Among the given compounds I, II, III, the cor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$II > I > III$

Vedclass · Answer

(A) The bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after homolytic fission of the $C-H$ bond.$(1)$ In $I$ (benzene), the $C-H$ bond is $sp^2$ hybridized. The resulting phenyl radical is highly unstable.$(2)$ In $II$ (phenylacetylene), the $C-H$ bond is $sp$ hybridized. The resulting ethynyl radical is even less stable than the phenyl radical due to high $s$-character.$(3)$ In $III$ (cyclopropene), the $C-H$ bond is $sp^3$ hybridized. The resulting cyclopropenyl radical is resonance-stabilized (allylic radical).Stability order of radicals: $III > I > II$.Therefore, the $BDE$ order is: $II > I > III$.

Among the given compounds $I, II, III$, the correct order of bond dissociation energy of the $C-H$ bond marked with $*$ is:

Similar Questions

Which one of the following carbocations is the most stable?

Which of the following intermediates has a complete octet around the carbon atom?

The correct order of the relative stability of the following carbanions is:
$I$. $HC \equiv C^{-}$
$II$. $CH_2=CH-CH_2^{-}$
$III$. $H_2C=CH^{-}$
$IV$. $(C_6H_5)_2CH^{-}$

The most stable carbocation among the following is:

Which of the following is the most stable free radical?

Vedclass Test Series

Exam Paper Generator

Online Exam Module