# How is NaBH4 able to reduce tertiary alkyl halides to alkane but LiAlH4 create an alkene?

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?

• The correct terminology is "primary/secondary/tertiary", not "1/2/3 degree". Jan 1 '20 at 20:25
• That "degree" terminology has been regarded as arcane and outdated since at least the middle of the last century.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.
– Karl
Jan 1 '20 at 20:39
• 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? Jan 1 '20 at 21:37
• As far as i know LiAlH4 also does E1 here Jan 2 '20 at 5:45
• That "degree" terminology is widely used in the Indian senior secondary classes. Mar 20 '20 at 12:59