Comparing C–H Bond Energies

The C–H bond energy depends on the hybridization state of the carbon atom and the stability of the resulting radical after bond dissociation. Higher s-character in hybridization leads to stronger bonds due to better orbital overlap and shorter bond lengths.

Step 1: Identify Hybridization States

Different carbon atoms have different hybridizations:

• sp³ hybridization (25% s-character) - alkanes
• sp² hybridization (33% s-character) - alkenes/aromatics
• sp hybridization (50% s-character) - alkynes

Step 2: Analyze Each Structure

From the image (assuming standard representations):

• a: Alkyne (sp carbon) - highest bond energy
• b: Alkane (sp³ carbon) - lowest bond energy
• c: Alkene/aromatic (sp² carbon)
• d: Benzene (sp² with resonance stabilization)
• e: Allylic position (slightly stabilized radical)
• f: Benzylic position (resonance stabilized radical)

Step 3: Order by Bond Strength

Bond energy order: sp > sp² > sp³, with modifications for radical stability:

Alkyne (a) > Benzene (d) > Alkene (c) > Allylic (e) > Alkane (b) > Benzylic (f)

Benzylic has lowest bond energy due to maximum radical stabilization: $E\propto \frac{1}{s\text{-character}}$

Final Answer

Correct order: a > d > c > e > b > f

Related Concepts

Bond Dissociation Energy: Energy required to break a bond homolytically. Affected by hybridization and radical stability.

Radical Stability: Tertiary > Secondary > Primary > Methyl, with additional stabilization through resonance.

Key Formulae

Bond energy generally correlates with s-character: $\mathrm{BE}\propto \%s$

For benzylic/allylic systems: $\mathrm{BE}\approx 88-5\times \left(\text{radical stabilization energy}\right)\text{kcal/mol}$