Which of the following compounds is formed as | vedclass

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(C) The reaction of $2-	ext{methylbut-}2-	ext{ene}$ with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition mechanism (Kharasch

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$2-	ext{Bromo-}3-	ext{methylbutane}$

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(C) The reaction of $2-	ext{methylbut-}2-	ext{ene}$ with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition mechanism (Kharasch effect).In this mechanism,the $Br^\bullet$ radical attacks the double bond to form the more stable free radical intermediate.For $CH_3-C(CH_3)=CH-CH_3$,the $Br$ radical adds to the $CH$ carbon to form a tertiary radical intermediate $(CH_3-C^\bullet(CH_3)-CH(Br)-CH_3)$,which is more stable than the secondary radical.However,in the presence of peroxide,the addition of $HBr$ to alkenes typically follows anti-Markovnikov's rule,where the $Br$ atom attaches to the carbon with more hydrogen atoms.For $2-	ext{methylbut-}2-	ext{ene}$,the structure is $CH_3-C(CH_3)=CH-CH_3$. The carbon atoms of the double bond are $C_2$ (which has no $H$) and $C_3$ (which has one $H$).According to the anti-Markovnikov rule,the $Br$ atom adds to the $C_3$ position,resulting in $2-	ext{bromo-}3-	ext{methylbutane}$ as the major product.

Which of the following compounds is formed as the major product in the following reaction?
$2-\text{methylbut-}2-\text{ene} \xrightarrow[\text{Peroxide}]{HBr} \text{Product}$

Similar Questions

Which hydrogen in compound $(E)$ is easily replaceable during bromination reaction in the presence of light?
$CH_3 - CH_2 - CH = CH_2$
$(E)$

Find the major product for the reaction shown below.

Ethylene reacts with Baeyer's reagent to give

The correct statement about $C$ and $D$ is:
$CH_3-C \equiv C-CH_3$ $\xrightarrow{H_2/Pd-BaSO_4} A$ $\xrightarrow{Br_2/CCl_4} C$
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Write the main product of the given reactions:
$(1)$ $CH_3-CH=C(CH_3)_2 + HCl \to$
$(2)$ $3-\text{Methylbut-}1-\text{ene} + HBr \to$

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Which of the following compounds is formed as the major product in the following reaction?$2-\text{methylbut-}2-\text{ene} \xrightarrow[\text{Peroxide}]{HBr} \text{Product}$