Matching Organic Reaction Processes to Major Reaction Products

To match the reactions in LIST-I with the corresponding major products in LIST-II, we analyze the mechanism for each process step-by-step:

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1. Reaction P

$\text{tert-BuONa} + \text{tert-BuBr} \longrightarrow$

• Mechanism: The tert-butoxide ion ($\text{tert-BuO}^-$) is a strong, sterically hindered base, and tert-butyl bromide ($\text{tert-BuBr}$) is a $3^\circ$ alkyl halide.
• Outcome: Steric hindrance completely blocks $S_N2$ substitution. Instead, an $E2$ elimination occurs: $\text{tert-BuO}^- + \text{(CH}_3)_3\text{C-Br} \xrightarrow{E2} \text{tert-BuOH (1)} + \text{CH}_2=\text{C(CH}_3)_2 \text{ (4)} + \text{Br}^-$
• Products:
  • $\text{tert-BuOH}$ (Compound 1)
  • Isobutene / 2-Methylpropene (Compound 4)

$\mathbf{P} \rightarrow 1, 4$

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2. Reaction Q

$\text{tert-BuOMe} + \text{HBr} \longrightarrow$

• Mechanism: Protonation of oxygen by $\text{HBr}$ gives the oxonium ion $[\text{tert-Bu}-\overset{+}{\text{O}}\text{H}-\text{Me}]$. Since the tert-butyl group forms a highly stable $3^\circ$ carbocation, cleavage occurs via an $S_N1$ mechanism.
• Outcome: Loss of methanol ($\text{MeOH}$) forms $\text{tert-Bu}^+$, which is subsequently attacked by $\text{Br}^-$: $\text{tert-Bu}-\overset{+}{\text{O}}\text{H}-\text{Me} \xrightarrow{-\text{MeOH}} \text{tert-Bu}^+ \xrightarrow{\text{Br}^-} \text{tert-BuBr (2)}$
• Products: $\text{tert-BuBr}$ (Compound 2)

$\mathbf{Q} \rightarrow 2$

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3. Reaction R

$\text{tert-BuBr} + \text{NaOMe} \longrightarrow$

• Mechanism: Sodium methoxide ($\text{NaOMe}$) is a strong base. Reaction of a $3^\circ$ alkyl halide ($\text{tert-BuBr}$) with a strong base proceeds almost exclusively via $E2$ elimination because $S_N2$ is sterically impossible on a tertiary carbon.
• Outcome: $\text{tert-BuBr} + \text{MeO}^- \xrightarrow{E2} \text{CH}_2=\text{C(CH}_3)_2 \text{ (4)} + \text{MeOH} + \text{Br}^-$
• Products: Isobutene / 2-Methylpropene (Compound 4)

$\mathbf{R} \rightarrow 4$

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4. Reaction S

$\text{tert-BuONa} + \text{MeBr} \longrightarrow$

• Mechanism: Methyl bromide ($\text{MeBr}$) has no $\beta$-hydrogens, so elimination is impossible. This is a classic Williamson ether synthesis where the bulky alkoxide ($\text{tert-BuO}^-$) acts as a nucleophile attacking the unhindered methyl carbon via an $S_N2$ pathway.
• Outcome: $\text{tert-BuO}^- + \text{CH}_3\text{Br} \xrightarrow{S_N2} \text{tert-BuOMe (3)} + \text{Br}^-$
• Products: tert-Butyl methyl ether (Compound 3)

$\mathbf{S} \rightarrow 3$

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Conclusion:

• $\mathbf{P} \rightarrow 1, 4$
• $\mathbf{Q} \rightarrow 2$
• $\mathbf{R} \rightarrow 4$
• $\mathbf{S} \rightarrow 3$

This matches Option (B).