The decreasing order of $-\mathrm{I}$ effect of the following is: $\begin{array}{llllll}\text { I. COOH } &…

The decreasing order of $-\mathrm{I}$ effect of the following is:
$\begin{array}{llllll}\text { I. COOH } & \text { II. F } & \text { III. OR } & \text { IV. } \mathrm{NH}_{2} & \text { V. OH } & \text { VI. Ph- }\end{array}$
  1. $\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}>\mathrm{V}>\mathrm{VI}$
  2. $\mathrm{II}>\mathrm{I}>\mathrm{III}>\mathrm{IV}>\mathrm{V}>\mathrm{VI}$
  3. $\mathrm{I}>\mathrm{II}>\mathrm{V}>\mathrm{III}>\mathrm{IV}>\mathrm{VI}$
  4. $\mathrm{II}>\mathrm{I}>\mathrm{V}>\mathrm{II}>\mathrm{IV}>\mathrm{VI}$

Solution

The decreasing order of -I effect is I. \(-\mathrm{COOH}>\) II. \(-\mathrm{F}>\mathrm{V} .-\mathrm{OH}>\) III. \(-\mathrm{OR}>\mathrm{IV} .-\mathrm{NH}_{2}>\) VI. Ph- The electron withdrawing nature of the atoms/species is called -I Effect. The correct order is- \(\mathrm{I}>\mathrm{II}>\mathrm{V}>\mathrm{III}>\mathrm{IV}>\mathrm{VI}\) .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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