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Delocalisation of p-electrons in conjugation is known as resonance.
(resonating structures) (Resonance hybrid)
in this form
Condition for showing resonance
1. Molecule should be planar, nearly planar or a part of it is planar
Q.1 Which are planar?
(A) (B) *(C) *(D)
Because all carbon atoms are sp2 hybridised.
2. Molecule should possess conjugated system.
Conjugated system :-
Continuous unhybridised p-orbital parallel to each-other.
Types of conjugated system:-
(1) p-bond alternate to p-bond
CH2 = CH - CH = CH2
(2) p-bond alternate to charge
CH2 = CH - CH2
(6) CH2 = CH - BH2
1. Resonance takes place due to delocalization of p e-s.
(b) Resonance absent
2. Position of the atoms remains the same, only delocalization of p e-s takes place.
Note:- [They are not resonating structure rather they are tautomers].
3. Bond pair gets converted into lone pair and l.p. gets converted into b.p.
4. In Resonance No. of unpaired e-s remains the same.
CH2 = CH - CH = CH2
(They are not resonating structure)
Resonating structure :
(1) Hypothetical strtucture exist on paper.
(2) The energy difference b/w different resonating structure is very small.
(3) All R. S. contribute towards the formation of resonance hybrid (Their contribution may be different).
(4) A single R. S. can't explain each & every property of that particular compound.
Draw the resonating structures : -
Q.2 -m of NO2 group
Resonance hybrid : -
It is a real structure which explains all the properties of a compound formed by the contribution of different R. S.. It has got maximum stability as compared to R. S.
Resonance Energy : -
It is the difference b/w theoretical value of H.O.H & experimental value.
It is the difference b/w more stable R.S. & R. H.
* More the resonance energy, more stable will be the molecule.
* Cyclohexane is thermodynamically more stable than benzene, even though resonance energy of benzene is more.
* Resonance energy is a absolute term.
Contribution of different R. S. towards resonance hybrid
(1) Non-polar R. S. contribute more than polar R. S.
(a) CH2 = CH - CH = CH2 (b) CH2-CH = CH - CH2 (c) CH2 - CH = CH - CH2
a > b = c stability
(2) Polar R. S. with complete octet will contribute more as compared with the one with incomplete octet
CH3 - CH - OCH3 CH3 - CH = :O - CH3
Incomplete octet Complete octet
(3) In polar R. S. the -ve charge should be on more electro - ve atom & +ve charge should be on more electro +ve atom.
(a) (more stable )
(b) (less stable )
(4) Compound with more covalent bonds will contribute more.
(5) Unlike charges should be closer to each other whereas like charges should be isolated.
(6) Extended conjugation contribute more than cross conjugation.
Cross conjugation < Extended conjugation
Fries Rule :-
Compound with more benzenoid structure are more stable as their Resonance energy is greater than those in which lesser no. of benzenoid structures are present.
R. E. is <
* If double bond is participating in resonance then it will acquire partial single bond character as a result of which bond length increase & bond strength decreases.
If a single bond is involved in resonance then it will acquire partial double bond character. As a result of which bond length decreases & bond strength increase.
a = e > b = d > c
a > b > c > d
Q.4 (a) (b) CH2 - CH = F a > b (stability)
Note:-When lone pair as well as double bond is present in some atom then only p bond will be participating in resonance. Whereas lone pair remains sp2 hybridised orbital. When an atom has two or more then two lone pair then only one lone pair will participate in resonance and the other one remains in sp2 hybridised orbital.
Permanent polarisation caused by displacement of s-electrons into p-molecular orbital is known as hyperconjugation.
Hyper conjugation is called No bond Resonance.
* More the C - H bond, more will be the no bond resonating structure (Hyper conjugation)
More the (C - H) bond, more will be the stability of free radical.
Properties of Free Radical:
1. It is a neutral species.
2. It has one un-paired electron that's why paramagnetic in nature.
→ methyl free Radical
→ ethyl free radical
3. Its hydridisation is sp2 and triangular planar shape.
Note : unpaired electron is not counted while calculating the hybridisation state.
(unpaired electron stay perpendicular to the plane)
Stability of free Radical :
Its stability can be determined with the help of hyperconjugation as well as Resonance effect.
Allylic Free Radical
(Free Radical is on next carbon to doubly bonded carbon atoms)
Effect of Resonance > Hyper conjugation
Benzylic Free Radical
* More Resonating structure, more will be the stability of the free Radical.
(di-benzylic free Radical)
No. of Resonating structures = 7
(Tri-benzylic free Radical)
No. of Resonating structures = 10
Stability Order :
Ex.1 Compare the stability of the following free Radicals
(Therefore this resonating structure is not possible)
Sol. b > a > c
Ex.2 (a) (b) (c) (d)
Compare the bond energy of the above compounds.
Sol. After forming free radical from the compound
(3°) (2°) (1°) methyl free radical
(a) (b) (c) (d)
Ex.3 Compare the potential energy of the following compounds (above compounds)
Sol. If compound after being in free Radical form is very stable (i.e., less energy) it means it would have possessed more energy initially i.e. its potential energy will be the most.
a < b < c < d
* Potential energy ≈ stability of free Radical
Ex.4 Compare the bond energies of C - H bond
(at a, b, c, d, e and f position)
e > b > a > f > c = d
Stability order of free Radical that might be formed after removal of H (Homolytically) from the given carbon.
→ e < b < a < f < c = d
(C - H bond energies)
In the above compound compare 2° benzylic allylic stability at two given positions:
while drawing the resonating structure of the above,
(Here inspite of Resonance three (C - H) bond are available for no bond Resonance.
→ Therefore extra stable than which have only two (C - H) bond for Hyper conjugation.
Therefore 2° benzylic allylic corresponding to structure (a) is more stable than that of structure (b)
Ex.5 Compare the stability of the following free Radical
⇒ c > b > a > d
Ex.6 Compare the potential energy of CH3 - CH3, CH2 = CH2, CH≡CH
Sol. After making free Radical of the above compounds:
a > b > c