Ethylene can be prepared in good yield by

Ethylene can be prepared in good yield by
  1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~N}^{+}\left(\mathrm{CH}_{3}ight)_{3} \mathrm{I}^{-} \stackrel{\text { heat }}{\longrightarrow}$ $\mathrm{CH}_{2}=\mathrm{CH}_{2}+\left(\mathrm{CH}_{3}ight)_{3} \mathrm{~N}+\mathrm{HI}$
  2. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~N}^{+}\left(\mathrm{CH}_{3}ight)_{3} \mathrm{OH}^{-} \stackrel{\text { heat }}{\longrightarrow}$ $\mathrm{CH}_{2}=\mathrm{CH}_{2}+\left(\mathrm{CH}_{3}ight)_{3} \mathrm{~N}+\mathrm{H}_{2} \mathrm{O}$
  3. Both (a) and (b)
  4. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} \stackrel{\text { heat }}{\longrightarrow}$ $\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{NH}_{3}$

Solution

This is an example of Hofmann elimination which generally takes place by E2 mechanism and the latter requires a strong base (recall that $\mathrm{OH}^{-}$ is a strong base than $\left.\mathrm{I}^{-}ight)$. The $\mathrm{NH}_{2}{ }^{-}$, being a strong base can't be eliminiated easily. ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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