The relation between (A) and (B) is: | vedclass

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(B) In the first reaction,$sec-butyl$ bromide undergoes $E2$ elimination with $EtO^-$ to form the more stable alkene,$but-2-ene$ (Saytzeff product),as

Vedclass · Accepted Answer

positional isomer

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(B) In the first reaction,$sec-butyl$ bromide undergoes $E2$ elimination with $EtO^-$ to form the more stable alkene,$but-2-ene$ (Saytzeff product),as the major product $(A)$.In the second reaction,$sec-butyl$ trimethylammonium ion undergoes $E2$ elimination with $EtO^-$. Due to the poor leaving group ability of the $-NMe_3^+$ group and the steric hindrance,the transition state $(T.S.)$ develops significant carbanion character. According to the Hofmann rule,the less substituted alkene,$but-1-ene$,is formed as the major product $(B)$.Since $(A)$ is $but-2-ene$ and $(B)$ is $but-1-ene$,they are positional isomers.

The relation between $(A)$ and $(B)$ is:

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