# Why is the ether linkage breaking here?

I have been asked to find the major product of the following reaction:

This is my thought process: the lone pairs of the oxygen atom connected to $$\ce{-Et}$$ are less delocalised than the other oxygen atom. So, that will abstract the $$\ce{H^+}$$ from $$\ce{HI}$$ leaving behind $$\ce{I^-}$$ and producing a positive charge on the oxygen atom. Now the oxygen atom will take up the bond pair electrons and $$\ce{I^-}$$ will form a bond in $$\ce{S_{N}2}$$ mechanism, producing $$\ce{I-Ph-O-Ph}$$ and $$\ce{EtOH}$$ as products.

But, the answer in the book states that:

The ether linkage will break as it does not have a double bond character due to the absence of resonance.

Why is the ether linkage breaking here? What does it mean by "it does not have a double bond character"?

• Having double bond character makes a bond stronger so the O-CH2R bond is weaker than the Ar-O bond. There are also steric factors to consider - Iodide is a bulky nucleophile and the approach to the O-CH2R is less hindered. Apr 19 '21 at 7:49
• $\mathrm{S_N2}$ reaction does not happen on $\mathrm{sp^2}$ centres, that is why your proposed reaction scheme would not work. See the accepted answer for this question: chemistry.stackexchange.com/questions/14482/… Apr 19 '21 at 9:05

I have used the image you used $$2nd$$, see carefully the markings.
The $$C-O$$ bond with benzene ring is stronger than the $$C-O$$ bond of ethyl, as there is resonance with benzene making it partially a double bond. So, that $$C-O$$ is difficult to break compared to the $$C-O$$ bond in ethyl branch.
• Yes C-O bond has partial double bond character, but I don't believe that is the main reason. Even if you have a leaving group like $\ce{I-}$ you would be hard pressed to achieve the reaction, because $\mathrm{S_N2}$ doesn't work with $\mathrm{sp^2}$ carbon centres. On aromatic rings, only $\mathrm{S_NAr}$ can happen. Apr 19 '21 at 15:04