Identify Reaction Not Yielding Major Product Via Rearrangement

To determine which reaction does not form its major product via a rearrangement reaction, we analyze the reaction mechanism for each option:

1. Option A: Friedel-Crafts alkylation of benzene with $1$-chloropropane in the presence of anhydrous $\text{AlCl}_3$: $\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \xrightarrow{\text{Anhyd. AlCl}_3} \text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2$

   • The primary carbocation intermediate $\text{CH}_3\text{CH}_2\text{CH}_2^+$ generated initially undergoes a $1,2$-hydride shift to form the more stable secondary carbocation $\text{CH}_3\overset{+}{\text{C}}\text{HCH}_3$.
   • Electrophilic aromatic substitution by this rearranged carbocation yields isopropylbenzene (cumene) as the major product.
   • Involves rearrangement.

2. Option B: Acid-catalyzed dehydration of 2-methylpropan-2-ol ($tert$-butanol): $(\text{CH}_3)_3\text{C-OH} \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_2\text{C}=\text{CH}_2$

   • Protonation of the alcohol followed by the loss of a water molecule directly forms a stable $3^\circ$ carbocation intermediate: $(\text{CH}_3)_3\text{C-OH} \xrightarrow{\text{H}^+} (\text{CH}_3)_3\text{C-OH}_2^+ \xrightarrow{-\text{H}_2\text{O}} (\text{CH}_3)_3\text{C}^+$
   • Loss of a proton ($\text{H}^+$) from this stable $3^\circ$ carbocation forms 2-methylpropene without requiring any structural or hydride rearrangement.
   • Does NOT involve rearrangement.

3. Option C: Isomerization of $n$-hexane using anhydrous $\text{AlCl}_3$ and $\text{HCl}$: $\text{CH}_3(\text{CH}_2)_4\text{CH}_3 \xrightarrow[\text{HCl}]{\text{Anhyd. AlCl}_3} \text{CH}_3\text{CH}(\text{CH}_3)(\text{CH}_2)_2\text{CH}_3$

   • The conversion of straight-chain $n$-hexane into branched 2-methylpentane involves carbocation generation followed by alkyl/hydride skeletal rearrangements.
   • Involves rearrangement.

4. Option D: Acid-catalyzed dehydration of 3,3-dimethylbutan-2-ol: $(\text{CH}_3)_3\text{C-CH(OH)-CH}_3 \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2$

   • Loss of water generates a $2^\circ$ carbocation, $(\text{CH}_3)_3\text{C-}\overset{+}{\text{C}}\text{H-CH}_3$.
   • A $1,2$-methyl shift takes place to convert it into a more stable $3^\circ$ carbocation, $(\text{CH}_3)_2\overset{+}{\text{C}}\text{-CH}(\text{CH}_3)_2$.
   • Subsequent deprotonation yields 2,3-dimethylbut-2-ene as the major product.
   • Involves rearrangement.

Thus, the reaction in Option B does not proceed through a rearrangement.

Correct Answer: Option B