# Why is the enthalpy of formation of 2,2-dimethylpentane larger than that of 2,2,3-trimethylbutane?

Firstly, it is commonly known and taught that branched alkanes are more stable than linear ones, and so have larger enthalpies of formation. But why is that the case? Are there some steric interactions at play?

Secondly, even if we just accept that relationship to be true, why are there instances where this relationship does not hold? Specifically, when I was exploring the data on NIST WebBook and I noticed that $$\Delta_\mathrm fH^⦵$$ is larger for 2,2-dimethylpentane than 2,2,3-trimethylbutane even though the latter is more branched than the former.

• Are you sure it is not a value sign issue ? As ΔfH⦵ should be larger for 2,2-dimethyl Pentane than 2,2,3-trimethyl Butane if the latter is more TD stable because of branching. ( Unless entropy takes decisive role, having lower formation deltaG even with higher formation deltaH ). Jul 13 at 10:50
• The NIST values are (kcal/mol): 2,2-dimethylpentane, -49.29 +/- 0.32; 2,2,3-trimethylbutane, -48.96 +/- 0.27. Considering the error in measurement, the latter may have a more negative heat of formation than the former. @Poutnik Jul 13 at 19:00
• Sorry for the confusion. As user55119 said in his comment, I meant the absolute value of ΔfH⦵ is larger for 2,2,3-trimethyl Butane compared to 2,2-dimethyl Pentane. @Poutnik Jul 14 at 5:47