Heating of Ethers with HI and Cleavage of Ethers

**Question:**



> 
> Give the major product formed by the heating of the following ether with HI.
> 
> 
> 
> ```
>         CH3
>         |
> CH3-CH2-CH-CH2-O-CH2-CH3
> 
> ```
> 
> 


**My Attempt:**


According to me, the product should be this due to delocalization of positive charge among the neighboring carbon atoms.



> 
> 
> ```
>         CH3
>         |    
> CH3-CH2-CH-CH2-I + CH3-CH2-OH
> 
> ```
> 
> 


**Answer:**


I checked the answer in my book and it is completely different:



> 
> 
> ```
>         CH3
>         |
> CH3-CH2-CH-CH2-OH + CH3-CH2-I
> 
> ```
> 
> 


Now, I think they may have considered **steric hindrance** to be the deciding factor of site of attack of OH but will not **delocalization** be dominant over steric effect?


Also, what will be the effect of hyperconjugation on the reaction?




Your ether has $\ce{CH2}$ groups at both sides of the oxygen atom. 


I would exclude a $S\_N1$ reaction here! Instead, cleavage of the ether proceeds via $S\_N2$. The oxonium ion resulting from addition of a proton is attacked by iodide at the least hindered carbon atom. Iodide is quite big ;-)


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> 
> According to me, The product should be this due to delocalisation of positive charge among the neighbouring Carbon atoms.
> 
> 
> 


Even if you consider an $S\_N1$ reaction pathway for this reaction, the positive charge could be more stable on the $\ce{-CH2CH3}$ branch as it has three hyperconjugation structures stabilizing the positive charge against one in the other branch. And gives the same products.


Moreover, as Klaus Warzecha said, steric hindrance comes into the play while considering $S\_N2$ pathway which looks a more plausible pathway due to the relatively unstable primary carbocations that form in $S\_N1$


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