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I have seen the following reaction in my textbook: Pent-3-en-2-ol + HBr

This was quite easy to understand, though I wonder why 2-bromopent-3-ene is not a possible result:

2-bromopent-3-ene

Isn't this how $\ce{-OH^}$ groups and hydrogen halides interact? Any clarification would be appreciated.

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It is very rare for a reaction to have no side products, and this reaction is certainly not exceptional in that regard. I'm actually a bit surprised your book chose this reaction to illustrate the regiochemistry of $\ce{HBr}$ addition to an alkene, as this would compete with the formation of an allylic cation, which are generally more stable than alkyl carbocations. Being more stable, these cations will form more quickly:

enter image description here

giving the product you correctly assumed would also be formed. You probably wouldn't obtain this though, as it would continue to react, forming a dibromoalkane (as would the products listed by your book):

enter image description here

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Although what ringo said is true, it can be explained quite simply.

Br- is more stable than OH- owing to it's larger size(and deceased electron density) and thus is a better leaving group than OH-

Therefore, OH- has a tendency to stay which results in the reaction you have shown.

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