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\(\frac{\text { (i) } \mathrm{BH}_{3}}{\text { (ii) } \mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}_{2}}\) Identify reactant

Short Answer

Expert verified
The reactant is an alkene.

Step by step solution

01

Understanding the Reaction

The reaction in question involves the use of two reagents. We have Borane, \( \text{BH}_3 \), as the organic reagent, which is typically used in hydroboration reactions. The second reagents are \( \text{NaOH} \) and \( \text{H}_2\text{O}_2 \), which are used in the oxidation step that follows hydroboration.
02

Identify the Typical Reaction Type

The sequence \( \text{(i) } \text{BH}_3 \) and \( \text{(ii) } \text{NaOH}, \text{H}_2\text{O}_2 \) suggests a two-step reaction sequence commonly used to convert alkenes to alcohols. This process is known as hydroboration-oxidation.
03

Association of Reagents with Hydroboration-Oxidation

In hydroboration-oxidation, the reactant is typically an alkene. \( \text{BH}_3 \) is used to add across the double bond in a syn-addition manner, and \( \text{NaOH} \) with \( \text{H}_2\text{O}_2 \) oxidizes the resulting organoborane to form an alcohol.
04

Determine the Initial Reactant

Given that hydroboration-oxidation involves an alkene as a starting material, the reactant in this two-step reaction is an alkene. This is because the alkene undergoes addition reactions with \( \text{BH}_3 \), followed by oxidation to form an alcohol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alkene Chemistry
Alkenes are hydrocarbons containing at least one carbon-carbon double bond. This double bond imparts unique reactivity to alkenes, as it can easily interact with various reagents, resulting in different types of addition reactions. Alkenes serve as crucial starting materials in several organic synthesis processes due to their ability to form reactive intermediates.
One notable feature of alkenes is their ability to undergo addition reactions, where the double bond opens up to allow new atoms or groups to attach, forming single bonds. This characteristic is harnessed in hydroboration-oxidation reactions, which we will explore in upcoming sections. The versatility of alkenes makes them valuable in the production of alcohols, polymers, and other organic compounds through these reactions.
Organic Reagents
In hydroboration-oxidation reactions, specific organic reagents play a key role. The process involves using borane (\( \text{BH}_3 \)) as the initial reagent for the hydroboration part. Borane is a molecule with a boron atom bonded to three hydrogen atoms, and due to its electron-deficient nature, it readily adds across the double bond of alkenes.
This reaction is followed by the use of sodium hydroxide (\( \text{NaOH} \)) and hydrogen peroxide (\( \text{H}_2\text{O}_2 \)) in the oxidation step. These reagents work together to convert the intermediate formed from the reaction with borane into a stable alcohol. The specificity and effectiveness of these reagents make them ideal for transforming alkenes into alcohols.
Reaction Mechanism
Understanding the reaction mechanism of hydroboration-oxidation is crucial for mastering this transformation. Hydroboration is a concerted reaction where borane adds to the alkene in a manner that both the boron and hydrogen atoms attach simultaneously. This attachment is called syn-addition, meaning both atoms add to the same side of the alkene double bond. As a result, this step produces an organoborane intermediate.
In the oxidation step, the organoborane is treated with \( \text{NaOH} \) and \( \text{H}_2\text{O}_2 \). This reaction converts the organoborane into an alcohol. The process involves replacing the boron atom with a hydroxyl group, typically retaining the stereochemistry established during hydroboration. This makes the hydroboration-oxidation not only region-specific but also stereospecific.
Alcohol Formation
The final product of the hydroboration-oxidation reaction sequence is an alcohol. This reaction is significant in organic chemistry due to its reliability in creating anti-Markovnikov alcohols, where the hydroxyl group attaches to the less substituted carbon atom of the double bond. This contrasts with acid-catalyzed hydration reactions, which typically form Markovnikov alcohols.
The formation of the alcohol product showcases the transformation of a simple alkene into a functionalized alcohol compound. This is a fundamental transformation, crucial for further synthesis in organic chemistry since alcohols serve as versatile intermediates for other functional groups. By following the hydroboration-oxidation procedure, chemists can efficiently create a broad array of alcohols tailored for further chemical transformations.

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Most popular questions from this chapter

The reaction of a Grignard reagent with a carboxylic acid does not give a secondary alcohol. This is because (a) Grignard reagents only react with the aldehydes, ketones, esters and epoxides. (b) The carboxylic acid is too sterically hindered to react. (c) The carboxylic acid is not electrophilic enough to react. (d) The Grignard reagent is a base, so an acid-base reaction occurs.

How many products are possible when ethanal and phenyl ethanal (mixture) is treated with dil. \(\mathrm{NaOH}\) at about \(10^{\circ} \mathrm{C}\).

Match the columns. Column I (a) PCC (Pyridiniumchlorochromate) (b) NBS (N-Bromosuccinimide) (c) \(\mathrm{AlPO}_{4}\) (d) Li/Liquid \(\mathrm{NH}_{3}\) Column II (p) C1=CCc2ccccc2C1 BrC1C=CCc2ccccc21 (q) \(\mathrm{MeC} \equiv \mathrm{CMe} \longrightarrow\) (r) (s) \(\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow \mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\)

In general, ketones are more reactive towards nucleophiles than esters because (a) The \(\alpha\) -protons of a ketone are more acidic than those of an ester. (b) The alkyl group in a ketone is an electron donating group due to hyperconjugation. (c) Alkoxy (RO-) groups are sterically larger than the related alkyl group. (d) Alkoxy (RO-) groups are stronger electron donating than alkyl groups via resonance.

Identify correct method of preparation of acetaldehyde from reaction of cyanide (a) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) DIBAL }}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}}\) (b) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) } \mathrm{SnCl}_{2}+\mathrm{HCl}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}}\) (c) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\text { (ii) dill. } \mathrm{NaOH}}\) (d) \(\mathrm{Me}-\mathrm{C} \equiv \mathrm{N} \frac{\text { (i) } \mathrm{Pd} / \mathrm{BaSo}_{4} / \mathrm{H}_{2}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{\oplus}}\)

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