Question

Choose the CORRECT option regarding given scheme?

• A. [P] is major product of reaction between 2-bromo 3-methyl butane and alc. KOH
• B. [Q] reacts with Cu at 573K to form [P]
• C. [Q] can be oxidized with PCC
• D. [S] gives positive iodoform test

Answer 1

Answer: (A), (B), (D)

The CORRECT options are: [P] is major product of reaction between 2-bromo 3-methyl butane and alc. KOH, [Q] reacts with Cu at 573K to form [P], [S] gives positive iodoform test

Answer 2

1. Determine [P]: The reaction of 2-bromo-3-methylbutane with alcoholic KOH is an elimination reaction. The possible alkenes formed are 2-methylbut-2-ene (trisubstituted) and 3-methylbut-1-ene (monosubstituted). According to Zaitsev's rule, the major product is the more substituted alkene, which is 2-methylbut-2-ene. Thus, the first option is correct.

2. Determine [Q] and [R]: Assuming [P] is 2-methylbut-2-ene ($\text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3$).

   • Hydroboration-oxidation of [P] yields 2-methylbutan-1-ol ([R]).
   • Oxymercuration-demercuration of [P] yields 2-methylbutan-2-ol ([Q]).

3. Evaluate Option 2: [Q] is 2-methylbutan-2-ol, a tertiary alcohol. Dehydration of tertiary alcohols with copper at 573K produces alkenes. Dehydration of 2-methylbutan-2-ol yields 2-methylbut-2-ene, which is [P]. Thus, the second option is correct.

4. Evaluate Option 3: [Q] is a tertiary alcohol. Tertiary alcohols are resistant to oxidation by mild oxidizing agents like PCC. Thus, the third option is incorrect.

5. Determine [S] and Evaluate Option 4: Ozonolysis of [P] ($\text{CH}_3\text{-C(CH}_3\text{)=CH-CH}_3$) followed by reductive workup (e.g., Zn/H$_2$O) yields acetone ($\text{CH}_3\text{-CO-CH}_3$) and acetaldehyde ($\text{CH}_3\text{-CHO}$). From the scheme, [S] is the aldehyde product. Acetaldehyde ($\text{CH}_3\text{-CHO}$) contains the $\text{CH}_3\text{-CO-}$ moiety and thus gives a positive iodoform test. Thus, the fourth option is correct.