CH_3-CH(CH_3)-CH=CH_2 + HBr to X (product)Wha | vedclass

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(B) The reaction of $3$-methyl-$1$-butene with $HBr$ follows the electrophilic addition mechanism.$1.$ Protonation of the double bond forms a secondar

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$CH_3-C(Br)(CH_3)-CH_2-CH_3$

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(B) The reaction of $3$-methyl-$1$-butene with $HBr$ follows the electrophilic addition mechanism.$1.$ Protonation of the double bond forms a secondary carbocation: $CH_3-CH(CH_3)-CH=CH_2 + H^{+} 	o CH_3-CH(CH_3)-C^{+}H-CH_3$ ($2^{\circ}$ carbocation).$2.$ $A$ $1,2$-hydride shift occurs to form a more stable tertiary carbocation: $CH_3-CH(CH_3)-C^{+}H-CH_3 \xrightarrow{1,2-H 	ext{ shift}} CH_3-C^{+}(CH_3)-CH_2-CH_3$ ($3^{\circ}$ carbocation).$3.$ Nucleophilic attack by $Br^{-}$ gives the major product: $CH_3-C^{+}(CH_3)-CH_2-CH_3 + Br^{-} 	o CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2$-bromo-$2$-methylbutane).

$CH_3-CH(CH_3)-CH=CH_2 + HBr \to X$ (product)
What is the major product $X$ in the reaction?

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$CH_3-CH(CH_3)-CH=CH_2 + HBr \to X$ (product)What is the major product $X$ in the reaction?