Question

Explain why methyl propyl ether forms both methyl iodide and propyl iodide when it is heated with excess HI.

Answer 1

Answer:

Methyl propyl ether protonates at oxygen, and this makes the oxygen a good leaving group for  $SN2$ attack by iodide ion which is a good nucleophile.

We are aware that  $SN2$ prefers 1 degree carbocation and here both methyl , propyl group is 1 degree.  

So we have two possible carbon atoms to react with iodide which are methyl group and  propyl group. They will react with iodide at similar rates, and so there is no great preference of one over the other.

As $HI$  is present in excess the hydrogen of $HI$  protonates the oxygen which leads to cleavage  that can happen on both sides either methyl or propyl side leading to different products.

The reaction proceeds in the manner

$CH3-O-CH2CH2CH3$  + $HI$   → CH3I + CH3CH2CH2OH

OR

$CH3-O-CH2CH2CH3$ + $HI$   → CH3OH + CH3CH2CH2I