The correct order of stability of the given c | vedclass

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(A) The stability of carbocations depends on factors like resonance, hyperconjugation, and the electronegativity of the carbon atom bearing the positi

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$A > C > B > D$

Vedclass · Answer

(A) The stability of carbocations depends on factors like resonance, hyperconjugation, and the electronegativity of the carbon atom bearing the positive charge.$A$ (Benzyl carbocation) is the most stable due to resonance stabilization by the benzene ring.$C$ $(CH_3-CH_2^+)$ is a primary alkyl carbocation, which is stabilized by the inductive effect $(+I)$ of the methyl group.$B$ $(CH_2=CH^+)$ is a vinyl carbocation where the positive charge is on an $sp^2$ hybridized carbon, which is more electronegative than an $sp^3$ carbon.$D$ $(HC \equiv C^+)$ is an acetylenic carbocation where the positive charge is on an $sp$ hybridized carbon, which is the most electronegative among the three $(sp > sp^2 > sp^3)$.Since a positive charge is less stable on a more electronegative atom, the order of stability for $B, C,$ and $D$ is $C > B > D$.Combining these, the overall order of stability is $A > C > B > D$.

The correct order of stability of the given carbocations is:

Similar Questions

The decreasing order of stability of the following carbocations is:
$A$. $m-CH_3O-C_6H_4-CH_2^+$
$B$. $p-CH_3O-C_6H_4-CH_2^+$
$C$. $C_6H_5-CH_2^+$
$D$. $p-NO_2-C_6H_4-CH_2^+$

What is the hybridisation state of the benzyl carbocation shown in the image?

Which of the following is the correct decreasing order of stability of carbanions?
$(1) (CH_3)_3C^-$
$(2) (CH_3)_2CH^-$
$(3) CH_3CH_2^-$
$(4) C_6H_5CH_2^-$

Compare the stability of the following free radicals:
$(1)$ $CH_3 - \dot{CH} - CH_3$
$(2)$ $C_6H_5 - \dot{CH_2}$
$(3)$ $\dot{CH_2} - CH(CH_3)_2$
$(4)$ $\dot{CH_2} - CH_3$

The decreasing order of stability of the given carbocations is:

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