Match the ions given in Column-$I$ with their nature given in Column-$II$.
Column-$I$ Column-$II$ $A$. $CH_3-O^{+}=CH-CH_3$ $1$. Stable due to resonance $B$. $F_3C^{+}$ $2$. Destabilised due to inductive effect $C$. $(CH_3)_3C^{-}$ $3$. Stabilised by hyperconjugation $D$. $CH_3-CH^{+}-CH_3$ $4$. $A$ secondary carbocation
| Column-$I$ | Column-$II$ |
|---|---|
| $A$. $CH_3-O^{+}=CH-CH_3$ | $1$. Stable due to resonance |
| $B$. $F_3C^{+}$ | $2$. Destabilised due to inductive effect |
| $C$. $(CH_3)_3C^{-}$ | $3$. Stabilised by hyperconjugation |
| $D$. $CH_3-CH^{+}-CH_3$ | $4$. $A$ secondary carbocation |
(A-1,2; B-2; C-2; D-3,4) $A-1, 2$; $B-2$; $C-2$; $D-3, 4$
$A$. $CH_3-O^{+}=CH-CH_3$: This is an oxonium ion. It is stable due to resonance: $CH_3-\ddot{O}^{+}-CH-CH_3 \leftrightarrow CH_3-\ddot{O}=CH^{+}-CH_3$. It is also destabilised by the $-I$ effect of the oxygen atom.
$B$. $F_3C^{+}$: Destabilised due to the strong $-I$ inductive effect of three fluorine atoms,which withdraw electron density from the positively charged carbon.
$C$. $(CH_3)_3C^{-}$: Destabilised due to the $+I$ inductive effect of three methyl groups,which increase electron density on the already negatively charged carbon.
$D$. $CH_3-CH^{+}-CH_3$: This is a secondary $(2^{\circ})$ carbocation. It is stabilised by hyperconjugation from the six $\alpha$-hydrogens.
$A$. $CH_3-O^{+}=CH-CH_3$: This is an oxonium ion. It is stable due to resonance: $CH_3-\ddot{O}^{+}-CH-CH_3 \leftrightarrow CH_3-\ddot{O}=CH^{+}-CH_3$. It is also destabilised by the $-I$ effect of the oxygen atom.
$B$. $F_3C^{+}$: Destabilised due to the strong $-I$ inductive effect of three fluorine atoms,which withdraw electron density from the positively charged carbon.
$C$. $(CH_3)_3C^{-}$: Destabilised due to the $+I$ inductive effect of three methyl groups,which increase electron density on the already negatively charged carbon.
$D$. $CH_3-CH^{+}-CH_3$: This is a secondary $(2^{\circ})$ carbocation. It is stabilised by hyperconjugation from the six $\alpha$-hydrogens.
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Similar Questions
The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$
MediumWBJEE 2023
View SolutionGiven below are two statements:
Statement $I$: $(CH_3)_3C^{+}$ is more stable than $CH_3^{+}$ as nine hyperconjugation interactions are possible in $(CH_3)_3C^{+}$.
Statement $II$: $CH_3^{+}$ is less stable than $(CH_3)_3C^{+}$ as only three hyperconjugation interactions are possible in $CH_3^{+}$.
In the light of the above statements,choose the correct answer from the options given below.
Statement $I$: $(CH_3)_3C^{+}$ is more stable than $CH_3^{+}$ as nine hyperconjugation interactions are possible in $(CH_3)_3C^{+}$.
Statement $II$: $CH_3^{+}$ is less stable than $(CH_3)_3C^{+}$ as only three hyperconjugation interactions are possible in $CH_3^{+}$.
In the light of the above statements,choose the correct answer from the options given below.
DifficultJEE MAIN 2026
View SolutionWhat are free radicals,carbanions,and carbocations? How are they formed?
Medium
View SolutionThe correct stability order for the following species is:
$(I)$ $o-CN-C_6H_4-CH_2^-$
$(II)$ $m-CN-C_6H_4-CH_2^-$
$(III)$ $p-CN-C_6H_4-CH_2^-$
$(IV)$ $C_6H_5-CH_2^-$
(Note: The image shows the structures of these carbanions.)
$(I)$ $o-CN-C_6H_4-CH_2^-$
$(II)$ $m-CN-C_6H_4-CH_2^-$
$(III)$ $p-CN-C_6H_4-CH_2^-$
$(IV)$ $C_6H_5-CH_2^-$
(Note: The image shows the structures of these carbanions.)
Medium
View SolutionThe most stable carbocation is
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