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Why is the 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?
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+ Chlorine
+ Chemistry
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Khari Slaughter
Why is the 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?
Well, given such conditions, radical chlorination of pentane can LEAD to THREE isomeric alkyl halides, i.e. [math]H_{3}C-CH_{2}CH_{2}CH_{2}CH_{2}Cl[/math], [math]H_{3}C-CH_{2}CH_{2}CH(Cl)CH_{3}[/math], and [math]H_{3}C-CH_{2}CH(Cl)CH_{3}CH_{3}[/math]. (We would expect that the two latter isomers are the major products of the reaction…)
On the other hand, radical halogenation of neopentane can LEAD to ONE halide, i.e. neopentyl chloride…i.e. all the methyl groups are equivalent by symmetry…
Well, given such conditions, radical chlorination of pentane can LEAD to THREE isomeric alkyl halides, i.e. [math]H_{3}C-CH_{2}CH_{2}CH_{2}CH_{2}Cl[/math], [math]H_{3}C-CH_{2}CH_{2}CH(Cl)CH_{3}[/math], and [math]H_{3}C-CH_{2}CH(Cl)CH_{3}CH_{3}[/math]. (We would expect that the two latter isomers are the major products of the reaction…)
On the other hand, radical halogenation of neopentane can LEAD to ONE halide, i.e. neopentyl chloride…i.e. all the methyl groups are equivalent by symmetry…
In case of neopentane, all the methyl groups are equivalent and hence, on radical chlorination, gives single monochlorinated product, that is, neopentylchloride. But in case of n-pentane, we get more than one monochlorinated product.
In case of neopentane, all the methyl groups are equivalent and hence, on radical chlorination, gives single monochlorinated product, that is, neopentylchloride. But in case of n-pentane, we get more than one monochlorinated product.
Well, given such conditions, radical chlorination of pentane can LEAD to THREE isomeric alkyl halides, i.e. [math]H_{3}C-CH_{2}CH_{2}CH_{2}CH_{2}Cl[/math], [math]H_{3}C-CH_{2}CH_{2}CH(Cl)CH_{3}[/math], and [math]H_{3}C-CH_{2}CH(Cl)CH_{3}CH_{3}[/math]. (We would expect that the two latter isomers are the major products of the reaction…)
On the other hand, radical halogenation of neopentane can LEAD to ONE halide, i.e. neopentyl chloride…i.e. all the methyl groups are equivalent by symmetry…
Well, given such conditions, radical chlorination of pentane can LEAD to THREE isomeric alkyl halides, i.e. [math]H_{3}C-CH_{2}CH_{2}CH_{2}CH_{2}Cl[/math], [math]H_{3}C-CH_{2}CH_{2}CH(Cl)CH_{3}[/math], and [math]H_{3}C-CH_{2}CH(Cl)CH_{3}CH_{3}[/math]. (We would expect that the two latter isomers are the major products of the reaction…)
On the other hand, radical halogenation of neopentane can LEAD to ONE halide, i.e. neopentyl chloride…i.e. all the methyl groups are equivalent by symmetry…
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In case of neopentane, all the methyl groups are equivalent and hence, on radical chlorination, gives single monochlorinated product, that is, neopentylchloride. But in case of n-pentane, we get more than one monochlorinated product.
In case of neopentane, all the methyl groups are equivalent and hence, on radical chlorination, gives single monochlorinated product, that is, neopentylchloride. But in case of n-pentane, we get more than one monochlorinated product.
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