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Primary alcohol can easily be prepared from primary alkyl halide via S_N2 reaction with aqueous NaOH. However, similar method does not work for the preparation of tertiary alcohol (tertiary butanol) from tertiary butyl bromide except?

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${({CH}_{3})}_{3} CBr + {CH}_{3} COOAg (aq) \to \overset{H_{3} O^{+}}{\to}$

${({CH}_{3})}_{3} CBr + KOH \to \overset{C_{2} H_{5} OH}{\to} \to_{Heat}^{dil . H_{2} {SO}_{4}}$

${({CH}_{3})}_{3} CBr \to_{{(C_{2} H_{5})}_{2} O}^{Mg} \to_{(ii) H_{3} O^{+}}^{(i) O_{2}}$

${({CH}_{3})}_{3} CBr \overset{KOH/Ethanol}{\to} \to_{(ii) H_{2} O_{2} / NaOH}^{(i) B_{2} H_{6}}$

answer is D.

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Detailed Solution

${({CH}_{3})}_{3} C B r \overset{K O H / E t h a n o l}{\to} {(C H_{3})}_{2} C = {CH}_{2} \to_{(2) H_{2} O_{2} / NaOH}^{(1) B_{2} H_{6}} {({CH}_{3})}_{2} {CHCH}_{2} OH$

A primary alocohol is obtained.

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