(D) The reaction of $2-methylpropene$ $(CH_2=C(CH_3)-CH_3)$ with $HBr$ follows Markovnikov's rule.$H^{+}$ adds to the carbon with more hydrogen atoms,

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(D) The reaction of $2-methylpropene$ $(CH_2=C(CH_3)-CH_3)$ with $HBr$ follows Markovnikov's rule.$H^{+}$ adds to the carbon with more hydrogen atoms,forming a stable tertiary carbocation $(CH_3-C^{+}(CH_3)-CH_3)$.Subsequently,$Br^{-}$ attacks this carbocation to yield $2-bromo-2-methylpropane$ $(CH_3-C(Br)(CH_3)-CH_3)$.

Class 11 Chemistry Hydrocarbons Alkene

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What is the main product of the reaction between $2-methylpropene$ and $HBr$?

A
$1-bromobutane$
B
$1-bromo-2-methylpropane$
C
$2-bromobutane$
D
$2-bromo-2-methylpropane$

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Which of the following alkenes on oxidation by $KMnO_4$ in dil. $H_2SO_4$ forms adipic acid?

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Assertion : $1$-Butene on reaction with $HBr$ in the presence of a peroxide produces $1$-bromobutane.
Reason : It involves the free radical mechanism.

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Aqueous solution of sodium salt of $2-$methylbutanedioic acid on Kolbe's electrolysis gives:

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In organic reactions, sodium in liquid ammonia is used as:

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In which of the following species,addition of $H^{+}$ will be faster in comparison to other three species?

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What is the main product of the reaction between $2-methylpropene$ and $HBr$?

Similar Questions

Which of the following alkenes on oxidation by $KMnO_4$ in dil. $H_2SO_4$ forms adipic acid?

Assertion : $1$-Butene on reaction with $HBr$ in the presence of a peroxide produces $1$-bromobutane.Reason : It involves the free radical mechanism.

Aqueous solution of sodium salt of $2-$methylbutanedioic acid on Kolbe's electrolysis gives:

In organic reactions, sodium in liquid ammonia is used as:

In which of the following species,addition of $H^{+}$ will be faster in comparison to other three species?

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Assertion : $1$-Butene on reaction with $HBr$ in the presence of a peroxide produces $1$-bromobutane.
Reason : It involves the free radical mechanism.