
Arrange the (C-H) bonds x, y and z in decreasing order of their bond dissociation energies in homolysis.
(A)- $y>x>z$
(B)- $z>x>y$
(C)- $z>y>x$
(D)- $y>z>x$
Answer
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Hint: In the homolytic cleavage, on the carbon radical, the hybridisation, substitutions and the resonance effect contribute to its stability, which further decides the energy required for the bond dissociation.
Complete step by step answer:
In the homolytic cleavage, there is splitting of the shared electron pair between the two covalently bonded atoms, such that each acquires one of the two shared electrons. Thus, forming two new species which have an unpaired electron present in them. This is called the free radical.
The energy required in the splitting of the covalent bond homolytically is the bond dissociation energy, which depends on the stability of the radical formed after splitting. This can further be determined from the hybridisation involved and the substitutions on the carbon atom.
At position x, the radical forms over an $s{{p}^{3}}$ hybridised carbon atom and secondary radical due to the presence of two alkyl groups attached to it. This leads to hyperconjugation as there are four hyperconjugable hydrogens present, making it slightly stable, as there is no partial double bond formed.
Now at position y, it is an allylic carbon, with the occurrence of both resonance and hyperconjugation. As there is both double- bond on adjacent carbon and also an alkyl group on another side. Thus, the free radical is dispersed over a larger space, showing partial $s{{p}^{2}}$ character and is highly stable.
Whereas, at position z, the radical is present over the $s{{p}^{2}}$ hybridised carbon or the vinylic carbon and is highly unstable due to the greater s-character, making the C-H the strongest.
Thus, the stability of the radical is related to the bond dissociation energy. As stronger the C-H bond, unstable the radical then more is the dissociation energy.
Therefore, the order of the bond dissociation energy of the homolytic cleavage of the C-H bond
So, the correct answer is “Option B”.
Note: Firstly, the hybridisation of the carbon radical, followed by the resonance or hyperconjugation effect contributes to its stability. As more the stability, lower the bond dissociation energy.
Complete step by step answer:
In the homolytic cleavage, there is splitting of the shared electron pair between the two covalently bonded atoms, such that each acquires one of the two shared electrons. Thus, forming two new species which have an unpaired electron present in them. This is called the free radical.
The energy required in the splitting of the covalent bond homolytically is the bond dissociation energy, which depends on the stability of the radical formed after splitting. This can further be determined from the hybridisation involved and the substitutions on the carbon atom.
At position x, the radical forms over an $s{{p}^{3}}$ hybridised carbon atom and secondary radical due to the presence of two alkyl groups attached to it. This leads to hyperconjugation as there are four hyperconjugable hydrogens present, making it slightly stable, as there is no partial double bond formed.
Now at position y, it is an allylic carbon, with the occurrence of both resonance and hyperconjugation. As there is both double- bond on adjacent carbon and also an alkyl group on another side. Thus, the free radical is dispersed over a larger space, showing partial $s{{p}^{2}}$ character and is highly stable.
Whereas, at position z, the radical is present over the $s{{p}^{2}}$ hybridised carbon or the vinylic carbon and is highly unstable due to the greater s-character, making the C-H the strongest.
Thus, the stability of the radical is related to the bond dissociation energy. As stronger the C-H bond, unstable the radical then more is the dissociation energy.
Therefore, the order of the bond dissociation energy of the homolytic cleavage of the C-H bond
So, the correct answer is “Option B”.
Note: Firstly, the hybridisation of the carbon radical, followed by the resonance or hyperconjugation effect contributes to its stability. As more the stability, lower the bond dissociation energy.
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