Question

Predict the major product of the following reaction sequence: 

• A.
• B.
• C.
• D.

Hint:

Key Points: \begin{itemize} \item Radical bromination prefers allylic position \item Alcoholic KOH causes elimination \item ROOR reverses normal addition orientation \end{itemize}

Answer 1

The Correct Option is B

The given sequence starts with methylcyclohexane and involves the following reagents:

(1) \( \mathrm{Br_2}/h\nu \)
(2) Alcoholic \( \mathrm{KOH} \)
(3) \( \mathrm{HBr} \) / \( \mathrm{ROOR} \), \( h\nu \)

Concept Used:

The reaction proceeds through three distinct stages:

1. Free-radical bromination at the most substituted carbon (tertiary position) under photochemical conditions.
2. β-Elimination (E2) by alcoholic \( \mathrm{KOH} \) to form an alkene (Zaitsev product).
3. Anti-Markovnikov addition of \( \mathrm{HBr} \) in the presence of peroxides (\( \mathrm{ROOR} \)), where bromine adds to the less substituted carbon.

Step-by-Step Solution:

Step 1: Radical bromination.

\[ \text{CH}_3\text{-C}_6\text{H}_{11} \xrightarrow{\mathrm{Br_2},\,h\nu} \text{1-Bromo-1-methylcyclohexane} \]

Bromine substitutes the tertiary hydrogen (attached to the carbon bearing the methyl group), forming a tertiary alkyl bromide.

Step 2: Elimination using alcoholic \( \mathrm{KOH} \).

\[ \text{1-Bromo-1-methylcyclohexane} \xrightarrow[\Delta]{\mathrm{alc.\ KOH}} \text{1-Methylcyclohexene} \]

Dehydrohalogenation gives the more substituted alkene (Zaitsev product).

Step 3: Radical addition of \( \mathrm{HBr} \) in the presence of peroxide.

\[ \text{1-Methylcyclohexene} \xrightarrow[\mathrm{ROOR},\,h\nu]{\mathrm{HBr}} \text{2-Bromo-1-methylcyclohexane} \]

Under peroxide conditions, the addition follows the anti-Markovnikov rule. Bromine attaches to the less substituted carbon of the double bond, forming 2-bromo-1-methylcyclohexane.

Final Computation & Result:

The major product is 2-bromo-1-methylcyclohexane.

Correct Option: (2)