200 Marks in JEE Mains Percentile 2025 - Expected Percentile and Rank

Hydroboration and Oxidation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • 16 Questions around this concept.

Solve by difficulty

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}}

Find out the major products from the following reaction

The final product A, formed in the following reaction sequence is :

Products A and B formed in the following set of reactions are

The product of the reaction given below is :

In the following reaction :

 

The major product is:

The major products A and B, respectively, are

UPES B.Tech Admissions 2025

Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements

Amrita Vishwa Vidyapeetham | B.Tech Admissions 2025

Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships

Concepts Covered - 2

Hydroboration and Oxidation

Hydroboration-oxidation serves as an important method for 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 double bond and H is transferred from boron atom to the other carbon atom of double bond. Thus it follows Anti-Markonikoff’s addition.

In the second step on reaction with \mathrm{H_2O_2,OH^{-}}, the \mathrm{OH} group of \mathrm{H_2O_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 mechanism of the reaction is outside the scope of your Class XI/XII syllabus.

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.


 

Upon heating the above product with boiling \mathrm{H_2O}\mathrm{OH^-} group replaces the \mathrm{HSO_4^-} leading to the formation of Alcohol

Study it with Videos

Hydroboration and Oxidation
Reaction of Alkene with Dilute H2SO4

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Reaction of Alkene with Dilute H2SO4

Chemistry Part II Textbook for Class XI

Page No. : 390

Line : 41

E-books & Sample Papers

Get Answer to all your questions

Back to top