For the following reactions,the major product | vedclass

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(A) At low temperature $(0^\circ C)$,the $1,2$-addition product $(CH_2=CH-CH(Br)-CH_3)$ is the major product because it forms faster due to a lower ac

Vedclass · Accepted Answer

$1 = 	ext{kinetic}, 2 = 	ext{thermodynamic}$

Vedclass · Answer

(A) At low temperature $(0^\circ C)$,the $1,2$-addition product $(CH_2=CH-CH(Br)-CH_3)$ is the major product because it forms faster due to a lower activation energy; this is known as kinetic control.At higher temperatures $(25^\circ C)$,the reaction is reversible and reaches equilibrium,favoring the more stable $1,4$-addition product $(Br-CH_2-CH=CH-CH_3)$ which has a more substituted double bond; this is known as thermodynamic control.

For the following reactions,the major products are shown: $H_2C=CH-CH=CH_2$ $\xrightarrow[0^\circ C]{HBr} CH_2=CH-CH(Br)-CH_3$ $\xrightarrow{25^\circ C} Br-CH_2-CH=CH-CH_3$. These provide an example of $1$ control at low temperature and $2$ control at higher temperature.

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