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(D) $(1)$ In the addition of $HCl$ to $2-	ext{methylbut-}2-	ext{ene}$ $(CH_3-CH=C(CH_3)_2)$,the reaction follows Markovnikov's rule. The $H^+$ adds

Vedclass · Accepted Answer

$(1)$ $2-	ext{chloro-}2-	ext{methylbutane}$,$(2)$ $2-	ext{bromo-}2-	ext{methylbutane}$

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(D) $(1)$ In the addition of $HCl$ to $2-	ext{methylbut-}2-	ext{ene}$ $(CH_3-CH=C(CH_3)_2)$,the reaction follows Markovnikov's rule. The $H^+$ adds to the carbon with more hydrogens,and $Cl^-$ adds to the more substituted carbon,forming $2-	ext{chloro-}2-	ext{methylbutane}$. $(2)$ In the addition of $HBr$ to $3-	ext{methylbut-}1-	ext{ene}$ $(CH_3-CH(CH_3)-CH=CH_2)$,a secondary carbocation $(CH_3-CH(CH_3)-C^+H-CH_3)$ is initially formed. This undergoes a $1,2-	ext{hydride shift}$ to form a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH_2-CH_3)$. The $Br^-$ then attacks the tertiary carbocation to form $2-	ext{bromo-}2-	ext{methylbutane}$.

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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Write the main product of the given reactions: $(1)$ $CH_3-CH=C(CH_3)_2 + HCl \to$ $(2)$ $3-\text{Methylbut-}1-\text{ene} + HBr \to$