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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". $\endgroup$ Jan 1, 2020 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    $\begingroup$ That "degree" terminology has been regarded as arcane and outdated since at least the middle of the last century.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, 2020 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, 2020 at 21:37
  • $\begingroup$ As far as i know LiAlH4 also does E1 here $\endgroup$ Jan 2, 2020 at 5:45
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    $\begingroup$ That "degree" terminology is widely used in the Indian senior secondary classes. $\endgroup$ Mar 20, 2020 at 12:59

1 Answer 1

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This is a classic case of $\ce{S_N2}$ nucleophilic substitution versus elimination. Sodium borohydride gives the former reaction, using a transferred hydride ion as the nuceophile, while lithium aluminum hydride produces elimination.

Two factors explain the different courses of the two reactions.

Getting down to basics. Generally, stronger proton bases favor elimination because they cleave the carbon-hydrogen bond more readily. Lithium aluminum hydride (which will deprotonate water and other protic solvents even under strongly basic conditions), is clearly a stronger proton base than sodium borohyidide (which is reasonably stable in water at sufficiently alkaline pH).

Elbow room. $\ce{S_N2}$ on tertiary substrates is made difficult by steric hindrance. A more compact nucleophile then becomes more favorable to overcome this barrier. Borohydride ion with its boron atom fits that compared with aluminohydride having the larger aluminum atom. The larger size of the aluminum atom also leads to weaker, more nearly ionic bonding to the (hydride-like) hydrogen and thus stronger proton basicity, which goes back to the basicity factor.

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