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If we react phenol with $\ce{C2H5I}$ in $\ce{C2H5O- Na+}$ (excess), $\ce{C2H5OH}$ (anhydrous) solutions .

In this reaction why is diethyl ether not formed and the actual product is phenyl ethyl ether? enter image description here

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You are adding excess sodium ethanolate so there is the possibility of deprotonation of phenol: $$\ce{C6H5OH + C2H5O- <=> C6H5O- + C2H5OH}$$

If we look at the $\text{p}K_\text{a}$ of ethanol and phenol we have 16 and 9.95 respectively and so the equilibrium above will strongly favour the right hand side. Therefore, the phenoxide ion will dominate as a nucleophile in the mixture, and the major product will be ethyl phenyl ether.

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    $\begingroup$ Yes, that's what I thought too, until I saw that the EtO- is apparently "in excess"... Ethoxide is definitely more nucleophilic than phenoxide and if both are present I would not be surprised if lots of Et2O was formed. $\endgroup$ – orthocresol Jun 4 '16 at 19:28
  • $\begingroup$ @orthocresol Hmm. You are probably right. $\endgroup$ – bon Jun 4 '16 at 19:33
  • $\begingroup$ So, you mean to say that phenoxide is not a better nucleophile than ethoxide (on the basis of resonating structures), but only happens to be the dominant nucleophile in this case just because it is present in more amount than ethoxide? $\endgroup$ – Gaurang Tandon Feb 2 '18 at 14:46
  • $\begingroup$ @GaurangTandon Yes that is what I am saying. $\endgroup$ – bon Feb 2 '18 at 15:17

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