0
$\begingroup$

enter image description here

Boron has an empty p-orbital and is highly electron deficient because of an incomplete octet.

So shouldn't this intermediate be more stable than $\ce{BR3}$ (whose electron deficiency is only somewhat fulfilled through hyperconjugation) since boron's electron deficiency is satisfied by the increased negative charge (although the majority of negative charge is still on oxygen due to its high electronegativity).

$\endgroup$
  • 1
    $\begingroup$ Sure, this thing feels somewhat more satisfied than BR3, but it has a suitable decomposition pathway, while BR3 has none. $\endgroup$ – Ivan Neretin Aug 8 '17 at 16:12
  • $\begingroup$ But why would it decompose into something less stable? Could it be because of the favorable entropy change? $\endgroup$ – xasthor Aug 8 '17 at 16:14
  • 1
    $\begingroup$ Who said "less stable"? The ultimate product is even more satisfied (and hence more stable). $\endgroup$ – Ivan Neretin Aug 8 '17 at 16:15
  • 1
    $\begingroup$ Why are you comparing a borate and a borane? Those don't even have the same number of atoms, let along the same number of electrons. $\endgroup$ – Zhe Aug 8 '17 at 16:39
3
$\begingroup$

Basically an intermediate form, that fulfills the octet rule, would be more stable than the trialkylborane, but only if we suppose that there are no other effects! And if there were no other ffects, it wouldn't be an intermediate form but an isolable compound.

In your reaction scheme you see that $\ce{OH-}$ will leave. You maybe know that $\ce{OH-}$ is - under basic conditions - a good leaving group. Moreover a peroxide is mostly pretty unstable, so it's happy to react and let $\ce{OH-}$ leave. The hydroboration–oxidation reaction gives you the tetrahydroxyborane-anion ($\ce{B(OH)4-}$) and three alcohols ($\ce{ROH}$) as final product. So you'll get an anion that fulfills the octet rule and is more stable than your reactant.


Edit: More details on the hydroxide as leaving group

The reaction finds place under strongly basic conditions. Usually you take $\ce{NaOH}$ to deprotonate the $\ce{H2O2}$ and form $\ce{OOH-}$, so in your solution there are plenty of $\ce{OH-}$ ions. The intermediate also splits off an hydroxide ion which is a good leaving group under this conditions. This means that the ion is stabilized in solution and is not less stable as the intermediate. Even if you can't really compare the stability of those two things because they are not the same... So let's only say, that the hydroxide ion is pretty stable in a basic solution.

As I also said in my original answer above, peroxides are pretty unstable. This is because the two oxygens are right next to each other. As you may know, oxygen is one of the most electronegative elements of the periodic table, so it "fights to get electrons". If you connect two oxygens like in a peroxide, they both fight to get the electron of the other one, so it resembles rope pulling. This isn't the best case scenario for a stable bond.

If you look at hydrogen peroxide, especially in concentrated solutions, it tends to decompose without any external influence.

$\ce{2H2O2 -> 2H2O + O2}$

This reaction releases 98,2 kJ/mol [1] of energy. This means that the products (oxygen and water) are way more stable than the hydrogen peroxide itself. This is mainly due to the very unstable rope pulling-bond between the two oxygens in $\ce{H2O2}$: $\ce{RO-OR}$.

This should explain why the intermediate group is so unstable.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ I understand that the final product is thermodynamically favored compared to the reactants. What I don't understand is why this intermediate is less stable than the first two reactants. Not only is boron's electron deficiency fulfilled but the reactive negative charge on oxygen is gone in the intermediate. $\endgroup$ – xasthor Aug 9 '17 at 3:53
  • $\begingroup$ Why would the intermediate decompose into two parts which are less stable than itself(one part having a deficient boron and the other part having a reactive negative charge) $\endgroup$ – xasthor Aug 9 '17 at 3:56
  • $\begingroup$ @xasthor Please consider my edit. $\endgroup$ – Sam Aug 9 '17 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.