In which of the following compounds,the C-Cl | vedclass

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Question

(A) The stability of a carbocation is determined by factors like resonance,hyperconjugation,and the inductive effect. $1$. $C_6H_5CH_2Cl$ ionizes to f

Vedclass · Accepted Answer

$C_6H_5CH_2Cl$

Vedclass · Answer

(A) The stability of a carbocation is determined by factors like resonance,hyperconjugation,and the inductive effect. $1$. $C_6H_5CH_2Cl$ ionizes to form the benzyl carbocation $(C_6H_5CH_2^+)$,which is highly stabilized by resonance with the benzene ring. $2$. $(CH_3)_3C-Cl$ forms a tertiary carbocation,which is stabilized by hyperconjugation and the inductive effect of three methyl groups. $3$. $(CH_3)_2CH-Cl$ forms a secondary carbocation,which is less stable than the tertiary one. $4$. $O_2NCH_2CH_2Cl$ forms a primary carbocation,which is further destabilized by the electron-withdrawing nitro group. Comparing the benzyl carbocation and the tertiary carbocation,the resonance stabilization provided by the phenyl ring in the benzyl carbocation makes it significantly more stable than the simple alkyl-substituted tertiary carbocation. Therefore,$C_6H_5CH_2Cl$ gives the most stable carbonium ion.

In which of the following compounds,the $C-Cl$ bond ionisation shall give the most stable carbonium ion?

Similar Questions

Consider the following carbocations:
$I. C_6H_5CH_2^+$
$II. CH_2=CH^+$
$III. CH_3-CH^+(CH_3)$
$IV. CH_3-CH_2^+$
$V. HC \equiv C^+$
Arrange the above carbocations in the order of decreasing stability.

Which of the following options given below is incorrect?

Which of the following carbanions is most stable?

The decreasing order of hydride affinity for the following carbocations is:
$A: CH_2=CH-C^+(CH_3)_2$
$B: (C_6H_5)_3C^+$
$C: (CH_3)_3C^+$
$D: (Cyclopropyl)_3C^+$
Choose the correct answer from the options given below:

The reaction of methyltrichloroacetate $(Cl_{3}CCO_{2}Me)$ with sodium methoxide $(NaOMe)$ generates

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