Problem: Starting from ethanol and 1-bromopropane, show the Williamson synthesis of ethyl propyl ether. Explain why you cannot use 2-bromopropane instead.
Step 1 — Prepare the alkoxide:
CH 3 CH 2 OH + NaH → CH 3 CH 2 O − Na + + H 2 \text{CH}_3\text{CH}_2\text{OH} + \text{NaH} \rightarrow \text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2 CH 3 CH 2 OH + NaH → CH 3 CH 2 O − Na + + H 2
Sodium ethoxide: SMILES:CC[O-].[Na+]
Step 2 — SN2 attack on 1-bromopropane:
CH 3 CH 2 O − Na + + CH 3 CH 2 CH 2 Br → SN2 CH 3 CH 2 OCH 2 CH 2 CH 3 + NaBr \text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{SN2}} \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_2\text{CH}_3 + \text{NaBr} CH 3 CH 2 O − Na + + CH 3 CH 2 CH 2 Br SN2 CH 3 CH 2 OCH 2 CH 2 CH 3 + NaBr
Product (ethyl propyl ether): SMILES:CCOCCC
Step 3 — Why 2-bromopropane FAILS:
2-Bromopropane is a 2° halide. The strong alkoxide (NaOEt) is also a strong base.
With a 2° halide and strong base, E2 elimination dominates:
CH 3 CH 2 O − Na + + (CH 3 ) 2 CHBr → E2 CH 3 CH = CH 2 + NaBr + CH 3 CH 2 OH \text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{(CH}_3)_2\text{CHBr} \xrightarrow{\text{E2}} \text{CH}_3\text{CH}=\text{CH}_2 + \text{NaBr} + \text{CH}_3\text{CH}_2\text{OH} CH 3 CH 2 O − Na + + (CH 3 ) 2 CHBr E2 CH 3 CH = CH 2 + NaBr + CH 3 CH 2 OH
Conclusion: Always use 1° alkyl halide in Williamson synthesis. Alkoxide provides -O-; alkyl halide provides the carbon.