Bromination of but-3-en-1-ol in basic medium and Baldwin's rules


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> [![enter image description here](https://i.stack.imgur.com/5UyEx.jpg)](https://i.stack.imgur.com/5UyEx.jpg) 
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After formation of the bromocarbocation the alkoxide anion attacks the bromocarbocation which leads to ring formation and can proceed through two pathways [![enter image description here](https://i.stack.imgur.com/z3ISHm.jpg)](https://i.stack.imgur.com/z3ISHm.jpg)


5 endo tet cyclization is disfavored but 4 exo tet cyclization is favored according to Baldwin's rules; however, the answer given in my text book is the 5 membered ring.




Your reasoning is totally correct according to Baldwin's Rule - the 4-exo-product would be more stable than the 5-endo-product ( see : [This](http://www.reading.ac.uk/acadepts/sd/pharmacy/public_html/staff/andre/BaldwinsRules-RAJ.pdf) for more details).


But your method may be wrong in the given question the product A was first formed and then a ring closure step was performed using $\ce{KOH}$ as shown below :
[![enter image description here](https://i.stack.imgur.com/A9O4i.jpg)](https://i.stack.imgur.com/A9O4i.jpg)
[![enter image description here](https://i.stack.imgur.com/4SoOy.jpg)](https://i.stack.imgur.com/4SoOy.jpg)


here the 5-exo-trig ring closure is favoured according to Baldwin's rule .


Let me also address something that may be a doubt ( It was mine ) i.e. why doesn't the oxygen atom attack the bromonium Ion in the first step ? this is because even if it will attack you will also form a acid $\ce{HBr}$ which will cause the cleavage in the cyclic ester we just formed returning us to the original reagent hence it doesn't attack.


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