Which C–H bond has the smallest bond dissociation enthalpy?
14% of national qualifiers got this right
Explanation
draftBond dissociation enthalpy (BDE) reflects the stability of the radical formed when the C–H bond is broken homolytically. The more stable (more delocalized or more substituted) the resulting radical, the weaker the bond and the lower the BDE.
- Vinylic C–H bonds (H on an carbon of a C=C, not allylic) are very strong because the radical formed is not stabilized and the C–H bond has more s-character.
- Primary C–H bonds give simple primary radicals, moderately strong.
- Tertiary C–H bonds give more stable radicals (hyperconjugation from three alkyl groups), so they are weaker than primary/secondary bonds.
- Allylic C–H bonds (adjacent to a C=C double bond) give a radical that is stabilized by resonance/delocalization into the π system, making it the most stable radical of all the choices. This means the allylic C–H bond requires the least energy to break.
Since is the allylic hydrogen, breaking it produces a resonance-stabilized allylic radical, giving it the smallest BDE among the four choices.
This is the vinylic hydrogen; students may think a bond on the double-bond carbon is 'reactive' and thus weak, but in fact the vinyl radical is unstabilized and this C–H bond has one of the highest BDEs.
Correct: is allylic, so breaking it forms a resonance-stabilized radical, giving the lowest bond dissociation enthalpy.
This is a primary C–H bond; students may mistakenly assume terminal/primary positions are always weakest, but primary radicals lack the resonance or hyperconjugative stabilization that lowers BDE the most.
This is a tertiary (or secondary) C–H bond; students may pick this thinking more alkyl substitution always gives the lowest BDE, but resonance stabilization in the allylic position lowers the BDE even further than simple hyperconjugation does.
USNCO 2019 National Part I exam, question 58. © American Chemical Society; reproduced for non-commercial practice.