Question

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

• A. $(C{H}_3{)}_{3}CBr+C{H}_{3}COOAg(aq)⟶\stackrel{H_2{O}^{+}}{\to }$
• B. $(C{H}_3{)}_{3}CBr+KOH\stackrel{C_2{H}_{2}OH}{\to }\underset{Heat}{\overset{H_{2}S{O}_4}{\to }}$
• C. $(C{H}_3{)}_{3}CBr\underset{(C_2{H}_5{)}_{2}O}{\overset{Mg}{\to }}\underset{\left(ii\right)H_3{O}^{+}}{\overset{\left(i\right)O_2}{\to }}$
• D. $(C{H}_3{)}_{3}CBr\stackrel{KOH/Ethanol}{\to }\underset{\left(ii\right)H_2{O}_2/NaOH}{\overset{\left(i\right)B_2{H}_6}{\to }}$

Answer 1

The correct option is A $(C{H}_3{)}_{3}CBr+C{H}_{3}COOAg(aq)⟶\stackrel{H_2{O}^{+}}{\to }$

1. Alcoholic $KOH$ will give elimination reaction with $3^{rd}$ degree alcohol so it can't be used to form tertiary alcohol.
   
2. Grignard reagent with $3^o$ carbanion part will not any reaction with $O_2,H_3{O}^{+}$