Question

For the following reaction:
$I.C{H}_{3}C{H}_{2}C{H}_{2}Br+KOH\to C{H}_{3}CH=C{H}_2+KBr+H_{2}O$

Which of the following statements is correct?

• A. $I$ and $II$ are elimination reactions, $III$ is addition reaction
  
• B. $I$ is elimination, $II$ is substitution, $III$ is addition reaction
  
• C. $I$ is elimination reaction, $II$ and $III$ are substitution reactions
  
• D. $I$ is substitution reaction, $II$ and $III$ are addition reactions
  

Answer 1

The correct option is B $I$ is elimination, $II$ is substitution, $III$ is addition reaction


$I.C{H}_{3}C{H}_{2}C{H}_{2}Br+KOH\to C{H}_{3}CH=C{H}_2+KBr+H_{2}O$
Here, breaking of 2 $\sigma$ bonds and formation of 1 $\pi$ bond. So it is an example of elimination reaction.
Here, replacement of $B{r}^{-}$ by $O{H}^{-}$. So it is an example of substitution reaction.
Here, breaking of 1 $\pi$ bond and formation of 2 $\sigma$ bonds. So it is an example of addition reaction.