Question

Can sodium propoxide and t-butyl chloride be used for the preparation of t-butyl propyl ether? Give suitable explanation. Justify your answer by suggesting the appropriate starting material required for the preparation of t-butyl propyl ether.

Hint:

A strong alkoxide with a tertiary halide gives elimination, not ether. Put the t-butyl group on the alkoxide and use a primary halide, 1-chloropropane.

Answer 1

Step 1: Break the ether bond on paper.
t-Butyl propyl ether has the skeleton $\mathrm{(CH_3)_3C{-}O{-}CH_2CH_2CH_3}$. Williamson synthesis builds it by joining one alkoxide ion to one alkyl halide, so there are two ways to cut the C-O bond.

Step 2: Route I, propoxide plus t-butyl chloride (the pair in the question).
Here the nucleophile $\mathrm{CH_3CH_2CH_2O^-}$ must attack a tertiary carbon. The carbon is shielded by three methyl groups, so $S_N2$ is not possible.
The same ion is also a strong base, and the tertiary halide has nine easily reached beta hydrogens. So E2 takes over:
\[ \mathrm{(CH_3)_3C{-}Cl + CH_3CH_2CH_2O^- \to (CH_3)_2C{=}CH_2 + CH_3CH_2CH_2OH + Cl^-} \]
The product is 2-methylpropene, so Route I fails.

Step 3: Route II, t-butoxide plus a primary propyl halide.
Now the nucleophile is $\mathrm{(CH_3)_3C{-}O^-}$ and the electrophilic carbon is the $\mathrm{CH_2}$ of 1-chloropropane. This carbon is open to attack and the halide leaves smoothly.
\[ \mathrm{(CH_3)_3C{-}ONa + Cl{-}CH_2CH_2CH_3 \to (CH_3)_3C{-}O{-}CH_2CH_2CH_3 + NaCl} \]
Elimination is only a minor path with a primary halide, so Route II gives the ether.

Final Answer:
Sodium propoxide with t-butyl chloride cannot be used because it gives 2-methylpropene by elimination. The right starting materials are sodium t-butoxide and 1-chloropropane (or 1-bromopropane).
\[ \boxed{\text{Sodium t-butoxide} + \text{1-chloropropane}} \]