The order of circled C-H bond dissociation en | vedclass

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Question

(C) The bond dissociation energy $(BDE)$ of a $C-H$ bond is inversely proportional to the stability of the free radical formed after homolytic cleavag

Vedclass · Accepted Answer

$I > II > IV > III$

Vedclass · Answer

(C) The bond dissociation energy $(BDE)$ of a $C-H$ bond is inversely proportional to the stability of the free radical formed after homolytic cleavage of that bond.$1.$ Ethylene $(I)$: The radical formed is a vinyl radical $(CH_2=CH^{\bullet})$,where the unpaired electron is in an $sp^2$ orbital. This is highly unstable.$2.$ Methane $(II)$: The radical formed is a methyl radical $(CH_3^{\bullet})$.$3.$ Propene $(III)$: The radical formed is an allyl radical $(CH_2=CH-CH_2^{\bullet})$,which is resonance stabilized and thus the most stable.$4.$ Propane $(IV)$: The radical formed is an isopropyl radical $((CH_3)_2CH^{\bullet})$,which is stabilized by hyperconjugation.Stability order of radicals: Allyl $(III)$ > Isopropyl $(IV)$ > Methyl $(II)$ > Vinyl $(I)$.Since $BDE \propto \frac{1}{	ext{Stability of radical}}$,the order of $BDE$ is: $I > II > IV > III$.

Column $-I$	Column $-II$
$A$. Free radical	$1$. Trigonal planar
$B$. Carbocation	$2$. Pyramidal
$C$. Carbanion	$3$. Linear

The order of circled $C-H$ bond dissociation energy in the following compounds is:

Similar Questions

The incorrect statement about carbene $(CH_2)$ is:

Arrange the following free radicals in the correct order of their stability:
$(i)$ $CH_2=\dot{C}H$
$(ii)$ $\dot{C}H_3$
$(iii)$ $CH_3-\dot{C}H-CH_3$
$(iv)$ $(CH_3)_3\dot{C}$

Match the intermediates given in Column $-I$ with their probable structure in Column $-II$.

Column $-I$ Column $-II$

$A$. Free radical $1$. Trigonal planar

$B$. Carbocation $2$. Pyramidal

$C$. Carbanion $3$. Linear

Which is the most stable carbocation?

Find out the number of $1-2-$shifts during the conversion of the given carbocation to the final alkene product.

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