For the molecule with the carbon skeleton CH2=CH-CH2-CH3 type chain bearing four distinct C-H bonds labelled C1-H, C2-H, C3-H and C4-H (where C1-H and C4-H are allylic/secondary sp3 C-H, C2-H is vinylic sp2 C-H, and C3-H is a primary alkyl sp3 C-H), arrange these C-H bonds in the correct o

Question

For the molecule with the carbon skeleton CH2=CH-CH2-CH3 type chain bearing four distinct C-H bonds labelled C1-H, C2-H, C3-H and C4-H (where C1-H and C4-H are allylic/secondary sp3 C-H, C2-H is vinylic sp2 C-H, and C3-H is a primary alkyl sp3 C-H), arrange these C-H bonds in the correct order of bond dissociation energy assuming homolytic cleavage. (The radical formed that is most stabilised corresponds to the weakest, lowest-BDE bond.)

Accepted Answer

C2-H > C3-H > C1-H > C4-H. Bond dissociation energy is highest when the radical formed is least stable. A vinylic C-H (C2-H) gives an unstable, high-energy vinyl radical, so it has the highest BDE. A primary alkyl C-H (C3-H) gives an unstabilised primary radical (next highest). The allylic positions (C1-H, C4-H) form resonance-stabilised allylic radicals, so they have the lowest BDE; the more substituted/better-stabilised allylic radical has the lowest BDE of all. This gives the order C2-H > C3-H > C1-H > C4-H.

Solution

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