Question

Radical chlorination of pentane is a poor way to prepare 1-chloropentane, but radical chlorination of neopentane, (CH3)4C, is a good way to prepare neopentyl chloride, (CH3)3CCH2Cl.
Why?

Answer 1

I would say that the intermediate formed, where a radical of chlorine attacks one of the hydrogens on the neopentane, is more stable than its linear pentane counterpart. This due to the good amount resonance that is attributed to the neopentane through its neighbouring methyl groups. (a total of 4 for neopentane but only 1 for primary pentane). The charge formed by the radical attack in neopentane has more room to propagate and is thus more stable. And reactions tend to favour a more stable state of bond dissociation energies.

The general rule for radical chlorination is tertiary radical species are more stable than secondary radical species, and secondary radical species are more stable than primary radical species—thus any single chlorination will favor substitution at the most substituted carbon.

Answer 2

The reaction will not be selective with pentane. It is more likely that the chlorine will go to the 2 or 3 positions than the 1 position. The neopentane is automatically selective since all the positions are the same.

Answer 3

the chlorine radical can attack the 1* or 2* carbon. In the neopentane you only have primary carbons as in pentane you have primary and secondary. I know chlorination i think doesnt prefer 1* or 2* but bromonation does.