Correct order for reaction with alcoholic $\mathrm{KOH}$ is

• A

  i $>$ ii $>$ iii $>$ iv
• Correct answer
  B

  i $>$ iii $>$ ii $>$ iv
• C

  iv $>$ ii $>$ iii $>\mathrm{i}$
• D

  i $>$ iv $>$ ii $>$ iii

Solution

$-\mathrm{Br}$ is good leaving group than $-\mathrm{Cl}$ due to low bond dissociation energy of $\mathrm{C}-\mathrm{Br}$ bond. $2^{\circ}$ carbocation is more stable than $1^{\circ}$ carbocation. More substituted alkene is more stable (Saytzeff's rule).

Hence, the correct answer is option: 'B'.