Chapter 12.11 - Homolytic Cleavage

In the previous section, we saw the details about heterolytic cleavage. In this section, we will see homolytic cleavage.

Details about homolytic cleavage can be written in 9 steps:
1. In homolytic cleavage, the covalent bond breaks in such away that, each of the two fragments get one electron from the broken bond.
• This can be explained using an example. It can be written in 5 steps:
(i) Fig.12.74(a) below shows the Lewis dot structure of CH₃X
    ♦ The green dots indicate the one valence electron of H
    ♦ The red dots indicate the four valence electrons of C
    ♦ The blue dots indicate the valence electrons of X.
         ✰ (Number of this valence electrons depend on the species represented by X)

Fig.12.74

(ii) The two cyan dashed curves indicate that, the bond between C and X undergoes fission.
• After the fission, we get two species. They are shown in figs.12.74 (b) and (c)
(iii) Fig.b shows that, the C atom has retained the red dot which originally belonged to it.
   ♦ This species is written as: $\mathbf{\rm{{H_3}\overset{.}{C}}}$
   ♦ The C atom in this species has an unpaired electron.
         ✰ This is indicated by the dot over C
(iv) Fig.c shows that, the X atom has also retained the blue dot which originally belonged to it.
   ♦ This species is written as: $\mathbf{\rm{\overset{.}{X}}}$
(v) The process in fig.12.74 can be written in a condensed form as shown below:

• The two curved arrows are 'half-headed'. That is., the heads have the shape of a fish hook. Such a shape indicates that, each of the two electrons in the bond, moves symmetrically after the cleavage.
2. Based on the above example, we can write:
After the homolysis, each of the two fragments will have an unpaired electron.
3. A species having an unpaired electron is called a free radical.
• So what we have in figs.12.74 (b) and (c) above, are free radicals. In our present case, fig.b can be named as methyl free radical.
• Such radicals in general are called alkyl radicals.
4. Consider fig.12.75(a) below:

Fig.12.75

• The compound under went homolysis in such a way that, one of the two electrons in the bond stayed with the right side C atom.
   ♦ So the species as a whole is a free radical.
• Note the right side C atom. The bond attached to it was cleaved. It is the C atom with the unpaired electron.
   ♦ Only one other C atom is directly attached to this C atom.
   ♦ So this species is known as primary alkyl radical.
   ♦ It is named as ethyl free radical.
   ♦ It is written as $\mathbf{\rm{CH_3\overset{.}{C}H_2}}$.
5. Consider fig.12.75(b) above.
• It has three C atoms. So it is related to propane. The middle C atom originally had two H atoms.
• But fig.b shows that, the middle C atom has retained it's electron after cleavage.
• Fig.c shows the structural formula.
   ♦ The middle C atom has an unpaired electron
   ♦ It is clear that, the species as a whole is an alkyl radical.
   ♦ Two other C atoms are directly attached to the middle C atom.
   ♦ So this species is known as secondary alkyl radical.
   ♦ It is named as isopropyl free radical.
   ♦ It is written as $\mathbf{\rm{(CH_3)_2\overset{.}{C}\,H}}$.
6. Consider the species in fig.12.69(d). It also has three C atoms. The right most C atom has retained it's electron after cleavage. It is the C atom with the unpaired electron. But this species cannot be called a secondary alkyl radical. This is because, the right most C atom is directly attached to only one other C atom. It is a primary alkyl radical.
7. Consider fig.12.76(a) below:

Fig.12.76

• The compound under went homolysis in such a way that, the central C atom has retained one of the two electrons in the cleaved bond.
   ♦ So the species as a whole is an alkyl free radical.
• Note the central C atom whose bond was cleaved.
   ♦ It is the C atom with the unpaired electron.
   ♦ Three other C atoms are directly attached to this C atom.
   ♦ So this species is known as tertiary alkyl radical.
   ♦ It is named as tert-butyl free radical.
   ♦ It is written as $\mathbf{\rm{(CH_3)_3\overset{.}{C}}}$.
8. In the above steps, we counted the ‘number of C atoms’ which are directly attached to the C atom with the unpaired electron.
• Instead of the ‘number of C atoms’, we can use the ‘number of alkyl groups’ also. This can be explained in 3 steps:
(i) In fig.12.75(a), the C atom is directly attached to one C atom. This one C atom is from a methyl group. So we can write:
The C atom is directly attached to one alkyl group.
(ii) In fig.12.75(b) and (c), the C atom is directly attached to two C atoms. These two C atoms are from methyl groups. So we can write:
The C atom is directly attached to two alkyl groups.
(iii) In fig.12.76, the C atom is directly attached to three C atoms. These three C atoms are from methyl groups. So we can write:
The C atom is directly attached to three alkyl groups.
9. Alkyl free radicals are highly unstable.
• To attain stability, they tend to enter into reactions with other species. That means, alkyl free radicals are very reactive.
• However, the alkyl groups which are directly attached to the C atom with unpaired electron, help to stabilize the radical. We will see the mechanism in later sections.
• Among the four radicals that we saw above, the tertiary butyl free radical has the greatest stability. Followed by isopropyl free radical, followed by ethyl free radical, followed by methyl free radical.

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• The organic reactions which proceed through homolytic bond cleavage are called free radical reactions.
• They are also known as homopolar reactions.
• Some times they are just called nonpolar reactions.

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Now we will see some solved examples:
Solved example 12.13
Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
(a) CH₃ㅡSCH₃ (b) CH₃ㅡCN (c) CH₃ㅡCu
Solution:
Part (a):
(i) Fig.12.77(a) below shows the Lewis dot structure of CH₃ㅡSCH₃
    ♦ The green dots indicate the one valence electron of H
    ♦ The red dots indicate the four valence electrons of C
    ♦ The blue dots indicate the seven valence electrons of S.

Fig.12.77

(ii) The cyan dashed curve indicates that, the bond between C and S undergoes fission.
• After the fission, we get two species. They are shown in figs.12.77 (b) and (c)
(iii) Fig.b shows that, the C atom now has only six electrons around it.
   ♦ Note that, the C atom has lost one red dot.
   ♦ So C has now a +ve charge.
   ♦ This species is written as: $\mathbf{\rm{{H_3}\overset{+}{C}}}$
(iv) Fig.c shows that, the S atom now has eight electrons around it.
   ♦ Note that, one red dot which originally belonged to the C, is now with S.
   ♦ So S now has a -ve charge.
   ♦ Also there are three lone pairs.
   ♦ This species is written as: $\mathbf{\rm{\overset{-}{S}CH_3}}$
(v) The process in fig.12.77 can be written in a condensed form as shown below:

• The curved arrow indicates that, both the electrons in the bond are transferred to the S atom.

Part (b):
(i) Fig.12.78(a) below shows the Lewis dot structure of CH₃ㅡCN
    ♦ The green dots indicate the one valence electron of H
    ♦ The red dots indicate the four valence electrons of left side C
    ♦ The yellow dots indicate the four valence electrons of right side C
    ♦ The blue dots indicate the seven valence electrons of N.

Fig.12.78

(ii) The cyan dashed curve indicates that, the bond between the two C atoms undergoes fission.
• After the fission, we get two species. They are shown in figs.12.78 (b) and (c)
(iii) Fig.b shows that, the left side C atom now has only six electrons around it.
   ♦ Note that, the C atom has lost one red dot.
   ♦ So C has now a +ve charge.
   ♦ This species is written as: $\mathbf{\rm{{H_3}\overset{+}{C}}}$
(iv) Fig.c shows that, the right side C now has eight electrons around it.
   ♦ Note that, one red dot which originally belonged to the left side C, is now with the right side C.
   ♦ So right side C now has a -ve charge.
   ♦ Also there is a lone pair.
   ♦ This species is written as: $\mathbf{\rm{\overset{-}{C}N}}$
(v) The process in fig.12.78 can be written in a condensed form as shown below:

• The curved arrow indicates that, both the electrons in the bond are transferred to the right side C atom.

Part (c):
(i) Fig.12.79(a) below shows the Lewis dot structure of CH₃ㅡCu
    ♦ The green dots indicate the one valence electron of H
    ♦ The red dots indicate the four valence electrons of C
    ♦ The blue dot indicate the valence electron of Cu.

Fig.12.79

(ii) The cyan dashed curve indicates that, the bond between C and Cu undergoes fission.
• After the fission, we get two species. They are shown in figs.12.79 (b) and (c)
(iii) Fig.b shows that, the C atom now has eight electrons around it. It has the extra blue dot which originally belonged to the Cu atom.
   ♦ So C has now a -ve charge.
   ♦ This species is written as: $\mathbf{\rm{{H_3}\overset{-}{C}}}$
(iv) Fig.c shows that, the Cu now has lost the blue dot.
   ♦ This species is written as: $\mathbf{\rm{\overset{+}{Cu}}}$
(v) The process in fig.12.79 can be written in a condensed form as shown below:

• The curved arrow indicates that, both the electrons in the bond are transferred to the C atom.

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In the next section, we will see nucleophiles and electrophiles.


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