What results from the reaction of 5-bromopentan-1-ol and NaOH?

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


1. pentane-(1,5)-diol


[![pentane-(1,5)-diol](https://i.stack.imgur.com/mIfpJ.gif)](https://i.stack.imgur.com/mIfpJ.gif)


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


2. (1,5)-epoxypentane


[![(1,5)-epoxypentane](https://i.stack.imgur.com/9Qquom.png)](https://i.stack.imgur.com/9Qquom.png)


~~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**


3. 4-penten-1-ol


[![4-penten-1-ol](https://i.stack.imgur.com/xkLbNm.gif)](https://i.stack.imgur.com/xkLbNm.gif)


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




The three reactions are:


An $\mathrm{S\_N2}$:


![](https://i.stack.imgur.com/yu4tx.png)


Deprotonation followed by an intramolecular Williamson ether synthesis ($\mathrm{S\_N2}$):


![](https://i.stack.imgur.com/0UqiE.png)


An $\mathrm{E2}$:


![](https://i.stack.imgur.com/tLIXk.png)


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