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 HA, HB and HC will come successively in alignment with the empty p orbital of C2.
4. When HA comes into alignment, hyperconjugation takes place between the C1ㅡHA 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 HA loses the bond with C1. Due to the loss of an electron, HA 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 HB comes into alignment with the empty p-orbital of C2. Hyperconjugation takes place between the C1ㅡHB 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.
◼ 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.
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 HA, HB and HC will come successively in alignment with the p-orbital of C2.
3. Hyperconjugation takes place between the C1ㅡHA σ bond and the p-orbital of C2.
• That is., the two electrons in the C1ㅡHA 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 HA loses the bond with C1. Due to the loss of an electron, HA 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 HB comes into
alignment with the p-orbital of C2. Hyperconjugation takes place
between the C1ㅡHB σ bond and the p orbital.
• The red curved-arrow indicates that, HA 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.
• 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.
• 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.
• The link below gives the folder containing additional solved examples on this chapter.
• Part 2 is related to reaction mechanism.
• In the next section we will see the methods of purification of organic compounds.
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