Question

Identify the sets in which \(A\), \(B\) and \(C\) of the following reaction sequence are in correct order \[ \mathrm{(CH_3)_3COH} \xrightarrow{A} X \xrightarrow[\text{(ii) }C]{\text{(i) }B} \mathrm{(CH_3)_2CHCH_2OH} \] (Major product) \[ \begin{aligned} \text{I. }&\mathrm{H_2SO_4;\ HBr;\ OH^-} \text{II. }&\mathrm{Cu/573\,K;\ HBr/(C_6H_5CO)_2O_2;\ OH^-} \text{III. }& 20\%\mathrm{H_3PO_4}/358\,K;\ (\mathrm{BH_3})_2;\ \mathrm{H_2O_2/OH^-} \text{IV. }& 20\%\mathrm{H_3PO_4}/358\,K;\ \mathrm{HBr;\ OH^-} \end{aligned} \] (Major product = major product) The correct answer is

• A. III, IV only
• B. II, III, IV only
• C. I, II, III only
• D. II, III only

Hint:

For alkenes, \[ \boxed{ \mathrm{HBr/peroxide} } \] and \[ \boxed{ \mathrm{BH_3,\ H_2O_2/OH^-} } \] both give the \[ \boxed{\text{anti-Markovnikov product}.} \]

Answer 1

The Correct Option is D

Step 1: Identify intermediate \(X\). Tert-butyl alcohol on dehydration gives \[ \boxed{\mathrm{2\mbox{-}methylpropene}.} \] Both \[ \mathrm{Cu/573\,K} \] and \[ 20\%\mathrm{H_3PO_4}/358\,K \] produce this alkene.

Step 2: Check the conversion to the final product. Set II: \[ \mathrm{HBr/(C_6H_5CO)_2O_2} \] adds HBr by the peroxide effect to form the anti-Markovnikov product, which on hydrolysis gives \[ \mathrm{(CH_3)_2CHCH_2OH}. \] Hence, Set II is correct.

Step 3: Check Set III. Hydroboration-oxidation, \[ (\mathrm{BH_3})_2,\ \mathrm{H_2O_2/OH^-}, \] directly gives the anti-Markovnikov alcohol \[ \mathrm{(CH_3)_2CHCH_2OH}. \] Hence, Set III is correct.

Step 4: Reject the remaining sets. Sets I and IV involve Markovnikov addition of HBr followed by hydrolysis, leading to tert-butyl alcohol rather than the required major product. Hence, they are incorrect. Therefore, \[ \boxed{\text{II and III only}} \] are correct. Thus, \[ \boxed{(D)} \] is the correct answer.