Which of the following is not expected to be | vedclass

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(D) The reaction is a solvolysis of $1$-iodo-$1$-methylcyclohexane in water. Step $1$: The leaving group $I^-$ departs to form a secondary carbocation

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(D) The reaction is a solvolysis of $1$-iodo-$1$-methylcyclohexane in water. Step $1$: The leaving group $I^-$ departs to form a secondary carbocation at the $C_2$ position. Step $2$: $A$ $1,2$-hydride shift occurs to transform the secondary carbocation into a more stable tertiary carbocation at the $C_1$ position (where the methyl group is attached). Step $3$: Water acts as a nucleophile and attacks the tertiary carbocation to form a protonated alcohol (oxonium ion). Step $4$: Deprotonation yields the final product,$1$-methylcyclohexanol. Comparing this with the options,the primary carbocation with a positive charge on the carbon bearing the methyl group is not formed,as the initial carbocation is secondary and the final one is tertiary.

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 is not expected to be an intermediate of the following reaction?

Similar Questions

Which of the following is the most stable carbanion?

The decreasing order of stability of the given carbocations is

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

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.

The stability of the carbocations follows the order:

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