Acidic cleavage of ether with a 2° alkyl group

Question

I know that an ether with a 3° alkyl group and a 1° alkyl group is cleaved by $\mathrm{S_N1}$ mechanism to give a tertiary alkyl halide and a 1° alcohol.

And an ether with a two 1° alkyl groups is cleaved by $\mathrm{S_N2}$ mechanism to give a 1° alkyl halide and a 1° alcohol.

But what happens if a ether with a 2° alkyl group and a 1° alkyl group is cleaved with a halogen acid?

Answer 1

what happens if a ether with a 2° alkyl group and a 1° alkyl group is cleaved with a halogen acid?

To this question I have drawn a scheme below. The given ether is $\ce{2^0}$ group to the left and methyl to right. Acid cleavage with a halogenic acid involves an initial protonation of ether. In an $\ce{S_N^2}$ reaction of this ether, nucleophile $\ce{I^-}$ attacks via path 1 and path2.

• In path 1, $\ce{I^-}$ approach is sterically hindered by phenyl and Methyl groups. This leads to a high energy transition state.
• In path 2, $\ce{I^-}$ approach is from right of protonated ether where steric hindrance is comparatively less( only $\ce{H}$). A more stable transition state favors $\ce{S_N^2}$ reaction .

Answer 2

In each of these cases, the ether will generally be cleaved to produce 2 alkyl halides. Acidic clevage can occur via Sn2 or Sn1 depending on the nature of the substrate. A bulky group, such as tert-butyl, will undergo an Sn1 mechanism in which the protonated ether group leaves before the nucleophile attacks. The same process is generally consistant with secondary substrates. I have drawn a mechanism involving both a primary and tertiary substrate to demonstrate this point; however, it is important to note that in the presence of most concentrated halogen acids and heat, ethers will almost always undergo complete acidic clevage in which 2 alkyl halides are generated. I have also included another mechanism involving a secondary and primary substrate. The primary substrate will undergo an Sn2 reaction due to the lack of steric hindrance; however, it is a little less clear with the secondary substrate. Will it undergo Sn1 since Sn2 would be more sterically hindered or will it undergo Sn2 since the carbocation intermediate in Sn1 is secondary and thus less stable? Since we are using an aqueous halogen acid in this case, it will most likely undergo Sn1 since it is in the presence of a polar protic solvent, which will thus stabilize the carbocation intermediate in the process of forming.