[sun725]

I have the following reaction that I have to show the reaction mechanism of. I don't know who to handle this type of problem and where to start.

Hex-5-en-ol is treated with an aqueous solution of bromine, and the compound "E" is formed.

https://imgur.com/a/Mpd5CiA

I hope for some guidance here.

Thanks in advance.

[Babcock_Hall]

It is a forum rule (see red link above) that you must show your attempt or provide your thoughts before we can help you.  What reactions have you learned that involve Br₂?

[sun725]

I believe it's halohydrins formation, where the alkene reacts in this case with Br₂ in the presence of water. I just don't get what happens since a cyclization occurs.

[Babcock_Hall]

Besides water, is there another nucleophile present?  It would probably help you if you drew the first intermediate in this reaction.

[sun725]

Yes, the double bond/alkene on the molecule acts as the nucleophile as well. I just don't know how to start regarding the mechanism. Reaction with the alkene with Br₂, should yield a bromonium ion intermediate of some sort, right?

But we are talking about halohydrins from alkenes?

[sun725]

Also I wonder if the lone pair on the water or the molecule itself that would attack the formed intermediate?

[Babcock_Hall]

Once the bromonium ion forms, there are actually three possible nucleophiles.  One is bromide ion, and a second is water.  It's the third nucleophile that is the key to this problem.  Draw this nucleophile attacking the bromonium ion, and that is most of the rest of the problem.

[sun725]

I tried and this is what I came up with. Please give me any advice and corrections.

https://imgur.com/a/ycytuJb

[Babcock_Hall]

I think that one of your arrows is not drawn correctly in the step in which the bromonium ion is the reactant.  One problem with this arrow is that the way that it is drawn suggests that the bromine and the oxygen will end up on the same carbon. With respect to the last step, there is one minor issue left to discuss.  Some textbooks show a water molecule acting as a base to remove the proton.  Some instructors may insist on showing the water picking up the proton, while others may not.

[sun725]

I am bit confused. Is it the arrow that points from the lone pair on the OH group towards the carbon on the bromonium ion intermediate? Also is the first step correct where the alkene attacks the Br?

Many thanks.

[Babcock_Hall]

Some authors draw the first step exactly as you have drawn it; others use an additional arrow from bromine atom that stays to one of the carbon atoms.  I slightly prefer the latter way because it allows one to see the formation of both C-Br bonds more easily IMO.  With respect to the second step, it's the other arrow.  Another way to look at it is that what you drew would make one carbon deficient in electrons and one carbon would have too many electrons.

[PKa]

Also instead of showing loss of H⁺, think about what is the other ion floating in the reaction? Clue: What happens to Br^-

[Babcock_Hall]

With respect to the loss of the proton, there is also water to consider.