Rank the bond dissociation energies of the bo | vedclass

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Question

(D) The bond dissociation energy is inversely proportional to the stability of the free radical formed after homolytic cleavage of the $C-H$ bond.$1$.

Vedclass · Accepted Answer

$3 < 1 < 2$

Vedclass · Answer

(D) The bond dissociation energy is inversely proportional to the stability of the free radical formed after homolytic cleavage of the $C-H$ bond.$1$. Bond $3$ leads to a tertiary $(3^{\circ})$ radical on the cyclohexane ring,which is the most stable due to hyperconjugation and inductive effects.$2$. Bond $1$ leads to a secondary $(2^{\circ})$ radical on the cyclohexane ring,which is less stable than the $3^{\circ}$ radical.$3$. Bond $2$ leads to a primary $(1^{\circ})$ radical on the methyl group,which is the least stable.Since the stability of radicals is $3^{\circ} > 2^{\circ} > 1^{\circ}$,the bond dissociation energy order is $3 < 1 < 2$.

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

Rank the bond dissociation energies of the bonds indicated with the arrows (from smallest to largest).

Similar Questions

In which pair is the second ion more stable than the first?

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

The stability of the carbocations follows the order:

Which of the following carbocations lacks hyperconjugative stability?

Which of the following is the most stable among the given ions?

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