HO - CH_2 - CH_2 - CH_2 - CH_2 - CH = CH_2 xr | vedclass

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(A) The reaction of an alkene with $Br_2$ typically leads to the formation of a vicinal dibromide. However,in this molecule,there is an internal nucle

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(A) The reaction of an alkene with $Br_2$ typically leads to the formation of a vicinal dibromide. However,in this molecule,there is an internal nucleophile (the hydroxyl group,$-OH$) present at the end of the chain.When $Br_2$ reacts with the double bond,a cyclic bromonium ion intermediate is formed.The oxygen atom of the $-OH$ group then performs an intramolecular nucleophilic attack on the more substituted carbon of the bromonium ion (or the carbon that best stabilizes the developing positive charge),leading to the formation of a cyclic ether (a ring closure reaction).Specifically,for $HO-(CH_2)_4-CH=CH_2$,the oxygen attacks the carbon at the $2$-position of the original double bond,resulting in a $6$-membered ring containing an oxygen atom (a tetrahydropyran derivative) with a $-CH_2Br$ group attached.This is an example of halocyclization,where the internal nucleophile traps the bromonium ion to form a cyclic ether.

$HO - CH_2 - CH_2 - CH_2 - CH_2 - CH = CH_2 \xrightarrow{Br_2} \text{Product ?}$

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