Question

Ethers are prepared by the dehydration of alcohols in presence of protic acids at 413 K. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves \( \mathrm{S_N2} \) attack of an alkoxide ion on primary alkyl halide. If a tertiary alkyl halide is used, elimination reaction occurs and an alkene is formed, no ether is formed. C - O bond in ethers are cleaved under drastic conditions with excess of HI. When unsymmetrical ethers react with HI, the alkyl halide is formed from smaller alkyl group. If one of the alkyl group is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary alkyl carbocation is more stable than the primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl-oxygen bond due to more stable aryl-oxygen bond. The order of reactivity of hydrogen halides is HI > HBr > HCl. Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring towards electrophilic substitution and directs the incoming group to ortho- and para-positions.

Answer the following questions :

(b) Write the names of alkyl halide and sodium alkoxide used to prepare tert-butyl ethyl ether.

Hint:

The halide must be primary for the Williamson synthesis. So the ethyl part comes from the halide and the tert-butyl part from the alkoxide.

Answer 1

Step 1: Write the target ether:
tert-Butyl ethyl ether is \( \mathrm{(CH_3)_3C{-}O{-}C_2H_5} \).
Williamson synthesis joins an alkoxide ion with an alkyl halide. There are two ways to split this ether.

Step 2: Test the first way:
Sodium ethoxide with tert-butyl chloride would need an \( \mathrm{S_N2} \) attack on a tertiary carbon.
That is not possible. The strongly basic ethoxide removes an H and gives 2-methylpropene (an alkene) by elimination.
The passage says exactly this: a tertiary alkyl halide gives an alkene and no ether.

Step 3: Test the second way:
Sodium tert-butoxide with a primary halide such as bromoethane works well.
The primary carbon is open to \( \mathrm{S_N2} \) attack.
\[ \mathrm{(CH_3)_3C{-}ONa + C_2H_5Br \longrightarrow (CH_3)_3C{-}O{-}C_2H_5 + NaBr} \]

Final Answer:
The alkyl halide is bromoethane (or chloroethane, a primary halide) and the sodium alkoxide is sodium tert-butoxide.