Question
$$tert$$ - Butyl ethyl ether can’t be prepared by which reaction ?
A.
$$tert$$ - Butanol + ethanol \[\xrightarrow{{{H}^{+}}}\]
B.
$$tert$$ - Butyl bromide + sodium ethoxide \[\to \]
C.
Sodium $$tert$$ - butoxide + ethyl bromide \[\to \]
D.
Isobutene + ethanol \[\xrightarrow{{{H}^{+}}}\]
Answer :
$$tert$$ - Butyl bromide + sodium ethoxide \[\to \]
Solution :
\[{{\left( C{{H}_{3}} \right)}_{3}}CBr+NaO{{C}_{2}}{{H}_{5}}\] can’t be applied for synthesising the ether because sod. ethoxide, being a strong base, will preferentially cause elimination reaction.
\[{{\left( C{{H}_{3}} \right)}_{3}}CBr\xrightarrow{^{-}O{{C}_{2}}{{H}_{5}}}{{\left( C{{H}_{3}} \right)}_{2}}C=C{{H}_{2}}+HBr\]
In isobutene + ethanol, isobutene will form $$tert$$ - butyl cation which reacts with ethanol, a nucleophile to form ether.
\[{{\left( C{{H}_{3}} \right)}_{2}}C=C{{H}_{2}}\xrightarrow{{{H}^{+}}}{{\left( C{{H}_{3}} \right)}_{2}}\overset{+}{\mathop{C}}\,C{{H}_{3}}\xrightarrow[\left( ii \right)\,-{{H}^{+}}]{\left( i \right)\,C{{H}_{3}}C{{H}_{2}}OH}{{\left( C{{H}_{3}} \right)}_{3}}COC{{H}_{2}}C{{H}_{3}}\]