@Hippalectryon (cc @PhMgBr ) Yeah that's one of the potential canon. ones. Even though the question itself is not that awesome written... but if I find the time I would do it ;)
> Alcohols are amphoteric; they can act both as acid or base. The lone pair of electrons on oxygen atom makes the –OH group weakly basic.
But there are two lone pairs, aren't there?
> Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion +OH2 (The pKa value of a tertiary protonated alcohol can go as low as -3.8).
@CowperKettle Actually, R-OH2(+) can release one H+, but that just brings us back to the original R-OH. So there is an equilibrium of R-OH + H+ <=> R-OH2(+). Only R-OH2(+) can release H2O, which is a nice stable molecule.
@CowperKettle What reaction of propene with H2SO4 do you expect?
> It is much easier to form the secondary carbocation because it is more energetically stable. The activation energy for the reaction will be less, and so most of the reaction happens via that mechanism.
@CowperKettle about that picture you posted before (about an alcohol reacting with an acid)... Why doesnt a H+ migrate from carbon-2 after H2O leaves so that a more stable alkene can be formed?
Found it. Markovnikov's reaction goes that way because in the secondary ion, the electron-pushing effect from alkyl groups neutralizes the positive charge in the middle.
His name derives from the male name "mark", and "ov" is the suffix
@AdityaDev "Morkovnick" was a nickname for a guy who grew carrots. That was one of his ancient ancestors. That ancestor's kids attached the "ov" suffix and created the surname
"ov" means "belonging to" (belonging to his genus)
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@roscoe_casita Different reactions are interesting for different reasons. In the explosion of nitroglycerine, the amount of gas released per mole and the heat released are intriguing.
Also sometimes some bizarre things happen in compounds that make them interesting, like that bent benzene Manish had a question about.
(It wasn't bent anyway, but that was a cool complex)
@roscoe_casita Reactions are at times equilibriums. That means the reactants wouldn't all react to give the products, but some will remain. A way to quantify this is the equilibrium constant, K_eq.
It's a measure of how much the reactants react to give products.
~ but my chemistry is pitiful. I'm seeking to correct this. The computational side is basically complete if I can demonstrate some 'simple' reaction information
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