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: $I > II > IV > III$.

Label	Carbocation
$1$	$CH_3-CH^+-CH_3$
$2$	$CH_3^+$
$3$	$(CH_3)_3C^+$
$4$	$CH_3-CH_2^+$

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

Similar Questions

Arrange the following carbocations in the increasing order of stability with respect to their labels:
Label Carbocation
$1$ $CH_3-CH^+-CH_3$
$2$ $CH_3^+$
$3$ $(CH_3)_3C^+$
$4$ $CH_3-CH_2^+$

Rank the following radicals in order of increasing stability (least $ < < < $ most).

In the organic reaction by fission of bond,which intermediate species are formed? Name these species and explain their stability.

Which is the correct order of stability of the following three carbonium ions?
$(I)$ $CH_2 = CH - CH^+ - CH_3$
$(II)$ $CH_2 = C(CH_3) - CH_2^+$
$(III)$ $CH_3 - CH = CH - CH_2^+$

What is the minimum energy required for the homolytic bond fission of the $C-H$ bond in the following molecule?

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