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Stability of Free Radical

Arrange the b...

Question

Arrange the bond energy of $H_{a}, H_{b}, H_{c} and H_{d}$ in decreasing order.

A

H_{b} > H_{d} > H_{c} > H_{a}

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B

H_{b} > H_{d} > H_{a} > H_{c}

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C

H_{c} > H_{a} > H_{d} > H_{b}

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D

H_{c} > H_{d} > H_{a} > H_{b}

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Solution

The correct option is C $H_{c} > H_{a} > H_{d} > H_{b}$
(i) By dissociation of $H_{a}$
The radical formed has $2 α - H y d r o g e n s$ , hence it is stabilised by hyperconjugation effect.

(ii) By dissociation of $H_{b}$
The radical formed has $3 α - H y d r o g e n s$ and it is also stabilised by resonance effect. Hence it is the most stabilised radical.

(iii) By dissociation of $H_{c}$
Here the radical is generated on vinylic carbon. Hence it is the least stable radical.

(iv) By dissociation of $H_{d}$
The radical formed has no $a l p h a - H y d r o g e n s$ but it is stabilised by resonance effect.

Stability of intermediate,
Resonance effect > Hyperconjugation effect
Hence the order of stability of free radical is
$H_{b} > H_{d} > H_{a} > H_{c}$
$B o n d e n e r g y \propto \frac{1}{S t a b i l i t y o f f r e e r a d i c a l}$
Hence the order of bond energy is
$H_{c} > H_{a} > H_{d} > H_{b}$

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Stability of Free Radical

Standard XII Chemistry

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