Which C-C bond would break first, the one in ethane, or 2,2-dimethylpropane?

Question

I have drawn the figures and determined intuitively that the larger molecule's $\ce{C-C}$ would break first. I made this assumption that ethane has the stronger $\ce{C-C}$ bond and I am correct, however I do not understand the underlying reasons why.

In both molecules we have single bonds, but in ethane we have more space for the hydrogen atoms to fill. This was my initial reasoning, that the larger molecule is less compact and will have longer $\ce{C-C}$ bonds. Am I correct in my reasoning, or does it have something to do with the radical stabilities? My book is saying "radical stability increases along the series from primary to secondary to tertiary; consequently, the energy required to create them decreases."

In ethane we would have primary carbons because they bond to only one other carbon. In 2,2-dimethylpropane the central carbon is bonded to four other carbons, making it quaternary. Is this the underlying reason, and if so then what is the reason for this trend?

Answer 1

It does not have to do with steric effects as you suggested. Rather, the reason behind this trend is the inductive effect. Figuratively speaking, carbon side-chains attached to an central atom "push" electron density towards this atom, while hydrogen atoms don't. If you break a C-C bond and produce a carbon radical, it will be stabilized if electron density is donated towards the site of the unpaired electron, which is the case in the tert-butyl radical formed from dimethylpropane, but not in the methyl radical from ethane.