The reaction is 5-bromo-butanol with Sodium Hydroxide (NaOH) would give you a cyclic ether. I understand the mechanism but I would like an explanation why the base (OH) picks up the H on the alcohol (OH) instead of a back side attack on the carbon with the bromine. Would appreciate the help, thanks!

You will actuallyget two competing reactions here; the one deprotonation of the alcohol followed by cyclisation and the secondwill be a S_N2 substitution of the bromide with hydroxide. This sounds like a typical early degree level question because they often just pick a reaction to make a point and not worry about the side reactions.

 

In practise, I would actually guess that the substitution would actually go much fast; hydroxide isn't really strong enough to completely deprotonate an alcohol. You usually use something much more basic (e.g. sodium hydride (NaH) or potassium t-butoxide (KO-tBu)). Both of these bases are actually non-nucleophilic so you would get minor susbtitution.

 

Also, there is no such compound as 5-bromobutanol....the number refers to which carbon atom the bromoide is attatched and butanyl means there are only four carbons. I assume you mean either 5-bromopentanol or 4-bromobutanol.

yeah was an error with the butanol, was typing this at 12am. anyways thanks for the input.