Hydration of an unsymmetrical alkene in anti-Markownikoff fashion can be effected by an indirect, convenient and high yielding process known as hydroboration-oxidation. The process involves addition of borane (BH3) to the double bond followed by oxidation and hydrolysis of the resulting alkyl borane with alkaline solution of hydrogen peroxide. For example:
3
2 2 2
3 2 2 / 3 2 2 2 3
3 2 2 2 3 3
BH THF 0 C H O NaOH/H O
3CH CH CH CH (CH CH CH CH ) B 1-Butene
3CH CH CH CH OH Na BO 1-Butanol
== ổổổổổ∞ ặ
ổổổổổổặ +
Mechanism of hydroboration: Because of the tendency of boron to accept an electron pair for completing its octet and also, because H is more electronegative than B, the B—H bond is polarized to give boron a partial positive charge (H—B),
d- d+
making BH2 the electrophile and H the nucleophile in hydroboration reaction. As boron accepts the p electrons from the alkene, it eliminates a hydride ion that also adds to alkene. However, unlike the addition of Br2, HBr, H2O, etc. the addition of borane to an alkene takes place in one step, i.e., it is a concerted reaction in which all the bond-breaking and bond-making processes occur in a single step. The electrophilic boron atom adds to the sp2 carbon that is bonded to most hydrogens. The addition, in fact, occurs through a four-centre transition state. A p-complex is formed initially by a donation of p electrons of the double bond to the vacant p orbital on boron. The p-complex then forms the transition state in which the boron atom is partially bonded to the less substituted doubly bonded carbon atom and the H atom is partially bonded to the more substituted doubly bonded carbon atom. As the transition state is approached, shifting of electrons in the direction of B atom and away from the more substituted C atom of the double bond occurs. As a consequence, the more substituted carbon atom, which is better able to accommodate positive charge, acquires a partial positive charge. This results in a relatively stable transition state that leads to the formation of Markownikoff addition product (because the negative part of the addendum adds to the carbon containing least number of hydrogens) predominantly.
Also, steric hindrance associated with alkylborane and particularly with dialkylborane causes the addition of boron to occur at the sp2 carbon that is bonded to the most hydrogens because that is the less sterically hindered of the two sp2 carbons. Therefore, the orientation of BH3 addition is favoured not only by electronic factor but also by steric factor.
The resulting alkylborane in which boron contains two more hydrogens reacts in a similar way with 2nd and 3rd equivalent of 1-butene to form di- and tributylborane respectively.
Mechanism of oxidation: When the hydroboration reaction is over, aqueous sodium hydroxide and hydrogen peroxide (H2O2) are added to the reaction mixture. Hydrogen peroxide reacts with alkali to form hydroperoxide ion
: . HOO@
The end result is the replacement of boron in alkyl borane by an OH group. Because the replacement of boron by an OH group is an oxidation reaction (oxygen is being added to the compound), the overall reaction is called hydroboration-oxidation.
The mechanism of oxidation of the trialkylborane involves the steps as follows:
Step 1: This involves nucleophilic attack by the hydroperoxide ion on the electron- deficient boron atom of the trialkylborane molecule to give an unstable tetracovalent boron intermediate.
(CH CH CH CH ) B + :OOH3 2 2 2 3
–
(CH CH CH CH ) B—O—OH3 2 2 2 2
CH CH CH CH2 2 2 3
Unstable intermediate
–
Step 2: The resulting compound undergoes rearrangement. An alkyl group migrates with its bonding electrons from boron to oxygen (a 1,2-anionic shift) and thereby displaces a hydroxide ion to form a borinic ester. The driving force for the rearrangement is the formation of the strong boron–oxygen bond.
(CH CH CH CH ) B—O—OH3 2 2 2 2 CH CH CH CH2 2 2 3
–
(CH CH CH CH ) B—O—CH CH CH + :OH3 2 2 2 2 2 2 3
–
The sequence is repeated twice to yield a trialkyl borate ester.
The borate ester is finally hydrolyzed by alkali. The reaction involves the steps as follows:
Step 1: The borate ester undergoes nucleophilic attack by the OH① ion.
(CH CH CH CH3 2 2 2O) B + :OH3 (CH CH CH CH O) B—OH3 2 2 2 3
– –
Step 2: An alkoxide ion departs from the borate anion, reducing the formal charge on boron to zero.
Step 3: A molecule of 1-butanol is obtained as a result of proton transfer.
(CH CH CH CH3 2 2 2O) B—O—H +2 CH CH CH CH O:3 2 2 2 – (CH CH CH CH O) B—O:3 2 2 2 2 + CH CH CH CH OH3 2 2 2
–
1-Butanol
The three preceding steps are repeated two times to give two more equivalent of 1-butanol.