Question: 76. Among the given compounds I-III, the correct order of bond dissociation energy of \mathrm{C-H} bond marked with * is:
(1) II > I > III
(2) I > II > III
(3) III > II > I
(4) II > III > I
Answer: Option (1)
Explanation:
Bond dissociation energy of a \mathrm{C-H} bond depends mainly on two factors:
(i) hybridisation of the carbon (more s-character gives a stronger \mathrm{C-H} bond,
hence higher bond dissociation energy) and
(ii) stability of the radical formed after homolytic cleavage
(more stable radical means lower bond dissociation energy).
In compound II, the hydrogen marked with * is attached to an
\mathrm{sp} hybridised carbon (terminal alkyne type
\mathrm{C\equiv C-H}).
An \mathrm{sp} carbon has the highest s-character (50\%),
so the \mathrm{C-H} bond is the strongest.
Therefore, compound II has the highest bond dissociation energy.
In compound I, the hydrogen marked with * is attached to an aromatic (benzene ring) \mathrm{sp^2} carbon.
An \mathrm{sp^2} carbon has less s-character (33\%)
than an \mathrm{sp} carbon,
so the \mathrm{C-H} bond is weaker than in II.
Hence, bond dissociation*