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I am trying to do a reaction between acetyl chloride $\ce{CH3COCl}$ with diethyl ether. I know that upon using anhydrous $\ce{ZnCl2}$ and heat, I'll get an ester and an alkyl chloride, but I'm not able to guess the mechanism.

Does $\ce{ZnCl2}$ break the $\ce{C-O}$ bond in ether as it would've done to alcohol? Please help me with the mechanism.

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    $\begingroup$ Note that on Chemistry SE site, it is adviced not to use MathJax in titles.When it is used in the body, use \ce{} structure for chemical formulas and equations. $\endgroup$ – Poutnik May 27 '19 at 7:15
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    $\begingroup$ Yes I believe the ZnCl2 is used to cleave the ether first to give the alkyl chloride and an alkoxide. This alkoxide then goes on to attack the acyl chloride, giving the ester. $\endgroup$ – Tan Yong Boon May 27 '19 at 7:45
  • $\begingroup$ I agree, the Zn2+ binds to the oxygen polarising the C-O bond enough for nucleophilic attack by Cl- at C. $\endgroup$ – Waylander May 27 '19 at 9:19
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I think that it reminds Fischer esterification at the beginning. Zinc dichloride acts as a Lewis acid by coordination at the carbonyl oxygen in AcCl. It activates C=O bond because of larger contribution of carbocation form. And this cation undergoes coordination of Et2O with formation of zwitterionic tetrahedral intermediate. This particle eliminates our catalyst and cloride-anion. The product of this step is acetyl diethyl oxonium. Cloride-anion is quite nucleophilic, but we have an excellent leaving group without charge (AcOEt), so Sn2-reaction occurs and lead to the final productsPossible mechanism of this reaction

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