Which one of the following compounds will be | vedclass

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(B) The elimination of $HBr$ from a haloalkane (dehydrohalogenation) typically proceeds via an $E2$ mechanism,which is facilitated by the presence of

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$Br-CH_2-CH_2-CH_3$

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(B) The elimination of $HBr$ from a haloalkane (dehydrohalogenation) typically proceeds via an $E2$ mechanism,which is facilitated by the presence of an electron-withdrawing group ($-M$ or $-I$ effect) at the $\beta$-carbon. These groups increase the acidity of the $\beta$-hydrogen,making it easier to remove. In the given options: $(a)$ $-NO_2$ is a strong electron-withdrawing group. $(c)$ $-CN$ is a strong electron-withdrawing group. $(d)$ $-CO_2Et$ is an electron-withdrawing group. $(b)$ $-CH_3$ is an electron-donating group (via inductive effect),which decreases the acidity of the $\beta$-hydrogen. Therefore,$Br-CH_2-CH_2-CH_3$ is the least susceptible to elimination.

Which one of the following compounds will be least susceptible to the elimination of hydrogen bromide $(HBr)$?

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