The correct order of stability of the carboca | vedclass

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Question

(A) The stability of the given carbocations is determined by resonance,aromaticity,and hybridization:$(i)$ Cyclopropenyl cation is aromatic ($2\pi$ el

Vedclass · Accepted Answer

$(i) > (ii) > (iii) > (iv)$

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(A) The stability of the given carbocations is determined by resonance,aromaticity,and hybridization:$(i)$ Cyclopropenyl cation is aromatic ($2\pi$ electrons,Huckel's rule),making it highly stable.(ii) Benzyl carbocation $(C_6H_5CH_2^+)$ is stabilized by resonance with the benzene ring.(iii) Allyl carbocation $(CH_2=CH-CH_2^+)$ is stabilized by resonance with the double bond.(iv) Vinyl carbocation $(CH_2=CH^+)$ is highly unstable because the positive charge is on an $sp$-hybridized carbon atom,which is more electronegative.Comparing their stabilities:- $(i)$ is aromatic,so it is the most stable.- $(ii)$ is resonance-stabilized by a benzene ring,which is more stable than the simple allyl cation $(iii)$.- $(iii)$ is resonance-stabilized.- $(iv)$ is the least stable due to the $sp$-hybridized carbon.Thus,the correct order is $(i) > (ii) > (iii) > (iv)$.

List-$I$ (Carbocation)	List-$II$ (Type)
$A$. $CH_3-C^{+}(CH_3)-CH_3$	$I$. Secondary carbocation
$B$. $CH_3-C^{+}H-CH_3$	$II$. Methyl carbocation
$C$. $CH_3-CH_2^+$	$III$. Primary carbocation
$D$. $CH_3^+$	$IV$. Tertiary carbocation

The correct order of stability of the carbocations given below is:

Similar Questions

Which of the following compounds forms the most stable carbocation: $(CH_3)_3C-Br$,$(C_6H_5)_3CBr$,$(C_6H_5)_2CHBr$,and $C_6H_5CH_2Br$?

Which of the following carbocations will undergo rearrangement?

Match the following columns and select the correct option.

List-$I$ (Carbocation) List-$II$ (Type)

$A$. $CH_3-C^{+}(CH_3)-CH_3$ $I$. Secondary carbocation

$B$. $CH_3-C^{+}H-CH_3$ $II$. Methyl carbocation

$C$. $CH_3-CH_2^+$ $III$. Primary carbocation

$D$. $CH_3^+$ $IV$. Tertiary carbocation

The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$

The compound $(X)$ will be:

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