What reagents are used to selectively reduce 2,2,3‐trimethyloxirane to secondary and tertiary alcohols?

Question

Problem

Answer

(c) $x = \ce{LiAlH4},~y = \ce{LiAlH4}/\ce{AlCl3}$

Questions

I know $\ce {LiAlH4}$ is a reducing agent because it gives four $\ce{H-}$ and will attack on our reactant on carbon with less substituents, i.e. in this case on $\ce{-CH-}$ and on attack the $\ce{-O-}$ will go to other carbon and on protonation we get 3° alcohol.

But what about 2° alcohol? How can we get that?

What does $\ce{AlCl3}$ do such that we get 2° alcohol?

I thought maybe the oxygen is giving its lone pair to $\ce{Al}$ being deficient in $\ce{e-}.$ Then what happens if it does so? Will now $\ce{H-}$ attack 3° carbon? Does that mean it's following S_N1 mechanism? Why?

Answer 1

As Waylander has pointed out in the comments, the mechanism would proceed as follows in case of $\ce{LiAlH4/AlCl3}$ as a reagent to produce a secondary alcohol (S_N1), whereas in case of just $\ce{LiAlH4}$, the reaction proceeds via S_N2 to give the tertiary alcohol by attacking the less hindered carbon.

Alane has more Lewis acid character than $\ce{LiAlH4}$ so when it binds to the oxygen a cationic centre develops at the tertiary carbon which is then reduced by hydride from the alane. In contrast with $\ce{LiAlH4}$ the hydride attack is a more nucleophilic process so it attacks the least hindred centre

Answer 2

Reaction with $\ce{LiAlH4}$ is base catalyzed ring opening while that with $\ce{LiAlH4/AlCl3}$ is acid catalyzed ring opening. In base catalyzed ring opening the nucleophile attacks less substituted (and therefore less sterically hindered) carbon atom of an unsymmetrical epoxide, while in acid catalyzed ring opening the nucleophile attacks at the more substituted carbon atom of an unsymmetrical epoxide.