For the reaction: CH_2=CH-CH=CH_2 + HBr (1 te | vedclass

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(B) The reaction of $1,3$-butadiene with $HBr$ at $80^{\circ}C$ (higher temperature) leads to the formation of the thermodynamically controlled produc

Vedclass · Accepted Answer

$1-$Bromobut$-2-$ene

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(B) The reaction of $1,3$-butadiene with $HBr$ at $80^{\circ}C$ (higher temperature) leads to the formation of the thermodynamically controlled product.In $1,3$-butadiene,the $1,2$-addition product is the kinetically controlled product (formed at lower temperatures),while the $1,4$-addition product is the thermodynamically controlled product.The $1,4$-addition product is $1$-bromobut-$2$-ene,which is more stable due to the more substituted double bond.Thus,$P_1$ (the major product at $80^{\circ}C$) is $1$-bromobut-$2$-ene.

For the reaction: $CH_2=CH-CH=CH_2 + HBr (1 \text{ eq}) \xrightarrow{80^{\circ}C} P_1 + P_2$ (where $P_1$ is the major product),identify $P_1$.

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