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.

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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? ...?

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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? ...?