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(C) Upon ionization,the $Br^-$ ion leaves,generating a carbocation at the carbon atom previously bonded to the bromine.Compound $1$ is $3$-bromo$-2-$e

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$1$ and $2$

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(C) Upon ionization,the $Br^-$ ion leaves,generating a carbocation at the carbon atom previously bonded to the bromine.Compound $1$ is $3$-bromo$-2-$ethylbut$-1-$ene. Ionization gives a secondary allylic carbocation: $CH_3-CH^+-C(=CH_2)-CH_2-CH_3$.Compound $2$ is $1$-bromo$-2-$ethylbut$-2-$ene. Ionization gives a primary allylic carbocation: $CH_3-CH=C(CH_2CH_3)-CH_2^+$.These two carbocations are resonance structures of each other,meaning they represent the same delocalized allylic system.Therefore,compounds $1$ and $2$ yield the same resonance-stabilized carbocation.

Which of the following compounds gives the same carbocation on ionization?

Similar Questions

Which of the following options given below is incorrect?

What is the increasing order of stability for the carbanions $P: \bar{C}H_3$,$Q: C_6H_5\bar{C}H_2$,$R: (CH_3)_2\bar{C}H$,and $S: H_2\bar{C}^{-}CH=CH_2$?

Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable):
$1. CH_3-OH_2^+ \to CH_3^+ + H_2O$
$2. (CH_3)_3C-OH_2^+ \to (CH_3)_3C^+ + H_2O$
$3. (CH_3)_2CH-OH_2^+ \to (CH_3)_2CH^+ + H_2O$

What are free radicals,carbanions,and carbocations? How are they formed?

$A$ carbanion acts as a:

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