I have been asked to find the major product of the following reaction: enter image description here

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. enter image description here

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. enter image description here

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

  • 3
    $\begingroup$ 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. $\endgroup$
    – Waylander
    Apr 19, 2021 at 7:49
  • 1
    $\begingroup$ $\mathrm{S_N2}$ 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/… $\endgroup$
    – S R Maiti
    Apr 19, 2021 at 9:05

1 Answer 1


enter image description here

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.

  • $\begingroup$ 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. $\endgroup$
    – S R Maiti
    Apr 19, 2021 at 15:04
  • $\begingroup$ Yes, SN2 do not occur at sp2 Carbon because vinylic substitution is hindered due to π electrons but with benzene there is more, which is lone pair donation to benzene ring from Oxygen, that is why i focused on double bond character $\endgroup$ Apr 19, 2021 at 15:21

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