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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?
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.
What's important to note here is that the substitution of 1° and 2° alkyl halides is an SN2 reaction. Like most SN2 reactions, it exists in equilibrium with the E2 mechanism in most circumstances. Because of steric hinderance, SN2 reactions are extremely poorly favored on 3° alkyl halides, so the substituted product will be created in only trace levels, with E2 products dominating the product mix.
