Among the given compounds,choose the two that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Upon ionization,compound $A$ ($3$-bromo$-1-$methylcyclohex$-1-$ene) loses a bromide ion to form the $1-$methylcyclohex$-2-$en$-1-$yl cation.Compou

Vedclass · Accepted Answer

$A, B$

Vedclass · Answer

(C) Upon ionization,compound $A$ ($3$-bromo$-1-$methylcyclohex$-1-$ene) loses a bromide ion to form the $1-$methylcyclohex$-2-$en$-1-$yl cation.Compound $B$ ($4$-bromo$-1-$methylcyclohex$-1-$ene) loses a bromide ion to form a secondary carbocation at the $C4$ position.This secondary carbocation undergoes a hydride shift to form the more stable allylic $1-$methylcyclohex$-2-$en$-1-$yl cation,which is the same intermediate as formed from compound $A$.Therefore,compounds $A$ and $B$ yield the same carbocation.

Column-$I$	Column-$II$
$A$. $CH_3-O^{+}=CH-CH_3$	$1$. Stable due to resonance
$B$. $F_3C^{+}$	$2$. Destabilised due to inductive effect
$C$. $(CH_3)_3C^{-}$	$3$. Stabilised by hyperconjugation
$D$. $CH_3-CH^{+}-CH_3$	$4$. $A$ secondary carbocation

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

Among the given compounds,choose the two that yield the same carbocation on ionization.

Similar Questions

Which of the following is the most stable carbanion?

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

Which of the following carbocations is the least stable?

The correct order of relative stability for the given free radicals is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module