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In our chemistry class, we noted three molecules that result from the reaction:

  1. pentane-(1,5)-diol

pentane-(1,5)-diol

This one is simple to understand, simply a nucleophilic substitution between $\ce{Br^-}$ and $\ce{OH^-}$.

  1. (1,5)-epoxypentane

(1,5)-epoxypentane

This is the main result, resulting from the intramolecular esterification of the two $\ce{OH^-}$ groups of the previous molecule. This is wrong, please see the explanation by ringo

  1. 4-penten-1-ol

4-penten-1-ol

This one, I don't understand. How is this a possible result? Could anyone help me understand? Thank you for your time.

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The three reactions are:

An $\mathrm{S_N2}$:

Deprotonation followed by an intramolecular Williamson ether synthesis ($\mathrm{S_N2}$):

An $\mathrm{E2}$:

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  • $\begingroup$ Thank you for the illustration, that made it much easier to understand. Would you mind me asking what software do you use to draw the equations? $\endgroup$ – Glycerius Feb 15 '17 at 20:45
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    $\begingroup$ ChemDraw. If you're at a university, they likely have a subscription and can get you a copy for free. $\endgroup$ – ringo Feb 15 '17 at 20:49

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