CycloEicosanone Posted July 1, 2008 Posted July 1, 2008 "When propene is treated with HCl in ethanol, one of the products of the reaction is ethyl isopropyl ether. Write a reasonable mechanism showing how this product forms. For each step of the mechanism, use curved arrows to show the direction that an electron pair moves when a bond is made or broken." Here's the reaction that I sketched out. It looks ok to me, but if someone could double check it for me I'd sincerely appreciate it! Just wanna make sure I got each subsequent transfer of electrons/breaking of bonds correct. Thanks!
ecoli Posted July 1, 2008 Posted July 1, 2008 I'm not sure, but you could be wrong, because I'm not sure the chloride ion would pull off the proton from the alcohol (it's pretty stable as is). It's possible the alcohol attacks the carbocation first, and the proton gets pulled off after the C - O bond has already formed.
Ozone Posted July 2, 2008 Posted July 2, 2008 Remember to push the electrons. Following this convention, the proton will not "go" over to the chloride anion. Then consider which would be the stronger base. In this case, the alkoxide...(it will not likely exist under these conditions). The hydroxyl species is present in enormous excess; it is feasible that it will add to any carbocation that is formed, which will undoubtedly be the 2° moiety (look up the order of stability of carbocations). The Cl- will serve to regenerate your catalyst. (try to get your HCl back). Cheers, O3
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