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?

Because it’s random that CH3CH2CH2CH2CH3 losing a (·H), so by reaction with Cl2, it can form many isomers, such as CH3CH2CH2CHClCH3, CH3CH2CHClCH2CH3, CH3CH2CH2CH2CH2Cl, so a poor way to prepare 1-chloropentane. While for (CH3)4C, each of the H atom are equivalent. So no matter which one be substituted, only one product, (CH3)CCH2Cl, can be prepared.

The correct answer is c b c a d

I'm not sure but i will try to find out