Carbon-carbon bond dissociation enthalpies have been measured for many alkanes. Without re...

Carbon-carbon bond dissociation enthalpies have been measured for many alkanes. Without referring to Table 4.3, identify the alkane in each of the following pairs that has the lower carbon-carbon bond dissociation enthalpy, and explain the reason for your choice.

(a) Ethane or propane

(b) Propane or 2-methylpropane

(c) 2-Methylpropane or 2,2-dimethylpropane

Sample Solution (a) First write the equations that describe homolytic carbon-carbon bond cleavage in each alkane.

Cleavage of the carbon-carbon bond in ethane yields two methyl radicals, whereas propane yields an ethyl radical and one methyl radical. Ethyl radical is more stable than methyl, and so less energy is required to break the carbon-carbon bond in propane than in ethane. The measured carbon-carbon bond dissociation enthalpy in ethane is 368 kJ/mol (88 kcal/mol), and that in propane is 355 kJ/mol (85 kcal/mol).