Electronic Displacement Effects | Chemistry Class 11 - NEET PDF Download

• The displacement of electrons within the same molecule is known as electronic displacement. These effects affect the stability of a species or compound and it also affects the acidic & basic strength.

Electronic Displacement Effect is divided into two parts:
(i) Permanent effect

• Inductive effect
• Mesomeric (resonance) effect
• Hyperconjugation

(ii) Temporary effect

• Electromeric effect
• Inductomeric effect

Inductive effect

• It is an effect in which permanent polarization arises due to partial displacement of s-electrons along the carbon chain or partial displacement of sigma-bonded electrons toward more electronegative atoms in the carbon chain.
  

• Magnitude of partial positive charge: δ₁ > δ₂ > δ₃ = δ^- (net charge remains constant in a molecule having inductive effect)
• ⇒ -I effect of X, if X is more electronegative.
•  ⇒ +I effect of Y, if Y is less electronegative.
•   (- I effect order)
• O^- < O < O  (-I effect order)

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➢ Characteristic of Inductive Effect

• It is a permanent effect.
• It is caused due to electronegativity difference.
• It operates via an s-bonded electron.
• It is a distance-dependent effect.
• As distance increases, its effect decreases.
• It can be neglected after the third carbon.
• It has a destabilizing effect.

➢ Types of Inductive Effect

• It is divided into 2 parts on the basis of electronegativity w.r.t. hydrogen atom:
  (i) + I effect
  (ii) - I effect

• If any atom or group having electronegativity greater than that of hydrogen, then it is considered as - I effect and vice-versa.
• Example:
  (i) CH₃ – CH₂ – Cl i.e. - I effect of Cl
  (ii) CH₃ – CH₂ = CH₂ i.e. - I effect of – CH₂ and + I effect of – CH₃
  (iii) CH₃ – CH₂ – C ≡ CH i.e. - I effect of – C ≡ CH and + I effect of – CH₂ – CH₃
  (iv) I – Cl i.e. + I of – I
• Order of -I effect showing group:
• Order of+ I effect showing group:

• Bond Strength: CT₃ > CD₃ > CH₃ 
  Thus, Order of +I effect: T > D > H

Q. Why carbon - hydrogen bond is longer than C - T bond?

Ans. As the mass increases, vibration decreases as a result of which the heavier isotope will be more closer to the C-atom for a longer time. Therefore C - T bond is stronger. 

C - T > C - D > C - H which implies that C - H bond has longest bond length.

➢ Application of Inductive Effect

• To compare the stability of intermediates.
  Intermediates: These are real separable species having measurable stability formed during the conversion of reactant to product. After bond cleavage and before bond formation.They are formed by homolytical and heterolytical cleavage.
  6 types of intermediates:
  (i) Free radical
  (ii) Carbocation
  (iii) Carbanion
  (iv) Carbene
  (v) Nitrene
  (vi) Benzyne

Electromeric Effect

• It involves the complete transfer of electrons of a multiple bond to one of the bonded atom in presence of an electron attacking reagent. It is called the E effect.
• This effect is temporary and takes place only in the presence of a reagent. As soon as the reagent is removed, the molecule reverts back to its original position.

➢ Types of Electromeric Effect

• +E effect: If the electrons of the π-bond are transferred to that atom of the double bond to which the reagent gets finally attached, the effect is called the +E effect.
  Example: Addition of acids to alkenes.
  
• -E Effect: If the electrons of the double bond are transferred to an atom of the double bonds other than the one to which the reagent gets finally attached the effect is called -E Effect.
  Example: Addition of Cyanide ion to the carbonyl group.

Table: Differences between Inductive and Electromeric Effect

Resonating Structure

➢ Conjugated System 

• Compounds having continuous unhybridized p-orbital parallel to each other such systems are known as conjugated system.
• Types of the conjugated system:
  (i) p-bond alternate to p-bond
  Example: CH₂ = CH - CH = CH₂ 
  Explantion:
  
  
  (ii) p-bond alternate to a positive charge
  Example: CH₂ = CH - CH₂
  Explanation:
• Examples:
  (i)
  (ii)
  (iii)
  (iv) CH₂ = CH - BH₂   
  (v)

➢ Resonance

• Delocalisation of p-electrons in conjugation is known as resonance.

• When one structure is not sufficient to explain each and every property (chemical & physical) then, a different structure has been drawn which is known as Resonating Structure (canonical structure).

• All these structures contribute to the formation of a Real structure, known as Resonating Hybrid.

•  (Actual Structure)
  (resonating structures)        (Resonance hybrid)
   in this form

• Condition for showing resonance:
  (i) Molecule should be planar, nearly planar or a part of it is planar
  (ii) Molecule should possess conjugated system.

Q. Which are planar?

 (A)  (B)  (C)  (D) 
Sol. (C) and (D), Because all carbon atoms are sp² hybridised.
Not (B) because it can be converted into

• Resonance takes place due to the delocalization of p e^-s.
  Example:
  (a)  ⇒ Resonance
  (b)  ⇒ Resonance absent
  (c) ⇒ Resonance
  (d) ⇒ Resonance
• The position of the atoms remains the same, only delocalization of p e^-s takes place.
• Bond pair gets converted into lone pair and l.p. gets converted into b.p.
• In Resonance, No. of unpaired e^-s remains the same.
  CH₂ = CH - CH = CH₂  (They are not resonating structure)

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➢ Resonating Structure

• Hypothetical structure existing on paper.
• The energy difference b/w different resonating structure is very small.
• All R.S. contribute towards the formation of resonance hybrid (Their contribution may be different).
• A single R.S. can't explain each & every property of that particular compound.

Q. Draw the resonating structures:

Sol. (i)
(ii) 

• Resonance hybrid: It is a real structure that explains all the properties of a compound formed by the contribution of different R.S.
  It has got maximum stability as compared to any R.S.
• Resonance Energy: It is the difference b/w theoretical value of H.O.H & experimental value. Or it can be defined as the difference b/w more stable R.S. & R. H.
  The more resonance energy, the more stable will be the molecule.
  Resonance energy is an absolute term.

  View Answer

Contribution of different R. S. towards resonance hybrid:

• Non-polar R.S. contributes more than polar R.S.
  Example: 
  (a) CH₂ = CH - CH = CH₂ (b)  CH₂-CH = CH - CH₂ (c) CH₂ - CH = CH - CH₂ 
  Order of stability of given R.S. ⇒ a > b = c
• Polar R. S. with complete octet will contribute more as compared with the one with an incomplete octet
  CH₃ - CH  - OCH₃ (Incomplete octet) CH₃ - CH = :O - CH₃ (Complete octet)
• In polar R. S. the -ve charge should be on more electro - ve atom & +ve charge should be on more electro +ve atom.
  Example: 
  (a)  (more stable)
  (b)  (less stable)
• Compound with more covalent bonds will contribute more.
• Unlike charges should be closer to each other whereas like charges should be isolated.
• Extended conjugation contributes more than cross conjugation.
  Example: < (Cross conjugation < Extended conjugation)

➢ Fries Rule:

• Compounds with more benzenoid structures are more stable. 
• As the Resonance energy is greater than those in which lesser no. of benzenoid structures are present.
• R. E. is  <
• If a double bond is participating in resonance then it will acquire a partial single bond character as a result of which bond length increases & 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.

Q. Find the order of Stability in the following:

(i) 

(ii) (a)  (b) (c) (d) 

(iii)
(iv) (a) (b) CH₂ - CH = F                         

Sol. (i) a = e > b = d > c
(ii)  a > b > c > d, c and are in complete
(iii) +CCl3 < +CF₃, due to back bonding in +CF₃

(iv) a > b (stability)

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Mesomeric Effect (Resonance effect)

• The mesomeric effect is valid only for the conjugated system.

➢ Types of Mesomeric Effect

• It is divided into 2 parts: 
  (i) +M effect (-R)
  (ii) - M Effect (-R)
• Consider the following conjugated system:
  
• Consider another conjugated system:
  

➢ Mesomeric Effect in the Compounds Having a Benzene Ring

• If the movement of electrons is towards the ring, it will show +M effect. This effect increases the electron density over the benzene ring.
• +M effect in phenol
  
• +M effect in aniline
  
• -M effect in Benzaldehyde
  
• +M group increases the electron density of ring while - M decreases the electron density of benzene ring.
• If NO₂ is present on the ortho or para position then along with its -I effect, It will also show -M effect.
  
  Above compound have M of -OH and -M of NO₂ group.
• 
  As we can easily see that -NO₂ at meta position is not attracting e^- density towards itself and that's why it will not show -M effect at m-position.

Q. Identify the compound showing +M or -M separately?

(a) (b)  (c)

Sol. (a) -M (b) -M (c) +M

Hyperconjugation

• Permanent polarisation caused by the 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 (Hyperconjugation).
  More the (C - H) bond, the more will be the stability of free radicals.

➢ Properties of Free Radical:

• It is a neutral species.
• It has one unpaired electron, therefore, why paramagnetic in nature.
• Structure
   → methyl free Radical
   → ethyl free radical
• Its hybridization is sp² and triangular planar shape.

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➢ Stability of Free Radical

Its stability can be determined with the help of hyperconjugation as well as the Resonance effect.

• Allylic Free Radical:
   (Free Radical is on next carbon to doubly bonded carbon atoms)
  Effect of Resonance > Hyperconjugation
  
• Benzylic Free Radical:
  
  More Resonating structure, more will be the stability of the free Radical.
  Example:
  (i) Di-Benzylic free Radical:
  No. of Resonating structures = 7
  (ii) Tri-benzylic free Radical: 
  No. of Resonating structures = 10
• Stability Order:
  

Solved Examples

Q.1 Compare the stability of the following free Radicals

(a) (b) (c) 

Sol. b > a > c, Since the resonating structure is not possible for (c)

 
Q.2. Compare the  bond energy of the following compounds.
(a)  (b)  (c)  (d)

Sol. After forming free radical from the compound:
Due to 3° free radical, (a) will be most stable, therefore will have more tendency to come in this form, and C–H bond will break very readily because bond energy will be very less.

Thus, Bond Energy Order: a<b<c<d

  View Answer

Q.3 Compare the potential energy of the following compounds.
(a)  (b)  (c)  (d)

Sol. If the 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. 
Potential energy ≈ stability of free Radical

Thus, Potential Energy Order: a < b < c < d

Q.4 (i) Compare the bond energies of the C - H bond at a, b, c, d, e, and f position in the compound below.

(ii) In the above compound compare 2° benzylic allylic stability at two given positions:

Sol. (i) Stability order of free Radical that might be formed after removal of H (Homolytically) from the given carbon: e > b > a > f > c = d

Thus, C - H bond energy order: e < b < a < f < c = d

(ii) Inspite of Resonance, three (C - H) bond are available for no bond Resonance.

Therefore, 2° benzylic allylic corresponding to structure (a) will be extra stable than (b) which have only two (C - H) bond for Hyperconjugation.

Q.5 Compare the stability of the following free Radicals.

Sol.  

 
Thus, Stability of the following free Radical: c > b > a > d

Q.6 Compare the potential energy of the following compounds.
(a) CH₃ - CH₃ (b) CH₂ = CH₂ (c) CH≡CH

Sol. After making free Radical of the above compounds:
(a),
(b)
(c)

Therefore,

Thus, Potential Energy Order: a > b > c