Question 3 of 13: Two Routes to Each Ether

Question 3: Two Routes to Each Ether (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Both targets are made by the Williamson ether synthesis: an alkoxide (S_N2 nucleophile) displaces a halide from an alkyl halide. Because the mechanism is S_N2, the alkyl halide half of the disconnection must be as unhindered as possible — whenever one side of the target ether is a secondary (or worse) carbon, the alkoxide should be built from that side and the halide should be the other, less-hindered partner, never the reverse.

a) Methyl cyclohexyl ether. Route 1: cyclohexoxide (from cyclohexanol + NaH) + CH₃Br/CH₃I (S_N2 on a methyl carbon — essentially unhinderable). Route 2: methoxide (CH₃O^–) + bromocyclohexane (S_N2 on a secondary ring carbon, competing with E2 elimination to cyclohexene since methoxide is also a base). Route 1 is preferred: attacking the methyl halide keeps the difficult, sterically-hindered S_N2 step off the secondary ring carbon entirely.

b) Ethyl isopropyl ether. Route 1: isopropoxide + CH₃CH₂Br (S_N2 on a primary ethyl carbon). Route 2: ethoxide + 2-bromopropane (S_N2 on a secondary isopropyl carbon, again competing with E2). Route 1 is preferred for the same reason as (a): put the secondary carbon on the alkoxide (where it does not need to undergo backside attack) and keep the halide primary.