[limonade]

I'm not sure that I understand the mechanism for this problem.

If you have 2 methylpropene and it undergoes hydroboration, you get anti markovnikov addition of BH2 to the less substituted carbon. I get this part.


The next part is what I don't get. If you react the resulting product with a carboxylic acid containing tritium instead of hydrogen, how does the tritium atom just replace the borane group?

[magician4]

if you take a look at electronegativity, you will realise that the carbon-boron bond is to be denoted with a certain carbanionic character at the carbon:

R₃-C-BH₂ equals R₃-C^{([itex] _{\delta}[/itex]-)} - B^{([itex] _{\delta}[/itex]+)}H₂

hence, adding a "dipolar" material for splitting the C-B bond like CH₃COO^--T⁺, the (negative) acetate-part will join with the (positive) boron, and the (positive) tritium will join with the (carbanionic) carbon


regards

Ingo

[limonade]

That makes so much sense now. Thank you

[orgopete]

if you take a look at electronegativity, you will realise that the carbon-boron bond is to be denoted with a certain carbanionic character at the carbon:

R₃-C-BH₂ equals R₃-C^{([itex] _{\delta}[/itex]-)} - B^{([itex] _{\delta}[/itex]+)}H₂

hence, adding a "dipolar" material for splitting the C-B bond like CH₃COO^--T⁺, the (negative) acetate-part will join with the (positive) boron, and the (positive) tritium will join with the (carbanionic) carbon


regards

Ingo

While that could be the case, I am reluctant to agree with it. I prefer to keep mechanisms as similar as possible. Boron is nearly a special case. It prefers to have a completed octet, but when it does, it hold those electrons very weakly. Your can anticipate this as it has the longest bonds of the second row octets. This tells us two things, trivalent boron is a Lewis acid. That is how it reacts with an alkene. Tetravalent boron can donate electrons, as it can with borohydride.

If I were to write a mechanism consistent with this, then I anticipate boron will first act as a Lewis acid to form a negatively charged tetravalent boron. Protonation of an electron pair then occurs. Now, because of the smaller nuclear charge, the electrons will remain with carbon and boron will leave as a neutral trivalent boron. This process can be repeated until all of the carbons (and hydrogens) are gone.

You should see a similarity in writing mechanisms for BH4(-) reductions of ketones using all four hydrogen atoms or the formation of H2 from borohydride upon acidification.