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When attacking tertiary alkyl halides, aren't both of them just supposed to follow $\mathrm{S_N2}$ and substitute the leaving group with Hydride ion? Apparently only $\ce{NaBH4}$ does it and $\ce{LiAlH4}$ causes an $\mathrm{E1}$ elimination resulting in an alkene. Why does that happen? $\ce{LiAlH4}$ does secondary and primary alkyl halides to alkanes, so why not tertiary as well?

Reaction scheme

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    $\begingroup$ The correct terminology is "primary/secondary/tertiary", not "1/2/3 degree". $\endgroup$
    – orthocresol
    Jan 1 '20 at 20:25
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    $\begingroup$ That "degree" terminology has been regarded as arcane and outdated since at least the middle of the last century. Whoever still teaches that should not be approached during a full moon, or without a handful of garlic. $\endgroup$
    – Karl
    Jan 1 '20 at 20:39
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    $\begingroup$ Do you think that the solvent in the NaBH4 reaction may play a role in the mechanism of the reduction? Do you have a reference for the data you have presented? $\endgroup$
    – user55119
    Jan 1 '20 at 21:37
  • $\begingroup$ As far as i know LiAlH4 also does E1 here $\endgroup$ Jan 2 '20 at 5:45
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    $\begingroup$ That "degree" terminology is widely used in the Indian senior secondary classes. $\endgroup$ Mar 20 '20 at 12:59

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