Imagine being in a kitchen and watching a chef who takes a set of raw ingredients to put together a gourmet dish. Each move is precise, each step calculated, and finally, in that finished product, the flavors and textures blend to delight the senses. Thus, in the kingdom of chemistry, there are also reactions able to convert these simple molecules into very complex and very important compounds. One of those processes that really changed the face of organic synthesis is, according to him, the hydroboration-oxidation reaction, whereby the simple alkenes have been easily converted into alcohols, a very essential building block in some various industrial and pharmaceutical applications
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Hydroboration-oxidation is a two-step procedure for converting alkenes to alcohols. First is hydroboration: the reaction of an alkene with BH3 to give an organoborane intermediate.This rather unusual process takes place in syn-addition—that is, the boron and hydrogen atoms add to the same side of the double bond of the alkene. The addition is regioselective.
This is followed by the oxidation of this organoborane intermediate with hydrogen peroxide in the presence of a base, usually sodium hydroxide. This gives alcohol as the product. Hence, clearly the net reaction can be written as follows:
Alkene + BH3 $\rightarrow$ Organoborane
Organoborane $\mathrm{H} _2 \mathrm{O} _2 \mathrm{NaOH} \rightarrow$ Alcohol
This is a pretty useful two-step procedure within the area of organic chemistry, as one can get mild reaction conditions and high yields of products that turn out to be alcohols. Hence, it becomes important to appreciate the mechanism in order to know its application and importance.
Hydroboration-oxidation serves as an important method for the synthesis of alcohol(1o & 2o). The reaction occurs as follows:
The addition of boron hydride is syn-addition. It is generally carried out by BH3 (boron hydride) or B2H6 (diborane) in THF. In each addition, the boron atom becomes attached to the less substituted carbon atom of the double bond and H is transferred from the boron atom to the other carbon atom of the double bond. Thus it follows Anti-Markonikoff’s addition.
In the second step on reaction with $\mathrm{H}_2 \mathrm{O}_2$, OH $\mathrm{H}_2$ $\mathrm{O}_2$ replaces the $\mathrm{BH}_2$from the less substituted carbon initially containing the double bond.
It is to be noted that there is no formation of carbocation during the reaction and hence no rearrangement occurs in the reaction.
The hydroboration-oxidation reaction turns out to be rather general and may be applied to most alkenes. The most peculiar feature of the reaction is that it is regioselective; the boron atom fixes itself to the least substituted carbon of the alkene. A reaction so useful in synthesizing alcohols of specific structural arrangement.
Another important parameter of stereoselectivity of hydroboration is that since the addition of the boron and hydrogen is coming from the same side of the alkene, this organoborane intermediate maintains its stereochemistry throughout. This stereospecificity will also play an important role in forming complex compounds with well-defined three-dimensional structures.
What's more, through the use of substituted boranes or other oxidizing reagents, the hydroboration-oxidation reaction can be modified according to need. It may be like using disiamylborane or 9-BBN, which increases process selectivity and efficiency. Most of the variations developed simply enhance the power associated with hydroboration-oxidation in complicated syntheses of organic compounds.
Cold-concentrated sulphuric acid is added to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction.
For example:
$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OSO}_2 \mathrm{OH}$
Mechanism: In this reaction, the carbon-carbon double bond is broken first. Then one of the H+ is released and combined with one of the carbon. Now, a carbocation is already formed after the breaking of the double bond. Now, if the carbocation can achieve more stability, then first it becomes more stable either by hydride shift or methyl shift. Then HSO-4 binds with carbocation and forms the final product as given below.
Upon heating the above product with boiling$\mathrm{H}_2 \mathrm{OH}_2$ group replaces the $\mathrm{HSO}_4$ leading to the formation of Alcohol
In real life, hydroboration-oxidation is of essential application mainly in the pharmaceutical industry. It offers a great opportunity to convert alkenes into alcohols with exceptionally high regio and stereoselectivity. It turns out to be very instrumental during API synthesis. This is necessary to get the proper molecular structure and activity required in some anti-inflammatory drugs and antibiotic preparation.
Hydroboration-oxidation has been one of the classic reactions in most college organic chemistry courses used in academic research. It has always been an extraordinary example of regioselective and stereoselective addition, giving a taste of the mechanisms of reactions and how synthetic strategies could be built. Simple yet efficient, it is forever a favorite in the laboratory to see students witness in awe how alkenes change into alcohol.
Moreover, hydroboration-oxidation has an even higher utility potential in the synthesis of new materials and chemical processes. Applications of the above reactions range from polymer synthesis through agrochemicals to perfumes, thus underlining the versatility of the reaction and the amplitude of its impact.
Example 1:
But-2-ene on reaction with alkaline KMnO4 at elevated temperature followed by acidification will give:
1) (correct) 2 molecules of CH3COOH
2)2 molecules of CH3CHO
3)one molecule of CH3CHO and one molecule of CH3COOH
4)
Solution
As we have learned,
Hot alkaline KMnO4 is a strong oxidizing agent and it converts alkenes into carbonyl compounds or carboxylic acids depending upon the alkylation around the double bond.
The given reaction occurs as
Hence, the correct answer is Option (1)
Example 2:
The product formed in the following multistep reaction is:
1) (correct)
2)
3)
4)
Solution
Example 3:
Acetone $\left(\mathrm{CH}_3 \mathrm{COCH}_3\right)$ is the major product in :
I$\mathrm{I} \mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2 \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}}$
II $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH} \xrightarrow{\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{HgSO}_4 / \mathrm{H}_2 \mathrm{O}}$
III $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH} \xrightarrow[\mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}]{\mathrm{BH}_3 \mathrm{THF}}$
1) I only
2)II only
3) III only
4) (correct)Both I and II
Solution
The given reactions occur as
Note: Protonation of allenes form vinylic cation and not allylic cation (FACT)
Thus, reactions (I) and (II) form Acetone as major products.
Hence, the correct answer is Option (4).
The hydroboration-oxidation reaction, put in simple terms, is one of the simplest yet most important chemical reactions of organic chemistry, by which an alkene is efficiently converted into an alcohol. The reaction is regioselective, meaning it adds on the less substituted carbon of the alkene. This amount of addition is stereoselective in the sense that it takes place on the same side of the alkene. Mild conditions, high yield, and a quite straightforward mechanism make the reaction quite helpful to both the academic and the industrial sectors. It has several real-life applications, mainly in the pharmaceutical industry.
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