Suppose 1,5-pentanediol were exposed to something like hydroiodic acid. Would the alkoxonium end of the diol be exposed to a nucleophilic attack by the remaining hydroxide at the other end for a ring formation? Or would the newly-free iodide ion be more competitive for substitution? Thanks!

edit: It was pointed out to me "newly-free iodide" was quite confusing. I meant it as though HI dissociated, the Hydrogen ion attacked one end of the diol, and the iodide from the HI being "newly-free". Now that one end of the diol has a good leaving group (H2O), is it more susceptible to be attacked by the iodide, or by the other end (OH) of the diol.

  • $\begingroup$ Unclear as to the meaning of "newly-free iodide". Did you mean 5-iod0pentane-1-ol? Depending on conditions, the cyclic ether (tetrahydropyran) would be susceptible to HI cleavage with formation of 1,5-diiodopentane. $\endgroup$ – user55119 Dec 17 '17 at 19:21
  • $\begingroup$ Thank you so much for the help/feedback! I did my best to clarify in the edit. $\endgroup$ – Bob Dec 18 '17 at 2:26
  • $\begingroup$ Basically, you’re asking whether the product will be 1,5-diiodopentane or tetrahydropyrane (oxane), aren’t you? $\endgroup$ – Jan Dec 18 '17 at 2:41

Bob: Here is the likely process. The alcohol 1 is protonated to provide intermediate 2 which undergoes SN2 displacement. The closure of 3 to tetrahydropyran 6 is unlikely under the reaction conditions. Rather a second displacement via 4 would lead to diiodide 5. If the tetrahydropyran did form, ethers are cleaved by HI. The conversion of 3 to 6 is normally accomplished in the presence of base (NaH/THF).

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