total number of optical isomers should be = 2n.
where n is number of chiral centre.
The fischer projections of the two isomers are
Br Br
H II H
Cl Cl
(i) (ii)
These are optically active, which are non-super imposable mirror image of each other and can rotate plane
of polarized light.
Stereoisomers which are mirror-image of each other are called enantiomers (or) enantimorphs. Thus (i) and
(ii) are enantiomers. All the physical and chemical properties of enantimoers are same except two :
(i) They rotate PPL to the same extent but in opposite direction. One which rotates PPL in clockwise
direction is called dextro-rotatory [dextro is latin word meaning thereby right] and is designated by d (or)
(+). One which rotates PPL in anti-clockwise direction is called laevo rotatory [means towards left] and
designated by l (or) (–).
(ii) They react with optically active compounds with different rates.
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Properties of enantiomers :
S .No. Properties R em a r k s
Same
1 Molecular formula Same
2 Structural formula D iffe r e nt
S te r e oche m ica l for m u la (str . For m u la with Same
3 orientation)
Same
P hysica l p r op e r tie s (m .p ., b .p . d e nsity, solu b ility), D iffe r e nt
4 refractive index etc.
Che m ica l p r op e tie s
5 (a) with optically inactive compound
(b ) with op tica lly a ctive c om p ou nd
Optical isomerism in compounds having more than one chiral carbons :
If an organic molecule cotains more than one chiral carbons then the molecule may be chiral (or) achiral
depending whether it has element of symmetry or not.
Elements of symmetry : If a molecule have either.
(a) a plane of symmetry, and/or
(b) centre of symmetry, and /or
(c) n-fold alternating axis of symmetry.
If an object is super imposable on its mirror image ; it can not rotate PPL and hence optically inactive. If an
object can be cut exactly into two equal halves so that half of its become mirror image of other half, it has
plane of symmetry.
COOH
H—C—OH (I)
Mirror plane
H—C—OH (II)
COOH
(I) and (II) are mirror images, hence plane of symmetry is present in the molecule.
COOH
H—C—OH (I)
Mirror plane
HO—C—H (II)
COOH
(I) and (II) are not mirror images, hence plane of symmetry is absent in the molecule.
Centre of symmetry : It is a point inside a molecule from which on travelling equal distance in opposite
directions one takes equal time.
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A centre of symmetry is usually present only in an even numbered ring. For example, the mole of trans-2,4-
dimethyl-cyclobutane-trans-1, 3 dicarboxylic acid has a centre of symmetry.
CH3 COOH
H •H H
H
COOH CH3
Thus, if an organic molecule contains more than one chiral carbon but also have any elements of symmetry,
it is super imposable on its mirror-image, cannot rotate PPL and optically inactive. If the molecule have more
than one chiral centres but not have any element of symmetry, it must be chiral.
Stereoisomerism in Tar taric Acid :
Compounds which contain two asymmetric carbon atoms and are the type Cabc-Cabc exist in only three
isomeric forms. Two of these are non-superimposable mirror images of each other and are optically active
and the third, a diastereomer of the first two, contains a plane of symmetry, is super imposable on its mirror
image and is not optically active e.g.,
OH OH OH CO2H
CO2H CO2H H—C—OH
CO2H HO2C
HO2C HO2C
OH OH OH
H—C—OH
-(–)- tartaric acid d-(+)-tartaric acid
Mesotartaric
acid CO2H
The inactive diastereomer is usually described as a meso form. As with other examples of diastereomers, the
properties of meso forms are different from those of the isomeric mirror-image pairs ; for example, mesotartaric
acid melts at a lower temperature [140°C] than the d and isomer [170°C] and is less dense, less soluble in
water, and a weaker acid.
Compounds with two asymmetric carbon atoms in which at least one substituent is not common to both
carbons occur in four optically isomeric forms, e.g., the four isomers of structure.
HH
H3C—C*—C*—CH2CH3
Br Br
2,3-dibromopentane
CH3 CH3 CH3 CH3
H—C—Br Br—C—H H—C—Br Br—C—H
H—C—Br Br—C—H Br—C—H H—C—Br
C2H5 C2H5
C2H5 (ii) (iii) C2H5
(i) (iv)
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In general, a compound possessing n distinct asymmetric carbon atoms exists in 2n optically active forms.
It should be noted that, whereas (i) and (ii) and (iii) and (iv) are mirror-image pairs and therefore have identical
properties in a symmetric environment, neither of the (i) and (ii) bears a mirror-image relationship to either
of those (iii) and (iv). These and other stereoisomers which are not enantiomers are described as diastereomers
(or diastereo isomers) and unlike enantiomers, they differ in physical and chemical properties.
Calculation of number of optical isomers :
The number of optical isomers of an organic compound depends on its structure and number of asymmetric
carbon atoms. Thus, the number of optical isomer may be determined from the knowledge of the structure
of the compound as follows :
(a) When the molecule is unsymmetrical
No. of optically active isomers, a = 2n
Number of meso forms (m) = 0
Number of racemic mixtures, r = a/2
Total no. of optically active isomers = (a + m) = 2n
(b) When the molecule is symmetrical and has even no. of asymmetric carbon atoms.
No. of optically active isomers, a = 2(n –1)
No. of meso forms, m = n 1
22
No. of racemic mixtures, r = a/2
Total no. of optically active isomers = a + m
(c) When the molecule is symmetrical and has an odd no. of asymmetrical carbon atoms.
no. of optically active isomers, a = 2(n–1) – (n / 21 )
2
2
(n /21)
No. of meso forms, m = 2 2
Total no. of optically active isomer = (a + m) = 2(n–1)
Optically active compounds having no asymmetric carbon :
1 . Allenes : An sp-hybridized carbon atom possess one electron in each of two mutually perpendicular p orbitals.
When it is joined to two sp2-hybridized carbon atoms, as in allene, two mutually perpendicualr -bonds are
formed and consequently the -bonds to the sp2-carbons are in perpendicular planes. Allenes of the type
abC=C=Cab (a b) are therefore not superimposable on their mirror images and despite the absence of any
asymmetric atoms, exist as enantiomers and several optically active compounds have been obtained.
(Ex. a = phenyl, b=1-naphthyl)
ba ab
C C C C C C
ab ba
2 . Any molecule containing an atom that has four bonds pointing to the corners of a tetrahedron will be optically
active if the groups are different.
16O —CH3
—CH2-S—
18O
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3 . Alkylidene cyclo alkanes : The replacement of one double bond in an allene by a ring does not alter the
basic geometry of the system and appropriately substituted compounds exist in optically active forms.
H CO2H
H3C H
Related compounds in which sp2-carbon is replaced by nitrogen have also been obtained as optical isomers.
H OH
N
CO2H
Difference between Racemic mixture and Meso compound :
A racemic mixture contains equimolar amounts of enantiomers. It is optically inactive due to external
compensation.
External compensation :
If equimolar amounts of d and -isomers are mixed in a solvent, the solution is inactive. The rotation of each
isomer is balanced (or) compensated by the equal but opposite rotation of the other. Optical inactivity having
this origin is described as due to external compensation. Such mixtures of (+) and
(–) isomer (Racemic mixtures) can be separated into the active components.
A meso compound is optically inactive due to internal compensation
Internal compensation :
In meso tartaric acid the inactivity is due to effects within the molecule and not external. The force of rotation
due to one of the molecule is balanced by the opposite and equal force due to the other half. The optical
inactivity so produced is said to be due to internal compensation. It occurs whenever a compound containing
two (or) more asymmetric carbon atoms has a plane (or) point of symmetry. Since the optical inactivity of such
a compound arises within the molecule, the question of separating into active components does not arise.
Example : CH3 CH3 CH3
CH3 Cl H H Cl Cl H
H Cl
H Cl Cl H Cl H H Cl
CH2CH3 CH2CH3 CH2CH3 CH2CH3
(I) (II) (III) (IV)
I, II = Enantiomers, III, IV = Enantiomers
I, III = Diastereomers, II, IV = Diastereomers
I, IV
II, III = Diastereomers, = Diastereomers
COOH COOH
H OH HO H
Example : H OH HO H
I and II are identical COOH (II)
COOH (I)
Achiral
Absolute Configuration (R, S configuration) :
The actual three dimensional arrangement of groups in a molecule containing asymmetric carbon is termed
absolute configuration.
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System which indicates the absolute configurastion was given by three chemists R.S. Cahn, C.K. Ingold and V.
Prelog. This system is known as (R) and (S) system or the Cahn–Ingold system. The letter (R) comes from the
latin rectus (means right) while (S) comes from the latin sinister (means left).
It is better system because in manycases configuration to a compound cannot be assigned by D, L method.
(R) (S) nomenclature is assigned as follows :
1 . Each group attached to stereocentre is assigned a priority on the basis of atomic number. The group with the
directly attached atom with highest atomic number out of the four groups gets top priority while the group with
the atom of least atomic number gets the least priority.
Cl 3 F Cl Br I
4C 1 Increa sing At.No.
Example : F C I
2 4 3 21
Br Increa sin g Pr iority
2 . If out of the four attached atoms in consideration, two are isotopic (like H and D), then priority goes to higher
atomic mass i.e. D.
3 . If out of the four attached atoms in consideration, two or more are same, then priority is decided on the basis
of the atom attached next to it in its group. e.g. out of CH3 and COOH, COOH gets priority.
4 . While deciding the priority, if the atom in consideration is attached is further to an atom through a double bond
then it is treated as if it is attached to two such atoms. For example :
CC CC OC NC
CC CO C N CN
CC NC
CO
5 . After assigning priorties, the least priority group is written at remotest valency (going away), while the top priority
group is written at the top directed valency (towards viewer).
(2) (2) COOH
COOH (1) R (3)
(4) H
NH2 CH3
NH2 C (C3)H3
H(4)
(1)
If lowest priority group is not on the dash position then,
Step 1 : Bring the lowest priority group to dash by even simultaneous exchanges.
Step 2 : Draw an arrow from first priority group to second priority group till third priority group.
Step 3 : If the direction of arrow is clockwise the configuration is R and if anticlockwise it is S.
Step 4 : Draw the Fisher projection formula having equivalent configuration to the wedge-dash formula.
(1) H C2H5
OH CH3 S CH3 S
(2)C2H5 C C OH
H CH3
OH C2H5
(4) (3)
H
Now the order from top priority to the one of second priority and then to the one of third priority is determined.
If this gives a clockwise direction then it is termed R configuration and if the anticlockwise direction is obtained
then it is assigned S configuration.
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Important : Note that the designation of a compound as R or S has nothing to do with the sign of rotation.
the Cahn-Ingold rule can be applied to any three dimensional representation of a chiral compound to determine
whether it is R or S only. For example in above case (i.e. lactic acid), R configuration is laevo rotatory is
designated as R-(–)-lactic acid. Now the other configuration of it will have opposite sign of rotation i.e. S-(+)-
lactic acid.
Special Point :
Chiral nitrogen containing tetra alkyl ammonium ion show optical isomerism.
R1 R1
N N
R4 R3 R2 R2 R3 R4
(I) (II)
I and II enatiomers
Chiral nitrogen containig tertiary amine do not show optical isomerism
Reason :- Rapid umbrella inversion.
N Room temperature N
R1 R2 R3 R3 R2 R1
(I) (II)
Energy required for this interconversion is available at room temperature so I and II are identical
Chiral C containing carbanion do not show optical isomerism.
Reason :- Rapid umbrella inversion.
C Room temperature C
R1 R2 R3 R3 R2 R1
(I) (II)
Energy required for this interconversion is available at room temperature.So, I and II are identical.
Substituted Allenes do not have chiral carbons but molecule is chiral, so show optical isomerism.
CH2 C CH2 XX
Allene CCC
YY
No chiral C but molecule is chiral
–bond –bond
Ha C –bond Hc
C –bond –bond C
Hb Hd
–bond
Only those substituted allenes will be optically active in which "each sp2 C have different atoms or group".
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E x . Which of the following is optically active –
(A) CH3—CH C CH—CH3 (B) CH2 C CH2
Cl Cl Me Cl
(C) C C C (D) C C C
Br Br Et Br
Sol. (A), (C) and (D)
Ortho substituted biphenyl compounds do not have any chiral carbon but due to chiral molecule, they are
optically active.
NO2 HOOC SO3H H3C
COOHO2N CH3 HO3S
Optically active
Horizontal Vertical
plane plane
CONFORMATIONAL ISOMERISM :
The different arrangement of atoms in space that result from the free rotation arround C—C bond axis
are called conformations. The phenomenon is called conformational isomerism
H HH
H
H
C C
CH H C H
HH HH
Here (60°) is dihedral angle, angle between two planes.
Ex. Which of the following show conformational isomerism.
HH
C O O
(1) HH
(2) C H (3)
HH
H HH
HH H HH
OO C CH C CC
(4) H (5) (6)
H
HH HH H
H HH
S o l . 2, 4, 5, 6 [Hint : Condition to show conformational isomerism There should be at least three continuous
sigma bonds.]
Conformers of ethane [CH3—CH3] :
HH H HH
C H H 60° H H
HH C H HH
H H
HH
(I) (II)
(Saw horse projection)
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HH
H
HH
H 60° H
H
(III) HH
H
H
(IV)
(Newman projection)
I = III (Eclipsed form) in this form distance between 2C–H bonds is minimum so maximum repulsion or
minimum stable.
II = IV (Staggered form) in this form distance between 2C–H bonds is maximum so minimum repulsion so
maximum stable.
There are infinite conformers between eclipsed and staggered forms which are called as skew forms
Stability order : Staggered > Skew > Eclipsed.
Dihedral Angle : Dihedral angle in eclipsed form of ethane is 0°.
Dihedral angle in staggered form of ethane is 60°.
The variation of energy with rotation about the C–C bond in ethane has been shown in figure below :
HH
HH
HH
Eclipsed
Potential Energy 12.5 kJ mol–1
HHH HHH
HHH HHH
Staggered Staggered
Rotation
Changes in energy during rotation about C–C bond in ethane
The difference in the energy of various conformers constitutes an energy barrier to rotation. The energy
required to rotate the ethane molecule about carbon-carbon single bond is called torsional energy. But
this energy barrier is not large enough to prevent the rotation. Even at ordinary temperature the molecules
possess sufficient thermal and kinetic energy to overcome the energy barrier through molecular collisions.
Thus, conformations keep on changing form one form to another very rapidly and cannot be isolated as
separate conformers.
Torsional energy : The energy required to rotate the ethane molecule about ‘C–C’ bond is called torsional
energy.
1CH3 4
CH3
Conformation of Butane [CH3—CH2—CH2—CH3] 2C 3
HH C
HH
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MeMe
60° 60° 60°
H
Me
(I) (II) (III) (IV)
I (Fully eclipsed form) : In this form distance between 2 methyl groups is minimum so maximum repulsion
or minimum stable.
IV (Anti or antistaggered) : In this form distance between 2 methyl groups is maximum so minimum repulsion
or maximum stable.
Stability order : IV > II > III > I
Dihedral angle : Angle between two planes.
Angle of rotation to get minimum stable to maximum stable form in butane is 180°.
Angle of rotation to get maximum stable to maximum stable form in butane is 360°.
The energy profile diagram for the conformation of butane is given below along with the difference of
energy between various conformation of butane.
H CH3 H3CCH3 H CH3
HH HH HH
H3C H HH H3C CH3
Eclipsed IV
Eclipsed II Eclipsed VI
Potential Energy 19 kJ mol–1
16 kJ mol–1 16 kJ mol–1
3.8 kJ mol–1 3.8 kJ mol–1
H CH3 H H3C CH3 H H CH3 H CH3
CH3 H
HH HH HH HH
CH3 H H CH3
Anti I
Gauche III Gauche V Anti I
0° 60° 120° 180° 240° 300° 360°
Rotation
Energy changes that arises from rotation about the C2–C3 bond of butane.
Conformational analysis of cyclohexane :
(A ) Chair form :
Experimental evidences show that cyclohexane is non-planar. If we look to the models of the different
conformations that are free of angle strain first there is chair conformation. If we sight along each of the
carbon-carbon bonds in turn we see in every case perfectly staggered bonds.
5 61 1 32
43 6
5
2 4
Chair conformation
Chair conformation(all staggered bonds)
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The conformation is thus free of all the strains, it lies at energy minimum and is therefore a conformational
isomer. The chair form is the most stable conformation of cyclohexane.
Axial and equatorial bonds in chair form of cyclohexane :
HH H H H H HH
H H H H H H
H H
HH H
Axial C–H bonds Equatorial C–H bonds H
HH
H
Axial & equatorial
bonds together
The 12 hydrogen atoms of chair conformation of cyclohexane can be divided into two groups. Six of the
hydrogens, called axial hydrogens, hence their bonds parallel to a vertical axis that passes through the rings
centre. These axial bonds are directed up & down on adjacent carbons. The second set of six hydrogens called
equatorial hydrogens are located approximately along the equator of the molecule.
( B ) Boat form :
Another conformation which is known as boat conformation has exactly eclipsed conformations.
Flagpole
hydrogens
41 6 1 2 HH
56 5 4 3
HH
32 Boat conformation
Boat conformation (all eclipsed bonds) e H H e
e H H e
HHa HaH
aa
In boat form of cyclohexane 6 hydrogens are equatorial, 4 hydrogens are axial and two hydrogens are flagpoles.
It is an unstable conformation of cyclohexane due to torsional strain among axial hydrogens and due to van
der waals strain caused by crowding between the "flagpole" hydrogens.
Conformational inversion (Ring flipping) in cyclohexane :
Like alkanes cyclohexane too is conformationally mobile. Through a process known as ring inversion, Chair-
chair interconversion, or more simple ring flipping one chair conformation is converted to another chair.
6 1 5
6
5
4 23
4
12
3
axial a'
a
equatorial a'
a
56 4 32 56 4 2 3
b 1 b'
equatorial
b b'
axial
By ring flipping all axial bonds convert to equatorial and vice-versa. The activation energy for cyclohexane ring
inversion is 45 kJ/mol. It is a very rapid process with a half-life of about 10–5 sec at 25°C.
Conformational analysis of monosubstituted cyclohexanes :
In ring inversion in methylcyclohexane the two chair conformations are not equivalent. In one chair the methyl
group is axial ; in the other it is equatorial. At room temperature 95% of the methylcyclohexane exist in
equatorial methyl group whereas only 5% of the molecule have an axial methyl group.
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CH3 CH3
H
(95%)
(5%) H
1,3-diaxial repulsion :
A methyl group is less crowded when it is equatorial than when it is axial. The distance between the axial
methyl groups at C-1 and two hydrogens at C-3 and C-5 is less than the sum of their vander waal radii which
causes vander waal strain in the axial conformation this type of crowding is called 1,3-diaxial repulsions. When
the methyl group is equatorial, it experience no significant crowding.
HH
HC H H H H
H5 H
5H 1 6
H6 H3 H
H C
4H 2
21
H
Vander waals strain between No vander waals strain between
hydrogen of axial CH3and axial
hydrogen at C-1 and axial
hydrogens at C-3 and C-5
hydrogens at C–3 and C–5
Difference between conformation and configuration
S .N. Conformation Configuration
It refers to different arrangement of atoms or It refers to different arrangement of atoms or
1. groups relative to each other and raised due to groups in space about a central atom.
fr e e r ota tion r ou nd a sig m a b ond
The energy different between two conformers is The energy difference between two configuration
2. lower
for m s is la r g e
3. Conformers are not isomers and they can not be These are optical isomers and can be separated
se p a r a te d fr om e a ch othe r . fr om e a ch othe r .
4. These are easily inter converted to one another. These are not easily inter converted to one
another.
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SOLVED EXA MPLE
Ex.1 Evaporation of an aqueous solution of ammonium cyanate gives urea. This reaction follows the class of-
Sol.
Ex.2 (A) Polymerization (B) Isomerization (C) Association (D) Dissociation
Sol. NH4CNO heat H2N–CO–NH2 Ans.(B)
The possible number of alkynes with the formula C5H8 is -
(A) 2 (B) 3 (C) 4 (D) 5
CH3CH2CH2C CH CH3 CH3CH2C CCH3 Ans.(B)
CH–C CH
CH3
Ex.3 How many isomers of C5H11OH will be primary alcohols -
Sol.
(A) 2 (B) 3 (C) 4 (D) 5
CH3CH2CH2CH2CH2OH CH3CHCH2CH2OH Ans.(C)
(i) CH3
(ii)
HOCH2 CHCH2CH3 CH3
CH3 CH3 C CH2OH
CH3
(iii) (iv)
Ex.4 Number of isomeric forms of C7H9N having benzene ring will be -
(A) 7 (B) 6 (C) 5 (D) 4
CH2NH2 CH3 NHCH3
NH2
Sol. Ans.(C)
Ex.5
Sol. (i) o-, m-, p- (v)
Ex.6 (iiii) (iv)
Sol. Which of the following is an isomer of diethyl ether-
(A) (CH3)3 COH (B) CH3CHO (C) C3H7OH (D) (C2H5)2CHOH
Diethyl ether has 4 carbon atoms, among different alternative alcohols only (CH3)3COH has 4 carbon
atoms.
Ans.(A)
Total number of isomeric alcohols with formula C4H10O are -
(A) 1 (B) 2 (C) 3 (D) 4
(i) CH3CH2CH2CH2OH CH3 CH3
(ii) CH3CH2 CH CH3 (iii) CH3 CH CH2OH (iv) CH3 C OH
CAHn3s. ( D )
OH
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Ex.7 The molecular formula of a saturated compound is C2H4Cl2. The formula permits the existence of two
Sol.
Ex.8 (A) functional isomers (B) Position isomers (C) Optical isomers (D) cis-trans isomers
Sol.
HH HH
(i) H C C Cl 1,1-dichloro ethane (ii) H C C H 1,2-dichloro ethane Ans .( B )
H Cl Cl Cl
Both are position isomers
The type of isomerism found in urea molecule is -
(A) Chain (B) Position (C) Tautomerism (D) None of these
NH2 C NH2 NH2 C NH Ans.(C)
O OH
urea Isourea
Ex.9 An alkane can show structural isomerism if it has ......... number of minimum carbon atoms -
Sol.
(A) 1 (B) 2 (C) 3 (D) 4
CH4,CH3– CH3, CH3–CH2–CH3 exist only in one structural form, while CH3CH2CH2CH3 can exist in
more than one structural form. Ans.(D)
E x . 1 0 How many chain isomers can be obtained form the alkane C6H14 is :-
(A) 4 (B) 5 (C) 6 (D) 7
Sol. CH3CH2CH2CH2CH2CH3 CH3CHCH2 CH2CH3 CH3CH CHCH3 Ans.(B)
(i) CH3 CH3 CH3
CH3CH2CH CH2CH3
(ii) (iii)
CH3
(iv) CH3
CH3 C CH2CH3
CH3
(v)
E x . 1 1 Keto - enol tautomerism is observed in -
O O
(A) C6H5 C H (B) C6H5 C CH3
O O CH3
(C) C6H5 C C6H5 (D) C6H5 C C C6H5
CH3
S o l . Only compound (B) contains hydrogen atom for showing keto enol tautomerism. Ans.(B)
Ex.12 The number of geometrical isomers in case of a compound with the structure :
Sol.
CH3–CH=CH–CH=CH–C2H5 is -
(A) 4 (B) 3 (C) 2 (D) 5
Given alkene is unsymmetrical one and has two double bonds, the number of geometrical isomers is given
by 2n. (n = 2) therefore number of geometrical isomers will be 22 = 4 Ans.(A)
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ABYSS Learning
Ex. 13 Which one of the following will show geometrical isomerism -
(A) CH2Cl CH3 (B) CH3 H
C C
H CH C(CH3)2 CH2Cl CH C(CH3)2
(C) CH2 CH C CH2Cl (D) CH3CH2CH=CHCH2CH3 Ans.(D)
H CH CH2
Ex. 14 In the reaction
CH3CHO + HCN CH3CH(OH)CN
a chiral centre is produced. Thus product would be -
(A) Meso compound (B) Racemic mixture (C) Laevorotatory (D) Dextrorotatory
S o l . Synthesis of a chiral compound from a chiral compound in the absence of optically active agent always
produces a racemic modification :
CN O CN
HCN HCN H OH
HO H C CH3
d – lactonitrile
CH3 H Ans.(B)
CH3
acetaldehyde
Ex.15 The molecule 3-penten–2–ol can exhibit -
(a) Optical isomerism (b) Geometrical isomerism (c) Metamerism (d) Tautomerism
The correct answer is -
(A) a and b (B) a and c (C) b and c (D) a and d
OH
Sol. CH3 *CH CH CHCH3 Ans.(A)
CH3
3-penten-2-ol
As given compound contains a assymmetric carbon atom and a double bond (with sufficient conditions
for geometrical isomerism). Therefore it can shown both optical and geometrical isomerism.
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ABYSS Learning
NOMENCLATURE OF ORGANIC COMPOUNDS
HISTORY :
In 1675, Nicholas Lemery had devided chemical substances into 3 parts.
(i) Mineral substance : which are obtained from minerals. eg. gold, silver, iron etc.
(ii) Vegetable substance : which are obtained from vegetables. eg. sugar, citric acid etc.
(iii) Animal substance : which are obtained from animals. eg. albumin, gilatin etc.
After some time when many of the chemical substance were discovered, it was found that some of them
can be obtained from both vegetables and animals. So this classification was failed. So chemical substance
were then divided into two parts :
(i) Organic compounds : which are obtained from living organism.
(ii) Inorganic compounds : compounds which are obtained from any other sources except living organisms.
VFT(Vital Force Theor y) : By Berzelius in 1815. Upto 1815, any organic compound could not be
synthesized in lab. So Berzelius suggested that there is a mysterious force in living organisms which was named
as Vital Force and said that organic compounds cannot be synthesized in lab. This theory was called as VFT.
But in 1828 a German scientist Wholar synthesized an organic compound in lab. Which was 'urea'. So VFT
was failed. Urea was synthesized in lab by heating of Ammonium cyanate (NH4CNO).
NH4 CNO Rearrangement NH2 C NH2
O
Ammonium cyanate Urea
Organic Compounds : Hydrocarbons and their derivatives are called as organic compounds.
Ex : Why are organic compounds found in larger no.? or Why are they studied as a separate subject ?
Sol. (i) Catenation Property : Carbon atom has a property by which it can join with other C–atoms and form
a long chain or a ring of different size and shapes.If covalency of atom is more, then catenation property
is also more.
(i i ) Organic compound shows isomerism.
(iii) Exhibits Homologous Series.
(i v) Same Empirical Formulae.
(v) Polymerisation.
Characteristics of C-Atoms
(a) Tetra valency : Atomic number of carbon atom is 6 and it have four valency electrons so C-Atom is
tetravalent. It is explained by promotion rule
2s 2p
In ground state (here covalency of carbon is 2)
First excited state (here covalency of carbon is 4)
Available for bond formation
(b) Tendency to form multiple bonds : Carbon atom forms following type of bonds, such as
C C , C C , C C , C C C
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ABYSS Learning
(c) Tetrahedral shape : T he four covalent bond are directed 109°28'
towards the four corners of a regular tetrahedron C
Bond angle 109028' or 109.5'
(d) Catenation : Self linking property of C-atom is known as catenation. It is responsible for the variety
and large number of organic compounds. It may also give rise to open chain and closed chain nature
of compounds. Bond energy for catenation of C is maximum.
Bond energy in Kcal : C—C Si—Si N—N P—P
85 54 39 50
(e) Hybridisation : The orbitals of different shape but almost of equal energies blend up to give the
same number of new orbitals of another shape and of identical energies.
Str ucture & bonds Hybridisation Bond angle Shape
C 4,0 sp3 109°28' Tetrahedral
C 3,1 sp2 120° Planar (Trigonal)
C 2,2 sp 180° Linear
C 2,2 sp 180° Linear
- (sigma) bonds : The molecular orbital formed by the overlapping of two-s atomic orbtials or one s
and one p atomic orbitals or co-axial overlapping of p-orbitals is called a bond.
or s p or p p
Note : (i) Overlapping of hybrid orbitals also give bonds. bonds are stronger, as they are resulted
from the effective axial overlapping.
(ii) More the directional character (p) in covalent bond more is the strength of the bond.
sp3 - sp3 > sp3 - sp2 > sp2 - sp2 > sp - sp
(Pi) bonds : bond is formed by the
lateral overlapping of two p-atomic +
orbitals. It is weaker than bond, as there
is only partial overlapping.
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ABYSS Learning
Ex. 1 sp sp sp2 sp2 sp3
Ex. 2
Note : HC C CH CH CH3
H
C Flat hexagonal structure due to sp2 hybridised
C-atom in benzene
H C C H
120°
H C C C H
H
(A) electrons are mobile hence bond is more reactive. bond is formed by the collateral overlapping
of sp2 orbitals.
(B) sp2 hybridised orbitals overlap with each other and with s orbitals of six H-atoms forming C–C and
C-H bonds.
(C) Six 2p unhybridised orbitals of 6 C-atom in benzene form 3 bonds by lateral overlapping with each
other. These six electrons are free to move over all the six carbon atoms. Since delocalised electrons
have lower energy than localised.
(D) The relative sizes of hybrid orbital follows the order sp3 > sp2 > sp
(E) The electronegativity of hybrid orbitals follows the order sp > sp2 > sp3
H H sp 3– sp3 H
sp3– s s – sp3 sp3– s
H
HH
s – sp3 HH
HH Ethane
Bond
Methane
p pz pz pz
z py
py
H sp2 sp2 H H sp sp H
sp sp
sp2 sp2
Ethene
sp2 sp2
H H
Ethyne
(Orbital diagram of methane, ethane, ethene and ethyne)
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ORGANIC COMPOUND
Open – chain, acyclic Closed – Chain or
or aliphatic compounds Cyclic compound
Saturated Unsaturated Homocyclic Heterocyclic
compound compound
ABYSS Learning
alkane
CLASSIFICATION OF ORGANIC COMPOUNDS
CH3–CH2–CH3 alkene alkyne Alicyclic *A r o m a t i c
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Propane CH3–CH=CH2 HCCH com pou nd s Alicyclic Aromatic
compounds compounds compounds
Propene Ethyne
..
Cyclo o N
hexane Pentahydropyran
Pyridine
Cyclo Benzene
butene
annulene [18]
*A r o m a t i c Non benzenoid annulene [18]
(Without any Benzene ring
compounds containing compounds)
Benzenoid Naphthalene
(Benzene ring
containing compounds) Benzene
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Organic Chemistry Some Basic Principles and Techniques
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