MEDIUMAmrita Vishwa Vidyapeetham | B.Tech Admissions 2025
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships
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 $\left(\mathrm{CH}_3 \mathrm{COCH}_3\right)$ is the major product in :
$\mathrm{I}: \mathrm{CH}=\mathrm{C}=\mathrm{CH}_2 \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}}$
II : $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH} \xrightarrow{\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{HgSO}_4 / \mathrm{H}_2 \mathrm{O}}$
III : $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH} \xrightarrow[\mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH} \oplus]{\mathrm{BH}_3 \mathrm{THF}}$
But-2-ene on reaction with alkaline $\mathrm{KMnO}_4$ at elevated temperature followed by acidification will give:
The major product (P) obtained in the above reaction sequence is
New: JEE Main 2025 Session 1 Result OUT; Check Now | Rank Predictor
JEE Main 2025: College Predictor | Marks vs Rank vs Percentile | Top NITs Cutoff
JEE Main 2025: January Session Official Question Paper for All Shifts | Exam Analysis
JEE Main 2025: High Scoring Topics | Sample Papers | Mock Tests | PYQs
The major product [B] in the following sequence of reaction is :
Find out the major products from the following reaction
Identify the product $\mathrm{A}$ and product $\mathrm{B}$ in the following set of reactions.
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{H}_2 \mathrm{SO}_4} \mathrm{~A} \xrightarrow[\text { Heat }]{\mathrm{H}_2 \mathrm{O}} \mathrm{B}$
What is B in this reaction?
An alkene(X) is passed into concentrated sulphuric acid then is diluted with water and the product formed is ethanol.What is X?
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 , the
group of
replaces the
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:
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 ,
group replaces the
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
Clear your Basics with NCERT
E-books & Sample Papers
Get Answer to all your questions
JEE Main Articles
Back to top