I understand the reason for why Markovnikov product is favoured in the addition of acidic reagents (HX) where the electrophilic addition reaction is brought out by the attack of a proton on the double bond. This enables the most stable carbocation (often the Markovnikov product) to form which gets attacked by the nucleophilic part of the reagent (X-).

But I've not been able to find any explanation for why Markovnikov products are favoured in the electrophilic addition of non acidic (or weakly acidic) reagents in non polar solvents.

Take the example of addition of Hypochlorous acid to propene in carbon tertrachlroide ($\ce{CCl4}$). enter image description here

In such reactions, a non classical carbocation is formed preventing any rearrangement to happen. Now the confusing part. Why does the water attack the middle carbon only? Moreover, my textbook (Chemistry of Organic Compounds - Chittaranjan Bhakta) suggests that the attack of the nucleophile (here water) on the cyclic chloronium ion is a SN2 type displacement, which means that the nucleophile should attack the primary carbon.

P.S - It has been notoriously difficult to find any explanation as to why this happens. I've searched all sites (even reputable ones such as chemistry libre texts) and all they can mention is only Markovnikov product forms and no one in the world knows why. This has left me thoroughly disgusted. Also I know that the diagram is a bit misleading but it was the best I could find. I'm new to SE so I don't know how to make my own diagrams yet.

  • $\begingroup$ The non classical carbocation is more stable on the middle one $\endgroup$
    – Baksish
    Commented Apr 21 at 5:04
  • $\begingroup$ @Baksish What do you mean the non classical carbocation is stable on the middle one? There is only one non classical carbocation possible in these types of reactions, that is where the double bond was. $\endgroup$
    – Nothing
    Commented Apr 21 at 9:16
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    $\begingroup$ It's not non classical, just a cyclic chloronium. It's not symmetric, so there has to be preference which side gets broken. $\endgroup$
    – Mithoron
    Commented Apr 21 at 16:11
  • $\begingroup$ @ Mithron Alright it is not non classical, but there is still a question as to why the Markovnikov product is formed. $\endgroup$
    – Nothing
    Commented Apr 21 at 17:18
  • $\begingroup$ how can the attack of nucleophile be SN2 if we have not taken a polar aprotic solvent? $\endgroup$
    – Baksish
    Commented Apr 23 at 12:43

3 Answers 3


The reason why water adds to the end that is best able to stabilize a carbocation and not the less sterically hindered end can be understood by looking at a similar heterocyclic 3 membered ring (just like the chloronium intermediate), the epoxide.

Depending on the reaction conditions the epoxide can be opened from either the more substituted end (acidic conditions, more $S_N1$ character (the leaving group starts to leave before the nucleophile comes in), also called "loose $S_N2$") or the less substituted end (basic conditions, more $S_N2$ character).* Regioselective opening of an epoxide

The mechanism for the opening of the epoxide in acidic conditions is as follows:Mechanism of acidic epoxide opening

As you can see, in an asymmetric protonated epoxide, the 2 carbon-oxygen bonds aren't identical, the bond to the end that can stabilize the partial positive charge better is longer and thus the $\sigma^*$ orbital is lower in energy, meaning it's easier for nucleophiles to donate into and break the bond.

Now, in the case of the chloronium ion, the Transition State (TS) resembles more the "loose $S_N2$" TS of the protonated epoxide, rather than a $S_N2$ TS, because the chlorine atom is positively charged, so we should expect so see similar reactivity, nucleophilic attack of a water molecule at the more substituted end. Mechanism of HOCl addition to an alkene

Also here is a cool animation on ChemTube3D, showing addition of Methanol to a bromonium ion (also similar to the chloronium ion).

*This only applies to small (usually "hard") nucleophiles, like oxygen and nitrogen based nucleophiles, for larger atoms like bromine the result is usually a mixture of the possible epoxide openings, favoring the less substituted opening. HBr addition to an epoxide


Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc. page 438


During the reaction, an intermediate product called a carbocation is formed. There are two types of carbocations that can be formed - primary and secondary. The secondary carbocation "requires" less activation energy to obtain it and, therefore, forms faster than the primary carbocation. This means that the reaction favours the formation of secondary alcohol. enter image description here

  • $\begingroup$ No, you're mistaken. An open carbocation (as you indicated) does not form in this reaction. The reaction intermediate is the cyclic chloronium ion whose stability plays no factor in the nature of the product formed (1 degree or 2 degree alcohol). Please refer to the diagram attached in the question. $\endgroup$
    – Nothing
    Commented Apr 23 at 16:07
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    $\begingroup$ I see what you mean, but I think you overcomplicate things, and sometimes, it's really good practice to simplify things when you do science. Anyway, please look at the page (link below) that provides the background to the answer to your question I have provided. See section Halohydrin Formation – with water and alcohols masterorganicchemistry.com/2013/03/15/…. $\endgroup$ Commented Apr 24 at 17:37
  • $\begingroup$ Thank you Samuil. This was the answer I was looking for. However in the paper it says that the cyclic halonium ion is a resonance hybrid of the two possible carbocations. I think this is a slightly wrong explanation. I think the more substituted C-X bond will be longer only because the electron density is more at that carbon (due to +I +H) making it less electronegative hence making its bond longer which ultimately results in its breaking. I was going to answer my own question after reading this but I remembered it's on a bounty. If you will answer it, I will reward you with it. $\endgroup$
    – Nothing
    Commented Apr 25 at 5:48

The preference for Markovnikov addition in non-polar solvents, especially in cases like the addition of hypochlorous acid to propene in carbon tetrachloride, involves several factors that are not always explicitly discussed in general chemistry texts but are crucial for understanding the reaction mechanism.

Steric Hindrance: Even though the formation of a non-classical carbocation prevents rearrangement, sterically hindered attack of the nucleophile (in this case, water) is still a factor. The middle carbon in the cyclic chloronium ion might be more accessible due to lower steric hindrance compared to the terminal carbons.

Solvent Effects: Non-polar solvents like carbon tetrachloride provide a medium where polar interactions are minimal. In such solvents, the reactivity might be governed more by the intrinsic electronic and steric factors than by solvent effects.

Nucleophilic Attack: Regarding the attack of water on the cyclic chloronium ion, even though it's described as an SN2-like displacement, the actual mechanism might involve some degree of carbocation-like character, especially considering the resonance stabilization in the cyclic chloronium ion. This could make the attack on the middle carbon more favorable due to electronic factors.

Thermodynamic vs. Kinetic Control: Even though the attack on the middle carbon might seem less intuitive from a simple electronic perspective, it's essential to consider whether the reaction is under kinetic or thermodynamic control. Kinetic control often leads to the formation of the less substituted product, while thermodynamic control favors the more stable product. In some cases, the reaction conditions might favor kinetic control, leading to the observed product distribution.

In essence, while the exact mechanism might not be explicitly discussed in general chemistry texts, the preference for Markovnikov addition in non-polar solvents involves a combination of steric, electronic, and kinetic factors, influenced by the nature of the reactants, solvent, and reaction conditions. This complexity might contribute to the difficulty in finding a straightforward explanation in introductory resources.

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    $\begingroup$ Please keep in mind that AI generated answers are frowned upon on this site. $\endgroup$ Commented Apr 27 at 5:54

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