Halohydrin, formation

Treatment of an alkene with a halogen X2 and H2O forms a halohydrin by addition of the elements of X and OH to the double bond. 

The mechanism for halohydrin formation is similar to the mechanism for halogenation addition of the electrophile X (from X2) to form a bridged halonium ion, followed by nucleophilic attack by H2O from the back side on the three-membered ring (Mechanism 10.4). Even though X is formed in Step [1] of the mechanism, its concentration is small compared to H2O (often the solvent), so H2O and not X is the nucleophile. 

Four bonds are broken or formed in this step the electron pair in the Ji bond and a lone pair on a halogen atom are used to form two new C-X bonds in the bridged halonium ion. The X-X bond is also cleaved heterolytically, forming X . This step is ratedetermining. 

Nucleophilic attack of H2O opens the halonium ion ring, forming a new C-X bond. Subsequent loss of a proton forms the neutral halohydrin. 

Recall from Section 7.8C that DMSO (dimethyl sulfoxide) is a polar aprotic solvent. 

Although the combination of Br2 and H2O effectively forms bromohydrins from alkenes, other reagents can also be used. Bromohydrins are also formed with A-bromosuccinimide (abbreviated as NBS) in aqueous DMSO [(CH3)2S=O]. NBS serves as a source of Br2, which then goes on to form a bromohydrin by the same reaction mechanism. 

Problem 7.3 What product would you expect to obtain from addition of CI2 to 1,2-dimethyIcyclo-hexene Show the stereochemistiy of the product. 

Problem 7.4 Unlike the reaction in Problem 7.3, addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the stereochemistry of each, and explain why a mixture is formed. 

Many different kinds of electrophilic additions to alkenes take place. For example, alkenes add HO-Cl or HO-Br under suitable conditions to yield 1,2-halo alcohols, called halohydrins. Halohydrin formation doesn t take place by direct reaction of an alkene with HOBr or HOCl, however. Rather, the addition is done indirectly by reaction of the alkene with either Br2 or CI2 in the presence of water. 

In practice, few alkenes are soluble in water, and bromohydrin formation is often carried out in a solvent such as aqueous dimethyl sulfoxide, CH ,SOCH j (DMSO), using a reagent called AT-bromosuccinimide (NBS) as a source of Bt2. NBS is a stable, easily handled compound that slowly decomposes in water to yield Br2 at a controlled rate. Bromine itself can also be used in the addition reaction, but it is more dangerous and more difficult to handle than NBS. 

Mechanism of bromohydrin formation by reaction of an aJkene with 8r2 in the presence of water. Water acts as a nucieophile to react with the intermediate bromonium ion. 

Explain which of these compounds has the faster rate of reaction with Br2  

The reactions of chlorine and bromine with alkenes described in the previous section are conducted in inert solvents such as CC14 and CH2C12. In these reactions the only nucleophile that is present to react with the halonium ion is the halide anion. However, if the reaction is performed in a nucleophilic solvent, such as water, then a water molecule can act as the nucleophile, resulting in the addition of a halogen and a hydroxy group to the double bond. 

The product is called a halohydrin. Because the concentration of the water molecules is so much higher than the concentration of the halide anions (water is the solvent), water wins the competition to act as the nucleophile, and only the halohydrin is formed in significant amounts. 

One of the major uses of these halohydrins is for the preparation of epoxides. Treatment of the halohydrin with base, such as NaOH or KOH, results in deprotonation of the alcohol followed by an intramolecular nucleophilic substitution (see Section 10.3), as shown in the following example. Remember that the nucleophilic oxygen must displace the chlorine from the opposite side, resulting in inversion of configuration at that carbon. 

CHAPTER I I ADDITIONS TO CARBON-CARBON DOUBLE AND TRIPLE BONDS 

The bromonium ion reacts with the bromide ions at equal rates by paths (a) and (b) to yield the two enantiomers in equal amounts (i.e., as the racemic form). 

When the bromonium ions react by either path (a) or path (b), they yield the same achiral meso compound. (Reaction of the enantiomer of the intermediate bromonium ion would produce the same result.) 

Molecules of water react with the halonium ion intermediate as the predominant nucleophile because they are in high concentration (as the solvent). The result is formation of a halo- 

This step is the same as for haiogen addition to an aikene (see Section 8.12A). 

however, a water molecule acts as the nucleophile and attacks a carbon of the proton (it is transferred to a ring, causing the formation of a molecule of water). This step 

---

Propylene oxide has found use in the preparation of polyether polyols from recycled poly(ethylene terephthalate) (264), haUde removal from amine salts via halohydrin formation (265), preparation of flame retardants (266), alkoxylation of amines (267,268), modification of catalysts (269), and preparation of cellulose ethers (270,271). 

Addition of Hypohalous Acids to Alkenes Halohydrin Formation... 

Figure 7.3 Mechanism of the oxymercuration of an alkene to yield an alcohol. The reaction involves a mercurinium ion intermediate and proceeds by a mechanism similar to that of halohydrin formation. The product of the reaction is the more highly substituted alcohol, corresponding to Markovnikov regiochemistry.

The final product is called a halohydrin (indicating that we have a halogen— Br— and an OH in the same compound). This reaction is commonly called halohydrin formation. 

The profile of halohydrin formation can be summarized in the following chart ... 

Product identification was carried out by NMR analysis using H, and H/ H correlation spectroscopy techniques. Under the given conditions, molar conversion yields were >90 % after 10 min, >90 % after 20 min and >60 % after 30 min for iodo-, bromo-and chloro-halohydrin formation respectively. 

The reagent may require acid activation depending on the type of transformation being attempted. The mechanism parallels that of halohydrin formation using an electrophilic source of halide in an aqueous medium ... 

Halohydrin formation, as depicted in Figure 6.13, is mechanistically related to halogen addition to alkenes. A halonium ion intermediate is formed, which is attacked by water in aqueous solution. 

The diastereoselective halohydrin formation, resulting from the reaction of chiral /V -enoyI -2-oxazoI idinones with Br2/l2 and water, promoted in the presence of silver , in aqueous organic solvents, has been found to occur with high regioselectivity and moderate to good diastereoselectivities. The alkenoyl, cinnamoyl, and electron-deficient cinnamoyl substrates readily produced the bromohydrin in aqueous acetone, but no iodohydrin formation was observed under these conditions. On the other hand, ( ) electron-rich cinnamoyl substrates preferred to afford iodohydrins in aqueous acetone with moderate diastereoselectivity enhanced diastereoselectivity was observed for aqueous THF.31... 

Halohydrin formation, anti addition, Markovnikov s rule... 

Halohydrin Formation Section 11.5 Figure 11.4 Hell-Volhard-Zelinsky Reaction Section 26.4 ... 

Fig. 6.8 Activation of halogens by CPO (a) reactions with 1,3-dicarbonyl compounds (b) halohydrins formation...

Several examples of halohydrin formation from styrene derivatives and saccharides catalyzed by CPO are reviewed by Adam and coworkers [23], Formation of bromohydrin derivatives of some saccharides can be of interest for the preparation of bioactive compounds [72]. 

Stereochemistry of Halohydrin Formation Because the mechanism involves a halonium ion, the stereochemistry of addition is anti, as in halogenation. For example, the addition of bromine water to cyclopentene gives fran.v-2-bromocyclopentanol, the product of anti addition across the double bond. 

Orientation of Halohydrin Formation Even though a halonium ion is involved, rather than a carbocation, the extended version of Markovnikov s rule applies to halohydrin formation. When propene reacts with chlorine water, the major product has the... 

Orientation of halohydrin formation The more substituted carbon of the chloronium ion bears more positive charge than the less substituted carbon. Attack by water occurs on the more substituted carbon to give the Markovnikov product. 

The Markovnikov orientation observed in halohydrin formation is explained by the structure of the halonium ion intermediate. The two carbon atoms bonded to the halogen have partial positive charges, with a larger charge (and a weaker bond to the halogen) on the more substituted carbon atom (Figure 8-5). The nucleophile (water) attacks this more substituted, more electrophilic carbon atom. The result is both anti stereochemistry and Markovnikov orientation. 

---