Chapter 12, Problem 12.22P

Interpretation Introduction

Interpretation:

The complete, detailed mechanism for the reaction, when the product in Problem 12.21 is treated with a basic solution of hydrogen peroxide, is to be drawn.

Concept introduction:

Hydroboration is addition of $\text{H}$ and ${\text{BH}}_{\text{2}}$ groups across the carbon-carbon double bond in an alkene. These are stereospecific reactions in which the two said groups are added in a syn fashion. Hydroboration is also regioselective, where the hydrogen atom adds to the carbon atom with the greater number of alkyl groups (i.e., the most substituted carbon atom in the double bond) whereas the ${\text{BH}}_{\text{2}}$ group adds to the carbon atom with the fewer number of alkyl groups (i.e., the least substituted carbon atom in the double bond). Thus, the addition is said to occur according to the anti-Markovnikov’s rule.

As with the addition of ${\text{BH}}_3$, each of these is a syn addition, and the bulkier B-containing portion adds preferentially to the less sterically hindered doubly bonded carbon atom. Once the trialkylborane is formed, it is treated with a basic solution of hydrogen peroxide to convert it into an alcohol. In this reaction, each $\text{C-B}$ bond in trialkylborane is replaced by a $\text{C-OH}$ bond, producing three equivalents of the alcohol. The $\text{OH}$ group in the product is syn to the hydrogen atom added from the previous hydroboration reaction. Thus, this oxidation takes place with retention of configuration at each carbon atom bonded to boron. The mechanism begins with coordination of ${\text{HOO}}^{-}$ (hydro peroxide ion) to the electron-deficient boron atom of the trialkylborane. This leads to the formation of an unstable tetrahedral intermediate. In step two, the breaking of a weak peroxide bond drives a 1, 2-alkyl shift that yields a borate ester. This pair of steps occurs twice more, resulting in a trialkylborate ester.

In step 7, the hydroxide ion coordinates to the boron atom of trialkylborate ester, followed by step 8, which is the hydrolysis that leads to the release of alkoxide ion as a leaving group. In step 9, the strongly basic alkoxide ion gains a proton from the water molecule to produce the first equivalent of the final alcohol. This trio of steps is then repeated twice as steps 10-15.