[xshadow]

Hi!!!

I have some doubt reading the electrophilic addition of alkenes on my textbook...

For example my book says that:

CH₃=CHCH₃  + Cl₂ | inert  solvent: CH₂Cl₂|  ----->  CH₃C-ClH-C-ClH-CH₃
(addition of two Cl)
                                                                                     
BUT if I have  as solvent H₂O , I have the addition of a single Cl and of  "OH" ,that works as  nucleophilic, (that comes from the solvent H₂O, )  + HCl.

So the product has as "nucleophilic" the H₂O and not the Cl^-  because water is the solvent, so is "more concentrated" than the solute ( so more odds to interact with the substrate)


Now does all this work also with Halogenidric acids??

For example:
I know that
H₂C=CH₂ + HCl  ---->  H₃C-CH₂Cl  (so  I suppose that in THIS reaction H₂O is not the solvent !!?! ) 

But if I have also the solvent H₂O (so I would have 2 nucleophilic,  Cl^- and H₂O ) what is the product??

H₂C=CH₂ + HCl + H₂O (solvent) ---->

H₂O is again more concentrated than Cl^- (a solute) so H₂O should be the nucleophilic?!!?
Thanks.

[Babcock_Hall]

Do the two kinds of reactions proceed via identical mechanisms, or are the mechanisms different?

[OCSaviour]

The addition of Cl2 to alkene takes place through the formation of a three-membered ring, kind of like epoxide. Now, after the electrophile attacks, we are left with Cl- ion and H2O. Now, you have to think which one is a stronger nucleophile.

If we have HCl and H2O in the same solution, the hydrogen from HCl protonates H2O to form H3O+ and the Cl- ions are free in solution. So, H2O can't act as a nucleophile.

[xshadow]

The addition of Cl2 to alkene takes place through the formation of a three-membered ring, kind of like epoxide. Now, after the electrophile attacks, we are left with Cl- ion and H2O. Now, you have to think which one is a stronger nucleophile.

If we have HCl and H2O in the same solution, the hydrogen from HCl protonates H2O to form H3O+ and the Cl- ions are free in solution. So, H2O can't act as a nucleophile.

I don't understand your reasoning...because also with HCl and H₂O at the end I always have [H2O] >>> [H₃O⁺] ...H₂O always exist  in the solution and it's the moSt concentrated specie


For example reading this: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/alkene/wasseraddition.vlu.html
HCl + Alkenes + H2O should be a acid catalyzed reaction of hydration of alkanes

And according to this:
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/addene1.htm

"The importance of choosing an appropriate solvent for these addition reactions should now be clear. If the addition of HCl, HBr or HI is desired, water and alcohols should not be used. These strong acids will ionize in such solvents to give ROH2(+) and the nucleophilic oxygen of the solvent will compete with the halide anions in the final step, giving alcohol and ether products. By using inert solvents such as hexane, benzene and methylene chloride, these competing solvent additions are avoided. "

[xshadow]

Do the two kinds of reactions proceed via identical mechanisms, or are the mechanisms different?

The intermediate is different...

In HCl + alkene there's a carbocation intermediate
In Cl₂,for example, a three membered-ring

[OCSaviour]

The addition of Cl2 to alkene takes place through the formation of a three-membered ring, kind of like epoxide. Now, after the electrophile attacks, we are left with Cl- ion and H2O. Now, you have to think which one is a stronger nucleophile.

If we have HCl and H2O in the same solution, the hydrogen from HCl protonates H2O to form H3O+ and the Cl- ions are free in solution. So, H2O can't act as a nucleophile.

I don't understand your reasoning...because also with HCl and H₂O at the end I always have [H2O] >>> [H₃O⁺] ...H₂O always exist  in the solution and it's the more concentrated specie


For example reading this: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/alkene/wasseraddition.vlu.html
HCl + Alkenes + H2O should be a acid catalyzed reaction of hydration of alkanes

And according to this:
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/addene1.htm

"The importance of choosing an appropriate solvent for these addition reactions should now be clear. If the addition of HCl, HBr or HI is desired, water and alcohols should not be used. These strong acids will ionize in such solvents to give ROH2(+) and the nucleophilic oxygen of the solvent will compete with the halide anions in the final step, giving alcohol and ether products. By using inert solvents such as hexane, benzene and methylene chloride, these competing solvent additions are avoided. "

Hi, I am sorry, I misread your statement. You are correct. In the solution, we'll have two nucleophiles, Cl^- and H₂O. Since, H₂O's concentration is more, the carbocation is much more likely to collide with H₂O.

I am sorry, if my previous statement caused you any connfusion.

[xshadow]

The addition of Cl2 to alkene takes place through the formation of a three-membered ring, kind of like epoxide. Now, after the electrophile attacks, we are left with Cl- ion and H2O. Now, you have to think which one is a stronger nucleophile.

If we have HCl and H2O in the same solution, the hydrogen from HCl protonates H2O to form H3O+ and the Cl- ions are free in solution. So, H2O can't act as a nucleophile.

I don't understand your reasoning...because also with HCl and H₂O at the end I always have [H2O] >>> [H₃O⁺] ...H₂O always exist  in the solution and it's the more concentrated specie


For example reading this: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/alkene/wasseraddition.vlu.html
HCl + Alkenes + H2O should be a acid catalyzed reaction of hydration of alkanes

And according to this:
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/addene1.htm

"The importance of choosing an appropriate solvent for these addition reactions should now be clear. If the addition of HCl, HBr or HI is desired, water and alcohols should not be used. These strong acids will ionize in such solvents to give ROH2(+) and the nucleophilic oxygen of the solvent will compete with the halide anions in the final step, giving alcohol and ether products. By using inert solvents such as hexane, benzene and methylene chloride, these competing solvent additions are avoided. "

Hi, I am sorry, I misread your statement. You are correct. In the solution, we'll have two nucleophiles, Cl^- and H₂O. Since, H₂O's concentration is more, the carbocation is much more likely to collide with H₂O.

I am sorry, if my previous statement caused you any connfusion.

No problem

Thanks

[orgopete]

Hi, I am sorry, I misread your statement. You are correct. In the solution, we'll have two nucleophiles, Cl^- and H₂O. Since, H₂O's concentration is more, the carbocation is much more likely to collide with H₂O.

I'd be cautious with this conclusion. An alcohol is a lot more basic than a halogen. Even if an alcohol may form as an intermediate, it may be reversible and lead to formation of a halide. Obviously this would depend on the alkene. A tri- or tetrasubstituted alkene can form a tertiary carbocation and will more readily result in a tertiary halide. A primary alkene is much less reactive and is also more difficult to undergo an acid catalyzed electrophilic addition. I'd look carefully at examples, conditions, and yields.