Question

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Answer 1

Acid dehydration of primary alcohols to ethers occurs by S_N2 mechanism in which the nucleophilic attack of the alcohol molecule takes place on the protonated alcohol molecule.

\[\ce{CH3CH2CH2\overset{\bullet\bullet}{\underset{\bullet\bullet}{O}}H + CH3CH2CH2 - \overset{+}{\underset{\bullet\bullet}{O}}\overset{+}{H} ->[S_N2][-H+, -H2O] CH3CH2CH2 - O - CH2CH2CH3}\]

Under these conditions, secondary and tertiary alcohols give alkenes instead of ethers. There is no nucleophilic attack of the alcohol molecule on the protonated alcohol molecule. Instead, protonated secondary and tertiary alcohols lose a molecule of water to form stable 2° and 3° carbocations. These carbocations preferentially lose H⁺ to form alkenes.

\[\begin{array}{cc}
\ce{CH3}\phantom{...........................}\ce{CH3}\phantom{.....................}\ce{CH3}\\\
|\phantom{.................................}|\phantom{...........................}|\phantom{.....}\\
\ce{\underset{\underset{(2^\circ alcohol)}{Propan-2-ol}}{CH3 - CH - OH} ->[H^-] \underset{Protonated 2^\circ alcohol}{CH3 - CH - \overset{+}{O}H2} ->[][-H2O] \underset{2^\circ Carbocation}{CH3 - CH^+}}
\end{array}\]

\[\begin{array}{cc}
\ce{CH3}\phantom{...}\ce{CH3}\phantom{..............................}\ce{CH3}\phantom{...........}\\
|\phantom{.........}|\phantom{....................................}|\phantom{..............}\\
\ce{\underset{2-propoxy-2-propane}{CH3 - CH - O - CH - CH3} ->[CH3CHOCH3][-H^+] CH3 - CH^+ ->[][-H^+] \underset{Propene}{CH3 - CH = CH2}}
\end{array}\]

Similarly, 3° alcohols give alkenes instead of ethers.

\[\ce{\underset{\underset{(3^\circ alcohol)}{2-Methylpropan-2-ol}}{(CH3)3C - OH} ->[H+] \underset{\underset{(3^\circ alcohol)}{Protonated 2-methylpropan-2-ol}}{(CH3)3C - \overset{+}{O}H2} ->[][-H2O] \underset{3^\circ butyl carbocation}{(CH3)3C^+}}\]

\[\begin{array}{cc}
\phantom{.......}\ce{CH3}\phantom{.}\ce{CH3}\phantom{....................................}\ce{CH3}\phantom{..................}\ce{CH3}\\
\phantom{.....}|\phantom{......}|\phantom{..........................................}|\phantom{.........................}|\\
\ce{CH3 - C - O - C - CH3 ->[(CH3)3COH][-H^+] CH3 - C^+ ->[][-H^+] \underset{2-Methylprop-1-ene}{CH3 - CH = CH2}}\\
|\phantom{.......}|\phantom{..........................................}|\phantom{......................}\\
\ce{\underset{2-Methyl-2-(2-methyl-2-propoxy)propane}{\phantom{..}CH3\phantom{..}CH3}}\phantom{...............}\ce{\underset{3^\circ butyl carbocation}{CH3}}\phantom{............................}
\end{array}\]

Answer 2

Consider the reaction between propan-2-ol molecules in the presence of an acid.

\[\begin{array}{cc}
\phantom{..............}\ce{CH3}\phantom{...........................}\ce{CH3}\\
\phantom{..........}|\phantom{.................................}|\\
\ce{CH3 - CH - OH ->[H+] CH3 - \underset{+}{C}H}\\
\end{array}\]

If an ether is to be formed, another alcohol molecule must carry out a nucleophilic attack on the carbocation as:

However, this does not happen because of:

1. the steric hindrance around the carbocation, and
2. the bulky size of the nucleophile would further cause crowding.

As a result, the carbocation prefers to lose a proton, forming an alkene.

For the same reason, 3° alcohols in the presence of acid do not form ethers since 3° alcohols are even more sterically hindered than 2° alcohols.