Question

Arrange the following alkyl halides in decreasing order of the rate of ~ elimination reaction with alcoholic KOH:

1. \[\begin{array}{cc}
   \ce{H}\phantom{...}\\
   |\phantom{...}\\
   \phantom{}\ce{CH3 - C - CH2Br}\\
   |\phantom{...}\\
   \ce{CH3}\phantom{}
   \end{array}\]
2. CH₃ – CH₂ – Br
3. CH₃ – CH₂ – CH₂ – Br

पर्याय

• (i) > (ii) > (iii)

• (iii) > (ii) > (i)

• (ii) > (iii) > (i)

• (i) > (iii) > (ii)

Answer 1

(i) > (iii) > (ii)

Explanation:

Only when there are a greater number of β - substituents will a more stable alkene be generated by β - elimination. Because I has two, (iii) has one - methyl substituent, and (ii) has none, the reactivity β - elimination decreases in order: (i) > (iii) > (ii)