8.4
Formation of Halohydrin from Alkenes
An alkene, such as propene, reacts with bromine in the presence of water to yield a halohydrin. Halohydrins contain a halogen and a hydroxyl group att…
Alkenes, such as propene, react with a halogen in the presence of water to yield a halohydrin, a compound having a halogen and a hydroxyl group on adjacent carbons.
The first step in the mechanism involves the electrophilic addition of bromine across the double bond giving a bromonium ion.
Water, present in excess, acts as a nucleophile and attacks the more substituted carbon of the bromonium ion to open the three-membered ring.
The steric crowding and non-availability of the bonding orbitals direct the attack of the nucleophile to an antibonding orbital on the opposite side of the carbon–bromine bond.
Since the protonated bromohydrin is a strong acid, it loses a proton to water giving the product 1-bromo-2-propanol.
The regioselective addition of a hydroxyl group at a more substituted carbon is explained based on the combination of two factors. The electrostatic potential map of the bromonium ion shows that the more substituted carbon possesses more carbocation character, facilitating the nucleophile’s attack.
Furthermore, the bond between bromine and the more substituted carbon is longer than the bond of the less substituted carbon, meaning that the ring-opening transition state requires lower energy when the nucleophile attacks the more substituted carbon.
The next step is to identify the stereochemical outcome when a chiral center is generated. The addition of a halogen to a propene produces enantiomerically bridged intermediates and the addition of a nucleophile from the opposite face leads to a pair of halohydrin enantiomers.
In another instance, 1-methylcyclohexene reacts with bromine to give a pair of enantiomeric bromonium ions. The anti addition of water delivers trans-2-bromo-1-methylcyclohexanol as a racemic mixture.
Because alkenes are insoluble in water, the reaction is usually carried out in solvents like aqueous dimethyl sulfoxide. N-Bromosuccinimide serves as a stable and less harmful source of bromine.
Interestingly, in biological systems, bromoperoxidase oxidizes the bromine ion to the corresponding hypobromous acid, which upon electrophilic addition to the substrate and subsequent reaction with water, yields 2-bromo-1-phenyl-1,3-propanediol, a bromohydrin.
Q1: What is a halohydrin and how does it form from an alkene?
A halohydrin is a compound containing both a halogen and a hydroxyl group attached to adjacent carbons. It forms when an alkene reacts with a halogen, such as bromine, in the presence of water. The reaction begins with electrophilic addition of bromine to the alkene's double bond, creating a bromonium ion intermediate. Water then acts as a nucleophile, attacking the bromonium ion to open the three-membered ring and form the halohydrin product.
Q2: Why does water attack the more substituted carbon of the bromonium ion?
Water preferentially attacks the more substituted carbon due to two factors. First, the electrostatic potential map shows the more substituted carbon has greater carbocation character, making it more susceptible to nucleophilic attack. Second, the bromine-carbon bond is longer at the more substituted carbon than at the less substituted carbon, requiring lower transition state energy for ring-opening at that position.
Q3: What is the stereochemical outcome when halohydrin formation creates a chiral center?
Halohydrin formation exhibits anti stereochemistry. When 1-methylcyclohexene reacts with bromine, enantiomeric bromonium ions form. Water's anti addition from the opposite face of the bromonium ion delivers the product as a racemic mixture. For example, the reaction yields trans-2-bromo-1-methylcyclohexanol as equal amounts of both enantiomers.
Q4: What is the mechanism of the nucleophilic attack in halohydrin formation?
The nucleophilic attack occurs through an SN2 process. Water uses a lone pair of electrons to attack the bromonium ion, opening the three-membered ring. The steric crowding and orbital availability direct the nucleophile's attack toward an antibonding orbital on the opposite side of the carbon-bromine bond, ensuring anti addition geometry.
Q5: How is the final halohydrin product formed after water attacks the bromonium ion?
After water attacks the bromonium ion, an oxonium ion intermediate forms. Since the protonated bromohydrin is a strong acid, it loses a proton to water in a deprotonation step. This yields a hydronium ion and the neutral bromohydrin product, completing the reaction sequence.
Q6: What practical considerations affect halohydrin formation reactions?
Since alkenes are insoluble in water, halohydrin reactions are typically conducted in solvents like aqueous dimethyl sulfoxide. N-Bromosuccinimide serves as a stable and less harmful source of bromine compared to molecular bromine. In biological systems, bromoperoxidase oxidizes bromide ions to hypobromous acid, which undergoes electrophilic addition to substrates.
Q7: How does the bromonium ion intermediate explain regioselectivity in halohydrin formation?
The bromonium ion intermediate is key to understanding regioselectivity. Its electrostatic potential map reveals that the more substituted carbon exhibits greater carbocation character. Combined with the longer bromine-carbon bond at the more substituted position, these factors make the ring-opening transition state more accessible when water attacks that carbon, explaining the regioselective hydroxyl group addition.