Arrange the following carbocations in decreas | vedclass

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(A) The stability of carbocations is determined by the inductive effect and hyperconjugation.Greater the number of alkyl groups attached to the positi

Vedclass · Accepted Answer

$(CH_3)_3\stackrel{+}{C} > (CH_3)_2\stackrel{+}{C}H > CH_3\stackrel{+}{C}H_2 > \stackrel{+}{C}H_3$

Vedclass · Answer

(A) The stability of carbocations is determined by the inductive effect and hyperconjugation.Greater the number of alkyl groups attached to the positively charged carbon atom,greater is the stability due to the $+I$ effect and hyperconjugation.$(CH_3)_3\stackrel{+}{C}$ is a tertiary carbocation (most stable).$(CH_3)_2\stackrel{+}{C}H$ is a secondary carbocation.$CH_3\stackrel{+}{C}H_2$ is a primary carbocation.$\stackrel{+}{C}H_3$ is a methyl carbocation (least stable).Therefore,the decreasing order of stability is: $(CH_3)_3\stackrel{+}{C} > (CH_3)_2\stackrel{+}{C}H > CH_3\stackrel{+}{C}H_2 > \stackrel{+}{C}H_3$.

Arrange the following carbocations in decreasing order of their stability:
$\stackrel{+}{C}H_3, (CH_3)_3\stackrel{+}{C}, CH_3\stackrel{+}{C}H_2, (CH_3)_2\stackrel{+}{C}H$

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Arrange the following carbocations in decreasing order of their stability:$\stackrel{+}{C}H_3, (CH_3)_3\stackrel{+}{C}, CH_3\stackrel{+}{C}H_2, (CH_3)_2\stackrel{+}{C}H$