Question

Primary alcohol can easily be prepared from primary alkyl halide via $S_{N} 2$ reaction with aqueous $NaOH$. However, similar method does not work for the preparation of tertiary alcohol (tertiary butanol) from tertiary butyl bromide except?

• A. $(CH_{3})_{3}CBr+CH_{3}COOAg(aq) \xrightarrow{H_{3}O^{+}}$
• B. $(CH_{3})_{3}CBr+KOH \xrightarrow{ C_{2}H_{5}OH} \xrightarrow[{\text{Heat}}] { H_{2}SO_{4}}$
• C. $(CH_{3})_{3} CBr \xrightarrow[{(C_{2}H_{5})_{2}O}] {Mg} \xrightarrow[{(ii)H_{3}O^{+}}]{(i)O_{2}}$
• D. $(CH_{3})_{3}CBr \xrightarrow{\text{KOH/Ethanol}} \xrightarrow[{(ii) H_{2}O_{2}/NaOH}] {(i) B_{2}H_{6}}$

Answer 1

Answer: (A)

$3^{\circ}$ alkylhalide undergoes elimination with base
$(CH_{3})_{3}CBr+CH_{3}COOAg (aq) \to$