Assertion (A): Reaction of but-1-ene with HBr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Markovnikov's rule,the addition of $HBr$ to an unsymmetrical alkene like $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$ results in the hydrogen at

Vedclass · Accepted Answer

$(A)$ is not correct but $(R)$ is correct

Vedclass · Answer

(D) According to Markovnikov's rule,the addition of $HBr$ to an unsymmetrical alkene like $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$ results in the hydrogen atom attaching to the carbon with more hydrogen atoms,and the bromine atom attaching to the carbon with fewer hydrogen atoms.Therefore,the major product formed is $2-bromobutane$ $(CH_3-CH_2-CH(Br)-CH_3)$,not $1-bromobutane$.Thus,Assertion $(A)$ is incorrect.However,the Reason $(R)$ stating that the addition of $HBr$ to unsymmetrical alkenes follows Markovnikov's rule is correct.Therefore,$(A)$ is incorrect but $(R)$ is correct.

Assertion $(A)$: Reaction of $but-1-ene$ with $HBr$ gives $1-bromobutane$ as the major product.
Reason $(R)$: Addition of $HBr$ to unsymmetrical alkenes proceeds according to Markovnikov's rule.
The correct answer is

Similar Questions

The major product $(P)$ formed in the below reaction is:
$CH_2$ attached to a cyclohexane ring $\xrightarrow[(ii) NaBH_4, OH^-]{(i) Hg(OAc)_2, H_2O-THF} P?$

An organic compound $'A'$ $(C_4H_8)$ on treatment with $KMnO_4 / H^{+}$ yields compound $'B'$ $(C_3H_6O)$. Compound $'A'$ also yields compound $'B'$ on ozonolysis. Compound $'A'$ is:

The correct order of rates of addition of $Br_2/\text{water}$ to the following alkenes is:
[$A$] $CH_2=CH_2$
[$B$] $CH_2=CH-NO_2$
[$C$] $CH_3-CH_2-CH=CH_2$
[$D$] $CH_3-CH=CH_2$

$CH_2 = CH - (CH_2)_5COOH \xrightarrow[HBr]{\text{Peroxide}} Z$,where $Z$ is:

In the given reaction,$(x)$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: Reaction of $but-1-ene$ with $HBr$ gives $1-bromobutane$ as the major product.Reason $(R)$: Addition of $HBr$ to unsymmetrical alkenes proceeds according to Markovnikov's rule.The correct answer is