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Hydroboration is a reaction in which boron hydride acts as an electrophile. \(\mathrm{R}_{2} \mathrm{BH}\) adds to a carbon-carbon double bond which acts as a nucleophile The organoborane compound then is oxidised by treatment with hydrogen peroxide in aqueous medium to form alcohol. The OH-group enters the carbon atom from the same side where the boron atom was present. Hence this reaction is highly regioselective and the boron atom attaches to that carbon atom which is less stearically hindered. \(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{NOC}}{\longrightarrow} \mathrm{X} ;\) Hence, \(\mathrm{X}\) is

Short Answer

Expert verified
The product \(\mathrm{X}\) is \(\mathrm{R}-\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}\).

Step by step solution

01

Understanding Hydroboration

Hydroboration involves the addition of boron hydride (\(\mathrm{R}_2\mathrm{BH}\)) to a carbon-carbon double bond, where the double bond acts as a nucleophile and the boron compound acts as an electrophile. The boron attaches to the less sterically hindered carbon atom in the alkene.
02

Identifying the Regioselectivity

In this reaction, the boron atom will add to the terminal carbon of the alkene (\(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_2\)), because it is less hindered compared to the other carbon, which might be attached to more substituents.
03

Oxidation of the Organoborane

Once the borane is added to the alkene, the organoborane intermediate is oxidized by hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_2\)) in an aqueous medium. This reaction converts the boron-containing compound into an alcohol where the OH group replaces the boron atom.
04

Understanding Syn-Addition and Markovnikov's Rule

The reaction is syn-addition, meaning both the boron and the incoming hydroxyl group add to the same side of the double bond. However, this reaction is anti-Markovnikov, which means the OH group ends up on the less substituted carbon, opposite to typical acid-catalyzed hydration reactions.
05

Identify Product X

With the less hindered terminal carbon receiving the boron initially and then the hydroxyl group, the product \(\mathrm{X}\) will be an alcohol in which the hydroxyl group is attached to the terminal carbon. Thus, \(\mathrm{X}\) is \(\mathrm{R}-\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}\).

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Key Concepts

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

Regioselectivity
In chemistry, regioselectivity is a key concept that helps us understand where exactly on a molecule a particular reaction will occur. When we talk about regioselectivity in the context of hydroboration, we're discussing which carbon atom in a carbon-carbon double bond will attract the boron atom. This selectivity is driven by the desire to add to the location that is least hindered.

For instance, in a molecule with the structure \(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_2\), the terminal carbon has fewer surrounding groups compared to its neighboring carbon. This makes it more accessible for the boron hydride addition. Because of this, the reaction exhibits high regioselectivity with the boron adding to the less crowded terminal carbon.
  • Important in determining the major product of a reaction.
  • Influences how molecules like alkenes react in the presence of electrophiles.
  • Crucial for creating specific desired alcohols through hydroboration-oxidation.
Electrophile
An electrophile is a species that seeks electrons, and thus reacts with regions of high electron density in other molecules. In the case of hydroboration, the boron hydride compound \(\mathrm{R}_2\mathrm{BH}\) behaves as the electrophile.

During this reaction, boron, which is electron-deficient, targets the electron-rich pi bond of the alkene. This interaction makes the double bond act as a nucleophile, offering up its electrons to satisfy the boron's electron need. This pairing is typical of electrophile-nucleophile interactions, which are fundamental to understanding chemical reactions.
  • Electrophiles are usually positively charged or neutral with empty orbitals.
  • They play a crucial role in many organic reactions, facilitating key transformations.
Nucleophile
A nucleophile is a molecule or ion that donates an electron pair to form a chemical bond in reaction with electrophiles. In the hydroboration of alkenes, the carbon-carbon double bond acts as the nucleophile.

This region of high electron density wants to form a stable bond, so it donates its electrons to the boron, which is electron-deficient. It is through this electron donation that the addition reaction progresses, making the nucleophile’s role critical.
  • Nucleophiles typically have pairs of electrons ready for bonding.
  • Common examples include anions, molecules with lone pairs, or pi bonds.
  • Nucleophilicity is influenced by factors such as charge, solvent, and the nature of the substituents attached to the nucleophile.
Anti-Markovnikov Addition
Anti-Markovnikov addition is a term describing a specific preference of addition reactions, in this case the placement of the hydroxyl group. Normally, in a typical acid-catalyzed hydration, the OH group would attach to the more substituted carbon atom, following Markovnikov's rule. But in hydroboration, we see the opposite, hence the term 'anti-Markovnikov'.

During hydroboration-oxidation, the boron initially attaches to the less substituted carbon. This leads to the final alcohol product having its OH group on the less substituted, often the terminal carbon.
  • This unexpected turn is due to the way in which the boron and transition states stabilize.
  • Anti-Markovnikov addition is crucial for creating certain alcohols.
  • Gives us insights into controlling the selectivity of chemical reactions toward desired outcomes.

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Most popular questions from this chapter

When ethane-1,2-dioic acid is heated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4^{\prime}}\) it gives (a) \(\mathrm{CO}+\mathrm{HCOOH}\) (b) \(\mathrm{CO}_{2}+\mathrm{HCOOH}\) (c) \(\mathrm{CO}+\mathrm{CO}_{2}+\mathrm{HCOOH}\) (d) \(\mathrm{CO}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

During the preparation of ethane by Kolb's electrolytic method using inert electrode the pH of the electrolyte (a) Decreases progressively as the reaction proceeds (b) Increases progressively as the reaction proceeds (c) Remains constant throughout the reaction (d) May decrease if concentration of the electrolytes is not very high

An optically active organic compound has the molecular formula \(\mathrm{C}_{7} \mathrm{H}_{10}\) (A). On reaction with \(\mathrm{H}_{2}+\mathrm{Pt}\) it forms an optically inactive compound (B). Then, compound (A) will be

When cyclohexadiene (A) reacts with \(\mathrm{Br}_{2^{\prime}}\) a mixture of cis- and trans-1, 2 -addition products is formed (in addition to other products). However, when cyclohexene (B) reacts with \(\mathrm{Br}_{2}\) under identical conditions, only trans product is observed. What is the best explanation for the observed difference in stereochemistry of the addition? (a) The cis and trans products are the result of aromaticity in the cyclic TS for reaction of A. In \(B\) there are only four electrons in TS, and cyclic TS is destabilised. (b) Reaction of A proceeds through an intermediate that has an \(\mathrm{sp}^{3}\) -hybridised carbocation, while the analogous intermediate in reaction of B has sp-hybridised carbocation. (c) Both reactions occur through bromonium ions, but because of planarity enforced by neighbouring double bond, cis addition is not sterically hindered in A. (d) B reacts through a bromonium ion intermediate, while A does through an allyl cation.

Compound having atleast one \(\pi\) -bond gives addition reaction. Alkene behaves as a nucleophile and hence it gives an electrophilic addition reaction. Electrophilic addition reaction in most of the cases takes place by formation of carbocation as reaction intermediate. Consider the following statements (i) Unsymmetrical alkene gives addition product according to Markovnikov's rule. (ii) Addition reaction is a regioselective reaction. (iii) Rearranged product is formed in addition reaction. (iv) Alkene gives mixed addition product with \(\mathrm{NaCl} / \mathrm{HOH} / \mathrm{H}^{+}\). Which one is/are correct? (a) i, ii, iii and iv (b) ii, iii and iv (c) i, iii and iv (d) i, ii and iv

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