Hydroboration and Oxidation - Practice Questions & MCQ

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MEDIUM

Acetone (CH₃COCH₃) is the major product in :

$\mathrm{I\; CH_{2}=C=CH_{2}\xrightarrow[]{H_{3}O^{+}}}$

$\mathrm{II\; CH_{3}C=CH\xrightarrow[]{H_{2}SO_{4}/HgSO_{4}/H_{2}O}}$

$\mathrm{III\; CH_{3}C=CH\xrightarrow[\mathrm{H_{2}O_{2}/OH^{\ominus }}]{BH_{3}THF}}$

I only

II only

III only

Both I and II

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In the following reaction :

The major product is:

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Concepts Covered - 2

Hydroboration and Oxidation

Hydroboration-oxidation serves as an important method for synthesis of alcohol(1^o & 2^o). The reaction occurs as follows:

The addition of boron hydride is syn-addition. It is generally carried out by BH₃ (boron hydride) B₂H₆ (diborane) in THF. In each addition, the boron atom becomes attached to the less substituted carbon atom of double bond and H is transferred from boron atom to the other carbon atom of double bond. Thus it follows Anti-Markonikoff’s addition.

Reaction of Alkene with Dilute H2SO4

Cold concentrated sulphuric acid adds 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, carbon-carbon double bond is broken first. Then one of the H⁺ is released and combines with one of the carbon. Now, a carbocation is already formed after the breaking of double bond. Now, if the carbocation has a possibility to 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.