The order of stability of the following carbo | vedclass

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(A) The stability of carbocations is determined by resonance and inductive effects.$III$ $(C_6H_5CH_2^+)$ is a benzyl carbocation,which is highly stab

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$III > I > II$

Vedclass · Answer

(A) The stability of carbocations is determined by resonance and inductive effects.$III$ $(C_6H_5CH_2^+)$ is a benzyl carbocation,which is highly stabilized by resonance with the benzene ring.$I$ $(CH_2 = CH - CH_2^+)$ is an allyl carbocation,which is stabilized by resonance with the double bond.$II$ $(CH_3 - CH_2 - CH_2^+)$ is a primary $(1^\circ)$ alkyl carbocation,which is the least stable among the three as it lacks resonance stabilization.Therefore,the order of stability is $III > I > II$.

The order of stability of the following carbocations is:
$I$. $CH_2 = CH - CH_2^+$
$II$. $CH_3 - CH_2 - CH_2^+$
$III$. $C_6H_5CH_2^+$

Similar Questions

Homolytic fission of the following alkanes forms free radicals: $CH_3-CH_3$,$CH_3-CH_2-CH_3$,$(CH_3)_2CH-CH_3$,and $CH_3-CH_2-CH(CH_3)_2$. The increasing order of stability of the resulting free radicals is:

Which of the following options given below is incorrect?

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

Which of the following is the most stable free radical?

In which case does the heterolysis of the carbon-chlorine bond lead to the most stable carbocation?

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The order of stability of the following carbocations is:$I$. $CH_2 = CH - CH_2^+$$II$. $CH_3 - CH_2 - CH_2^+$$III$. $C_6H_5CH_2^+$