Identification of Major Products in a Multi-step Organic Synthesis

To determine the correct option(s), we analyze each step in the reaction sequence:

1. Structure of Compound $Q$ and Alkene $R$:

• The starting compound $\text{C}_8\text{H}_{10}\text{O}$ reacts with Grignard reagent $\text{MeMgBr}$ followed by hydrolysis ($\text{H}_2\text{O}$) to form the alcohol $Q$, which is 1-methylcyclohexanol.
• Acid-catalyzed dehydration of $Q$ using $20\%\text{ H}_3\text{PO}_4$ at $360\text{ K}$ yields the most substituted, most stable alkene as the major product $R$, which is 1-methylcyclohexene.

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2. Formation of Product $S$:

• Reaction of $Q$ (1-methylcyclohexanol) with concentrated $\text{HCl}$ proceeds via an $\text{S}_\text{N}1$ pathway:
  1. Protonation of the $-\text{OH}$ group forms $-\text{OH}_2^+$.
  2. Loss of a water molecule generates a highly stable tertiary ($3^\circ$) carbocation at position C-1 of the ring.
  3. Attack by nucleophilic $\text{Cl}^-$ gives 1-chloro-1-methylcyclohexane as the major product $S$.

$\text{1-methylcyclohexanol} \xrightarrow{\text{conc. HCl}} \text{1-chloro-1-methylcyclohexane } (S)$

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3. Formation of Product $T$:

• Catalytic hydrogenation of $R$ (1-methylcyclohexene) with $\text{H}_2/\text{Ni}$ reduces the double bond to form methylcyclohexane.
• Free radical bromination of methylcyclohexane using $\text{Br}_2/h\nu$ occurs with high regioselectivity at the tertiary ($3^\circ$) carbon due to the superior stability of the $3^\circ$ free radical intermediate.
• Thus, the major product $T$ is 1-bromo-1-methylcyclohexane.

$\text{1-methylcyclohexene} \xrightarrow{\text{i) H}_2\text{, Ni}} \text{methylcyclohexane} \xrightarrow{\text{ii) Br}_2\text{, } h\nu} \text{1-bromo-1-methylcyclohexane } (T)$

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4. Formation of Product $U$:

• Addition of $\text{HBr}$ to $R$ (1-methylcyclohexene) in the presence of benzoyl peroxide ($\text{PhCO-O-O-COPh}$) and heat ($\Delta$) proceeds via a radical mechanism (anti-Markovnikov addition or Kharasch effect):
  1. The bromine radical ($\text{Br}^\bullet$) adds to the less hindered C-2 carbon to generate the more stable tertiary ($3^\circ$) carbon radical at C-1.
  2. Hydrogen abstraction from $\text{HBr}$ yields 1-bromo-2-methylcyclohexane as the major product $U$.

$\text{1-methylcyclohexene} \xrightarrow{\text{HBr, benzoyl peroxide, } \Delta} \text{1-bromo-2-methylcyclohexane } (U)$

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Conclusion and Option Verification:

• Option (C) correctly depicts $S$ as 1-chloro-1-methylcyclohexane and $U$ as 1-bromo-2-methylcyclohexane.
• Option (D) correctly depicts $U$ as 1-bromo-2-methylcyclohexane and $T$ as 1-bromo-1-methylcyclohexane.

Thus, the correct options are (C) and (D).