Which of the following intermediates is not e | vedclass

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(B) The reaction is a solvolysis of $1-$iodo$-2-$methylcyclohexane in water.$1$. The leaving group $I^-$ departs to form a secondary carbocation at th

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(B) The reaction is a solvolysis of $1-$iodo$-2-$methylcyclohexane in water.$1$. The leaving group $I^-$ departs to form a secondary carbocation at the $C-2$ position (Image $D$).$2$. This secondary carbocation undergoes a $1,2-$hydride shift to form a more stable tertiary carbocation at the $C-1$ position (Image $C$).$3$. Water acts as a nucleophile and attacks the tertiary carbocation to form an oxonium ion intermediate (Image $A$).$4$. Finally, deprotonation yields $1-$methylcyclohexanol.Comparing the options:- Option $A$ is the oxonium ion intermediate.- Option $B$ is a primary carbocation, which is highly unstable and not expected to be formed in this mechanism.- Option $C$ is the tertiary carbocation.- Option $D$ is the initial secondary carbocation.Thus, the intermediate not expected to be formed is $B$.

Which of the following intermediates is not expected to be formed in the following reaction?

Similar Questions

Homolytic fission of the following alkanes forms free radicals: $CH_3-CH_3$,$CH_3-CH_2-CH_3$,$(CH_3)_2CH-CH_3$,and $CH_3-CH_2-CH(CH_3)_2$. The increasing order of stability of the resulting free radicals is:

The correct order of increasing stabilities of the following carbocations is:
$(i)$ Allyl carbocation
$(ii)$ Ethyl carbocation
$(iii)$ Benzyl carbocation
$(iv)$ Isobutyl carbocation

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

Which is the correct order of stability of the following three carbonium ions?
$(I)$ $CH_2 = CH - CH^+ - CH_3$
$(II)$ $CH_2 = C(CH_3) - CH_2^+$
$(III)$ $CH_3 - CH = CH - CH_2^+$

The geometry of the reaction intermediate in an $S_N1$ reaction is:

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