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(C) The stability of the transition state in these $S_N1$ type dissociation steps is directly related to the stability of the resulting carbocation in

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$1 < 3 < 2$

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(C) The stability of the transition state in these $S_N1$ type dissociation steps is directly related to the stability of the resulting carbocation intermediate.Step $1$ produces a methyl carbocation $(CH_3^+)$,which is the least stable.Step $3$ produces an isopropyl carbocation $((CH_3)_2CH^+)$,which is a secondary $(2^o)$ carbocation.Step $2$ produces a tert-butyl carbocation $((CH_3)_3C^+)$,which is a tertiary $(3^o)$ carbocation.Since the stability order of carbocations is $3^o > 2^o > 1^o$ (methyl),the stability of the transition states follows the same order: $1 < 3 < 2$.

Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable):
$1.$ $CH_3-O^+H_2 \to CH_3^+ + H_2O$
$2.$ $(CH_3)_3C-O^+H_2 \to (CH_3)_3C^+ + H_2O$
$3.$ $(CH_3)_2CH-O^+H_2 \to (CH_3)_2CH^+ + H_2O$

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The decreasing order of stability of the following carbocations is:
$A$. $m-CH_3O-C_6H_4-CH_2^+$
$B$. $p-CH_3O-C_6H_4-CH_2^+$
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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:

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Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable):$1.$ $CH_3-O^+H_2 \to CH_3^+ + H_2O$$2.$ $(CH_3)_3C-O^+H_2 \to (CH_3)_3C^+ + H_2O$$3.$ $(CH_3)_2CH-O^+H_2 \to (CH_3)_2CH^+ + H_2O$