What happens when prop-2-en-1-ol (allyl alcohol) reacts with concentrated hydrogen bromide?

My Idea: I thought that in the first step protonation of the hydroxyl group will take place followed by formation of a carbocation. Then the attack of bromide ion from hydrogen bromide takes place and that would give me 3-bromoprop-1-ene. Also, I thought since the double bond still remains Markovnikov addition at the double bond might also take place.

But the reaction results into something different Here is a picture of the above reaction:

enter image description here

The answer has been given as option C. Could someone point out where I might be going wrong or how should the reaction actually proceed?

  • $\begingroup$ Please don't use MathJax in titles. In particular you don't need it to type the name of a chemical. $\endgroup$ Nov 2, 2016 at 11:27
  • $\begingroup$ NOOO! Stop proposing primary (even allylic carboncations)! This is a normal hydrobromination of a alkene. It should follow anti-Markovnikov rules. $\endgroup$
    – Zhe
    Nov 2, 2016 at 15:28
  • 2
    $\begingroup$ @Zhe Hydrobromination of alkenes follow Markovnikov rules. Furthermore, $\ce{HBr}$ is a strong acid and perfectly capable of protonating hydroxy groups. Whether the mechanism happens to me $\mathrm{S_N2}$, $\mathrm{S_N2'}$ or $\mathrm{S_N1}$ is of little to no importance. $\endgroup$
    – Jan
    Nov 2, 2016 at 16:04
  • $\begingroup$ Ugh. Undo. I learned the history wrong. Markovnikov addition it is... en.wikipedia.org/wiki/… Thanks, @Jan. $\endgroup$
    – Zhe
    Nov 2, 2016 at 16:58
  • $\begingroup$ Organic Syntheses, Coll. Vol. I, pg. 27, Gilman & Blatt, ed., describes the conversion of allyl alcohol to allyl bromide with 48% HBr (from Br2, SO2 and water) and H2SO4 on a 4 mole scale with a 92-96% yield. $\endgroup$
    – user55119
    Dec 11, 2018 at 19:30

2 Answers 2


Rather than trying to rationalise what the question wants to know, I ran a quick SciFinder search. One of the first results was the following reaction, reported for the perdeuterated allyl alcohol by Thiem, Mohn and Heesing in 1985:[1]

reaction of perdeuterated allyl alcohol with concentrated HBr
Scheme 1: Reaction of perdeuterated allyl alcohol with $\ce{HBr}$. All implicit hydrogens in this scheme are deuterium atoms.[1]

Thiem, Mohn and Heesing do not go into details regarding the mechanism, but as I commented it really doesn’t matter whether the reaction follows a $\mathrm{S_N1}$, $\mathrm{S_N2}$ or $\mathrm{S_N2'}$ mechanism, since the final product will always be the same. In any case, the first step will be protonation of the hydroxy group to turn it into a better leaving group.

Another one of the first results was the following, reported by Karki and Magolan in 2015:[2] reaction of allyl alcohol with HBr in chloroform/DMSO mixtures
Scheme 2: Reaction of normal allylic alcohol with $48~\%\ \ce{HBr}$ in DMSO/chloroform.[2]

Their proposed mechanism includes the in situ creation of $\ce{Me2S^+-Br}$ which will then react with the double bond giving a bromonium ion; this will then be captured by a second bromide to give 2,3-dibromopropan-1-ol.

I was unable to find single-step reactions of allyl alcohol with hydrogen bromide leading to either 2-bromopropan-1-ol (which would imply a single Markovnikov addition of $\ce{HBr}$ onto the double bond), 1,2-dibromopropane (which would imply Markovnikov hydrobromination of the double bond of allyl bromide) or 1,3-dibromoprop-1-ene (where I have no clue how exactly that would be generated from allyl alcohol in the first place). Funnily enough, you state that this last product, C according to your scheme be the correct answer. I challenge the exam.


[1]: J. Thiem, H. Mohn, A. Heesing, Synthesis 1985, 775. DOI: 10.1055/s-1985-31344.

[2]: M. Karki, J. Magolan, J. Org. Chem. 2015, 80, 3701. DOI: 10.1021/acs.joc.5b00211.

  • $\begingroup$ Can you form 1,2,3-tribromopropane via a combination of the two reactions and then eliminate one of the bromides? Not sure why it wouldn't eliminate anything else though... $\endgroup$
    – Zhe
    Nov 2, 2016 at 17:10
  • $\begingroup$ @Zhe The elimination would require basic coditions to the best of my knowledge; they aren’t available in concentrated $\ce{HBr}$ … $\endgroup$
    – Jan
    Nov 2, 2016 at 17:12
  • $\begingroup$ Was thinking more $\mathrm{E}1$ type mechanism and that would probably require refluxing solvent like in example 1... $\endgroup$
    – Zhe
    Nov 2, 2016 at 18:11
  • $\begingroup$ @Zhe $\mathrm{E1}$ still requires a base of some sort; something you just don’t have in refluxing $\ce{HBr}$ … $\endgroup$
    – Jan
    Nov 2, 2016 at 18:12

I found a patent dating back to the 1930s in which the dichloro homologue of C was produced in almost quantitative yield via Hydrochlorination. Long reaction time (100 hours) at 50-70C. When it was discovered, it was a surprise (they say) but I presume people didn't leave hot-plate stirrers going for days....

With this in mind, I decided to find out if Hydroiodination would yield 1,3-diiodoprope-1-ene (C analogue). Will hydroiodination live up to the theory?

It turns out that it's a rather difficult compound to come by with only 1 synthesis offered and that was hard to find:

To a solution of (Z)-1-iodo-3-methanesulfonyloxy-1-propene (3.40 g, 13.0 mmol) in acetone (18.6 ml) was added sodium iodide (1.94 g. 13.0 mmol) at 0 OC. The mixtute was stirred at RT for 2h. diluted with ether, and filtered The filtrate was washed with 10% aqueous Na2S203 and brine, dried (NazS04). and concentrated. The product was purified by silica gel chromatography (hexane) to give (3.19 g, 83%) as a dark red oil.

If I have missed a simple synthesis then I can only apologize but the stuff doesn't have a CAS. Searching for (E/Z ignored) '1,3-diiodoprop-1-ene' yields almost nothing so try '1,3-diidopropene' which if you think about it, describes it! IUPAC can be no help.


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