Chapter 12.18 - Hyperconjugation in Isopropyl Cation

In the previous section, we saw the hyperconjugation in ethyl cation. In this section, we will see hyperconjugation in isopropyl cation. It can be written in steps:

1. Fig.12.109 below shows the 3D view of the isopropyl cation.

Fig.12.109

• C₁ and C₃ are sp³ hybridized. C₂ is sp² hybridized.
• In this situation, C₂ carries the +ve charge. This is shown in the resonance structure I in fig.12.110 further below.
2. Two types of rotation are possible:
    ♦ C₁ can rotate about the red axis.
    ♦ C₃ can rotate about the green axis.
• The red and green axes intersect at C₂. But those two axes have different directions.
3. Suppose that, C₃ remains stationary while C₁ rotates about the red axis.
• Then H_A, H_B and H_C will come successively in alignment with the empty p orbital of C₂.
4. When H_A comes into alignment, hyperconjugation takes place between the C₁ㅡH_A σ bond and the empty p orbital of C₂.
• This is indicated by the red curved-arrow in I.

Fig.12.110

• As a result, the C₁ㅡC₂ single bond becomes C₁=C₂ double bond. This double bond is shown in II. The H_A loses the bond with C₁. Due to the loss of an electron, H_A gets a +ve charge. This is also shown in II.
5. So now we know how II is obtained. We can analyze the curved-arrows in II.
• The magenta curved-arrow in II comes into play when H_B comes into alignment with the empty p-orbital of C₂. Hyperconjugation takes place between the C₁ㅡH_B σ bond and the empty p orbital. Thus we get the structure III.
6. In this way, by following the route shown by the double headed green arrows, we can understand all the seven resonance structures of the isopropyl cation. Note that, IV, V, VI and VII are obtained when C₁ remains stationary and C₃ rotates.
7. Now we can write about the stability of the isopropyl cation. It can be written in three steps:
(i) Originally, the isopropyl cation does not have any double bond. The + charge is carried by the C₂ atom. This is one of the seven resonance structures.
(ii) Due to hyperconjugation, the single bonds between the three C atoms become double bonds. The + charge is carried successively by six H atoms.
(iii) So in the hybrid structure, the + charge is distributed among seven atoms:
One C atom and six H atoms.
• Such a distribution of charge, gives greater stability to the isopropyl cation.
8. In an earlier section, we wrote about the stability of isopropyl cation. (step 9 below fig.12.70 of section 12.10). The above seven steps helps us to explain it's stability.
9. We can write a comparison between the stability of various cations. It can be written in 5 steps:
(i) In ethyl cation, the +ve charge is distributed among four atoms.
(ii) In isopropyl cation, the +ve charge is distributed among seven atoms.
(iii) So isopropyl cation will have greater stability when compared to ethyl cation.
(iv) In tertiary butyl cation (CH₃)₃C, the +ve charge is distributed among even more number of atoms.
(v) Thus we get the order:
Tertiary butyl cation is the most stable, followed by isopropyl cation, followed by ethyl cation, followed by methyl cation.

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◼ We see that, the methyl cation has the least stability. The reason can be written in 2 steps:
1. Fig.12.71 of section 12.10 shows the 3D view of methyl cation. We see that, all the CㅡH σ bonds lie in a plane. But the empty p-orbital is perpendicular to that plane.
2.So none of the CㅡH σ bonds can ever come into alignment with the empty p-orbital.
• We can write:
Hyperconjugation can never occur in methyl cation. So it has the least stability when compared to the other cations.

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Next we will see the hyperconjugation in propene. It can be written in 6 steps:
1. Fig.12.111 below shows the 3D view of propene.

Fig.12.111

• C₁ is sp³ hybridized. C₂ and C₃ are sp² hybridized.
• The p-orbitals of C₂ and C₃ are not empty. They contain one electron each. So a 𝜋 bond is formed between C₂ and C₃ by the lateral overlap of those p-orbitals. This lateral overlap is indicated by the two double headed yellow arrows.
• Thus we see the double bond C₂=C₃ in the resonance structure I in fig.12.112 further below.
(The p-orbital of C₂ of the isopropyl cation that we saw previously, was empty because, the cation has lost one electron. But in our present case of propene, no electron is lost)
2. C₁ can rotate about the red axis.
• Then H_A, H_B and H_C will come successively in alignment with the p-orbital of C₂.
3. Hyperconjugation takes place between the C₁ㅡH_A σ bond and the p-orbital of C₂.
• That is., the two electrons in the C₁ㅡH_A σ bond gets delocalized into the p-orbital.
• This is indicated by the red curved-arrow in I.

Fig.12.112

• As a result, the C₁ㅡC₂ single bond becomes C₁=C₂ double bond. This double bond is shown in II. The H_A loses the bond with C₁. Due to the loss of an electron, H_A gets a +ve charge. This is also shown in II.
• But due to the red arrow in I, C₂ has gained two extra electrons. It does not need two extra electrons. It has already octet. So the magenta arrow comes into play. Two electrons are transferred to C₃. The C₃ gains two electrons and a -ve charge.
4. So now we know how II is obtained. We can analyze the curved-arrows in II.
• The magenta curved-arrow in II comes into play when H_B comes into alignment with the p-orbital of C₂. Hyperconjugation takes place between the C₁ㅡH_B σ bond and the p orbital.
• The red curved-arrow indicates that, H_A regains it's two electrons. Thus we get the structure III.
5. In this way, by following the route shown by the double headed green arrows, we can understand all the four resonance structures of propene.
6. Note that, the +ve charge is distributed among three H atoms.
• The negative charge is not distributed. It permanently resides at C₃.
• The actual propene molecule is the hybrid of the four resonance structures. So the hybrid structure will be a polarized structure. One end of that structure has a +ve charge and the other end has a -ve charge.
• In other words, all molecules in a sample of propene will be in a polarized state.

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• We saw three examples for hyperconjugation. Ethyl cation, isopropyl cation and propene. Based on those examples, we can now write a definition for hyperconjugation. It can be written in 4 steps:
1. Hyperconjugation involves delocalization of the electrons in the CㅡH σ bond of an alkyl group.
[Recall that, in all the examples that we saw, the delocalization started from the CㅡH σ bond of the alkyl group  (methyl group: ㅡCH₃)]
2. In some cases, the delocalized electrons move into an unshared p-orbital.
[Recall that, this happened in ethyl cation and isopropyl cation]
3. In some other cases, the delocalized electrons move into a nearby unsaturated system (double or triple bond).
[Recall that, this happened in propene]
4. Hyperconjugation is a permanent effect.

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• We have completed a discussion on mechanisms of organic reactions. These mechanisms will help us to understand the following types of reactions:
    ♦ Substitution reactions
    ♦ Addition reactions
    ♦ Elimination reactions
    ♦ Rearrangement reactions
• We will see these reactions in later sections.

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• The link below gives the folder containing additional solved examples on this chapter.
• Part 2 is related to reaction mechanism.

Additional solved examples

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• In the next section we will see the methods of purification of organic compounds.

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Chapter 12.17 - Electromeric Effect And Hyperconjugation

In the previous section, we saw the details about resonance effect. In this section, we will see electromeric effect.

• The resonance effect that we saw in the previous section, is a permanent effect. For each case that we saw, there were various resonance structures. The molecule exists as the hybrid of those resonance structures.
• But electromeric effect is a temporary effect. This effect occurs in a substrate only when an attacking reagent comes to the vicinity of that substrate. The effect disappears if the attacking reagent is removed from the vicinity of the substrate.
This can be demonstrated using two examples.

Example 1:
This can be written in 4 steps:
1. The fig.12.103 below shows the reaction between ethene and H⁺ ion. All atoms of ethene have octet.

Fig.12.103

2. When the H⁺ ion approaches the ethene molecule, the two electrons in the 𝜋 bond moves to the right side C atom.
• The right side C atom uses those two electrons to form a bond with H⁺
• The left side C atom acquires a +ve charge.
3. Let us write the three salient features of this reaction:
(i) Transfer of electrons occurred due to the presence of the attacking reagent H⁺.
(ii) Electrons in the 𝜋 bond are transferred.
(iii) After the transfer, the electrons were acquired by one of the atoms present on either ends of the 𝜋 bond.
• The reagent H⁺ got attached to the atom which acquired the electrons.
(iv) Effect is temporary:
• The electron transfer indicated by the curved-arrow takes place when the H⁺ come into the vicinity of the ethene molecule.
• If the H⁺ is taken away from the vicinity of the ethene molecule, the electrons move back to their original positions.
4. If the four conditions mentioned in the above step are satisfied, it is known as positive electromeric effect (+E effect)

Example 2:
This can be written in 4 steps:
1. The fig.12.104 below shows the reaction between ethene and CN^- ion. All atoms of ethene have octet.

Fig.12.104

2. When the CN^- ion approaches the ethene molecule, the two electrons in the 𝜋 bond moves to the right side C atom.
• The left side C atom thus loses an electron.
• The incoming CN^- ion supplies the required electron and forms a bond with the left side C atom.
• The right side C atom acquires a - ve charge.
3. Let us write the three salient features of this reaction:
(i) Transfer of electrons occurred due to the presence of the attacking reagent CN^-.
(ii) Electrons in the 𝜋 bond are transferred.
(iii) After the transfer, the electrons were acquired by one of the atoms present on either ends of the original 𝜋 bond.
• The reagent CN^- got attached to the other atom which did not acquire electrons.
(iv) Effect is temporary:
• The electron transfer indicated by the curved-arrow takes place when the CN^- come into the vicinity of the ethene molecule.
• If the CN^- is taken away from the vicinity of the ethene molecule, the electrons move back to their original positions.
4. If the four conditions mentioned in the above step are satisfied, it is known as negative electromeric effect (-E effect)

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◼ What happens when inductive effect and electromeric effect act at the same time?
• The answer can be written in steps:
1. We have seen inductive effect in an earlier section. Inductive effect also involves pulling of electrons.
2. It may so happen that:
• Due to the inductive effect, electrons tend to move in a particular direction.
• But due to the electromeric effect, the electrons tend to move in the opposite direction.
3. Then there will be a competition between the two effects.
• If such a situation occur, electromeric effect will be the winner.

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Hyperconjugation

• Some basics of hyperconjugation can be written based on ethyl cation. It can be written in 9 steps:
1. We know the structure of ethane: CH₃ㅡCH₃. The two C atoms, carry three H atoms each.
• If one H atom leaves (taking both electrons in the bond with it), we get the ethyl cation. We write it as: $\mathbf{\rm{CH_3\overset{+}{C}H_2}}$
• We saw it’s structure when we discussed heterolytic cleavage. See fig.12.69(a) of section 12.10.
2. So the ethyl cation has two parts:
    ♦ $\mathbf{\rm{CH_3}}$ㅡ
    ♦ ㅡ$\mathbf{\rm{\overset{+}{C}H_2}}$
• The C atom in CH₃ part has octet. That C atom is sp³ hybridized. It is shown in fig.12.105(a) below. The H atoms in this part are named as H_A, H_B and H_C

Fig.12.105

• The C atom in CH₂ part has only sextet. That C atom is sp² hybridized. It is shown in fig.12.105(b) above. We already know how this structure is derived. See fig.12.71 of section 12.10.
3. Figs (a) and (b) show us the two parts separately. When they combine together, we get the ethyl cation. It is shown in fig.c
• They are combined together through a σ bond between the two C atoms.
4. Though the two parts are combined together, rotation can take place about the red axis. We saw this in the animation in fig.4.141 of section 4.25.
• Consider a particular instant during the rotation. At that instant, the H_A atom and the hybrid orbital carrying H_A, comes in alignment (same plane) with the empty p-orbital of the CH₂ part.
• In such a situation, the two electrons in the CㅡH_A bond will get delocalized into the empty p-orbital.
• This is shown in fig.12.106(a) below. This type of delocalization is known as hyperconjugation.

Fig.12.106

5. As a result, H_A will lose both electrons in the bond.
• Those two electrons will be used to form a 𝜋 bond between the two C atoms.
    ♦ It is a 𝜋 bond because, the overlap is lateral.
• Thus the original CㅡC single bond becomes a double bond C=C.
• The structural formula of the resulting structure will be as shown in fig.12.106(b) above. In that fig., we see that, H_A has no bond with C. Also, H_A has a + charge. This is due to the loss of one electron. The C has gained an electron. So it's original + charge disappears.
6. Now, as the rotation continues, at another instant, the H_B atom and the hybrid orbital carrying H_B, comes in alignment (same plane) with the empty p-orbital of the CH₂ part.
• At that instant, the two electrons in the CㅡH_B bond will get delocalized into the empty p-orbital.
• The structural formula of the resulting structure will be as shown in fig.12.107(a) below. In that fig., we see that, H_B has no bond with C. Also, H_B has a + charge. This is due to the loss of one electron.

Fig.12.107

7. Similarly, as the rotation continues, at another instant, the H_C atom and the hybrid orbital carrying H_C, comes in alignment (same plane) with the empty p-orbital of the CH₂ part.
• At that instant, the two electrons in the CㅡH_C bond will get delocalized into the empty p-orbital.
• The structural formula of the resulting structure will be as shown in fig.12.107(b) above. In that fig., we see that, H_C has no bond with C. Also, H_C has a + charge. This is due to the loss of one electron.
8. The structural formulas in figs.12.106 and 107 are all resonance structures of the same ethyl cation.
• The analysis of the resonance structures can be written in 3 steps:
(i) The structure I in fig.12.108 below shows the original structure.
• The red curved arrow indicates that, the two electrons in the CㅡH_A bond is transferred to the CㅡC single bond. That single bond thus becomes a double bond. This double bond is shown in II.
• Due to the action of the red curved arrow, the bond between C and H_A is lost. There is 'no bond' between C and H_A. This is also shown in II.
• The red curved arrow in I, comes into play when the H_A atom becomes aligned with the empty p orbital. It is a hyperconjugation.

Fig.12.108

 

(ii) From (i) above, we know how the structure II is obtained. So now we can analyze the curved-arrows in II.
• The red curved-arrow in II indicates that the CㅡH_A receives the two electrons back. So in III, we see the CㅡH_A bond.
• But the C=C double bond does not become a single bond. This is because, the H_B now becomes aligned with the empty p-orbital and hyperconjugation takes place.
• The transfer of electrons by this hyperconjugation is indicated by the magenta curved-arrow in II. As a result, there is ‘no bond’ between C and H_B. We see this in III
(iii) From (ii) above, we know how the structure III is obtained. So now we can analyze the curved arrows in III.
• The red curved-arrow in III indicates that the CㅡH_B receives the two electrons back. So in IV, we see the CㅡH_B bond.
• But the C=C double bond does not become a single bond. This is because, the H_C now becomes aligned with the empty p-orbital and hyperconjugation takes place.
• The transfer of electrons by this hyperconjugation is indicated by the magenta curved-arrow in III. As a result, there is ‘no bond’ between C and H_C. We see this in IV
9. Now we can write about the stability of the ethyl cation. It can be written in five steps:
(i) Originally, the ethyl cation does not have any double bond. The + charge is carried by one of the C atoms. This is one of the four resonance structures.
(ii) Due to hyperconjugation, the single bond between the two C atoms become a double bond. The + charge is now carried by H_A. This is another resonance structure.
(iii) Again due to hyperconjugation, the single bond between the two C atoms become a double bond. The + charge is now carried by H_B. This is another resonance structure.
(iv) Again due to hyperconjugation, the single bond between the two C atoms become a double bond. The + charge is now carried by H_C. This is yet another resonance structure.
(v) So in the hybrid structure, the + charge is distributed among four atoms:
One C atom and three H atoms.
• Such a distribution of charge, gives greater stability to the ethyl cation.

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In an earlier section, we wrote about the stability of ethyl cation. (step 9 below fig.12.70 of section 12.10). The above nine steps helps us to explain it's stability. In the next section we will see the stability of isopropyl cation.

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